Foundations of Euclidean Constructive Geometry

FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY

MICHAEL BEESON

Abstract. Euclidean geometry, as presented by Euclid, consists of straightedge-and- compass constructions and rigorous reasoning about the results of those constructions. A consideration of the relation of the Euclidean “constructions” to “constructive mathematics” leads to the development of a first-order theory ECG of “Euclidean Constructive Geometry”, which can serve as an axiomatization of Euclid rather close in spirit to the Elements of Euclid. Using Gentzen’s cut-elimination theorem, we show that when ECG proves an existential theorem, then the things proved to exist can be constructed by Eu- clidean ruler-and-compass constructions. In the second part of the paper we take up the formal relationships between three versions of Euclid’s parallel postulate: Euclid’s own formulation in his Postulate 5, Playfair’s 1795 version, which is the one usually used in modern axiomatizations, and the version used in ECG. We completely settle the questions about which versions imply which others using only constructive logic: ECG’s version implies Euclid 5, which implies Playfair, and none of the reverse implications are provable. The proofs use Kripke models based on carefully constructed rings of real-valued functions. “Points” in these models are real-valued functions. We also characterize these theories in terms of different constructive versions of the axioms for Euclidean fields.1

Contents 1. Introduction 5 1.1. Euclid 5 1.2. The collapsible vs. the rigid compass 7 1.3. Postulates vs. axioms in Euclid 9 1.4. The parallel postulate 9 1.5. Polygons in Euclid 10 2. Euclid’s reasoning considered constructively 10 2.1. Order on a line from the constructive viewpoint 10 2.2. Logical form of Euclid’s propositions and proofs 14 2.3. Case splits and reductio inessential in Euclid 15 3. Constructions in geometry 16 3.1. The 48 Euclidean constructions 16 3.2. The elementary constructions 16 3.3. Circles and lines eliminable when contructing points 19 3.4. Describing constructions by terms 20 4. Models of the elementary constructions 21 4.1. The standard plane 21 4.2. The recursive model 22 4.3. The algebraic model 23 4.4. The Tarski model 23

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5. The geometrization of arithmetic without case distinctions 23 5.1. Uniform perpendicular and projection 24 5.2. Rotation 27 5.3. Addition 28 5.4. Multiplication and division 28 5.5. Square roots 31 5.6. Formalizing the geometrization of arithmetic 32 6. The arithmetization of geometry 32 6.1. Euclidean ﬁelds in constructive mathematics 32 6.2. Line-circle and circle-circle continuity over Euclideanﬁelds 34 7. Otheraxiomatizationsofgeometry 36 7.1. Pasch 37 7.2. Pieri and Peano 37 7.3. Hilbert 37 7.4. Tarski and his students 38 7.5. Tarski’stheoryofEuclideangeometry 40 7.6. Borsuk-Szmielew 40 7.7. Avigad, Dean, and Mumma 40 7.8. Heyting and other constructive approaches 41 8. The theory ECG of Euclidean Constructive Geometry 41 8.1. Logic of Partial Terms (LPT) 41 8.2. Replacing LPT and sorts with predicates if desired 43 8.3. Language 44 8.4. Intuitionistic logic and stability 45 8.5. Incidence and intersection axioms 46 8.6. Constructorandaccessoraxioms 47 8.7. Meaning of equality 48 8.8. Betweenness axioms 48 8.9. Sides of a line 51 8.10. Right and Left : handedness and sides of lines 52 8.11. Dimension axioms 54 8.12. Rays and segments 54 8.13. Congruence axioms 55 8.14. Line-circle continuity 56 8.15. Intersections of circles 58 9. Development of neutral geometry in ECG 61 9.1. Why we cannot import negative theorems from Tarski’s theory 62 9.2. Congruence and betweenness lemmas 64 9.3. Ordering of segments 68 9.4. Angles and triangles 70 9.5. Perpendicular lines 74 9.6. Existence of midpoints and perpendiculars 75 9.7. Right angles 76 9.8. Various forms of the Pasch axiom 82 9.9. Inner Pasch implies Outer Pasch and plane separation 84 9.10. The crossbar theorem 85 9.11. Ordering of angles 86 FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 3

9.12. Triangle construction and uniqueness 87 9.13. Euclid Books I, II, and III 92 9.14. Consequencesoftheupperdimensionaxiom 94 9.15. Right and left turns and sides of lines 99 9.16. Definability of IntersectCirclesSame and IntersectCirclesOpp 101 9.17. Rotation and Uniform Reflection 104 9.18. The other intersection point 110 9.19. Defining the order of points on a line 111 10. The parallel postulate 113 10.1. Alternate Interior Angles 113 10.2. Euclid’s parallel postulate 115 10.3. The strong parallel postulate 116 10.4. Playfair’s axiom 118 10.5. Another version of the strong parallel postulate 120 10.6. Coordinates 121 10.7. The strong parallel postulate and stability of IntersectLines 123 10.8. Stability of definedness and ∼= 125 10.9. ECG proves Euclid’s parallel postulate 129 10.10. Variousformsoftheparallelaxiom 131 11. Comparisons of ECG to other formalizations 136 11.1. Comparison of ECG to Hilbert-style theories 136 11.2. Comparison of ECG to Tarski’s theories 136 12. Connections between geometry and Euclidean fields 138 12.1. Signed addition 139 12.2. Signed multiplication 144 12.3. The distributive law 149 12.4. Reciprocals 150 12.5. Interpreting ECG in field theory 151 12.6. Area in Euclid, Hilbert, and ECG 174 12.7. Handedness, cross product, and linear transformations 181 12.8. Interpreting field theory in ECG 192 12.9. Faithfulness of the interpretations 201 13. Classical geometry and constructive geometry compared 219 13.1. The double negation interpretation 219 13.2. The double-negation interpretation applied to ECG 221 14. The relation of ECG toTarski’stheoriesofgeometry 222 14.1. Interpretation of Tarksi’s theory in ECG 225 15. Interpretation of ECG in Hilbert’s and Tarski’s theories 227 15.1. Right and left handedness 227 15.2. Apartness 228 16. Metatheorems 230 16.1. Things proved to exist in ECG canbeconstructed 230 16.2. Disjunction properties 232 17. Independence results for the parallel axioms 233 17.1. Kripke models of ring theory 233 17.2. Euclid 5 does not imply Axiom 58 235 17.3. Playfair does not imply Euclid 5 239 4 MICHAEL BEESON

17.4. Independence of Markov’s principle 240 18. Apartness 240 18.1. Constructions and Apartness 240 18.2. Euclidean fields and apartness 240 18.3. Apartness and the parallel axioms 240 19. Appendix A: Roads not taken 242 19.1. Rigid compass undefinable from the collapsible compass 242 19.2. Rigid compass not definable, even with degenerate circles 244 19.3. Circles of zero radius, the rigid compass, and projection 246 19.4. Constructing the center of a circle 247 20. Appendix B: Possible reductions in the primitive construction mechanisms 248 20.1. TheMohr-Mascheronitheorem 248 20.2. Strommer’s theorem 248 FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 5

§1. Introduction. 1.1. Euclid. Euclid’s geometry, written down about 300 BCE, has been ex- traordinarily influential in the development of mathematics, and prior to the twentieth century was regarded as a paradigmatic example of pure reasoning.1 During those 2300 years, most people thought that Euclid’s theory was about something. What was it about? Some may have answered that it was about points, lines, and planes, and their relationships. Others may have said that it was about methods for constructing points, lines, and planes with certain specified relationships to given points, lines, and planes, for example, constructing an equilateral triangle with a given side. In these two answers, we see the viewpoints of pure (classical) mathematics and of algorithmic mathematics represented. Near the end of the nineteenth century, the Italian “Peano school”, and especially Mario Pieri, began to use the “hypothetical-deductive method”, today called the “axiomatic method.” At the same time, and apparently without communication, Hilbert took a similar approach, and published a famous book [15] in the last year of the century. According to the axiomatic method, Euclid’s theories were not about anything at all. Instead of “points, lines, and planes”, one should be able to read “tables, chairs, and beer mugs.” All the reasoning should still be valid. The names of the “entities” were just place holders. That was the viewpoint of twentieth-century axiomatics. In this paper, we re-examine Euclidean geometry from the viewpoint of constructive mathematics. The phrase “constructive geometry” suggests, on the one hand, that “constructive” refers to geometrical constructions with straightedge and compass. On the other hand, the word “constructive” may suggest the use of intuitionistic logic. We investigate the connections between these two meanings of the word. Our method is to keep the focus on the body of mathematics in Euclid’s Elements, and to examine what in Euclid is constructive, in the sense of “constructive mathematics”. Our aim in the first phase of this research was to formulate a suitable formal theory that would be faithful to both the ideas of Euclid and the constructive approach of Errett Bishop. We did achieve this aim, and the resulting theory ECG of “Euclidean constructive geometry” is presented in this paper. In constructive mathematics, if one proves something exists, one has to show how to construct it. In Euclid’s geometry, the means of construction are not arbitrary computer programs, but ruler and compass. Therefore it is natural to look for quantifier-free axioms, with function symbols for the basic ruler-and- compass constructions. The terms of such a theory correspond to ruler-and- compass constructions. In number theory, if one proves an existence theorem, then for a constructive version, one has to show how to compute the desired number as a function of the parameters. In analysis, if one proves an existence theorem, one has to be able to compute approximations to the desired number from approximations to

1Readers interested in the historical context of Euclid are recommended to read [6], where Max Dehn puts forward the hypothesis that Euclid’s rigor was a reaction to the first “foundational crisis”, the Pythagorean discovery of the irrationality of √2. 6 MICHAEL BEESON the parameters. In particular, the solution will depend continuously on parameters, at least locally. This feature of constructive analysis depends, in a way, on what we think it means “to be given” a number x. Whatever that may mean, it surely means that we have a way to get a rational approximation to x within any specified limit of accuracy.2 Geometry is more like analysis than number theory, in the sense that we do not want to assume in advance that points can be given to us all at once in a completely determined location; points are given only approximately, by dots on paper or a computer screen, or in Euclid’s case, by indentations in sand (the Greeks drew their diagrams in sand). It might be doubtful whether two such points coincide; in such a case one would have to ask the one who made the diagram to refine it. It follows that in constructive geometry, we should have local continuous dependence of constructions on parameters. We can see that dramatically in computer animations of Euclidean constructions, in which one can select some of the original points and drag them, and the entire construction “follows along.” Thus our theory should (i) be quantifier free, so as to allow terms for constructions, and (ii) allow only continuous constructions. There is a third consideration: the axiomatization should be disjunction-free. There are two reasons for this: First, disjunction does not occur in Euclid. Euclid did not use first- order logic; the logical structure of his theorems will be discussed below, but for now we just note that he never states or uses a disjunctive proposition. Second, disjunctive axioms would make trouble for our desired applications of Gentzen’s cut-elimination theorem. We want to have a theory in which things proved to exist can be constructed by ruler and compass, and that will be straightforward with a quantifier-free, disjunction-free axiomatization. That observation is the key: we need a theory that is faithful to Euclid’s Elements. • has terms for geometrical constructions, and those terms describe construc- • tions continuous in parameters. is quantifier-free. • is disjunction free. • We present here a theory ECG that has these properties. A version of this theory was presented in [3]. The version presented is slightly different, but is equivalent.3 Regarding the “faithfulness” to Euclid: we discovered only one serious non-constructivity in Euclid’s Elements, namely, Book I Prop. 2, in which Euclid shows that a collapsible compass can be simulated by a rigid compass. This issue is discussed in detail below; here we say only that the remedy is to adopt a rigid compass as primitive; essentially, we take Euclid I.2 as an axiom

2There are philosophical issues here: can we “be given” a number by using a random process to generate approximations? Or must the approximations be generated by a computer program? It doesn’t matter for the conclusion that functions are continuous, because it is a theorem that even if we assume numbers are generated only by programs, computable functions are still continuous. This theorem (of Kreisel-Lacombe-Shoenfield-Tseitin) is discussed in [2], p. 61. 3It is not completely obvious what the axioms of such a theory should be; in Appendix A and Appendix B we consider alternate choices and present results justifying the choices we made. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 7 instead of as a theorem.4 Although it was a shock to find that the second proposition in Euclid is not constructive, after that there are no more shocks. It was also necessary to make a slight strengthening of Euclid’s parallel postulate, but the necessary revision is relatively minor, and these are the only two points in which ECG varies from Euclid. The second phase of this research is to study the metamathematical properties of ECG and related theories. First we have the result alluded to above, that existential theorems are “realized” by ruler-and-compass constructions. Next there is Tarski’s “representation problem”. We want to characterize the models of ECG. Classically, the models of Euclidean geometry are well-known to be planes over Euclidean fields (ordered fields in which nonnegative elements have square roots). But constructively, what is a Euclidean field? It turns out that there are three natural ways of formulating that, and that planes over such fields correspond to three different versions of the parallel postulate. The proof of such a representation theorem (classically or constructively) involves both the “arithmetization of geometry” and the “geometrization of arithmetic.” The geometrization of arithmetic (originally due to Descartes) shows how to add, subtract, multiply, divide, and take square roots of segments, and how to introduce coordinates in an arbitrary plane. The arithmetization of geometry (also due to Descartes) is using coordinates and algebra to verify that the axioms of geometry hold in F 2, where F is a Euclidean field. Both of these offer some constructive difficulties; for example, Descartes’s constructions apply only to positive segments, but we need to define arithmetic on signed numbers, without making case distinctions whether the numbers are zero or not. These difficulties are overcome and the representation theorem proved in this paper. The third phase of this research is to investigate the provability or independence of other geometrical theorems, relative to some constructive base theory. Our most interesting result is to settle the relations between the three versions of the parallel postulate. As it turns out, no two are equivalent. We also obtain a few other independence results. The readers of this paper may include logicians (who may or may not have studied axiomatic geometry or intuitionistic logic); geometers (who may or may not have studied logic or proof theory or constructive mathematics); and constructive mathematicians (who may or may not have studied axiomatic geometry and may or may not have studied proof theory). The paper is written to be accessible to this entire audience, so experts may find they can skip certain sections. 1.2. The collapsible vs. the rigid compass. A “rigid compass” can be used to transfer a given distance, say BC, from one location in the plane to another, thus constructing at any point A another point D such that AD = BC. A “collapsible compass”, on the other hand, can only be used to draw a circle with a given center A passing through another given point B. The compass “collapses” as soon as you pick it up. There is a word in Dutch, passer, for a

4Euclid I.2 with the assumption that all the points in the diagram are distinct is still a theorem. Maybe that was what Euclid intended; but we need to take as an axiom the version without that assumption. 8 MICHAEL BEESON rigid compass, which was used in navigation in the seventeenth century. But there seems to be no single word in English that distinguishes either of the two types of compass from the other. Euclid’s compass is collapsible; but a rigid compass is important in geometry, so right at the outset, in Book I, Proposition 2, Euclid attempted to show that a rigid compass could be simulated by a collapsible compass. This famous construction shows how to use the collapsible compass to construct a point D = e(A, B, C) such that whenever B = C and A = B then AD = BC. The reader is advised to take out his or her6 Euclid and review6 the simple, but non- trivial and beautiful, construction.5 Our interest in this subject began with a computer animation of Euclid’s constructions that permits the user to drag the starting points, and see how the construction depends on the changed starting points. The results for Euclid I.2 were surprising and interesting. The first problem with this construction is that it does not work when A = C. Of course, in that case we can just take D = B; but that case distinction requires classical logic. And the computer animation reveals that when we drag point C close to A, and then around A in a small circle, then the constructed point D moves around A in a circle of radius close to BC. Hence D does not depend continuously on C. This discontinuity, together with the need for a case distinction just mentioned, makes it clear that from the intuitionistic point of view, there are two different versions of Euclid I.2: the version in which (in addition to B = C) we take A = C as a hypothesis, or the “uniform” version in which we do not6 assume A = C6 , but assert that D can be constructed, whatever the values of A, B, and C, as6 long as B = C. The uniform version corresponds to a rigid compass (with which we can cop6y the segment BC so it has one end at A). From the constructive point of view, what Euclid I.2 shows is that a rigid compass can be simulated by a collapsing compass plus a “decision function”, that enables us to decide wither two given points are equal or not. The latter is not a constructive principle, in the sense of intuitionistic logic. It is also not a constructive principle, in the sense of ruler and compass constructions, since, given two fixed points A and B (which are and remain unequal), and two more points P and Q (which may or may not be equal), we do not know how to construct a point C which is equal to P if A = B and to Q if A = B. Moreover, since such a construction could not be continuous in P and Q, it6 is not a matter of ignorance: no such construction exists. The uniform version of Euclid I.2 seems to be important in developing geometry; we use it in constructing the projection of a point on a line, which is in turn needed for defining addition and multiplication of segments (the geometrization of arithmetic). Therefore, from the constructive point of view, it is a defect that the rigid compass cannot be simulated by the collapsing compass. We repair this by requiring a rigid compass in ECG. That amounts to including Euclid I.2 as a new axiom.

5Should any reader not possess a copy of Euclid, we recommend [10] and [9]; or for those wishing a scholarly commentary as well, [8]. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 9

1.3. Postulates vs. axioms in Euclid. Euclid presents his readers with both “postulates” and “axioms”. Modern mathematicians often treat these words as synonyms. For Euclid and his contemporaries, however, they had quite different meanings. Here is the difference, as explained by Pambuccian [22], p. 130. For Proclus, who relates a view held by Geminus, a postulate prescribes that we construct or provide some simple or easily grasped object for the exhibition of a character, while an axiom asserts some inherent attribute that is known at once to one’s auditors. And just as a problem differs from a theorem, so a postulate differs from an axiom, even though both of them are undemonstrated; the one is assumed because it is easy to construct, the other accepted because it is easy to know. That is, postulates ask for the production, the poesis of something not yet given . . . , whereas axioms refer to the gnosis of a given, to insight into the validity of certain relationships that hold between given notions. 1.4. The parallel postulate. Euclid’s famous “parallel postulate” (Postu- late 5, henceforth referred to as Euclid 5) states that if two lines L and M are traversed by another line T , forming adjacent interior angles on one side of T adding up to less than two right angles, then L and M will intersect on that side of T . Stated this way, the postulate can be viewed as a construction method for producing certain triangles. In view of the remarks of Geminus and Proclus, it seems likely that Euclid viewd his postulate in this way, or he would have called it an axiom. In 1795, Playfair introduced the version that is usually used today, which is an axiom rather than a postulate: Given a line L, and a point P not on L, there exists exactly one parallel to L through P . (Parallel lines are by definition lines that do not meet.) Written this way, the parallel postulate does not directly assert the existence of any specific point. It is not immediately clear whether this version implies Euclid 5 using only intuitionistic (constructive) logic, although of course it does in classical logic. This question is resolved in this paper. The answer is in the negative: Playfair does not imply Euclid 5. A more subtle distinction arose in the formulation of ECG. There the following version of the parallel postulate is used: if K is a line parallel to line L through point P , and M is another line through P (different from K) then M meets P . This is Axiom 58 of ECG. We prove that Axiom 58 implies Euclid 5, and that the implication cannot be reversed (relative to the other axioms of ECG). The difference between Axiom 58 and Euclid 5, but is crucial for the ability of ECG to define (signed) arithmetic operations in a continuous way. We show below that the theory of Euclidean fields is related to geometry, just as in classical mathematics, by coordinatization. The difference between Axiom 58 and Euclid 5 corresponds, in the theory of ordered fields, to whether 1/x has to exist when x = 0, or only when x > 0. If it exists for x > 0, it exists for x < 0; so the question6 can also be put, whether 1/x has to exist when x = 0, or only when x > 0 x < 0, which is constructively not the same as x6 = 0. (Playfair’s axiom corresponds∨ to requiring that elements without reciprocals6 are zero.) In different sections below, we establish these results that characterize the models 10 MICHAEL BEESON of ECG, and show that the implications between versions of the parallel axioms (or Euclidean field axioms) cannot be reversed. 1.5. Polygons in Euclid. Euclid gives 48 two-dimensional constructions (we will ignore his three-dimensional constructions entirely). These mostly deal with configurations of a fixed number of points; for example, he constructs a regular pentagon, hexagon, and octagon, but does not address the general n-gon (which of course is now known to not be constructible with ruler and compass anyway). Some small fixed number of sides would suffice for the Euclidean constructions. In Euclid, no figure with more sides than an octagon is constructed, and no figure with more than four sides is an input to another construction, except for constructions that work on any “figure.” But, some of the later constructions do use the word “figure”, which apparently means something like what a modern mathematician would call a “closed polygon” (and has no connection to “figure” in the sense of “ilustrative drawing”.) The general concept of a closed polygon of any number of sides may be logically problematic as it drags the concept of integer into geometry. We regard such theorems of Euclid formally as theorem schemata, which become theorems for each particular number n of sides of the “figure.” In this we join everyone who previously considered first-order geometric theories.

§2. Euclid’s reasoning considered constructively. In the late twentieth century, contemporaneously with the flowering of computer science, there was a new surge of vigor in algorithmic, or constructive, mathematics, beginning with Bishop’s book [4]. In algorithmic mathematics, one tries to reduce every “existence theorem” to an assertion that a certain algorithm has a certain result. It was discovered by Brouwer that if one restricts the laws of logic suitably (to “intuitionistic” logic), then one only obtains algorithmic existence theorems, so there is a fundamental connection between methods of proof, and the existence of algorithms to construct the things that have been proved to exist. Brouwer thought it necessary to do more than just restrict logic; he also wanted to state some additional principles. Bishop renounced additional principles and worked by choosing his definitions very carefully, but using only a restricted form of logic. Results obtained this way are classically valid as well as constructively. What happens if we examine Euclid’s Elements from this point of view? Some- what surprisingly, we only had to make two modifications to Euclid, both of which we have already mentioned. Namely, we need to adopt the uniform version of Euclid I.2 as an axiom, since Euclid’s proof is valid only for the non-uniform version; and we had to strengthen Euclid 5 (the parallel postulate) as discussed above. In this section, we justify this conclusion in more detail. 2.1. Order on a line from the constructive viewpoint. In this section we explain how a constructivist views the relations x

This guarantees that the limit x of such a sequence satisﬁes x xn 1/n. Two | − |≤ such sequences are considered equal if xn yn 2/n; it is important that diﬀerent sequences can represent equal (or “the same”,− ≤ if you prefer) real numbers. Now observe that if x

Could we assert x y implies y < x? Subtracting y we arrive at an equivalent version of the question¬ ≤ with only one variable: can we assert x 0 implies 0 < x? Since x 0 is equivalent to 0 < x, the question is whether¬ ≤ we can assert ≤ ¬ x> 0 x> 0 (Markov’s principle) ¬¬ → In other words, is it legal to prove that a number is positive by contradiction? One could argue for this principle as follows: Suppose x > 0. Now compute ¬¬ the approximations xn one by one for n = 1, 2,... . Note that trichotomy does hold for xn and 1/n, both of which are rational. You must find an n such that xn > 1/n, since otherwise for all n, we have xn 1/n, which means x 0, contradicting x> 0. Well, this is a circular argument:≤ we have used Markov’s≤ principle in the¬¬ justification of Markov’s principle. Shall we settle it by looking at the recursive model? There it can easily be shown to boil down to this: if a Turing machine cannot fail to halt, then it halts. Again one sees no way to prove this, and some may feel is intuitively true, while others may not. Historically, the Russian constructivist school adopted Markov’s principle, and the Western constructivists did not. It reminds one of the split between the branches of the Catholic Church, which also took place along geographical lines. In any case, as we will see below, it seems appropriate to adopt this principle for a constructive treatment of Euclid. From the constructive viewpoint, the main difference between x

Euclidean proofs do little more than introduce objects satisfying lists of atomic (or negation atomic) assertions, and then draw further atomic (or negation atomic) conclusions from these, in a simple linear FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 15

fashion. There are two minor departures from this pattern. Some- times a Euclidean proof involves a case split; for example, if ab and cd are unequal segments, then one is longer than the other, and one can argue that a desired conclusion follows in either case. The other exception is that Euclid sometimes uses a reductio; for example, if the supposition that ab and cd are unequal yields a contradiction then one can conclude that ab and cd are equal. 2.3. Case splits and reductio inessential in Euclid. It is our purpose in this section to argue that Euclid’s reasoning can be supported in ECG, including the two types of apparently non-constructive reasoning just discussed. The reason for this is that the apparent non-constructivities come down to the two basic types mentioned just above, and these can be dealt with by assuming the “stability” of equality and betweenness. By the stability of equality, we mean x = y x = y. ¬¬ → This formula simply codifies the principle that it is legal to prove equality of two points by contradiction, and it is taken as an axiom in our theory. Similarly, if B(a,b,c) means that b is between a and c on a line, we take as an axiom the stability of betweenness: B(a,b,c) B(a,b,c). ¬¬ → While Euclid never explicitly mentions betweenness (which is, as is well-known, the main flaw in Euclid), the stability of betweenness and equality together account for all apparent instances of nonconstructive arguments in Euclid. A typical example of such an argument in Euclid is Prop. I.6, whose proof begins Let ABC be a triangle having the angle ABC equal to the angle ACB. I say that the side AB is also equal to the side AC. For, if AB is unequal to AC, one of them is greater. Let AB be greater, . . . To render this argument in first-order logic, we have to make sense of Euclid’s “common notion” of quantities being greater or less than other quantities. This is usually done formally using the betweenness relation. In the case at hand, we would lay off segment AB along ray AC, finding point D on ray AC with AD = AC. Then since AB is unequal to AC, D = C. Then “one of them is greater” becomes the disjunction B(A, D, C) B(A,C,D6 ). This disjunction is not constructively valid. But its double negation∨ is valid, since if the disjunction were false, we would have B(A, D, C) and B(A,C,D)), which would contradict D = C. The rest of¬ Euclid’s argument¬ shows that each of the disjuncts implies6 the desired conclusion. We therefore conclude that the double negation of the desired conclusion is valid. The conclusion of I.6, however, is negative (has no or ). Hence the double negation can be pushed inwards, and the stability of the∃ atomic∨ sentences applied to make the double negations disappear. In Euclid, disjunctions never appear, even implicitly, in the conclusions of propositions.6 The conclusions are simply conjunctions of literals. Sometimes

6An apparent exception to this rule is Prop. I.13, “If a straight line set up on a straight line make angles, it will make either two right angles or angles equal to two right angles.” Here the disjunction is superfluous: we can just say, “it will make angles equal to two right angles.” 16 MICHAEL BEESON there is an implicit existential quantifier, but if (as will always be the case in our theory) we can explicitly exhibit terms for the constructed points, the resulting explicit form of the proposition will be quantifier-free. Then the double negation can be pushed inwards as just illustrated. In this way, all arguments of the form beginning “For, if AB is unequal to AC, one of them is greater”, can be constructivized. Prop. I.26 gives an example of the use of the stability of equality: “...DE is not unequal to AB, and is therefore equal to it.” In the examples above, this principle is not really needed to reach Euclid’s desired conclusion. Since the conclusion concerns the equality of certain points, we can simply double- negate each step of the argument, and then add one application of the stability of equality at the end. In fact, this had to happen: we proved in [3], and reprove them in this paper, explaining why the principle in question, and indeed, any uses of classical logic whatsoever, are in principle eliminable from proofs of theorems of the form found in Euclid. We note that if the conclusion of the proposition does not itself involve Euclid’s “greater than” or “less than”, then only the stability of equality is needed at the end. Whether or not betweenness actually is needed in the conclusions (as opposed to the hypotheses) in Euclid, it certainly could occur in “the spirit of Euclid”, so we include the stability of betweenness in our theory.

§3. Constructions in geometry. 3.1. The 48 Euclidean constructions. We are dealing only with the two- dimensional part of Euclid, although in some sense the culmination of Euclid is the construction of the five Platonic solids. Thus we focus on Books I-VI. In those four books one finds a total of 172 propositions. Of those, 48 make a claim that something can be constructed. We do not take the space to list the propositions, but with your Euclid at hand, it should suffice to list the proposition numbers: Book I, propositions 1,2,3,9,10,11,12,22,23,31,42,44,45,46; Book II, propositions 11, 14; Book III, propositions 1,2,17,25,30,33,34; Book IV, all 16 propositions; Book VI, 9,10,11,12,18,25,28,29,30 These constructions are, so to speak, the “data” that our theory is to “explain.” 3.2. The elementary constructions. The Euclidean constructions are carried out by constructing lines and circles and marking certain intersection points as newly constructed points. Our aim is to give an account of this process with modern precision. We use a system of terms to denote the geometrical constructions. These terms can sometimes be “undefined”, e.g. if two lines are parallel, their intersection point is undefined. A model of such a theory can be regarded as a many-sorted algebra with partial functions representing the basic geometric constructions. Specifically, the sorts are Point, Line, and Circle. We have constants and variables of each sort. It is possible to define extensions of this theory with definitions of Arc and Segment. These are “conservative extensions”, which means that no additional theorems about lines and points are proved by reasoning about arcs and segments. Angles are treated as triples of points. While this conservative extension FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 17 result applies to logical theories, a similar result applies to constructions considered algebraically. If we replace rays and segments by lines, and angles by pairs of lines, then some terms may become defined that were not defined before, as lines may intersect where rays or segments did not, etc. But any point that was constructible with rays and angles will still be constructible when rays and angles are replaced by lines. Therefore it suffices to restrict attention to points, lines, and circles, which we do from now on.7 This collection of data types is adequate to cover the return types and argument types of the 48 plane Euclidean constructions (given the convention about polygons or “figures” discussed above). Lines are constructed by drawing a line through two distinct points; the resulting line is Line (A, B). Circles are constructed “by center and radius”; Circle 3(A, B, C) is the circle with center A and radius BC. This term corresponds to a “rigid compass”. Its necessity for constructive geometry is already suggested by the analysis of Euclid’s proof of Prop. I.2 above; and in Appendix A we give a proof that not only does Euclid’s simulation of the rigid compass by a collapsing compass fail to be everywhere defined, but any simulation will have that problem: the rigid compass cannot be defined from the collapsing compass. Therefore the construction of circles from center and radius must be taken as primitive. We allow circles of radius zero, which we call “degenerate circles”, i.e. we consider Circle 3(A, B, C) to be defined when B = C. Reasons for this are also given in Appendix A. Starting with at least three noncollinear points, we can draw lines and circles using these constructors, and then construct more points using the following six “elementary constructions”, each of which has return type Point: IntersectLines (Line K, Line L) IntersectLineCircle1 (Line L, Circle C) IntersectLineCircle2 (Line L, Circle C) IntersectCircles1 (Circle C, Circle K) IntersectCircles2 (Circle C, Circle K) IntersectCircles1 and IntersectCircles2 construct the intersection points of two circles; how the two points are distinguished will be explained below. Euclid does not use function symbols for the intersection points of two circles, and in particular did not worry about how to distinguish one from the other, although sometimes (as in Euclid I.9, the angle bisector theorem8 ) his proofs need to be repaired by selecting “the intersection point on the opposite side of the line joining the centers as the given point x”, or “on the same side.” To directly support this kind of argument we might consider adding function symbols IntersectCirclesSame (Circle C, Circle K, Point p) IntersectCirclesOpp (Circle C, Circle K, Point p)

7Of course, diagrams will look better with rays and segments, so in a computer program to draw diagrams, we would want to (and did) use them; but this is a theoretical paper. 8See Heath’s commentary on I.9 in [8]. 18 MICHAEL BEESON

These two constructions give the intersection point of C and K that is on the same side (or the opposite side) of the line joining their centers as p, when p is not on that line. It turns out that constructions meeting this specification can be defined, and hence it is not necessary to include them as primitive symbols. However, the constructions that define them are somewhat elaborate, and in the interest of having a system “faithful to Euclid”, in the sense that it is easy to translate Euclid directly into the formal system, we include symbols for these constructions. They will be shown redundant in Theorem 9.96. Another way in which Euclid avoids the need for function symbols to identify the intersection points is the phrase “the other intersection point.” We will show in Theorem 9.109 that there is a construction to get “the other intersection point”, given two circles and one intersection point. In this picture of constructions, every line comes equipped with the two points that were used to construct it. In hypothetical constructions, however, we may be “given” some lines, and we need to have “accessor functions” to recover points on those lines. Thus, pointOn1 (L) and pointOn2 (L) construct two points on a line; those will be a and b where L = Line (a,b). Similarly, since circles can only be constructed from a given center, every circle comes equipped with its center, so we should have an “accessor function” center(C) for the center of C, and pointOnCircle (C), to enable us to “construct” a point on a given circle.9 Think, for example, of always choosing the southern extremity of the circle. Note that lines and circles are treated differently here: pointOnCircle (C) will be the same for any two circles with the same points, but pointOn1 (L) will depend on the points from which L was constructed. This will be discussed more extensively below. There are three issues to decide: when there are two intersection points, which one is denoted by which term? • In degenerate situations, such as Line (P, P ), or • IntersectCirclesSame (C,K,p) when p is on the line joining the centers, what do we do? When the indicated lines and/or circles do not intersect, what do we do • about the term(s) for their intersection point(s)? We will answer these questions as follows: When the indicated lines and/or circles do not intersect, the term for their intersection points will be “undefined”. What this means for the logic will be spelled out in due course. In other words, the operations of these “algebras” do not have to be defined on all values of their arguments. The same issue, of course, arises in many other algebraic contexts, for example, division is not defined when the denominator is zero, and √x is not defined (when doing real arithmetic) when x is negative. In degenerate situations such as Line (P, P ), we will be guided by continuity. Thus Line (P, P ) will be undefined, since it cannot be defined in such a way as to make Line (P,Q) continuous as P approaches Q. Similarly, IntersectCirclesSame (C,K,p) and IntersectCirclesOpp (C,K,p) will be undefined when p is on the line joining the centers of C and K. On the other

9See Appendix A for a discussion of issues raised by Euclid III.1 in this connection. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 19 hand, degenerate circles such as Circle (A, A) are convenient, and do no harm to continuity, so we allow them. (That decision is my “final answer” after much vacillation; some reasons are spelled out in Appendix A.) The question about intersection points is more complex. We note that lines are treated intensionally in our theory; that is, Line (A, B) is not equal to Line (B, A), even though the two lines are extensionally equal, i.e. the same points lie on them. That is, a line “comes equipped” with two points that were used to construct it; to “be given” a line involves being given two points on that line, and order matters. It thus makes sense to define the two intersection points of Line (A, B) and a circle to be such that their ordering on the line is the same as that of A and B. While this definition works semantically in the standard plane, its axiomatic development is also possible, as we shall see. Our original development in [3] handled circle-circle intersections using only the two operations that take only the two circles as arguments; but the other two operations, that takes an extra argument of type “point” to specify that the intersection point should be on the same side of the center-to-center line as the specified point, seems more faithful to Euclid. We wish to consider all four operations, if only to clarify their relations. (It turns out that the second two can be defined, so from the point of view of minimizing the number of symbols in the language, they could be eliminated.) With the first two operations, we are faced with the problem of distinguishing the two intersection points, when given only the circles. Unlike lines, circles do not “come with” a built-in direction. Consider intersecting circles with centers at A and B. Let the two intersection points be C and D. Then we note that one of ABC is a “right turn” and the other is a “left turn”, except in the degenerate case A = B (when the terms given by IntersectCircles1 and IntersectCircles2 will be undefined) and in the case when the circles are tangent, and in that case there is no need to distinguish the (identical) intersection points. This observation enables us to define the semantics of IntersectCircles1 and IntersectCircles2 , but leaves us with the problem of giving definitions of the notions Left (A, B, C) and Right (A, B, C), whose intended interpretations are that ABC is a left-hand turn or a right-hand turn, respectively. In [3], we gave a complicated solution of this problem, involving many axioms. In this paper, we give a simpler axiomatization. 3.3. Circles and lines eliminable when contructing points. It is intuitively plausible that one does not really need to draw the lines and circles in a geometric diagram; one just uses the circles and lines to get their points of intersection with other circles and lines; sometimes Each of these has several “overloaded” variants, which can be defined from these using constructors and accessors. For example,

IntersectLines (Point A, Point B, Point C, Point D) = IntersectLines (Line (A, B), Line (C,D)) IntersectLineCircle1 (Point A, Point B, Point C, Point D) = IntersectLineCircle1 (Line (A, B), Circle (C,D)) IntersectLineCircle1 (Point A, Point B, Circle C) = IntersectLineCircle1 (Line (A, B), C) 20 MICHAEL BEESON

As these three examples illustrate, One can regard circles and lines as mere intermediaries; points are ultimately constructed from other points. In a lemma at the end of this section, we state and prove this principle precisely. On the other hand, the overloaded versions do not enable one to construct any more points, so for simplicity they are omitted from our axiomatic theories. There is a second constructor for circles, which we can describe for short as “circle from center and radius”, as opposed to the first constructor above, “circle from center and point.” Specifically Circle3 (A, B, C) constructs a circle of radius BC and center A, provided B = C. These two constructors for circles correspond to a “collapsible compass” and6 a “rigid compass” respectively. The compass of Euclid was a collapsible compass: you cannot use it to “hold” the length BC while you move one point of the compass to A. You can only use it to hold the radius AB while one point of the compass is fixed at A, so in that sense it corresponds to Circle (A, B). The second constructor Circle3 corresponds to a rigid compass. The theory ECG includes Circle3 . That is because (as we show in detail in Appendix A and in [3], without a rigid compass, one cannot project a point P onto a line L, without making a case distinction between the case when P lies on L and the case when it does not; and the ability to make such projections is crucial to defining a coordinate system and showing how to perform addition and multiplication on segments. We note in the following lemma that, as far as constructing points goes, the other types are mere conveniences. The elementary constructions can be expressed, as we have noted, in several ways using variables of different types. For example, we could have IntersectLines (K,L) where K and L have Line, or IntersectLines (A,B,C,D), where A, B, C, and D have type Point, and IntersectLines (Line(A, B),Line(C,D)) = IntersectLines (A,B,C,D). Lemma 3.1. Let t be a term of type Point, whose variables are all of type Point. Then there is a term t∗ with the same variables as t, also of type Point, such that in the standard plane t and t∗ determine the same function, and t∗ contains only function symbols of type Point having Point arguments. Proof. By induction on the complexity of t. Suppose that t has the form IntersectLines(r, s) where r and s are terms of type Line . None of the elementary constructions has type Line, and r and s cannot be variables (since all variables in t are of type Point), so r must have the form Line (p, q) for some terms p and q, and s must have the form Line (u, v) for some terms u and v. Then t∗ can be taken to be IntersectLines (p∗, q∗,u∗, v∗). The other elementary constructions are treated similarly. The basis case, when t is a variable or constant, is treated by taking t∗ = t. 3.4. Describing constructions by terms. It is customary to describe constructions by a sequence of elementary construction steps. In this section, for the benefit of readers not expert in logic, we show how terms in a logical system correspond to traditional descriptions. For example, we might describe bisecting a segment this way: Midpoint(A,B) { C = Circle(A,B); // center at A, passing through B FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 21

K = Circle(B,A); P = IntersectCircles1(C,K); Q = IntersectCircles2(C,K); L = Line(P,Q); // the perpendicular bisector of AB J = Line(A,B); E = IntersectLines(L,J); // the desired midpoint return E; } This description (which actually is in a precisely-deﬁned language used for computer animation of constructions) looks quite similar to textbook descriptions of constructions. We call such a description a “construction script” or just a “script.” For theoretical purposes, as in this paper, it is easier to describe constructions oﬃcially by terms as above. When the midpoint construction is described as a term, it looks like this:

IntersectLines (Line (IntersectCircles1 (Circle (A, B), Circle (B, A)), IntersectCircles2 (Circle (A, B), Circle (B, A))), Line (A, B)) Conversely, given a term, one can “unpack” it, introducing new variable names for subterms, and even, if desired, collapsing duplicate subterms (e.g., Circle (A, B) need not be drawn twice). Since we have chosen readable names for our function symbols, such as IntersectLines instead of f, our terms sometimes grow typographically cumbersome, and for human readability we will often use scripts to describe constructions; but we do not deﬁne scripts formally; they always just stand for terms.

§4. Models of the elementary constructions. There are several interesting models, of which we now mention four. To define these models, we assume there are three constants α, β, and γ of type Point. 4.1. The standard plane. The standard plane is R2, with the usual interpretation of points, lines, and planes, and three arbitrary points picked for α, β, and γ. The standard plane is R2. Points, lines, and circles (as well as segments, arcs, triangles, squares, etc., in extended algebras in which such objects are considered) are interpreted as the objects that usually bear those names in the Euclidean plane. More formally, the interpretation of the type symbol “Point ” is the set of points, the interpretation of “Line ” is the set of lines, etc. In particular we must choose three specific non-collinear points to serve as the interpretations of α, β, and γ. Let us choose α = (0, 1), β = (1, 0), and γ = (0, 0). The constructor and accessor functions listed above also have standard and obvious interpretations. It is when we come to the five operations for intersecting lines and circles that we must be more specific. As discussed above, the interpretations of the five function symbols for the elementary constructions, such as IntersectLineCircle1 , are partial functions, so that if the required intersection points do not exist, the term simply has no interpretation. In degenerate situations, terms are defined if and only if they can be defined so as to make the 22 MICHAEL BEESON interpreting function continuous; thus Line (P, P ) is undefined and Circle (P, P ) is the zero-radius circle, and the only point on it is P .10 To distinguish the intersection points of two circles: IntersectCircles1 (C,K) is the intersection point P such that the angle from center (C) to center (K) to P makes a “left turn”. This is defined as in computer graphics, using the sign of the cross product. Specifically, let A = center (C) and B = center (K). Then the sign of (A B) (P B) determines whether angle ABP is a “left turn” or a “right turn”.− Thus× αβγ−is a left turn and γβα is a right turn. In case the two intersection points are different, one of these cases must apply. This explanation has used a case distinction as well as the cross product; later we will show how to define “left turn” and “right turn” using only the axioms of Euclidean geometry and intuitionistic logic. For now we simply note that this notion is constructively appealing, because of continuity: there exists a unique continuous function of C and K that satisfies the stated handedness condition for IntersectCircles1 when C and K have two distinct intersection points, and is defined whenever C and K intersect (at all). The principle of continuity leads us to make IntersectCircles1 (C,K) and IntersectCircles2 (C,K) undefined in the “degenerate situation” when circles C and K coincide, i.e. have the same center and radius. Otherwise, as the center of C passes through the center of K, there is a discontinuity. It makes sense, anyway, to have them undefined when C andK coincide, as the usual formulas for computing them get zero denominators, and there is no natural way to select two of the infinitely many intersection points. IntersectCirclesSame (C,K,p) and IntersectCirclesOpp are undefined when p is on the line joining the centers of the circles; and we note that each of the two is continuous on its domain, but cannot be extended continuously to the case when p is on that line, because when p crosses the line joining the centers, the intersection point jumps to the “other” intersection point. 2 4.2. The recursive model. The recursive plane Rrec consists of points in the plane whose coordinates are given by “recursive reals”, which were defined above. It is a routine exercise to show that the recursive points in the plane are closed under the Euclidean constructions. In particular, given approximations to two circles (or to a circle and a line), we can compute approximations to their “intersection points”, even though it may turn out that when we compute still better approximations to the circles, we see that they do not intersect at all. In the case of IntersectCirclesSame (C,K,p) or IntersectCirclesOpp (C,K,p), we compute rational approximations pn to the point p, and rational approximations to the centers of C and K, until we see that p cannot lie on the line joining the centers. Then we can compute the sign of the cross product required to see which side of the line p is on. Then we compute IntersectCircles1 (C,K) or IntersectCircles2 (C,K), as appropriate. We showed above that in the recursive plane, there is no computable test-for- equality function. Recall that the proof made use of a recursive real number E(x) such that E(x) > 0 if x (x) halts and E(x) = 0 if x (x) does not halt. { } { }

10We shall also consider (brieﬂy) the model R2− in which degenerate circles are not allowed, so Circle (A,A) is undeﬁned. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 23

The same example shows that we cannot test computably for incidence of a point on a line; the number E(x) lies on the y-axis if and only if Turing machine x does not halt at x. In the recursive model, we cannot decide in a finite number of steps whether a circle and a line intersect or whether two circles intersect; if they intersect transversally we will find that out after computing them to a sufficient accuracy, but if they are tangential, we won’t know that at any finite approximation. Technically, the circle of center (0, 1) and radius 1 E(x) will meet the x-axis if and only if the x-th Turing machine does not halt− at x. Readers familiar with recursion theory may realize that there are several ways to define computable functions of real numbers. The model we have just described is essentially the plane version of the “effective operations”. It is a well-known theorem of Tseitin, Kreisel, LaCombe, and Shoenfield, known traditionally as KLS (and easily adapted to the plane) that effective operations are continuous. Of course, in the case at hand we can check the continuity of the elementary constructions directly. 4.3. The algebraic model. The algebraic plane consists of points in the plane whose coordinates are algebraic. Since intersection points of circles and lines are given by solutions of algebraic equations, the algebraic plane is also closed under these constructions. Note that the elements of this plane are “given” all at once. We assume algebraic numbers are given by means of a rational interval (a,b) and a square-free polynomial f Q[x] such that f has only one root in (a,b). ∈ In the algebraic plane, there is a computable test-for-equality function D. To determine if (a,b) and f determine the same or a different real number than (p, q) and g, first check if the two rational intervals overlap. If not, the two reals are different. If so, let (r, s) be the intersection. Now we have to determine if f and g have a common zero on (r, s). There is a simple recursive algorithm to do that: Say g has degree greater than or equal to that of f. Then write f = gh + r with r of lower degree than g. Then f and g have a common zero on (r, s) if and only if f and r have a common zero. Recurse until both polynomials are linear, when the decision is very easy to make. Similarly, we can compute whether two circles or a circle and a line intersect. Since algebraic numbers can be computed, the algebraic model is isomorphic to a submodel of the recursive plane. 4.4. The Tarski model. The Tarksi model is K K, where K = Q(√ ) is the least subfield of the reals containing the rationals× and closed under taking the square root of positive elements. This is a submodel of the algebraic plane. Its points are the points constructible with ruler and compass. Here α, β, and γ are interpreted as three fixed rational points. That the intersection points of circles can be computed using only the solution of quadratic equations is checked in detail in section 6.2.

§5. The geometrization of arithmetic without case distinctions. To- day we usually think of analytic geometry as coordinatizing a plane and translating geometrical relations between points and lines into algebraic equations and 24 MICHAEL BEESON inequalities. But the converse is also possible: translating algebra into geometry, and this is important for lower estimates on the power of geometric constructions, for example for showing that the models of the geometry of constructions are planes over Euclidean fields. In modern books (such as [5]) arithmetic is geometrized as operations on congruence classes of segments. We operate instead on points on some fixed line X = Line (0, 1), where 0 and 1 are two arbitrarily fixed points. As far as I can tell, past work on coordinatization has always assumed some discontinuous constructions, such as test-for-equality or at least apartness. Since coordinatization itself is patently computable and continuous, it is unesthetic to appeal to discontinuous and non-computable “constructions” to achieve coordinatization and arithmetization. Although coordinatization is standard, old, and not complicated, we need to check that it can in fact be done from the specified primitives, without using apartness or test-for-equality, by definitions that apply without (for example) case distinctions as to whether numbers being multiplied are equal to 0 or 1 or not. It is not a priori clear that this can be done, and it is definitely not old and standard. In this section, we give the definitions of constructions that serve to implement coordinatization and the arithmetic operations in a continuous way. We present this mathematics informally, in the same spirit that Descartes presented his original coordinatization. In the process we will isolate the geometric theorems that are needed, so that one can later show that the work of this section can be formalized in a specific geometric theory. Chief among those principles is the existence of a “uniform perpendicular” construction, i.e. a construction that produces from a point p and a line L, a line K perpendicular to L that contains p, without a case distinction whether p is or is not on L. Second among those principles is the notion of “right turn” as applied to a triple of distinct points abc (and of course “left turn” is similar). It follows from the informal discussion in this section, that in the particular models of geometry discussed above, signed addition, multiplication, and division are defined and behave as desired. In a later section, we will give an axiomatic theory capable of formalizing these correctness proofs without reference to models. Recall that the concepts of “right turn” and “left turn” have been introduced and discussed in Section 8.10 For example, in Fig. 2, AOZ is a “right turn”, because the sign of the cross product OA OB is positive. Intuitively, traveling from A to O to Z requires one to turn right× at O. This definition of “right turn” is adequate for this section, since we are only concerned with models where cross product makes sense. (Later, we will define “right” and “left” in an axiomatic context.) 5.1. Uniform perpendicular and projection. In this section we show that some fundamental constructions needed to define (signed) arithmetic can be defined from the elementary constructions. A very fundamental construction in constructive geometry is the uniform perpendicular Perp(x, L), which constructs a line perpendicular to L passing through FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 25 x, without a case distinction as to whether x lies on L or not. In classical treatments of geometry, this case distinction is made and a different construction is used for each case. Definition 5.1. The term Perp(x, L) is given by the following construction script (See Fig. 1.) Perp(Point x, Line L) { a = pointOn1(L) b = pointOn2(L) Q = Circle3(b,x,a) c = IntersectLineCircleTwo(L,Q) C = Circle3(x,a,c) p = IntersectLineCircle1(L,C) q = IntersectLineCircle2(L,C) K = Circle(p,q) R = Circle(q,p) d = IntersectCircles1(K,R) e = IntersectCircles2(K,R) return Line(d,e) } Later on, when we have given some axioms and a formal theory, we will prove the correctness of this script. For now we just give an informal argument. The point is that we need to construct two points p and q on line L such that the perpendicular bisector of pq will be the desired line; thus we must have p = q even if x is on L, and whether or not x = a or x = b. Well, we can get such6 points p and q if we can draw a circle with center x, whose radius r is “large enough.” It will suffice if we take r to be the length of x plus the length of ab. Since a = b, that will be a nonzero radius, and since it is more than ax, the circle with6 that radius will meet L twice, in points p and q. Then we just bisect segment pq by an ordinary method, drawing circles with centers p and q that pass through each other’s centers. The formal proof of the correctness of this script can be found below in Lemma 9.56. The construction Project(P,L) takes a point P and line L and produces a point Q on L such that P lies on the perpendicular to L at Q. The well-known Euclidean construction for the projection applies only if P is known not to be on L. To define Project using that construction, we would require a test-for- incidence that allows us to test whether point P is on line L or not. But no such test-for-incidence construction is computable over the computable plane, so the Euclidean projection construction does not lead in any obvious way to a definition of Project. (That does not, however, constitute a proof that Project is not definable in terms of the elementary constructions; below we give such a proof.) Lemma 5.2. Strong extension and projection are definable from Circle3 (with circles of zero radius allowed). Also one can, using those primitives, construct the perpendicular to line L passing through point P , without conditions as to whether P is or is not on L. That perpendicular meets L in a point Project(P,L), the 26 MICHAEL BEESON

Figure 1. M = Perp(x, L)) is constructed perpendicular to L without a case distinction whether x is on L or not. Note bc = xa so the radius ac of C is long enough to meet L twice. K R

b e

C Q x b

b b b b b L p a b c q

b d projection of P on L, where Project is a certain explicitly given term for a ruler- and-compass construction, and the perpendicular itself is given by another such term.

The following gives an explicit construction for extending a non-null segment on a given end by a (possibly null) segment:

Definition 5.3.

Extend (A,B,C,D)= IntersectLineCircle2 (Line (A, B), Circle3 (B,C,D))

Proof. To construct the projection of point P on line L, we just need some circle with center P that intersects L in two distinct points Q and R. Then the projection of P on L is the midpoint of segment QR. If L is Line (A, B), then a suitable radius would be the sum of the lengths of AB and P A. Thus the circle we need can be constructed as Circle3 (P, A, Extend (A,B,P,A)). Note FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 27 that circles of zero radius are crucial here, so that we can handle the case P = A. That completes the deﬁnition of projection. Now, to construct the perpendicular to L at P , we simply erect the perpendicular to L at the projection of P on L, using the usual Euclidean construction. That completes the proof of the lemma.

Lemma 5.4. We can deﬁne a construction Para such that, for any line L and any point P (which may or may not be on L), Para(P,L) passes through P , and if P is not on L then Para(P,L) is parallel to L, while if P is on L, then Para(P,L) has the same points as L.

Proof. The deﬁnition of Para is

Para(P,L)= Perp(P, Perp(P,L)).

In words: First find the perpendicular to L passing through P . Then erect the perpendicular to that line at P . It is a theorem of neutral geometry (“neutral” means no parallel postulate is used) that two lines with a common perpendicular are parallel. Since Perp is everywhere defined, regardless of whether P is or is not on L, the same is true of Para. That completes the proof of the lemma. 5.2. Rotation. We next define a construction Rotate , which requires as in- puts three distinct, non-collinear points P , O, and Q (think of angle POQ), as well as a point A on Line (O, P ). The desired result of Rotate (P,O,Q,A) is a point Z on Line (O,Q) such that OZ = OA and if A = O then AOZ is a right turn. (See Fig. 2, but note that A might also be on the6 other side of O.) The

Figure 2. Z = Rotate (P,O,Q,A)

L Z

O A P point is that Z is deﬁned even when A = O (in which case it is just O, of course), and if A moves along Line (O, P ) through O, then Z moves along Line (O,Q), passing through O as A does. To construct Z, we ﬁrst bisect the angle POQ (by the usual Euclidean construction, which is not problematic since the three points are not collinear). Let the angle bisector be line L. Then let line K be the perpendicular to Line (O, P ) at A, let B be the intersection point of K and 28 MICHAEL BEESON

L, and let Z be the projection of B on Line (O,Q) (which is defined no matter whether O = A or not). Note that there are, if A = O, two points Z on OQ such that OZ = OA. The one constructed by Rotate (P,O,Q,A6 ) is such that, if POQ is a right turn, then AOZ is a right turn when A = O, regardless of whether A is between O and P or not. Similarly, if POQ is a6 left turn, so is AOZ. 5.3. Addition. To perform addition geometrically we suppose given a line L = Line (R,S) and an “origin”, a point O on L with S between R and O. We need to define a construction Add(A, B), which also depends, of course, on S, R, and O, such that Add(A, B) is a point C on L representing the (signed) sum of A and B, with O considered as origin. Lemma 5.5. Given line L = Line(R,S), and a point 0 on L with S between R and 0, we can construct a point Add(A, B) on L representing the signed sum of A and B, with O considered as origin, using the elementary constructions and Circle3. Remark. In order to appreciate that this lemma is not trivial, consider the following obvious, but incorrect, attempted solution: Add(A, B) := IntersectLineCircle2 (Line (0,B),Circle3(A, 0,B)). This works fine for B = 0, whatever the “sign” of A and B, and it even works when A = 0, but when6 B = 0 it is undefined. Proof. With Rotate in hand, we can give a construction for Add(A, B) (depending also on R, S, and O). (The construction is illustrated in Fig. 3 and Fig. 4) First, we replace R with a new points on L = Line (R,S), farther away from O, so that O, A, and B are all on the same side of R, and the new R and S are in the same order on line L as R and S were before. (This can be done using Extend , as we will prove rigorously later.) Now erect the perpendicular K to L at O, and the perpendicular H to L at B. In the process of erecting these perpendiculars, we will have constructed a points C on K such that ROC is a right turn. Then let D be the projection of C on H and let U = Rotate (R,O,C,A) V = Project(U, H) W = Rotate (D,B,R,V ) We set Add(A, B) = W . Then Add(A, B) is defined for all A, B. Suppose A = O. Then UV is perpendicular to both K and H. Then U and V are on the same6 side of L, since if UV meets L at a point X, then XU and XO are both perpendicular to K, which implies U = O, which implies A = O, contradicting A = O. It then follows from the property of Rotate that B and W occur on line L in6 the same order that O and A occur. Refer to Fig. 4 for an illustration of the case when A is negative. This implies that Add(A, B) represents the algebraic sum of A and B, since in magnitude BW = OA. 5.4. Multiplication and division. Having defined addition, we now turn to multiplication, division, and square root. The geometrical definitions of these FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 29

Figure 3. Signed addition without test-for-equality. A is rotated to U, then projected to V , then rotated to W . C H

b b U V

b b b b R O A B W = A + B

Figure 4. Signed addition when A is negative

A 0 W = A + B B

U V operations go back (at least) to Descartes.11 On the second page of La Geometrie [7], he gives constructions for multiplication and square roots. We reproduce the drawings found on page 2 of his book [7] in Figures 2 and 3. Here is Descartes’ explanation of these ﬁgures:

Figure 5. La Multiplication according to Descartes

B A D

1. Let AB be taken as unity.

11Although Descartes is usually credited with this, compare Euclid VI.12, which is very similar to Descartes’s treatment of multiplication and division, and Euclid VI.13, which is very similar to Descartes’s construction of square roots. 30 MICHAEL BEESON

2. Let it be required to multiply BD by BC. I have only to join the points A and C, and draw DE parallel to CA; then BE is the product of BD and BC. 3. If it be required to divide BE by BD, I join E and D, and draw AC parallel to DE; then BC is the result of the division. 4. If the square root of GH is desired, I add, along the same straight line, F G equal to unity; then, bisecting F H at K, I describe the circle F IH about K as a center, and draw from G a perpendicular and extend it to I, and GI is the required root.

From the point of view of constructive geometry, there is a problem with the construction. Namely, Descartes has only told us how to multiply two segments with non-zero lengths, and at least one of whose lengths is not 1 (the length of unity–he needs this when constructing AC parallel to DE), while we want to be able to multiply in general, without a test-for-equality construction. Using the Para construction of Lemma 5.4 where Descartes calls for “drawing DE parallel to CA”, we no longer have a problem multiplying numbers near 1 or 0. In [3], we gave a construction that successfully generalizes Descartes’s multiplication method to signed arguments. That method uses Rotate . However, Hilbert ([15], p. 54) gives another construction, whose result is equivalent to that of Descartes for positive arguments, but which directly works correctly for signed arguments as well. It is illustrated in Fig. 6.

Figure 6. Multiplication according to Hilbert

ab b

I b

a b b b

The construction is as follows: Start with Line (0, 1). Erect a perpendicular to Line (0, 1) at 0, and let I be a point on it such that 0I = 01. Given points a and b on Line (0, 1), construct a circle passing through I, a, and b. This must be done uniformly, so that when a = b, the circle passes through I and is tangent to Line (0, 1) at a. Then the answer is “the other point of intersection” of the circle and Line (0, I), except that this point lies not on Line (0, 1) but on Line (0, I), FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 31 so it needs to be rotated clockwise back to Line (0, 1). This construction is more easily visualized (for signed arguments) than that of Descartes. In order to carry out the construction uniformly, we first must be able to uniformly construct the perpendicular bisector of ab, in such a way that it is defined and perpendicular to Line (0, 1) even when a = b. Then we construct the perpendicular bisector of aI (which is unproblematic since I does not lie on Line (0, 1)), and then the center of the circle will lie where those two lines meet. Here we must use the parallel axiom; the existence of a circle passing through three given noncollinear points (i.e., circumscribing a triangle) is classically equivalent to the parallel postulate. We will take up the details of both multiplication constructions in Section 12.2. Descartes’ division method is handled similarly, using Para where Descartes constructs a parallel. 5.5. Square roots. Now we take up the geometrical construction of square roots. Fig. 7 shows Descartes’ construction for finding the square root of HG. His answer is the length of segment IG. Here is a geometrical construction term (encoded as a program) to carry out Descartes’ construction uniformly, as long as G is not to the left of H in the illustration, regardless of whether G = H or not. We assume that H in the diagram is 0 and the horizontal line is Line (0, 1). The point 1 is not shown in the figure.

Figure 7. Square roots according to Descartes

H K GF

SquareRoot(Point G) { // H in Descartes’ diagram is 0 F = Add(0,G,0,1) // so FG has unit length K = Midpoint(F,0) C = Circle(K,F) L = Perp(G,Line(0,F)) I = IntersectLineCircle1(L,C) // next rotate unit length to line L U = IntersectLineCircle1(L,Circle(G,F)) R = Rotate(U,G,F,I) // so now RG = IG but R is on Line(H,G), on the same side of G as F // now we need N so that N0 = RG MinusOne = IntersectLineCircle2(1,0,Circle(0,1)) N = Extend(MinusOne,0,G,R) 32 MICHAEL BEESON

return N }

Descartes stops when he has constructed I. What we have to do extra is to construct a point N such that 0N = GI. In order to do that uniformly, we must not assume that I = G. In order to get a non-degenerate angle we cannot use IGF ; instead we lay6 off a unit length on the perpendicular GI, which has been correctly constructed even if 0 = H = G = I. Thus we do not need to assume G = 0 for this construction to work; we only need that 0 is not between G and 1,6 i.e. loosely speaking G 0. This works because Perp is total, i.e. everywhere defined. ≥ 5.6. Formalizing the geometrization of arithmetic. It remains, of course, to prove in some axiomatic geometrical theory that multiplication and addition satisfy the field laws. We do not take that up at this point since we have not yet discussed theories and axioms. In due course, after having presented an axiomatic theory, we will formalize these results in deatil. But the reason for presenting them informally is to emphasize that the ideas really do not depend on a particular formalization of constructive Euclidean geometry. All we need is that the formalization should support the basic constructions Para, Perp, and Rotate used above. Of these Perp is basic, since the other two are defined using Perp. Just as in classical geometry, one can define addition and prove its properties without using the parallel axiom, but one needs the parallel axiom to define and prove the properties of multiplication; and one needs line-circle continuity to verify Descartes’s square root construction. The details of these points will be taken up in Sections 12.1 and 12.2.

§6. The arithmetization of geometry. Our aim in this section is to lay the groundwork for a constructive version of the classical representation theorem: the models of Euclidean geometry are exactly planes F 2 where F is a Euclidean field, i.e. an ordered field in which positive elements have square roots. The groundwork in question consists in developing the theory of Euclidean fields constructively, and verifying constructively that the axioms of line-circle continuity and circle-circle continuity hold in each Euclidean field. 6.1. Euclidean fields in constructive mathematics. We consider the analogue for ECG, which requires us to define constructive Euclidean fields. We use a language with symbols + for addition and for multiplication, and a unary predicate P (x) for “x is positive”. · We first discuss the axiomatization of Euclidean fields with intuitionistic logic. We take the usual axioms for fields, except the axiom for multiplicative inverse, which says that positive elements have multiplicative inverses. If positive elements have inverses, it is an easy exercise to show that negative elements do too. We define a Euclidean field to be a commutative ring satisfying the following FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 33 additional axioms: 0 =1 EF0 6 x =0 y (x y =1) EF1 6 → ∃ · P (x) P (y) P (x + y) P (x y) EF2 ∧ → ∧ · x + y =0 (P (x) P (y)) EF3 → ¬ ∧ x + y =0 P (x) P (y) x =0 EF4 ∧ ¬ ∧ ¬ → x + y =0 P (y) z(z z = x) EF5 ∧ ¬ → ∃ · P (x) P (x) EF6, or Markov’s principle ¬¬ → Axiom EF5 says that non-negative elements have square roots. This is a stronger axiom, intuitionistically, than simply specifying that positive elements have square roots. We could conservatively extend field theory by a unary function symbol for negation, x, in which case EF5 could be more readably written P ( x) z (z2 = x). − ¬ − → ∃ As usual, we define x

( y(x y = 1)) x =0 EF9 ∀ · 6 → P (y) P (z) y + z = x (y v = 1) w (w x =1) EF10 ∧ ∧ ∧ · → ∃ · Fields that satisfy EF0, EF2–EF6, and EF9-10 are called “Playfair ﬁelds”, because (as we will see far below) they correspond to models of the Playfair parallel axiom. EF9 enables us to verify the Playfair axiom, and EF10 enables us to verify Pasch’s axiom. 34 MICHAEL BEESON

To recap: In a Euclidean field, all nonzero elements have multiplicative inverses, while in a weakly Euclidean field, it is only required that elements known to be positive or negative have multiplicative inverses. In a strongly Euclidean field, nonzero elements are either positive or negative, so the distinction doesn’t matter. In a Playfair field, elements without reciprocals are zero, and elements greater than an invertible element are invertible. Remark. Since the double negation of EF8 follows from EF4, every Euclidean field is not not strongly Euclidean; in other words, one cannot give an example of a Euclidean field that is not strongly Euclidean. The standard plane is strongly Euclidean if and only if Markov’s principle holds in the reals. (To see this, suppose x = 0; apply Markov’s principle to x we have x > 0; then by apartness, which holds6 in the standard plane, we have| x>| x /|2| or x< x /2, and in the first case x> 0 while in the second case x< 0.) −| | | | 6.2. Line-circle and circle-circle continuity over Euclidean fields. Let F be an ordered field. Then “the plane over F ”, denoted by F 2, is a geometrical structure determined by defining relations of betweenness and equidistance in F 2, using the given order of F . Namely, b is between a and c if it lies on the interior of the segment ac. That relation can be expressed more formally in various ways, for example (using the cross product) by (c b) (b a) = 0 and (b c) (a b) > 0. Similarly, non-strict betweenness means− × that−b lies on the closed− segment· − ac, or formally, (c b) (b a)=0 and(b c) (a b) 0. The equidistance relation E(a,b,c,d− ×), which− means that− segment· −ab ≥is congruent to segment cd, can be defined by (b a)2 = (d c)2. Note that no square roots were used, so these definitions are valid− in any− ordered field. Though we have not yet given specific axioms for betweenness and congruence, it is reason- able to demand that any constructive axioms for betweenness and congruence should hold in F 2, at least for Euclidean fields F . By line-circle continuity we understand this axiom: let point a be non-strictly inside circle C, and point b be non-strictly outside circle C. Then Line (a,b) meets circle C in a point non-strictly between a and b. By circle-circle continuity, we mean that if circle C has a point non-strictly inside circle K, and another point non-strictly outside circle K, then there is a point on both circle C and circle K.

Theorem 6.1 (proved constructively). Let F be an ordered field (satisfying any of the axioms about the existence of reciprocals). Then the following are equivalent: (i) F 2 satisfies circle-circle continuity. (ii) F 2 satisfies line-circle continuity. (iii) F is a Euclidean field (nonzero elements have reciprocals and nonnegative elements have square roots). Remark. We never need to divide in this proof, hence the constructive distinctions about the existence of reciprocals do not come into play. Proof. A similar theorem is proved classically in [14], p. 144. We follow that proof as closely as possible, but some details are missing there. The issue is that it is not enough to observe that the equations for the intersection points are FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 35 quadratic. One has to translate the hypothesis that one circle has a point inside and a point outside the other circle into algebra and show algebraically that this implies the equations for the intersection are solvable. First we show (i) implies (ii). If f = 0 is the equation of a circle C, and g =0 the equation of a line L, then f + g = 0 is the equation of another circle, whose intersections with C are the same as the intersections of C and L. Hence (i) implies (ii). Next we show (ii) implies (iii). Let b 0, let H be the origin and let G = (0,b). To show that b has a square root. Refer≥ to Fig. 7, which shows the geometrical construction of square roots that goes back to Euclid. We wish to show that the point I = (b, √b) exists, as the intersection point a line and circle. The circle has center K = ((1 + b)/2, 0) and radius (1 + b)/2), and the line is x = b. So the line has the point (0,b) non-strictly inside the circle, since (b (1+b)/2 (1+b)/2; and the line has points outside the circle, for example (|b, (1+− b). So|≤ by line-circle continuity, the intersection point I exists, and hence b has a square root. Note that the case b = 0 is allowed. Finally we have to show (iii) implies (i). Let C and K be non-concentric circles. By making linear transformations, we can assume that the two circles have their centers at (0, 0) and (1, 0) respectively; then the equations for the circles are x2 + y2 = r2 and (x 1)2 + y2 = s2. We must make use of the hypothesis− that C has a point (non-strictly) inside K and another (non-strictly) outside K. To do so we need the following lemma:

Lemma 6.2. With C and K as above, C has a point (non-strictly) inside K and another (non-strictly) outside K if and only if r + s 1 and (s r)2 1. ≥ − ≤ Proof. First suppose r + s 1 and (r s)2 1. Then we have ≥ − ≤

r 1 s since r + s 1 ≥ − ≥ r s 1 since (r s)2 1 − ≤ − ≤ r 1+ s. ≤

That is, 1 s r 1+s. Then s r 1 s. Squaring, we have (r 1)2 s2. Hence (r, 0)− is≤ a point≤ on C and− inside≤ −K. ≤ − ≤ Since (r s)2 1 we have s r 1; hence r 1 s. Since s 0, when we square this− inequality≤ we get (−r +≤ 1)2 s2. Therefore,− − ≤− the point (≤r, 0) (which is on C) is outside K. That proves the≥ easy half of the lemma. − Now suppose that the point (x, y) is on C and inside K. We will prove r+s 1. Since a b is equivalent to (a>b), we can prove r + s 1 by contradiction.≥ Suppose≤ that r + s < 1. Then¬ intuitively, the line x = (r≥+1 s)/2 separates circles C and K, contradiction, since (x, y) is inside both circles.− We prove this more rigorously: Since (x, y) is on C and inside K, the point (x, 0) is inside both circles. That is, x2 r2 and (x 1)2 s2. Since r + s< 1, we have s< 1, and since (x, 0) is inside≤K, we have−x > ≤0. Hence 0 < x r < 1 and 1 x

Now suppose in addition that the point (u, v) is on C and outside K. Then we claim the point ( r, 0) on C is also outside K. Here is the proof: − (u 1)2 + v2 s2 since (u, v) is outside K − ≥ u2 2u +1+ v2 s2 − ≥ r2 2u +1 s2 since u2 + v2 = r2 − ≥ r2 +2r +1 s2 +2r 2u ≥ − (r + 1)2 s2 + 2(r u) ≥ − But u r, since (u, v) is on C. Hence r u 0, and we have ≤ − ≥ (r + 1)2 s2 ≥ which shows that ( r, 0) is outside K as claimed. Hence s r + 1. Hence s r 1. Squaring,− we have the desired inequality (r s)2 1.≤ That completes the− proof≤ of the lemma. − ≤ Returning to the proof of the theorem, we want to show that the equations x2+ y2 = r2 and (x 1)2 +y2 = s2 have a common solution, under the hypothesis that C has a point− (non-strictly) inside K and another point (non-strictly) outside K. Subtracting the two equations we obtain a linear equation:

x2 (x 1)2 = r2 s2 − − − 2x 1= r2 s2 − 1 − x = (r2 s2 + 1) 2 − Putting that back into x2 + y2 = r2 we have 1 (r2 s2 + 1)2 + y2 = r2 4 − 4y2 =4r2 (r2 s2 + 1)2 − − =4r2 (r4 + s4 +1+2r2 2r2s2 2s2) − − − =2r2 r4 s4 1+2r2s2 +2s2 − − − = (r2 s2)2 + 2(r2 + s2) 1 − − − = ((r + s)2 1)(1 (r s)2) − − − By the lemma, both factors on the right are nonnegative; so their product is nonnegative, and we can take the square root in any Euclidean ﬁeld (or weakly Euclidean or even Playfair ﬁeld). That completes the proof of the theorem.

§7. Other axiomatizations of geometry. In the previous sections, we have presented constructive geometry informally, that is, without the use of the apparatus of formal logic, or even an informal axiomatic system for geometry. It is, however, our intention to present a formal theory ECG of Euclidean constructive geometry. Before doing so, we will review the existing formalizations of classical geometry. It is not our purpose to review in detail the history of axiomatic geometry, but some history is helpful in understanding the variety of geometrical axiom systems. In fact the modern subject of formal logic was born in the effort to understand the foundations of geometry; in particular to FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 37 understand the efforts to understand the relation of Euclid’s parallel postulate to the other postulates; in a sense non-Euclidean geometry was the mother of modern logic. 7.1. Pasch. Moritz Pasch’s 1882 book [23],[24] introduced the relation of betweenness that has been used in all axiomatizations of geometry since 1882. Betweenness is the relation B(a,b,c) that holds when points a, b, and c are distinct points on a line, “in that order”. Pasch pointed out the defects in Euclid that require the use of betweenness and supplied the basic axioms for betweenness. The most complicated of those axioms is known as Pasch’s axiom; it says that if line L meets side AC of triangle ABC, and none of A, B, or C lies on L, and L and does not meet side AB, then L meets side BC. 7.2. Pieri and Peano. The modern axiomatic method, or as Pieri called it, the “hypothetical-deductive method”, differed from Euclid not only in increased precision (the use of betweenness, for example) and the realization that some notions must be taken as undefined, but also in the important idea that geometry might have many models, rather than just “the intended model”. These three ideas arose in the second half of the nineteenth century in more than one place, and an accurate history of their development is far beyond the scope of these brief paragraphs. It is, however, clear that the “Italian school” made considerable progress in this direction. For example, Pieri’s Point and Motion [25] was a fully modern development, taking as primitive objects points and “motions”, and as a primitive relation the application of a point to a motion. Apparently neither Pieri nor Hilbert knew about the work of the other before the publication (see [20], pp. 150-151). FINISH THIS–what about Peano? 7.3. Hilbert. Hilbert’s influential book [15] used the notion of betweenness and the axioms studied by Pasch. His theory was what would today be called “second-order”, in that sets were freely used in the axioms. Segments, for example, were defined as sets of two points, so by definition AB = BA since the set A, B does not depend on the order. Of course, this is a trivial departure from{ first-order} language; but Hilbert’s last two axioms, Archimedes’s axiom and the continuity axiom, are not expressible in a first-order geometrical theory. On the other hand, lines and planes were regarded not as sets of points, but as (what today would be called) first-order objects, so incidence was an undefined relation, not set-theoretic membership. At the time (1899) the concept of first- order language had not yet been developed, and set theory was still fairly new. Congruence was treated by Hilbert as a binary relation on sets of two points, not as a 4-ary relation on points. Early geometers thought that the purpose of axioms was to set down the truth about space, so as to ensure accurate and correct reasoning about the one true (or as we now would say, “intended”) model of those axioms. Hilbert’s book promoted the idea that axioms may have many models; the axioms and deductions from them should make sense if we read “tables, chairs, and beer mugs” instead of “points, lines, and planes.” This is evident from the very first sentence of Hilbert’s book: 38 MICHAEL BEESON

Consider three distinct sets of objects. Let the objects of the first set be called points . . . ; let the objects of the second set be called lines . . . ; let the objects of the third set be callled planes. Hilbert defines segments as pairs of points (the endpoints),although lines are primitive objects. On the other hand, a ray is the set of all points on the ray, and angles are sets consisting of two rays. So an angle is a set of sets of points. Hence technically Hilbert’s theory, which is often described as second-order, is at least third-order. Hilbert’s language has a congruence relation for segments, and a separate congruence relation for angles. Hilbert’s congruence axioms involve the concept of angles: his fourth congruence axiom involves “angle transport” (constructing an angle on a given base equal to a given angle), and his fifth congruence axiom is the SAS triangle congruence principle. Hilbert’s Chapter VII discusses geometric constructions with a limited set of tools, a “segment transporter” and an “angle transporter”. These correspond to the betweenness and congruence axioms. Hilbert does not discuss the special cases of line-circle continuity and circle-circle continuity axioms that correspond to ruler-and-compass constructions, despite the mention of “compass” in the section titles of Chapter VII. 7.4. Tarski and his students. Hilbert’s geometry contained two axioms that go beyond first-order logic. First, the axiom of Archimedes (which requires the notion of natural number), and second, an axiom of continuity, which essentially says that Dedekind cuts are filled on any line. This axiom requires mentioning a set of points, so Hilbert’s theory with this axiom included is not a “first-order theory” in a language with variables only over points, lines, and circles. Later in the 20th century, when the concept of “first-order theory” was widely understood, Tarski formulated his theory of elementary geometry, in which Hilbert’s axiom of continuity was replaced with an axiom schemata. The set variable in the continuity axiom was replaced by any first-order formula. Tarski proved that this theory (unlike number theory) is complete: every statement in the first-order language can be proved or refuted from Tarski’s axioms. Tarski’s axioms are elegantly stated using only variables for points. He replaces Hilbert’s fourth and fifth congruence axioms (angle transport and SAS) with an elegant axiom, known as the five-segment axiom. We will eventually give the formal statement of this axiom, but for now, we refer to Fig. 8. The 5-segment axiom says that in Figure 8, the length of the dashed segment cd is determined by the lengths of the other four segments. Formally, if there is another figure like the one shown, and the four solid segments are pairwise congruent to the corresponding segments in the second figure, then the dashed segments are also congruent. Tarski’s 5-segment axiom is a thinly-disguised variant of the SAS criterion for triangle congruence. To see this, think of another copy of Fig. 8 labeled with upper-case letters. The triangles we are to prove congruent are dbc and DBC. We are given that bc = BC and db = DB. The congruence of angles dbc and DBC is expressed in Tarski’s axiom by the congruence of triangles abd and ABD, whose sides are pairwise equal. The conclusion, that cd = CD, give the congruence of triangles dbc and DBC. Later we will give a formal proof of the FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 39

Figure 8. Tarski’s 5-segment axiom. cd is determined.

ab c

SAS criterion from the 5-segment axiom. We note that Borsuk-Szmielew also took this as an axiom (see [5], p. 81, Axiom C-5). We note that an earlier version of Tarski’s theory included as an axiom the “triangle construction theorem”, which says that if we are given triangle abc, and segment AB congruent to ab, and a point x not on Line (A, B), then we can construct a point C on the same side of Line (A, B) as x such that triangle ABC is congruent to triangle abc. It was later realized12 that this axiom is provable. For example, one can drop a perpendicular from c to Line (a,b), whose foot is the point d on Line (a,b), and then find a corresponding point D on Line (A, B), and then lay off dc on the perpendicular to Line (A, B) at D on the same side of Line (A, B) as x, ending in the desired point C. Of course one must check that this construction can be done and proved correct on the basis of the other axioms. But as it stands, this construction demands a case distinction about the order and possible identity of the points d, a, and c on Line (a,b). Hence, at least this proof of the triangle construction theorem from the axioms of Tarski’s theory is non-constructive. (The construction itself is obvious: C is the intersection point of circles about A and B of radii ac and bc that lies on the same side of Line (A, B) as x. The problem is to prove that the circles actually intersect, without using a constructively invalid case distinction.) Tarski’s early axiom systems also included axioms about betweenness and congruence that were later shown [13] to be superfluous. The final version of this theory appeared in [26]; for the full history see [30].13 The achievement of [26] is to develop a really minimal set of axioms for betweenness and congruence.14 Hilbert’s intuitive axioms about betweenness disappeared, leaving only the axiom B(a,b,a) and the Pasch axiom and axioms to guarantee that congruence is an ¬equivalence relation. We shall adopt constructive versions of these axioms, and

12Acording to [30], Tarski included this principle as an axiom in his first two published axiom sets, but then discovered in 1956-57 with the aid of Eva Kallin and Scott Taylor, that it was derivable; so he did not include it in [29]. (See the footnote, p. 20 of [29].) But Tarski did not publish the proof, and Borsuk-Szmielew take the principle as their Axiom C-7 [5]. 13Note that the version mentioned in [1] is not the final version used in [26]; inner transitivity for betweenness was eliminated in [26]. 14We would like to emphasize the important contributions of Gupta, which are important to the development in [26], and are credited appropriately there, but without a careful study one might not realize how central Gupta’s results were. These results were apparently never published under Gupta’s own name, and still languish in the Berkeley math library in his doctoral dissertation. 40 MICHAEL BEESON we are able to show that negative (that is, disjunction-free and existence-free) statements that follow classically from those axioms also follow constructively. Thus, we will be able to draw freely on the labors of Tarski, Gupta, Szmielew, Schwabhauser, and others who contributed to this elegant axiom set. That will enable us to focus on the constructive aspects, rather than on derivations of (de- ceptively) simple-looking theorems, whose complications have nothing essential to do with constructivity. 7.5. Tarski’s theory of Euclidean geometry. But Tarski’s theory is “elementary” only in the sense that it is first-order. It still goes far beyond Euclid. To capture Euclid’s geometry, Tarski considered the sub-theory in which the continuity axiom is replaced by “line-circle continuity” and “circle-circle continuity”. These axioms assert the existence of the intersection points of lines and circles, if some point on the line lies inside the circle, or some point on one circle lies inside the other circle. In the language of ECG, these axioms are expressed without needing to say “there exists”; since we have terms to denote those intersection points, all we have to do is say that, for instance, IntersectLineCircle1 (L, C) lies on both L and C. ECG is essentially a constructive version of Tarski’s theory of the elementary constructions.15 7.6. Borsuk-Szmielew. This book follows Hilbert, with some axioms taken from Tarski, filling in many details. It treats both Euclidean and hyperbolic geometry. The formal systems are, like Hilbert’s, second-order, and the metamathematics is restricted to the construction of (second-order) models. It does introduce a four-place equidistance relation between points, instead of treating congruence as a relation on sets; but it still regards, as Hilbert does, a segment as a set of points. (Strangely, the segment ab is a,b , while the “open segment” ab is the set of all points between a and b.) For{ example,} the betweenness axioms are those Hilbert used, including the Pasch axiom. The first edition of this book was in Polish, and the English edition is not merely a translation, but also incorporated some changes in the system; although I have never seen the Polish edition, according to the introduction of the English edition, the original followed Hilbert more closely, in having primitive sorts for lines and planes; the English edition treats them as sets of points, in order to make the model theory easier. The book uses an axiom system based on Hilbert, but incorporating some of Tarski’s improvments, such as the five-segment axiom, which is to be expected as co-author Szmielew worked closely with Tarski; but the manuscript was prepared too early to benefit from the later improvements made by Gupta and others. Today this work is superceded by [26]. 7.7. Avigad, Dean, and Mumma. In [1], these three authors give a formal theory with classical logic that is intended to faithfully reflect Euclid’s reasoning. Their theory is many-sorted, and has many primitives and many axioms, partly in order to support educational applications (it has been used as the basis for mechanical theorem-proving support of a geometry course). They distinguish different kinds of inferences: diagrammatic, metric, diagram-metric transfer, and

15It is confusing that in axiomatic geometry, “elementary” sometimes refers to the elementary constructions, and sometimes to the full first-order theory of Tarski. In this paper we shall not refer again to the full first-order theory. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 41 superposition inferences. One has a sort for angles (that is, measures of angles) and a sort for lengths (measures of segments). One can perform addition and subtraction on these sorts (but not multiplication), and compare them. This is used to interpret Euclid’s comparison of segments and angles. The point of this is that there is an efficient decision procedure for this subtheory. The diagrammatic inferences involve betweenness, axioms about the primitive notion “same side of”, including versions of Pasch’s axiom, axioms about incidence and intersection, etc. The idea is that the diagrammatic inferences are the ones that Euclid does not state explicitly. Finally there are the diagram-metric transfer instances, in which diagrammatic information is used to infer metric information. For example, if b is between a and c, then ab + bc = ac. The fact that this theory uses classical logic is not very essential. There is nothing in the framework that would preclude a constructive version of the theory. It is already “constructive” in the sense that it does not have continuity axioms beyond line-circle and circle-circle. Such a theory would no doubt be a conservative extension of our theory ECG. 7.8. Heyting and other constructive approaches.

§8. The theory ECG of Euclidean Constructive Geometry. In this section we give and explain the axioms of a first order axiomatic theory of geometry as close as possible to Euclid. We call it ECG, for “Elementary Constructive Geometry”. We will take care to formulate our axioms without quantifiers and without disjunction, which will be key to our applications of proof theory. What we aim to do in this section is to formulate such a theory, which we feel is quite close in spirit to Euclid. In formulating this theory, we made use of the famous axioms of Hilbert [15], which have been given a more modern and detailed formulation in the textbook of Greenberg [12]. Of course, we do not take the full continuity axioms of Hilbert, but only the line-circle and circle-circle continuity axioms. Where possible, we formulate our axioms as correctness statements about the constructions; in that form they are automatically quantifier-free. Some axioms, which are not about constructions, are inherently quantifier-free. The only question of serious interest is whether disjunction can be completely avoided. It can, as it turns out. The details of this axiomatization may seem somewhat tedious, but the system must of course be specified in complete detail in order to use it in metamathematical proofs. Moreover, some of the details, as far as I can determine, are actually new. In particular, we show how to define the relations “ABC is a left turn” and “ABC is a right turn” in our theory; the experts we consulted thought this was new. We need this in order to distinguish the two intersection points of two circles. 8.1. Logic of Partial Terms (LPT). Since many of our function symbols denote “partial functions”, i.e. functions that are not always defined, we will use the “logic of partial terms” LPT in our theories. This is a modification of first-order logic, in which the formation rules for formulas are extended by adding the following rule: If t is a term then t is a formula. Then in addition the quantifier rules are modified so instead of x↓(A(x) A(t)) we have x(t ∀ → ∀ ↓ 42 MICHAEL BEESON

A(x) A(t)), and instead of A(t) x A(x) we have A(t) t x A(x). ∧Details→ of LPT can be found in [2], p.→ 97. ∃ ∧ ↓ → ∃ We could try to deal with partial terms, such as √x, by simply using an ordinary function symbol for √, but not saying anything in the axioms about √ of negative numbers. Thus √ 1 would some real number, but we would not − know or care which one, and we would not be able to prove that its square is 1. This approach rapidly becomes awkward when complicated terms involving square− roots of different quantities are used, and you must add extra hypotheses to every theorem asserting that what is under every square root is positive, and we choose to use LPT instead. LPT includes axioms c for all constants c of any theory formulated in LPT; this is in accordance with↓ the philosophy that terms denote things, and while terms may fail to denote (as in “the King of France”), there is no such thing as a non-existent thing. Thus 1/0 can be undefined, i.e. fail to denote, but if a constant is used in LPT, it must denote something. The meaning∞ of t = s is that t and s are both defined and they are equal. We write t ∼= s to express that if one of t or s is defined, then so is the other, and they are equal.

Definition . 8.1 For terms in any theory using the logic of partial terms, t ∼= q means t t = q q t = q. ↓ → ∧ ↓ → This is read t and q are equal if deﬁned.

Thus “∼=” is an abbreviation at the meta-level, rather than a symbol of the language. LPT contains the axioms of “strictness”, which are as follows (for each function symbol f and relation symbol R in the language):

f(t1,... ,tn) t1 . . . tn ↓ → ↓ ∧ ∧ ↓ R(t1,... ,tn) t1 . . . tn → ↓ ∧ ∧ ↓ Note that in LPT, under a given “valuation” (assignment of elements of a structure to variables), each formula has a definite truth value, i.e. we do not use three-valued logic in the semantics. For example, if P is a formula of field theory with a reciprocal operation 1/x, then P (1/0) is false, since 1/0 is undefined. For the same reason P (1/0) is false. Hence P (1/0) P (1/0) is false too; but that does not contradict¬ the classical validity of x(P∨(x ¬) P (x)) since we are required to prove t before deducing an instance∀P (t) ∨P ¬(t). As an example of↓ the use of LPT, and also for further∨ ¬use below, we reformu- late the theory of Euclidean fields using the logic of partial terms. The existential quantifiers associated with the reciprocal axioms, with the axiom of additive inverse, and with the square-root axiom of Euclidean field theory are replaced by a function symbol √, a unary minus , and a function symbol for “reciprocal”, − which we write as 1/x instead of reciprocal(x). The changed axioms are FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 43 x + ( x) = 0 (additive inverse) − x =0 x (1/x)=1 (EF1′) 6 → · P (x) x (1/x)=1 (EF7′) → · x + y =0 P (y) √x √x = x (EF5′) ∧ ¬ → · Definition 8.2. The theory EF is the theory of Euclidean fields, formulated using the logic of partial terms with function symbols for additive inverse, square root, and reciprocal, using the axioms EF1′ (nonzero elements have reciprocals), EF2 (sums and products of positive elements are positive), EF3 and EF4 (roughly, that x and x are not both positive and not both non-positive unless x − is zero), EF5′ (given just above), and EF6 (Markov’s principle). 8.2. Replacing LPT and sorts with predicates if desired. Since the usual theorem-provers do not understand the logic of partial terms, for purposes of experimentation with automated deduction, one may wish for a formulation without LPT. That can be done, and at the same time one can eliminate the use of multi-sorted logic. Add unary predicates Point, Line, and Circle, and then replace t by either Point (t), Circle (t), or Line (t), according to the sort of t, in all the axioms,↓ including the logical axioms of LPT that involve . The nonlogical axioms then have all the variables of a given sort relativized↓ to the corresponding unary predicate, e.g. point variables are relativized to Point. It may be easier to understand this procedure if we consider a more familiar theory, field theory. There is originally only one sort of variables, so we will add just one predicate symbol F , and the axiom x = 0 1/x would become x = 0 F (1/x), and the axiom (x 1/x) 6 →x (1/x↓) = 1 becomes F (6x (1/x→)) x (1/x) = 1. In any model· of this↓ → system,· the interpretation F¯ of· F is (the→ carrier· set of) a field, but there will be elements of the model, including an interpretation of 1/0, that are not in F¯. The reciprocal of zero is something, but it doesn’t matter what, as long as it is not in F¯. The elements of the model that are not in F¯ serve as the denotations of terms that would be undefined in LPT. One might try to not use the new symbol F , and just let 1/x be always defined, and in the field, but only assert that x = 0 x (1/x) = 1. In this system we can prove 1/0 = 0 (since if 1/0 = 0 then6 (1→/0) 0· = 1, but anything times 0 is 0 and 0 = 1). This equation is not6 true in LPT·, so the interpretation is not faithful from6 LPT to this system. Moreover, in LPT we can prove 1/0 = 0, since if 1/0 = 0 then 1/0 , so 1/0 0 = 1, but 1 = 0, contradiction; hence6 1/0 = 0 as claimed. Hence the↓ interpretation· is not even6 sound; one cannot just let 16 /x be an “unspecified number”; if it is defined, it has to be an “unspecified something” that is not a number. Thus eliminating LPT necessarily involves introducing unary predicates (or sorts) with some “undefined elements” lurking in purgatory, so to speak, e.g. 1/0, which cannot satisfy F . To make these considerations precise, consider any theory T in many-sorted predicate logic with the logic of partial terms. For each formula φ, let φˆ be the formula in ordinary one-sorted predicate logic obtained by replacing the sorts of T by unary predicate symbols (with the same names as the sorts), relativizing quantifiers over sort P to the predicate symbol P , and for each sort P and term 44 MICHAEL BEESON of sort P , replacing t by P (t). If T is a theory in predicate calculus, let Tˆ be ↓ the theory axiomatized by the φˆ for φ and axiom of T . Theorem 8.3. Let T be any theory in many-sorted predicate calculus with the logic of partial terms. Let φˆ be defined as just above. Then T proves φ if and only if φˆ is provable in ordinary first-order predicate logic. Proof. Note that the interpretation commutes with implication. Therefore, by the deduction theorem we may assume, without loss of generality, that T is the empty theory. (For readers with little background in logic, what we are saying here is that if T proves φ, then pure logic proves ψ φ, for some conjunction of axioms ψ of T , so we might as well consider ψ → φ with an empty theory T .) → We first prove the left-to-right direction, that if many-sorted logic with LPT proves φ, then predicate logic proves φˆ. We proceed by induction on the length of proofs. Consider the axiom of LPT, x φ t φ[x := t] ∀ ∧ ↓ → when x is a variable of sort P . This becomes x( (x) φˆ P (t) φ[x := t]. ∀ ¶ → ∧ → which is provable in first-order predicate calculus. Similarly, consider the axiom φ[x := t] t x φ. ∧ ↓ → ∃ This translates to φ[x := t] P (t) x (P (x) φ). ∧ → ∃ ∧ which is also provably in first-order predicate calculus. The other axioms and rules of inference do not involve the logic of partial terms and are straightforward. That completes the proof that if many-sorted LPT proves φ, then predicate logic proves φ. For the converse, we Strong Parallel Axiom1 8.3. Language. For foundational purposes, it seems simplest to use only Point, Line, and Circle, and that is what we do in ECG.16 We choose a multi- sorted theory, with “sorts” corresponding to those types. We use the words “sort” and “type” synonymously in this paper. It is, of course, not difficult to add sorts Ray, Arc, and Segment, and axioms making the extended theories conservative over ECG, but we do not do so here. Similarly, it is not difficult to eliminate all the sorts but Point. We take function symbols corresponding to constructors and accessors for those types, described in detail below. The relation symbols we use are standard in axiomatic geometry, B for (strict) betweenness and E for equidistance. We emphasize that B is used for strict betweenness; as Hilbert put it, if C is between A and B, then A, B, and C are three distinct points.

16Using more sorts makes the theory closer to practice, and using fewer sorts makes it easier to analyze metamathematically. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 45

We use on (P,L) for the incidence of point P on line L, and On (P, C) for the incidence of point P on circle C. There is a complete list of the axioms of ECG in the Appendix, for reference. ECG has six basic function symbols of type Point, shown here with arguments: IntersectLines (L,K) IntersectLineCircle1 (L, C) IntersectLineCircle2 (L, C) IntersectCircles1 (C,K) IntersectCircles2 (C,K) IntersectCirclesSame (C,K,p) IntersectCirclesOpp (C,K,p) The intuitive meaning of these symbols has been discussed above. ECG does not have “overloaded” versions of these functions; in other words, we just write IntersectLines (Line (A, B), Line (P,Q)) instead of having an overloaded version of IntersectLines that takes four points. 8.4. Intuitionistic logic and stability. Our underlying logic is intuitionistic. We first give the specifically intuitionistic parts of our theory, which are very few in number. We do not adopt decidable equality, nor even the substitute concept of “apartness” introduced by Heyting (and discussed above), primarily because we aim to develop a system in which definable terms (constructions) denote continuous functions, but also because we wish to keep our system closely related to Euclid’s geometry, which contains nothing like apartness. Our first four axioms express our intuition that there is nothing asserting existence in the meaning of equality or incidence; hence assertions of equality or incidence can be constructively proved by contradiction. Technically, the word “stable” is applied to a predicate Q if Q Q. The following axioms assert the stability of equality, equidistance,¬¬ and→ incidence: x = y x = y (Axiom S1) ¬¬E(a,b,c,d→ ) E(a,b,c,d) (Axiom S2) ¬¬on (p,L) →on (p,L) (Axiom S3) ¬¬On (p, C) → On (p, C) (Axiom S4) ¬¬ → We also want stability for betweenness. Since we use non-strict betweenness, this stability axiom has a different character, and it will be discussed as a betweenness axiom below. Nevertheless we group the axiom here as a “stability axiom” since it disappears with classical logic: B(x,y,z) B(x,y,z) (Axiom S5) ¬¬ → This axiom is also called “Markov’s principle for betweenness”, as we will discuss below. One might consider the following candidate for an axiom (schema): for every term t, t t (S6, not an axiom) ¬¬ ↓ → ↓ 46 MICHAEL BEESON

For example, if two lines cannot fail to intersect, then they do in fact intersect; or if two circles cannot fail to intersect, then they intersect. This seems somewhat similar to Markov’s principle: we just follow the lines until they intersect, and they “must”, just as a Turing machine that cannot fail to halt “must halt”. But of course, as with Markov’s principle, that argument is circular. If we could not prove S6, we would need to adopt it as an axiom. However, it turns out that we can prove S6 in ECG, so we do not need to add it as an axiom. See Theorem 10.16 for the proof; we cannot give it here since it relies on developing enough geometry in ECG to prove geometrical equivalents for the definedness of each kind of term. 8.5. Incidence and intersection axioms. There are two incidence relations, for on (P,L) for points lying on lines, and On (P, C) for points lying on and circles. We start with three “incidence axioms”: a = b on (a, Line (a,b)) (Axiom I1) a 6= b → on (b, Line (a,b)) (Axiom I2) on6 (p,L→) on (q,L) p = q on (r, Line (p, q)) on (r, L) (Axiom I3) ∧ ∧ 6 ∧ → Axiom I3 can be paraphrased,“two points determine a line.” The following axioms express the meaning of the five main function symbols of ECG. They do not, however, make any assertions of geometrical content, except that two lines intersect in at most one point. In particular they do not say that any intersection point of two circles is one of the two particular ones given by the function symbols; that would involve a disjunction, which we wish to avoid; and they do not say that a term meant to define an intersection point of two circles is actually defined, unless we already have a point on both circles. The axioms that say intersection points are defined are given in another section; in this section we just say that if these terms are defined, they denote intersection points. For readability, the first five of these axioms introduce a name “p” for an intersection point; this extra variable, and the implication symbol in the axioms, can easily be eliminated by substituting the term for p; one would wish to do this if using the axioms for automated deduction. p = IntersectLines (L,K) on(p,L) on (p,K) (Axiom I4) → ∧ IntersectLines (L,K) ∼= IntersectLines (K,L) (Axiom I5) (The symbol ∼= is defined in Definition 8.1.) p = IntersectLineCircle1 (L, C) on (p,L) On (p, C) (Axiom I6) p = IntersectLineCircle2 (L, C) → on (p,L) ∧ On (p, C) (Axiom I7) p = IntersectCircles1 (C,K) →On (p, C) On∧ (p,K) (Axiom I8) p = IntersectCircles2 (C,K) → On (p, C) ∧ On (p,K) (Axiom I9) p = IntersectCirclesSame (C,K,x→ ) On∧(p, C) On (p,K) (Axiom I10) (B(p,y,x) on(y, Line (center→(C), center∧(K)))) ∧¬ ∧ Axiom I10 says that the (open) segment px does not meet the line L joining the centers. The next axiom, about IntersectCirclesOpp , specifies that the segment px does meet L; but it is allowed to meet it at p, which we need in case the circles are tangent. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 47 p = IntersectCirclesOpp (C,K,x) On (p, C) On (p,K) (Axiom I11) ( B(p, IntersectLines ( → ∧ ∧ ¬ ¬ Line (p, x), Line (center(C), center(K)), x) p = IntersectLines (Line (p, x), Line (center(C),∧ center(K))) 6 on (p,L) on (p,K) on (q,L) on (q,K) (Axiom I12) ∧IntersectLines∧ (L,K) ∧ ¬ IntersectLines→ (L,K) on (p,L↓ ) (Axiom I13) on (p,K) p ↓= ∧IntersectLines (L,K) on (p,L∧ ) On (p,→ C) IntersectLineCircle1 (L, C) (Axiom I14) on (p,L) ∧ On (p, C) → IntersectLineCircle2 (L, C) ↓ (Axiom I15) On (p, C)∧ On (p,K)→ IntersectCircles1 (C,K) ↓ (Axiom I16) On (p, C) ∧ On (p,K) → IntersectCircles2 (C,K) ↓ (Axiom I17) ∧ → ↓ We want the intersection points of circles C and K to be undefined when circles C and K coincide (have the same points). The following axioms so specify. IntersectCircles1 (C,K) center(C) = center(K) (Axiom I17) IntersectCircles2 (C,K) ↓ → center(C) 6= center(K) (Axiom I18) IntersectCirclesSame (C,K,x↓ → )) center6 (C) = center(K) (Axiom I19) IntersectCirclesOpp (C,K,x)) ↓ →center(C) =6 center(K) (Axiom I20) ↓ → 6 Similarly, we want the intersection point of two lines to be undefined when the lines coincide. IntersectLines (L,K) (Axiom I21) (on (pointOn1 (K↓) →,L) on (pointOn2 (K),L)) ¬ ∧ 8.6. Constructor and accessor axioms. Since we allow Circle3 , we do not need Circle (x, y) as an official part of the language. We regard Circle (x, y) as an abbreviation:

Definition 8.4. Circle (x, y) := Circle3 (x, x, y) There are function symbols corresponding to the constructor and accessor functions for each of the sorts. The argument and value types of these symbols are obvious, and hence not speciﬁed here. Here are the axioms relating the constructors and accessors. Line (pointOn1 (L), pointOn2 (L)) = L (Axiom CA1) A = B pointOn1 (Line (A, B)) = A (Axiom CA2) A =6 B → pointOn2 (Line (A, B)) = B (Axiom CA3) Line6 (A,→ A) (Axiom CA4) center¬ (Circle3↓(a,b,c)) = a (Axiom CA5) Circle (center (C), pointOnCircle (C)) = C (Axiom CA6) Circle3 (A, B, C) (Axiom CA7) ↓ Note that Axiom CA7 allows for “degenerate” circles, i.e. circles of zero radius. We note that it is not necessary to have as an axiom that pointOn1 (L) = pointOn2 (L) 6 since it can be derived; namely if A = pointOn1 (L) and B = pointOn2 (L) then L = Line (A, B) by Axiom CA1, so Line (A, B) , so by Axiom CA4 , A = B. ↓ 6 48 MICHAEL BEESON

Note that we do not have pointOnCircle (Circle (A, B)) = B. Even though we have On(B, Circle (A, B)), not every circle is constructed as Circle (A, A, B); and in general we need pointOnCircle (C) to depend continuously on circle C, however C is given. For example, in models we could take pointOnCircle (C) to be the “southernmost point” of circle C, where “south” is given by a ray from the center of C parallel to a certain fixed ray. Since we want all our constructions to be continuous, we will need some axiom to guarantee that pointOnCircle (C) is continuous in the circle C; that axiom will involve the equidistance (or “congruence” relation) and will be given below. 8.7. Meaning of equality. Lines in ECG have direction. Line (A, B) is not in general equal to Line (B, A). We say they “have the same points”. Lo- gicians call this “extensional equality”. We use the equality symbol between points to mean “identically equal”. Between lines, equality means “intensional equality”. In the spirit of constructive mathematics, lines “come equipped” with two associated distinct points. Thus, Line (A, B) = Line (P,Q) if and only if A = B and P = Q. This does not need to be assumed, as it follows from the axioms given above for the accessor and constructor functions. It may, however, be confusing to those not accustomed to constructive mathematics. The notion of “extensional equality” refers to the defined relation between two lines, that the same points are on both lines. In practice, to avoid confusion, we rarely if ever mention equality between lines. It should be noted, however, that our theory does depend on this view of lines, since the order of the two intersection points of line Line (A, B) with circle C is opposite to the order of the two intersection points of Line (B, A) with C. In other words, when considering IntersectLineCircle1 (L, C), it is essential that L is given by two points. Unlike lines, circles do not come equipped with a direction. Two circles having the same points are considered equal. We will state this below, when the equidistance (or “congruence”) relation is introduced. This is not essential; we could introduce a new symbol for extensional equality of circles, and keep the equality symbol to mean intensional equality; but we have no need of intensional equality of circles for geometry; once a circle is constructed, we don’t care how it was constructed. By contrast, we do care about the direction of a line. We don’t care in general about the specific two points used to construct a line, so we could identify two lines with the same direction, but there is no pressing reason to do so. 8.8. Betweenness axioms. The basic relations of our theories are equidistance and betweenness, which have been recognized as fundamental at least since Hilbert’s famous 1899 book [15]. All the arguments of these two relations have sort Point. The (strict) betweenness relation is written B(a,b,c).17 We read this “b is between a and c”. The intended meaning is that that the three points are collinear and distinct, and b is the middle one of the three.

17Hilbert [15] and Greenberg [12] use strict betweenness, as we do. Tarski [30] used non- strict betweenness. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 49 B(a,b,c) on(b, Line (a,c)) (Axiom B1-i) B(a,b,c) → a = c (Axiom B1-ii) B(a,b,c) → a =6 b (Axiom B1-iii) B(a,b,c) → B6(c,b,a) (Axiom B1-iv) → Before giving a constructive version of the remaining betweenness axioms, we discuss a related principle, “Markov’s principle for betweenness”, which has already been listed as Axiom S5: B(A, B, C) B(A, B, C) ¬¬ → Markov’s principle expresses the idea that by computing two points P and Q etc. to greater and greater accuracy, if they are not identical we will eventually find that out. We want it to be provable in ECG for several reasons: it is used in Euclid (e.g. I.6 and I.26, as is discussed below); it is needed to permit argument by cases for a conclusion that is a conjunction of equalities and betweenness statements; and without such arguments we could not make significant headway in geometry. Besides, it makes for a smooth metatheory. While some may consider Markov’s principle in number theory to be of questionable constructivity, we consider that geometry without Markov’s principle is awkward. Specifically, it would not support Euclid. As we shall see in other sections, Euclid does use it, and including it does not harm our ability to construct things that are proved to exist. In terms of order, it expresses the principle that x 0 x> 0. Hilbert’s second betweenness axiom says that Line (P,Q¬ )≤ contains→ a point between P and Q. We call this the “density” axiom. Since we want a quantifier- free axiomatization, we would like to specify the point asserted to exist. The natural candidate for a point between P and Q is the result of Euclid’s segment- bisection construction. We therefore take the following axiom: B(P, IntersectLines (Line(P,Q),Line( (Axiom B2) IntersectCircles1 (Circle(P,Q),Circle(Q, P )), IntersectCircles2 (Circle(P,Q),Circle(Q, P )))),Q) Note that, in view of the strictness axioms of LPT, this axiom implies that the circles and intersection points involved are defined. Hilbert’s second axiom for betweenness says that between two distinct points on a line there is another point. His third axiom for betweenness is, “Of any three points on a line there exists no more than one that lies between the other two.”18 Here is a constructive version: a = b a = c b = c on(a,L) on(b,L) on(c,L) (Axiom B3) 6 ( ∧B(6a,b,c∧) 6 B∧(b,c,a) ∧ B(c,a,b∧ )) → (¬B(b,c,a) ∧ ¬B(c,a,b) → ¬¬B(a,b,c))∧ (¬B(c,a,b) ∧ ¬B(a,b,c) → ¬¬B(b,c,a))∧ ¬(B(a,b,c) ∧B ¬ (b,c,a)) →(B ¬¬(a,b,c) B(∧b,a,c)) ¬(B(b,c,a) ∧ B(b,a,c)) ∧ ¬ ∧ ∧ ¬ ∧

18In Greenberg’s system this becomes the slightly diﬀerent “given three distinct points on a line, one and only one of the points is between the other two.” That formulation is too strong, constructively. (For example its translation into the recursive plane is not provable in HA plus Markov’s principle.) 50 MICHAEL BEESON

In view of Markov’s principle for betweenness (Axiom S5), we could remove the double negations on the right. If we did so, then Axiom S5 would become superfluous, as is shown in the following lemma. But we prefer to keep Axiom S5 and the Axiom B3 just given, because it facilitates the discussion of different versions of Euclid’s parallel postulate, whose relations depend on Markov’s principle. Lemma 8.5. Let B3* denote Axiom B3 with the double negations removed. Markov’s principle for betweenness (Axiom S5) is provable from Axiom B3*. Proof. Suppose B(a,x,b). We want to prove B(a,x,b). From B(a,x,b) we immediately¬¬ have a = b and a = x and x = b. By Axiom B3* it¬¬ suffices to prove B(x,a,b) and B6 (a,b,x). To6 prove B6 (x,a,b), suppose B(x,a,b). Then B(a,x,b¬ ), by Axiom¬ B3*. But that contradicts¬ B(a,x,b). That proves ¬B(x,a,b). Similarly we have B(a,b,x). That completes¬¬ the proof of the lemma.¬ ¬ 19 Hilbert’s fourth betweenness axiom is often known as Pasch’s axiom. It says that if line L meets (the interior of) side AB of triangle ABC then it meets (the interior of) side AC or side BC as well. Constructively we had better say that if it meets AB and does not meet AC, then it meets BC. Note that the most natural way of writing this does not further specify (as Hilbert did) that the three points A, B, and C should not be collinear. We do not want to do that constructively since generally we can’t decide whether three points are or or not collinear, and the Pasch axiom is uniformly valid, so there is no pressing reason to require the three points to not be collinear. Tarski refined Pasch’s axiom, considering three forms, “inner Pasch”, “outer Pasch”, and “weak Pasch”, of which only the first is an axiom of ECG. The point of Tarski’s refinements seems to be twofold: (i) these restricted versions can be expressed in an AE form, which become quantifier-free when we have the function symbol IntersectLines (this is not true of the unrestricted Pasch axiom), and (ii) the restricted forms are valid even in three-dimensional space, so they do not make an implicit dimensional assertion, as the unrestricted Pasch axiom does (it fails in three-space). The version we take is similar to the “inner Pasch” Axiom 7 of [26]; it differs only in replacing an existential by a term using IntersectLines for the point asserted to exist. See Fig. 9. B(a,p,c) B(b,q,c) p = b q = a on(c, Line (a, q) IntersectLines∧ (Line∧ 6 (p,b∧), Line6 (∧a, ¬ q)) → B(p, IntersectLines (Line (p,b), Line (a,↓ q ∧)),b) B(q, IntersectLines (Line (p,b), Line (a, q)),a)∧ (Pasch’s Axiom B4)

This version diﬀers from [26] in that the point x such that B(p,x,b) B(q,x,a) is explicitly given, instead of having the axiom say that some x exists∧ with the

19Hilbert did not need this axiom because he had his “plane separation axiom.” It is derived from the plane separation axioms as Greenberg’s Prop. 3.3 [12], p. 112. Tarski used non-strict betweenness, here called T , and his version of this axiom was T(a, b, a) a = b; it is Axiom → 6 in [30] and [26]. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 51

Figure 9. Inner Pasch (left) and Outer Pasch (right). Line pb meets triangle acq in one side. The open circles show the points asserted to exist on the other side.

b b

b p b c

b b b q p a x b b b b a x q desired betweenness properties. Also, we use strict betweenness instead of non- strict betweenness, and the hypothesis means that the lines whose intersection point is x must be defined and must not coincide, since then x would not be defined. We will see that Tarski’s version, with non-strict betweenness, is not constructively valid. For now let it suffice to observe that as b goes to q and p goes to a, the position of x could be anywhere between a and q, depending how close b is to q and p is to a, so x cannot be constructed continuously as would be required for a constructive version of this axiom with non-strict betweenness. The inner Pasch axiom can be used to derive the outer and weak Pasch axioms, as we shall see; and as Tarski, Szmielew, Gupta, and Schwabhauser showed, Tarski’s version, with non-strict betweenness, can be used with the axiom B(a,b,a) to derive our axioms B3 and B1-iii and B1-iv; but that is not true¬ with strict betweenness. 8.9. Sides of a line. Two points a and b not on line L are on opposite sides of L if a = b and there is a point of L between a and b, i.e., the segment ab meets L. 6 Definition 8.6. OppositeSide (a,b,L) := B(a, IntersectLines (Line (a,b),L),b) The definition of being on the same side is less straightforward. One obvious definition of SameSide (a,b,L) would be that segment ab does not meet L. We have to ensure that a point not on L is on the same side of L as itself, though, without implying that we can test whether two points are equal, so we would have to say x (B(a,x,b) on(x, L)). ∀ ¬ ∧ If we try to make this quantifier-free by replacing x by IntersectLines (Line (a,b),L) we have to watch out for the case a = b. It does not work to say (a = b B(a, IntersectLines (Line (a,b),L),b)) ¬ 6 ∧ 52 MICHAEL BEESON because (intuitively) the truth value of B(a, IntersectLines (Line (a,b),L),b) is indeterminate when a = b. More precisely, when we put a = b in this formula, the result will be equivalent to B(a, IntersectLines (Line (a,a),L),a), ¬ which contains an undefined term and is not provable, even though x B(a,x,b) is derivable. ∀ ¬ One can get an existential quantifier instead of a univeral quantifier by using Tarski’s definition, as given in [26]. This says that a and b are on the same side of L if there is some c such that both a and b are on the opposite side of L from c. Another advantage of this definition is that it works in more than two dimensions. Definition 8.7. SameSide (a,b,L) := c (B(a, IntersectLines (Line (a,c),L,c) ∃ ∧ B(b, IntersectLines (Line (b,c),L),c)) The question of the equivalence of this definition with the first candidate will be addressed in Lemma 9.83 below. Evidently the upper dimension axiom will be needed. 8.10. Right and Left : handedness and sides of lines. Consider two circles C and K with different centers a and b respectively. They intersect (if at all) in two points p (which may coincide if the circles are tangent). If they do not coincide then we define abp to be a “right turn” if p = IntersectCircles1 (C,K):

Definition 8.8. Right (a,b,p) := p = IntersectCircles1 (Circle (a,p), Circle (b,p)) Left (a,b,p) := p = IntersectCircles2 (Circle (a,p), Circle (b,p)) Note that this counts abb as both a right turn and a left turn, but aab and aaa are neither right turns nor left turns, since the terms on the right are undefined in those cases by Axioms I17 and I18. We then take as axioms the statements that if abp is a right turn and abq is a right turn, and p and q are not on Line (a,b), then no point of that line is between p and q, and similarly for left turns: Right (a,b,p) Right (a,b,q) on(p, Line (a,b)) on(q, Line (a,b)) (on(x, Line∧ (a,b)) B(∧p, ¬x, q))∧ ¬ (Axiom→ H1) Left (¬a,b,p) Left (a,b,q∧) on(p, Line (a,b)) on(q, Line (a,b)) (on(x,∧Line (a,b)) ∧B ¬(p, x, q))∧ ¬ (Axiom→ H2) ¬ ∧ Similarly, if abp is a right turn and abq is a left turn, and p and q are not on Line (a,b), then p and q are on opposite sides of Line (a,b): Right (a,b,p) Left (a,b,q) on(p, Line (a,b)) on(q, Line (a,b)) B(p, IntersectLines∧ (Line (p,∧ q ¬), Line (a,b)), q)∧ ¬ (Axiom→ H3) Conversely, if p and q are on the same side of Line (a,b), then the turns abp and abq have the same sense. Here we use the definition of SameSide given FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 53 above. That will put an existential quantifier in the antecedent of an implication; writing it out more explicitly the axiom will be quantifier-free. Thus SameSide (p,q, Line (a,b)) φ becomes → B(p,x,y) B(q,z,y) on(x, Line (a,b)) on (z, Line (a,b) φ. ∧ ∧ ∧ → Applying this procedure to SameSide (p,q, Line (a,b)) Right (a,b,p) Right (a,b,q) ∧ → and to the similar formula with Left instead of Right, we have the next two axioms (see Fig. 10):

Figure 10. (Axiom H4) abq is a right turn if and only if abp is a right turn. Points x, y, and z witness that p and q are on the same side of Line (a,b).

b y

ab bb z b b x

b b q p

B(p,x,y) B(q,z,y) on(x, Line (a,b)) on (z, Line (a,b) Right∧(a,b,p) ∧Right (a,b,q)∧ ∧ (Axiom H4) B(p,x,y) B(q,z,y→) on(x, Line (a,b)) on (z, Line (a,b) Left (a,b,p∧ ) Left∧ (a,b,q)∧ ∧ (Axiom H5) → The next two axioms say that if p and q are on opposite sides of Line (a,b), then the turns have the opposite sense: B(p,q,x) on (x, Line (a,b)) Right (a,b,p) Left (a,b,q) (Axiom H6) B(p,q,x) ∧ on (x, Line (a,b)) ∧ Left (a,b,p) →Right (a,b,q) (Axiom H7) ∧ ∧ → The following axioms connects the turns in a triangle, or technically, in a list of three distinct points. Note that Right (a,b,c) implies a = b, as does Left (a,b,c), since it is deﬁned using the intersection points of circles6 with centers a and b, and Axiom I17 and A18 guarantee that these intersection points are not deﬁned for concentric circles. That is why we do not need the hypothesis a = b in the next four axioms. 6 Right (a,b,c) b = c Right (b,c,a) Right (c,a,b) (Axiom H8) Left (a,b,c) ∧b =6 c →Left (b,c,a) Left∧ (c,a,b) (Axiom H9) Right (a,b,c∧) 6 Left→(a,c,b)∧ (Axiom H10) Left (a,b,c) →Right (a,c,b) (Axiom H11) → 54 MICHAEL BEESON

Note that it is not necessary to specify that a, b, and c are not collinear in the above axioms, since in that case all the turns are both left and right. There are three constants α, β, and γ in the theory. The following axiom resolves an ambiguity that otherwise would exist (which one could think of as an ambiguity about from which side we are viewing “the plane”). Left (α,β,γ) (Axiom H12) Nothing in these axioms prevents the three points from being collinear, but that will be ruled out by the dimension axioms below. 8.11. Dimension axioms. There are three constants α, β, and γ with axioms saying that these three points are non-collinear. Specifically on (α, Line (β,γ)) (Axiom D1) ¬on (β, Line (α, γ)) (Axiom D2) ¬on (γ, Line (α, β)) (Axiom D3) α¬ = β α = γ β = γ (Axiom D4) 6 ∧ 6 ∧ 6 We do not have to say explicitly that α because it is part of the logic of partial terms that every constant is defined–it is nothing↓ special to any particular theory. Axiom D4 is needed since without it there is nothing to rule out the possibility that α = β = γ; in that case Line (β,γ) would be undefined, so on(α, Line (β,γ)) is false, and axiom D1 would still hold. One also needs an “upper dimension axiom” to ensure that models of the theory are planes.20 We take the one chosen by Tarski, which says that if three circles with different centers have a common chord, their centers are collinear:

On(a, C1)) On(b, C1) On(a, C2) On(b, C2) (Axiom D5) ∧ ∧ ∧ On(a, C3) On(b, C3) a = b center (C1) = center (C2) ∧ ∧ ∧ 6 ∧ 6 center (C1) = center (C3) center (C2) = center (C3) ∧ 6 ∧ 6 On(center (C3), Line (center (C1), center (C2))) → 8.12. Rays and segments. Although we have not included rays and segments in ECG, we do want to support our claim that a conservative extension including them can easily be introduced; and also, we sometimes make informal arguments using rays with the implication that they can be formalized in ECG. We now show that the use of intuitionistic logic does not cause a problem about incidence on rays or segments. Using betweenness, we can define incidence for rays. However, there is a technicality: the origin O of Ray (O,B) is considered to lie on the ray, i.e. “rays are closed”, while betweenness means “strictly between.” It is thus easier to define the “opposite ray”: P is on the opposite ray to Ray (O,B) if P is on Line (O,B) and O is between P and B. Then Q is on Ray (O,B) if it is on Line (O,B) but not on the opposite ray: on (Q,Ray(O,B)) := on (Q, Line(O,B)) B(P,O,B) (Definition 4) ∧ ¬ This definition can be used to express informal arguments about rays in ECG without needing to introduce an explicit sort and axioms for rays.

20With classical logic and full ﬁrst-order continuity in place of line-circle continuity, one can take for such an axiom essentially any formula that holds in R2 but not in Rn for n > 2, as shown by Scott [27]. Whether this is true also with only line-circle continuity and classical logic, let alone with intuitionistic logic, I do not know. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 55

In a similar way we could define incidence for segments, so that “segments are closed”, but we do not do so. 8.13. Congruence axioms. The equidistance relation is officially written E(A,B,C,D). Axioms for equidistance are sometimes called “congruence axioms” since equidistance can be thought of as congruence of segments. Fol- lowing a tradition that goes back to Euclid, we write AB = CD instead of E(A,B,C,D). Hilbert’s first congruence axiom, paraphrased from [12], is closely related to the uniform version of Euclid’s Book I, proposition 2: A = B C = D (not an axiom of ECG) 6 R∧(on (6 R,Ray→(A, B)) AR = CD ∃ ∧ This axiom permits us to “lay off” segment CD along Ray (A, B). Since we are seeking a quantifier-free axiomatization, we want to specify the point R. This we do by taking R(A,B,C,D)= IntersectLineCircle2 (Line (A, B), Circle3 (A,C,D)). It should be IntersectLineCircle2 rather than IntersectLineCircle1 because we intend that those two intersection section points should occur on Line(A, B) in the same order as A and B. Since we do not take Ray as a primitive of the language, we need to eliminate it, replacing on (R,Ray(A, B)) by on(R, Line (A, B) B(R,A,B). Since we allow degenerate circles, we drop the hypothesis C = D∧ as¬ well. Our version of the axiom is thus 6 a = b on (R(a,b,c,d),Line(a,b)) 6 → B(R(a,b,c,d),a,b) E(a, R(a,b,c,d),c,d) ∧¬ ∧ Note that the strictness axioms then will imply that R(a,b,c,d) is defined when a = b. It is allowed that c = d since we are allowing degenerate circles. The official6 version, with R replaced by is definition, is Axiom C1: a = b (Axiom C1) 6 → (on (IntersectLineCircle2 (Line (a,b), Circle3 (a,c,d)),Line(a,b)) B(IntersectLineCircle2 (Line (a,b), Circle3 (a,c,d)),a,b) ∧E ¬ (a, IntersectLineCircle2 (Line (a,b), Circle3 (a,c,d)),c,d) ∧ We do not need to require the uniqueness of R in the axiom, since we can prove the uniqueness from other axioms; this will be demonstrated in Lemma 9.4 below. Just below are four more congruence axioms, adapted from Tarski’s system. The first two turn out be enough to guarantee that congruence of segments is an equivalence relation (which we prove in a subsequent section); Hilbert stated that in his axioms. Note that Axiom C2 is not exactly transitivity of congruence, although it has a similar flavor. Hilbert did not need C3 (which was introduced by Tarski), since he did not allow degenerate segments; and he did not need C4 since he treated segments as (unordered) pairs of points. In our theory, ab = cd is a 4-ary relation on points, so we need this axiom. Axiom C3 can be thought of as saying that addition is well-defined on congruence classes of segments. We therefore refer to it as the “additive congruence axiom.” ab = cd ab = pq cd = pq (Axiom C2) ab = cc ∧ a = b → (Axiom C3) ab = ba → (Axiom C4) 56 MICHAEL BEESON

One might take the additivity of congruence B(a,b,c) B(p,q,r) ab = pq bc = qr ac = pr ∧ ∧ ∧ → as an axiom, except that it would be redundant to do so, as will be shown in Lemma 9.3 below. Using equidistance, we define incidence for circles: On (x, Circle3 (a,b,c)) ax = bc (Axiom C5) ↔ Since we are allowing circles of zero radius, we do not need to specify A = Q. We need Axiom C3 to guarantee that the only point on a circle of zero radius6 is its center. Now to complete the list of congruence axioms, we have to choose whether to follow Hilbert or Tarski. Hilbert’s axioms are perhaps closer in spirit to Euclid, though Euclid did not perceive the necessity of any congruence axioms beyond what has already been stated. (He tried to prove SAS, using an argument by “superposition”.) In [3] we followed Hilbert, but here we choose to follow Tarski, as that enables us to complete the list of axioms immediately, and separate the discussion of angles and triangles into another section; and it also makes it easier for us to make use of the results of [26], which is based on Tarski’s axioms. The crucial axiom is Tarski’s 5-segment axiom, which is illustrated in Fig. 8 and discussed near that figure. We take the following variant of Tarski’s 5-segment axiom as our last congruence axiom: a = b B(a,b,c) B(A, B, C) ab = AB bc = BC (Axiom C6) ad6 =∧AD bd =∧BD cd =∧CD ∧ ∧ ∧ → Note that Axiom C6 does not allow a = b or b = c, but it does allow d to lie on Line (a,b). Constructively this axiom may seem inadequate, because we would like the length of cd to be determined even when c is between a and b, and without having to make a case distinction about that; but we will show in Lemma 9.10 that an extended version of the axiom is already provable. In essence, one can argue by cases to prove a congruence relation, because congruence is stable. 8.14. Line-circle continuity. We now formulate the axioms of line-circle continuity. The first two of these just tell us when a line and a circle intersect– namely, when there is a point on the line closer (or equally close) to the center than the radius of the circle.21 But we have not defined inequalities for segments yet, so the formal statement is a bit more complex. Moreover, we have to include the case of a degenerate circle or a line tangent to a circle, without making a case distinction.22 Therefore we must find a way to express “P is inside the closed Circle (A, Y )”. For that it suffices that there should be some X non-strictly between A and Y such that AX = AP . Since this will appear in the antecedent of the axiom, the “some X” will not involve an existential quantifier. Nonstrict betweenness T can be defined, since we have Markov’s principle for betweenness:

21Note that in spite of the use of the word “circle” the axiom is valid in n-dimensional Euclidean space, where it refers to the intersections of lines and spheres. 22Avigad et. al. count only transverse intersection, not tangential intersection, as “intersection.” FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 57

Definition 8.9. Non-strict betweenness is defined by T(a,b,c) := (a = b b = c B(a,b,c)) ¬ 6 ∧ 6 ∧ ¬ This disjunction-free definition is equivalent to (a = b b = c B(a,b,c)) ¬¬ ∨ ∨ Definition 8.10. ab < cd (or cd > ab) means x(B(c, x, d) ax = ab). ab cd (or cd ab) means x(T(c, x, d) ax∃= ab), where∧ T is non-strict betweenness.≤ ≥ ∃ ∧ Definition 8.11. Let C be a circle with center a. Then point p is strictly inside C means there exists a point b on C such that ap < ab, and p is inside C, or non-strictly inside C, means ap ab. ≤ We could equivalently say “for every b on C” or “for b = pointOnCircle (C)”. The first axiom for line-circle continuity will then say that if line L has a point p inside circle C, then both intersection points of the line L and Circle (A, Y ) are defined. We first express this in terms of T. See Fig. 11

Figure 11. Line-circle continuity. p is inside the circle, so L meets the circle.

b p a b b b b x

on(p,L) ax = ap T(a,x,b) IntersectLineCircle1∧ ∧ (L, Circle→ (a,b)) IntersectLineCircle2 (L, Circle (a,b)) ↓ ↓ Now we express this in terms of the primitives of the language, eliminating the abbreviation T: on(p,L) ax = ap (a = x x = b B(a,x,b)) IntersectLineCircle1∧ ∧ ¬ (6 L, Circle∧ 6 (a,b∧)) ¬ (Axiom→ Cont 1) IntersectLineCircle2 (L, Circle (a,b)) ↓ ↓ 58 MICHAEL BEESON

Our next axiom says that if L contains a point strictly inside C then the two intersection points are distinct. on(p,L) ax = ap (a = x B(a,x,b)) ax = ap IntersectLineCircle1∧ ∧ ¬ (6 L, Circle∧ ¬ (a,b)) =∧ → (Axiom Cont 2) IntersectLineCircle2 (L, Circle (a,b)) 6 This axiom deserves some discussion. Consider the propositions (i) if L is a line and e is a point not on L then there is a perpendicular from e to L, and (ii) if L is a line containing a point x strictly inside circle C, then L meets C in at least two points. It is not hard to prove (i) from (ii), but if we follow Euclid, we need (ii) for (i). Also (i) can be proved from the “triangle existence theorem” (Lemma 9.67 below): let a and b be two points on L, and construct triangle abE congruent to abe with E on the other side of L from E; then eE L. The triangle existence theorem is an axiom in Hilbert’s system, so with Hi⊥lbert’s axioms we can dispense with Axiom Cont 2. Actually, Axiom Cont 2 is superfluous, since (i) can also be proved in Tarski’s system, which we are following here, without appealing to line-circle continuity at all. This development is presented in [26] and is completely constructive. However, it is long and complicated, and far from Euclid. We have therefore chosen to include this “harmless little axiom”, and follow Euclid in the construction of perpendiculars. Our next axiom says that the intersection points depend extensionally on the circle: A = center (C) A = center (K) (Axiom Cont 3) On (P, C) On∧(Q,K) AP = AQ∧ ∧ ∧ → IntersectLineCircle1 (L, C) = IntersectLineCircle1 (L,K) ∼ ∧ IntersectLineCircle2 (L, C) ∼= IntersectLineCircle2 (L,K) Note that these axioms do not guarantee the continuous dependence (on the point and circle) of the intersection points defined by each of the two function symbols. In other words, we have not yet specified which of the two intersection points is which. We want to say that the two intersection points occur in the same order on Line (a,b) as a and b occur. That is done by the following axioms:

on(z,L) Right (z,a,b) Right (z, IntersectLineCircle1 (Line (a,b), C), ¬ IntersectLineCircle2∧ (Line→ (a,b), C)) (Axiom Cont 4) on(z,L) Left (z,a,b) Left (z, IntersectLineCircle1 (Line (a,b), C), ¬ IntersectLineCircle2∧ →(Line (a,b), C)) (Axiom Cont 5) The order of the arguments is important here, because if the line is tangent to the circle, the two intersection points will coincide, so we may have Right (p,a,a), which is true, but Right (a,a,p) is always false. Note that the axiom is still correct if the two intersection points coincide, since pxx is both a right and a left turn. 8.15. Intersections of circles. We next give the axiom(s) known as line- circle continuity. The axioms in question should say that if point p on circle K lies (non-strictly) inside circle C, and point q lies (non-strictly) outside C, then both intersection points of the circles are deﬁned. (Incidence axioms already given then guarantee they are on both circles.) To express that p is non-strictly inside (or outside) Circle (A, Y ) we use the same technique as just above. Namely, p FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 59 is inside C if ap = ax for some X non-strictly between a and a point y on C. The situation is illustrated in Fig. 12. We need this axiom to apply even to degenerate circles, and to points that are on C rather than strictly inside, so we cannot say B(x,y,z) as we must allow x = y or y = z, and we must even allow a = x = y = z; hence the point w in the axiom is needed so that we deﬁnitely have a line connecting a and w on which x, y, and z must lie, non-strictly in that order. The axiom expresses this using the primitive relation B.

Figure 12. Circle-circle continuity. p is inside C and q is outside C, as witnessed by x, y, and z, so the intersection points 1 and 2 exist. ab1 is a right turn and ab2 is a left turn.

b w b b z 2 q y b

b b x p

b b a b

1 C K

ap = ax aq = az on (q,K) on (p,K) on (y, C) (Axiom Cont 6) B(y,x,z∧ ) B∧(x,z,y) B∧(a, x, w) ∧B(a,y,w) B(a,z,w) ∧ ¬ IntersectCircles1∧ ¬ (C,K)∧ IntersectCircles2∧ (C,K∧ ) → ↓ ∧ ↓ In Section 8.10, we have already provided axioms to distinguish the two points of intersection, according as the triple formed by the two centers and the point of intersection is a right turn or a left turn. The axiom for IntersectCirclesSame is similar (but does not involve Right and Left): On(P, C) On(Q, C) AP = AX AQ = AZ (Axiom Cont 7) on(X,∧Line (A, Y ))∧ on(Z, Line∧(A, Y )) ∧ B(X, A, Y ) B(∧A,Y,X) B(A, Z,∧ Y ) B(Z, A, Y ) Z = A ¬ B(A,Y,X∧) ¬ AX = AP ∧ ¬On(P, C) ∧ ¬ ∧ 6 ∧ ¬on(x, Line (∧center(C), center∧ (K)) ∧ ¬ IntersectCirclesSame (C,K,x) IntersectCirclesOpp (C,K,x) → ↓ ∧ ↓ 60 MICHAEL BEESON

We have already provided in Axiom I10 that once IntersectCirclesSame (C,K,x) is defined, it lies on the same (closed) side of the line joining the centers of C and K as x does, and IntersectCirclesOpp (C,K,x) lies on the opposite side. In other words, if the circles are tangent, and x is not on the line L joining the centers, then IntersectCirclesSame (C,K,x) is still defined. This axiom and I10 together do not specify anything about what happens when x lies on L (since nothing is specified, we cannot use IntersectCirclesSame unless we know x is not on L). Our next axiom specifies that the intersection points of two circles depend extensionally on the circles:

A = center (C1) A = center (C2) (Axiom Cont 8) ∧ ∧ On (P, C1) On (Q, C1) AP = AQ ∧ ∧ → IntersectCircles1 (C1,K) = IntersectCircles1 (C2,K) ∼ ∧ IntersectCircles2 (C1,K) = IntersectCircles2 (C2,K) ∼ ∧ IntersectCircles1 (K, C1) = IntersectCircles1 (K, C2) ∼ ∧ IntersectCircles2 (K, C1) ∼= IntersectCircles2 (K, C2) IntersectCirclesSame (C,K,x) ∼= IntersectCirclesSame (C,K,x) IntersectCirclesOpp (C,K,x) ∼= IntersectCirclesOpp (C,K,x) This contrasts with the line-circle intersection operation, where if we change Line (A, B) to Line (B, A), the two intersection points with a given circle switch labels. That is absolutely necessary, as the following example shows: let line L pass through the center of circle C. Then continuously rotate line L through 180 degrees. That switches the two points of intersection, and changes Line (A, B) to Line (B, A). We deﬁnitely want the points of intersection to vary continuously, so we cannot have them depend extensionally on the line. Consider what happens when we rotate circles instead. Start with two intersecting circles of the same size, and rotate both circles 180 degrees, around the midpoint of the segment connecting their centers center of one around the center of the other. Then the intersection point that was originally at the top of the diagram is now at the bottom. But the circles have also switched places. So there is also a label- switching property here, but it involves switching the labels of the centers of the circles as well as the labels of the intersection points. That is, IntersectCircles1 (Circle3 (a,x,y), Circle3 (b,x,y)) = IntersectCircles2 (Circle 3(b,x,y), Circle3 (a,x,y)) This does not contradict Axiom Cont 7. We need an axiom to guarantee the continuity of pointOnCircle . The intuitive idea is that pointOnCircle (C) should always be a point on C in a canonical direction (say “north”) from the center of C. But there are no “canonical directions”, so we have to express continuity indirectly. Consider two circles C and K, and let p = pointOnCircle(C) and q = pointOnCircle (K), and let a and b be the centers of C and K, respectively. If the two circles are the same size we could require that pq = ab. If they are not the same size, then if t is a point on Ray (b, q) with bt = ap, we require pt = ab. See Fig. 13 for an illustration. We want the axiom to guarantee that the ﬁgure is accurate, but to do that we also need to specify that p and t are on the same side of Line (a,b) (if a = b). Of course a = b is also allowed, and also all four points could be collinear.6 FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 61

Figure 13. Axiom Cont 9: if bt = ap then pt = ab and segment pt does not meet ab. q b K p b t b C

b b b a

a = center (C) b = center (K) (Axiom Cont 9) p = pointOnCircle∧ (C) q = pointOnCircle (K) ∧ ( B(b,t,q) t = q ∧b = t) ap = bt ∧B ¬(u,b,q¬ ) On∧(u,K6 ) ∧B(6 p,x,t∧) ∧ pt = ab∧ B(a,x,b∧) → ∧ ¬ In the ﬁgure, point v is IntersectLines (Line (p,u), Line (a,b)). The hypothesis B(u,b,q) means the axiom will not apply when K is a degenerate circle, but in that case there is only one possibility for q anyway. The formula ( B(b,t,q) t = q b = t) is just the translation into primitive symbols of T(b,t,q¬ ¬) (non-strict∧ betweenness).6 ∧ 6

§9. Development of neutral geometry in ECG. It will be necessary to prove some basic theorems in ECG, in spite of the existence of the double negation interpretation and the work in [26]; we will explain this point in more detail below. The word “neutral” here means that part of geometry not using the parallel axiom. In fact, for much of this development we also do not use the upper dimension axiom; it is used for the first time in section 9.14. In this version of ECG, we have followed Tarski’s lead in choosing a small set of short, simple axioms for betweenness and congruence, while keeping the larger language of points, lines, and incidence introduced by Hilbert. The detailed development of elementary geometry from Tarski’s axioms has been carried out in [26] in almost formal detail, so it would be nice if could use some of this work rather than going over similar territory but paying attention to the constructivity of the proofs. As mentioned above, we have used strict betweenness B instead of non-strict betweenness T as Tarski did, and consequently we have a few more betweenness axioms, and our other axioms, similar in form to Tarski’s, may be weaker or different because they use B instead of T. However, in carrying out proofs, one notices repeatedly that one argues by cases, for example on B(a,b,c) B(b,c,a) B(c,a,b) for three distinct points on the same line, and then one∨ notes that∨ since the double negation of this disjunction is valid, the 62 MICHAEL BEESON double negation of whatever one has proved by cases is valid. If that conclusion is, for example, an atomic formula or conjunction of atomic formulas, then its double negation implies itself, by the axioms of ECG. We would haved liked to formulate a metatheorem about this phenomenon that would have enabled the wholesale importation of a useful class of theorems from [26]. The first subsection below explains why this cannot be made to work. The rest of the section gives proofs of some basic theorems of constructive geometry in ECG(without using any parallel axiom). We avoid the parallel axiom so that these results can be used in our study of the provability relations between different versions of the parallel axiom, and so that they can be used (in subsequent work) to study constructive non-Euclidean geometry. Since [26] works with non-strict betweenness, we must be careful not to confuse our betweenness B with non-strict betweenness, which we here denote by T. Classically, T(a,b,c) means that a = b b = c B(a,b,c); constructively, we define it by taking the double negation and∨ pushing∨ one negation in: T(a,b,c) := (a = b b = c B(a,b,c). ¬ 6 ∧ 6 ∧ ¬ 9.1. Why we cannot import negative theorems from Tarski’s theory. Gödel introduced [11] his double-negation interpretation for this purpose. This interpretation assigns a formula A− to every formula A, by replacing by ∃ ¬∀¬ and replacing A B by ( A B). For atomic formulae, A− is defined to be A. The rules∨ of intuitionistic¬ ¬ ∧ ¬ logic are such that if A is classically provable ¬¬ (in predicate logic) then A− is intuitionistically provable. Hence, if we have a theory T with classical logic, and another theory S with intuitionistic logic, whose language includes that of T , and for every axiom A of T , S proves A−, then S also proves A− for every theorem A of T . In case the atomic formulas in the language of T are stable in S, i.e. equivalent to their double negations, then of course we can drop the double negations on atomic formulas in A−. One might like to try this with T taken to be Tarski’s theory as used in [26], and S taken to be ECG. Indeed that is straightforward, and we shall present the details in a later section, after developing some geometry in ECG. The question addressed here is whether we could possibly avoid some of that development. The short answer is “no”. We will explain the difficulties, in order to justify the necessity of going over the ground plowed by [26] again from a constructivist viewpoint. Having the double negation interpretation from Tarski’s theory into ECG only shows us that with classical logic the two theories are equivalent; that will make only negative theorems automatically constructively provable. But many of the theorems in which we are interested are not negative. To give a specific example, the “outer Pasch” theorem, Satz 9.6 of [26] asserts the existence of a point x, which (in the non-degenerate case) is given as the intersection of two lines, but the double negation interpretation is only going to tell us that such a point cannot fail to exist. If it does exist, it will be the intersection point of two specific lines, but the interpretation is only going to tell us that those two lines cannot fail to intersect, not that they actually do intersect. To conclude that they do intersect we need the stability principle t t , which we considered taking as Axiom S6, but did not, because¬¬ it can↓ → be p↓roved using FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 63 other axioms of ECG, in particular in this case the strong parallel axiom. But we want to prove that inner Pasch implies outer Pasch without using the strong parallel axiom; we need to work without the strong parallel axiom in order to study the independence relations of the various parallel axioms. We might have considered taking S6 as an axiom if it would enable us to import the negative theorems of [26] in one blow. But there is a second difficulty: Tarski’s axioms have many “degenerate” cases, in which the points that appear in the diagram are collinear or some of them coincide. These cases are very useful in developing Tarski’s theory and contributed to the small number of axioms required: for example, the symmetry property of betweenness is proved using the five-segment axiom in a degenerate case. (That is why ECG has to have the symmetry property as an axiom.) One cannot define explict Skolem functions in ECG to cover the degenerate cases. Therefore, one cannot avoid the need for S6 by first extending Tarski’s theory by Skolem functions. One can extend Tarski’s theory by Skolem functions, of course, but then one cannot interpret them by definable terms of ECG, unless one is also willing to change the axioms of Tarski’s theory to avoid degenerate cases. But if one makes such changes, then one is no longer able to import theorems wholesale from [26]. A second example of a theorem derived using degenerate cases of the axioms is the “inner transitivity” of betweenness, B(a,b,d) B(a,b,c) B(b,c,d). This was an axiom of early versions of Tarski’s theories,∧ but was→ proved in Gupta’s thesis, and occurs only halfway through [26], and the proof uses degenerate cases of the axioms. As it happens those particular degenerate cases are also allowed in ECG, so that proof can be constructivized. But there is no way to distinguish these two examples without looking at the details. The conclusion is that there is no cheap way to import the results of [26]. One must actually go through the mathematics and establish the constructive content. Moreover, there are other reasons why that is worth doing: one also has to constructivize the definitions, i.e. to make sure that the defined concepts can be constructivized, and to choose the correct way of doing so. For example, what does it mean to construct a triangle congruent to abc “on a given side of” segment AB (assuming AB is congruent to ab? One may ask for a specific counterexample to the conjecture that negative theorems of Tarski’s theory are provable in ECG. For example, the inner Pasch axiom of Tarski’s theory is negative, and its translation into ECG is also negative when non-strict betweenness is given a negative interpretation, and the result is not provable in ECG(because as the given points tend to a degenerate case, the dependence of the constructed point is not continuous). Details of this result are given in Theorem ??. In addition to these issues, there are other reasons for developing some geometry in ECG. There are a number of definitions to be made (ordering of segments, angles, congruence of angles, ordering of angles, etc.) that are necessary to support Euclid, and there are some interesting constructive distinctions to be made. In other words, the development of geometry in ECG has independent interest, and is not just a matter of taking the double negation interpretation of the classical development. On the other hand, much of the development in [26] is essentially constructive, and generally speaking, using strict betweenness 64 MICHAEL BEESON makes the proofs easier, not more difficult. Therefore we do not repeat lengthy arguments from [26] when there is no significant new twist require for the con- structivization. 9.2. Congruence and betweenness lemmas. In this section we prove some basic lemmas about congruence and betweenness. In some cases we have followed proofs from [26]. Note how the five-segment axiom is used, even when all the segments lie on the same line. All the lemmas in this section are proved in ECG. In Tarski’s theories, the betweenness axioms have almost disappeared, compared to Hilbert; only the axiom that T (x, y, x) x = y. One recovers the missing betweenness axioms by clever applications→ of Pasch’s axiom and the five-segment axiom, often applied in ‘degenerate cases” where three vertices of a “triangle” lie on a line, etc. Some of the degenerate cases introduced by using non-strict betweenness are not constructively valid, and we have chosen to use strict betweenness. Since Euclid did not mention betweenness at all, one cannot use “faithfulness to Euclid” to justify the choice, and no doubt the theory could be developed either way. But, for example, in ECG one can no longer (at least not by a similar proof) derive the symmetry axiom T(a,b,c) T(c,b,a) from Pasch’s axiom, is done in Tarski’s theories (see [26], Satz 3.2,→ p. 30). Lemma 9.1. Congruence is an equivalence relation on segments. Remark. For Hilbert and for Borsuk-Szmielew, this was not an issue, since segments were defined as pairs of points. Proof. Reflexivity. Suppose given points p and q. By Axiom C4 we have qp = pq. Now Axiom C2 yields pq = pq, taking both hypotheses of Axiom C2 to be the same congruence qp = pq. Symmetry. Suppose ab = cd. By reflexivity, which we have just proved, ab = ab. Then by Axiom C2, since ab is congruent to cd and to ab, we have cd = ab. Transitivity. Suppose ab = cd and cd = pq. By symmetry (just proved) we have cd = ab. Then cd is congruent to both ab and pq, so by Axiom C2 we have ab = pq. That completes the proof of the lemma. Lemma 9.2. All null segments are congruent, i.e. aa = bb. Proof. Extend segment ba to a point x on Ray (b,a) such that ax = bb, by Axiom C1. Then by Axiom C3, we have a = x. Hence aa = bb. That completes the proof. Lemma 9.3 (Additivity of congruence). Suppose B(a,b,c) and B(A, B, C) and ab = AB and bc = BC. Then ac = AC. Proof. Since B(a,b,c) we have a = b. We wish to apply the five-segment axiom, Axiom C6, with d = a and D = A6 . Then ad = AD means aa = AA, which was proved in Lemma 9.2. The condition db = DB becomes ab = AB, which is true by hypothesis. Hence Axiom C6 is applicable, and it yields cd = CD, which is ca = CA. By Axiom C4 and Lemma 9.1, we have ac = AC. That completes the proof of the lemma. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 65

Lemma 9.4 (Uniqueness of extension). B(q,a,x) B(q,a,y) ax = bc ay = bc x = y. ∧ ∧ ∧ → Proof. Since ax = bc and ay = bc, by Lemma 9.1 we have ax = ay. By Lemma 9.3 we have qx = qy. Now we apply the five-segment axiom, Axiom C6, with (a,b,c,d) = (q,a,x,x) and (A,B,C,D) = (q,a,x,y). We conclude xx = xy. By Lemma 9.1, xy = xx. By Axiom C3, x = y. That completes the proof. Sometimes it is more convenient to phrase the uniqueness of extension in terms of rays: Lemma 9.5. Suppose x and y are two points on Ray (a,b), which is to say that they are on Line (a,b) and B(x,a,b) and B(y,a,b), and suppose ax = ay. Then x = y. ¬ ¬ Proof. Let q be a point with B(E,A,B), and apply Lemma 9.4. Lemma 9.6 (Subtractivity of congruence). Suppose B(a,b,c) and B(A, B, C) and ac = AC and ab = AB. Then bc = BC. Similarly if instead of ab = AB we assume bc = BC, then ab = AB. Proof. It suffices to prove the first assertion, since the second follows from the first using the axiom, interchanging the roles of a and C and the roles of A and C using Axiom B1-iv, which says B(a,b,c) implies B(c,b,a). Assume B(a,b,c) and B(A, B, C), and ac = AC and ab = AB. Then there is a point x on Ray (A, B) such that Bx = bc, and by Lemma 9.4, x is the unique such point. By Lemma 9.3, we have Ax = ac. Since ac = AC, by Lemma 9.1, we have Ax = AC. Let R be a point on Ray (B, A) such that RA = AB. Then B(R,A,B), and x and C are two points on Ray (R, A) such that Ax = AC. Then by Lemma 9.4, x = C. That completes the proof. Lemma 9.7 (Inner transitivity). B(a,b,d) B(b,c,d) B(a,b,c) ∧ → Remark. This (with non-strict betweenness) was an axiom of Tarski’s theories until and including the thesis of Gupta [13], where many other earlier axioms were proved. It is listed as Ax. 15 in [30], who says that Szmielew used results of [13] to show that it could be replaced with the identity axiom for betweenness, which with non-strict betweenness is B(a,b,a). It appears as Satz 3.5 in [26], and we give the proof from there, presumably¬ due to Szmielew, to check that it still works with non-strict betweenness and constructive logic. Proof. We prove only the first claim. Suppose B(a,b,d) B(b,c,d). Applying Axiom B4 (Inner Pasch), we obtain a point x such that ∧B(b,x,b) B(c,x,a). (Note that it was important that Axiom B4 allows all the points to be∧ collinear.) But B(b,x,b) contradicts Axiom B1-ii. That completes the proof of the first claim. The other three are proved similarly. Lemma 9.8 (Outer transitivity). B(a,b,c) B(b,c,d) B(a,c,d) B(a,b,d). ∧ → ∧ 66 MICHAEL BEESON

Proof. Assume B(a,b,c) and B(b,c,d). It suﬃces to prove B(a,b,d), for then we can apply that lemma with (a,b,c,d) replaced by (d,c,b,a) to conclude B(d,c,a), and then by Axiom B1-iv we have B(a,c,d). Extend segment bc to a point x such that B(a,c,x) and cx = cd. Using the symmetry of betweenness (Axiom B1-iv) we have B(c,b,a) and B(x,c,a). By Lemma 9.7 we have B(x,c,b), and by Axiom B1-iv we have B(b,c,x). Then by Lemma 9.4 we have x = d. But since B(a,c,x) and x = d we have B(a,c,d) as required. That completes the proof. Lemma 9.9 (Transitivity extended). B(a,b,d) B(b,c,d) B(a,c,d) ∧ → B(a,b,c) B(a,c,d) B(b,c,d) B(a,b,d) ∧ → ∧ Proof. These are easy consequences of the previous two lemmas and the symmetry of betweenness. From a constructive viewpoint, the formulation of Tarski’s 5-segment axiom is inadequate, at least on the face of it, since it depends on the ordering of points b and c in Fig. 8. We want segment dc to be determined even if c is between a and b. A remedy for this is to require that cd be determined by ad, ab, bd, and then ac instead of bc. To say that a, b, and c are collinear without specifying the order of b and c, and to stay in Tarski’s language (i.e. without mentioning lines and incidence), we must mention another point e, which would be on the line to the right of b and c, as shown in Fig. 14:

Figure 14. cd is determined, but the order of b and c is not speciﬁed.

ab ce

This leads us to the following formulation, which turns out to be provable: Lemma 9.10 (6-segment lemma, see Fig. 14). a = b B(a,b,e) B(a,c,e) B(A, B, C) B(A, B, E) 6 ∧ ∧ ∧ ∧ ab = AB ac = AC ad = AD bd = BD ae = AE cd = CD ∧ ∧ ∧ ∧ ∧ → Proof. We ﬁrst show that in each of the three cases B(a,b,c) or b = c or B(a,c,b), the theorem holds. If b = c then dc = db, and db = DB by hypothesis, so dc = DB by Lemma 9.1. Then we claim C = B. We have AC = ac by hypothesis, and ab = ac since b = c, and ab = AB by hypothesis, so AB = AC by Lemma 9.1. Then B = C by Lemma 9.4. That disposes of the case b = c. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 67

Now assume B(a,b,c). We have ac = AC and ab = AB, so by Lemma 9.6, we have bc = BC. Then by Axiom C6 we have cd = CD as required, disposing of that case. Now we assume B(a,c,b). Then bc = BC by Lemma 9.6, since we have ab = AB and ac = AC. We have ae = ae = AE = EA by hypothesis and Axiom C4. Since ab = AB by hypothesis, we have be = BE by Lemma 9.6; so eb = EB by Axiom C4. Now we apply Axiom C6 with (a,b,c,d) = (a,b,e,d) and (A,B,C,D) = (A,B,E,D). The hypotheses have been verified, so we can conclude ED = ed. Next we apply Axiom C6 again, this time with (a,b,c,d)= (e,b,c,d) and (A,B,C,D) = (E,B,C,D). The hypotheses are B(e,b,c) and ec = EC and eb = EB and ed = ED. We have already verified ed = ED and eb = EB and bd = BD is given by hypothesis. We have ea = EA and B(a,c,e) by hypothesis, so ec = EC by Lemma 9.6. that leaves just B(e,b,c) to check. We have B(a,c,b) by the hypothesis of this case, and B(a,b,e) by hypothesis. By Axiom B1-iv (symmetry of betweenness) we have B(b,c,a) and B(e,a,b). Hence by the inner transitivity of betweenness, Lemma 9.7, with (a,b,c,d) in the lemma replaced by (e,b,c,a) here, we have B(e,b,c) as desired. Hence Axiom C6 is applicable as claimed. The conclusion is cd = CD, disposing of the third case. We have thus shown that if B(a,b,c) b = c B(a,c,b), then the theorem holds. But because of the stability of∨ equality∨ and betweenness, the double negation of the theorem implies the theorem. Hence it suffices to prove the double negation of B(a,b,c) b = c B(a,c,b), i.e. to prove that not all three cases fail. For proof by contradiction,∨ ∨ then, assume that all three cases fail. Then by Axiom B3 and Axiom B1-iv, we have B(b,a,c). Since B(a,c,e), by Lemma 9.8 we have B(b,a,e). Now we have B(a,b,e) by hypothesis. These two betweenness relations contradict Axiom B3. That completes the proof of the lemma. The following lemma shows that a segment cannot be congruent to a subsegment: Lemma 9.11. If B(a,b,c) then not ab = ac. Proof. Let d be a point with B(d,a,b). Suppose ab = ac; then b and c are on Ray (d, a) and by Lemma 9.4 we have b = c. That completes the proof.23 The following lemma shows that we did not need to provide for uniqueness in the congruence axiom about laying off a segment along a ray. Here is a corollary that surprises one by not coming earlier in the development. Lemma 9.12. The center of a nondegenerate circle is between the endpoints of any diameter; that is, if line L contains the center p of circle C and two distinct additional points a and b, then B(a,p,b). Proof. We have ap = bp by Axiom C5. Since C is nondegenerate, a = p and b = p. By hypothesis a = b. By Axiom B3 and Markov’s principle for between-6 ness,6 it suffices to rule6 out the possibilities B(b,a,p) and B(p,b,a). Suppose B(b,a,p). By axiom B1-iv we have B(p,a,b). Since ap = bp, Lemma 9.11

23Narboux [21] says that this lemma is needed to fill gaps in [26], Ch. 5. Since our Lemma 9.4 is Satz 2.12 in [26], the addition required to [26] is minimal. 68 MICHAEL BEESON yields a contradiction. On the other hand, if B(p,b,a), then by Axiom C1 and Lemma 9.11 we have b = a, contradiction. That completes the proof. 9.3. Ordering of segments. Euclid uses the notion of one segment being “greater than” another, without defining it. In [26], ab > cd means that there exists an x between a and b (or equal to a, because null segments are allowed) such that ax = cd. We adopt a similar definition. 24 Lemma 9.13. The points of an equilateral triangle do not all lie on a line. That is, we cannot have three distinct points on a line with ab = bc = ac. Proof. Suppose three such points exist. Suppose B(a,b,c). Then segment ac is congruent to its proper subsegment ab, contradicting Lemma 9.11. Hence B(a,b,c). Similarly, B(b,c,a) and B(c,a,b). Then by Axiom B3, the three points¬ do not all lie on¬ a line. That completes¬ the proof. Lemma 9.14. Suppose a and d lie on Ray (b,a) and are not equal to b. Then there is a point e such that B(b,a,e) and B(b,d,e). Proof. Extend segment ba by bd to find point x with B(b,a,x) and ax = bd. Extend segment bd by ba to find point y with B(b,a,y) and dy = ba. Then by Lemma 9.3 we have xb = by, since xa = bd and ab = dy. Hence bx = by. Since a and d both lie on Ray (b,a), x and y both lie on Ray (b,a) as well. Hence by Lemma 9.5, x = y. Take e = x. Then B(b,a,e) since B(b,a,x) and e = x; and B(b,d,e) since B(b,d,y) and e = y. That completes the proof of the lemma. Recall that ab < cd was defined in Definition 8.10. Lemma 9.15 (Stability of ab < cd). ab < cd ab < cd. ¬¬ → Proof. By definition of ab < cd, ab < cd means ¬¬ x(B(c, x, d) cx = ab). ¬¬∃ ∧ But by Lemma 9.5, there is one and only one possible point p on Ray (c, d) such that cx = ab. Hence ab < cd is equivalent to B(c,p,d), and ab < cd is equivalent to B(c,p,d). By Markov’s principle for betweenness,¬¬ we have ab < cd ab¬¬ < cd. That completes the proof. ¬¬ → Lemma 9.16 (Ordering of segments is transitive). If ab < cd and cd < ef then ab < ef. Proof. Suppose ab < cd and cd < ef. Let K be the circle with center e and radius cd; let it meet Ray (e,f) in point y; then since cd < ef we have B(e,y,f). Let C be the circle with center c and radius ab; let it meet Ray (c, d) in point z; then since ab < cd we have B(c,z,d). Let J be the circle with center e and radius ab; let it meet Ray (e,f) in point x; we must show that B(e,x,f). We have cz = ex since both are equal to ab, and cd = ey by construction of y, and

24This approach to the ordering of segments seems to be faithful to Euclid, in the sense that it accounts for Euclid’s use of the concept, and does not go much beyond it. In particular, we do not associate any number to a segment to “measure its length”, and we do not perform arithmetic, even addition, on the lengths of segments. The formalization of Euclid in [1] takes a diﬀerent approach, using linear arithmetic on lengths and on the measures of angles. That is convenient for computerizing proofs, since it permits the use of decision procedures for the limited arithmetic involved. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 69

B(c,z,d), and we wish to prove B(e,x,y). Since x lies on Ray (e,f), by Axiom B3, not not x =0 x = y g B(y,x,g) If we show that each of the three cases implies B(e,x,y),∨ or implies∨ ∃ the ﬁnal goal ab < ef (or is contradictory), then we can push the double negation in and apply Markov’s principle for betweenness (which implies the stability of ab < ef), to ﬁnish the proof. Therefore we can proceed to argue by cases, as we would classically. Case 1, e = x. Then because cz = ex we have cz = ee and hence by axiom C3, c = z, and hence by Axiom C3 again, a = b. But then ab is a null segment, so ab < cd is automatic. Case 2, x = y. Then cz = ex = ey = cd, so segment cd is congruent to its subsegment cz, contradicting Lemma 9.11. Case 3, B(y,x,g) for some g. Then let g′ lie on Ray (z, d) with dg′ = yx. Then g′ = z since if g′ = z then y = x, contradicting B(y,x,g). Then since cd = ey 6 we have cg′ = eg by Lemma 9.3. Then cz = ex = cg′. We have B(z,d,g′) by Axiom B3, since g′ lies on Ray (z, d) and is not equal to z. Then segment cg′ is congruent to its subsegment cz, contradicting Lemma 9.11. Thus each case has been shown to lead to the desired conclusion (remember a contradiction implies anything). That completes the proof of the lemma.

Lemma 9.17.

ab = AB ac = AC B(a,b,c) on(B, Line (A, C)) B(A, B, C). ∧ ∧ ∧ → Proof. Suppose ab = AB ac = AC B(a,b,c) on(B, Line (A, C)). We must show B(A, B, C). Suppose,∧ in order∧ to refute it,∧ that B(A,C,B). Extend segment ac by bc to find point x such that B(a,c,x) and cx = bc. Now apply the five-segment axiom with (a,b,c,d) replaced by (a,c,x,b) and (A,B,C,D) replaced by (A,C,B,B). The hypotheses hold since ac = AC, cx = CB = bc, ab = AB, and cb = CB. The conclusion is xb = BB. By Axiom C3, x = b. Now since B(a,c,x) and x = b we have B(a,c,b). But also we have B(a,b,c), contradicting Axiom B3. Hence B(A,C,B). If C = B then ab = AB = AC = ac, so segment ac is congruent¬ to its proper subsegment ab, contradicting Lemma 9.11. Hence C = B. If B = A then AA = AB = ab, so by Axiom C3, a = b, contradicting B(6a,b,c). Now assume B(B, A, C). Then extend segment ca to find point x with B(x,a,c) and xa = ba = BA. Now apply the five- segment axiom with (a,b,c,d) replaced by (c,a,x,b) and (A,B,C,D) replaced by (C,A,B,B). Checking the hypotheses, we have ac = AC, ax = AB, cb = CB, and ab = AB. The conclusion is xb = BB. Then by Axiom C3, x = b. Since B(x,a,c) we have B(b,a,c), which together with B(a,b,c) contradicts Axiom B3. This contradiction refutes B(B, A, C). We have now shown that B cannot be equal to A or C and we cannot have B(B, A, C) or B(A,C,B). By Axiom B3 we therefore have B(A, B, C). Then by Markov’s principle for betweenness, we have B(A, B, C¬¬) as desired. That completes the proof of the lemma.

Lemma 9.18 (Ordering respects congruence).

ab < cd ab = AB cd = CD AB < CD. ∧ ∧ → 70 MICHAEL BEESON

Proof. Define X = IntersectLineCircle2 (Line (C,D), Circle3 (C,A,B)) x = IntersectLineCircle2 (Line (c, d), Circle3 (c,a,b) In order to prove AB < CD, according to the definition of AB < CD, we must show B(C,X,D). Since ab < cd we have B(c, x, d). We have cx = ab = AB = CX, so by Lemma 9.1 we have cx = CX. Since cd = CD, we can apply Lemma 9.17 with (a,b,c) = (c, x, d) and (A, B, C) = (C,X,D). The conclusion is B(C,X,D). That completes the proof of the lemma. Lemma 9.19 (Extended transitivity). ab cd cd ef ab ef ≤ ∧ ≤ → ≤ ab cd cd < ef ab < ef ≤ ∧ → ab < cd cd ef ab < ef ∧ ≤ → Proof. Similar to the proof of Lemma 9.16. Lemma 9.20. ab cd (cd < ab). ≤ ↔¬ Proof. Suppose ab bc and cd < ab. By Lemma 9.19, ab < ab. Then by definition of <, there≤ is a point x between a and b with ax = ab. But that contradicts Lemma 9.5. That completes the proof. For those new to constructive mathematics, it is not true that (y < x) x

We take an angle, like a triangle, to be given by three distinct points, as in “∠ABC”, in which B is the vertex of the angle. However, if A′ is another point on Ray (B, A) (different from B) and C′ is another point on Ray (B, C) (different from B), then angle A′BC′ is “the same angle” as angle ABC. Classically we would define an angle to be an equivalence class under this equivalence relation. Constructively we continue to regard an angle as (given by) three points, but we have the notion of two triples of points “determining the same angle.” Since Euclid uses the phrase “equal angles” to mean “congruent angles”, it would be too confusing to use the word “equal” for this equivalence relation, or to use the symbol ‘=’, so we will just use the English “same angle”, as in “ABC and A′BC′ are the same angle.” Whenever we define a relation involving angles, we will check that it respects this equivalence relation, i.e. if angles are replaced by other triples of points defining the same angles, then the new relation is preserved. That way, when we are “given an angle ABC”, then we can “without loss of generality” replace A and B by any other more convenient points not equal to B on Ray (B, A) and Ray (B, C) respectively. Note, however, that abc is not the same angle as cba. We will prove below that they are congruent, but they are not the same angle. This decision is somewhat arbitrary; we could have chosen to define the relation “same angle” more broadly.

Definition 9.21. Angle abc is congruent to angle ABC if triangle a′bc′ is congruent to triangle A′BC′, where the primed points are constructed so that a′ lies on Ray (b,a), c′ lies on Ray (b,c), A′ lies on Ray (B, A), C′ lies on Ray (B, C), and all the segments ba , bc′, BA′, and BC′ are congruent to a fixed segment αβ. (Here α and β are′ two of the points mentioned in the lower dimension axiom.) This definition has the advantage that it is quantifier-free. There are other possible definitions we could take25 :

(i) We could require that there exist points a′, c′, A′, and C′ on Ray (b,a), Ray (b,c), Ray (B, A), and Ray (B, C) respectively, such that triangle a′bc′ is congruent to triangle A′BC′.

(ii) We could require that if a′ and c′ are on Ray (b,a) and Ray (b,c) respectively, and A′ and C′ are on Ray (B, A) and Ray (B, C) respectively, and ba′ = BA′ and bc′ = BC′, then a′c′ = A′C′. Note that choices (i) and (ii) use a quantifier (universal for (ii) and existential for (i)), while (iii) is quantifier-free. All the choices make it apparent that congruence is symmetric and respects the “same-angle” equivalence relation. The following lemma shows that the possible definitions are all equivalent. Lemma 9.22. The three possible definitions of angle congruence listed above, i.e. (i) and (ii) and the actual definition, are equivalent. Proof. Evidently the actual definition implies (i) and (ii) implies the actual definition, so it suffices to show that (i) implies (ii). Suppose we are given

25In [26], p. 95, a diﬀerent way of constructing canonical points is used, similar to the construction in the proof of Lemma 9.14. This deﬁnition implies (i) and is implied by (iii). 72 MICHAEL BEESON triangle abc congruent to triangle ABC, and points a′, c′, A′, and C′ on Ray (b,a), Ray (b,c), Ray (B, A), and Ray (B, C) respectively, and a′ = b, c′ = b, A′ = B, 6 6 6 and C′ = C, and ba′ = BA′ and bc′ = BC′. We must prove a′c′ = A′C′. By Lemma6 9.14, there is a point e on Ray (b,a) such that B(b,a,e) and B(b,a′,e), and a point f be a point on Ray (b,c) such that B(b,c,f) and B(b,c′,f). We intend to use these points in applying Lemma 9.10, so we need corresponding points E and F : let E and F be points on Ray (B, A) and Ray (B, C) respectively such that BE = be and BF = bf. Then we claim B(B, A, E) and B(B, A′, E) and B(B,C,F ) and B(B, C′, F ). We prove B(B, A, E) (the other three claims are similar). We claim A = E. Assume that A = E. Then ba = BA = BE = be, so ba = be, contradicting6 Lemma 9.11. Hence A = E. Next assume B(B,E,A). In that case we have BE

Definition 9.26. Suppose triangles ABC and abc are congruent. Then the corresponding angles of the triangles are the pairs ABC and abc, BCA and bca, and CAB and cab. Lemma 9.27. Corresponding angles of congruent triangles are congruent. Proof. Suppose triangle ABC is congruent to triangle abc. We show that angle ABC is congruent to angle abc. We have AB = ab and AC = ac and BC = bc by Lemma 9.22. Lemma 9.28. The base angles of an isosceles triangle are congruent. Proof. Let triangle ABC be isosceles; assume AB = BC. Then angle ABC is congruent to angle CBA, by Lemma 9.24. Then triangle ABC is congruent to triangle CBA by SAS. The base angles of triangle ABC are corresponding angles in this congruence, and hence are equal. That completes the proof. Definition 9.29. Angles ABC and CBE are supplements if B(A, B, E). Angles ABC and DBE are supplements if D is on Ray (B, C) and B(A, B, E). Lemma 9.30. Supplements of congruent angles are congruent. Proof. Let ABC and DBE be supplements. Suppose that angle ABC is congruent to angle abc, with B(A, B, E), and e is a point with B(a,b,e) and d a point on Ray (b,c), so angle abc and angle dbe are supplements. We must prove angle DBE is congruent to angle dbe. Without loss of generality we can assume D = C and d = c and ab = AB and bc = BC and be = BE. We can therefore apply the ﬁve-segment axiom, Axiom C6, to conclude that CE = ce. That implies that angle DBE is congruent to angle dbe. That completes the proof. Lemma 9.31. An angle congruent to a right angle is a right angle. Proof. Let abc be a right angle and let d be a point with B(a,b,d) and ab = bd. Since abc is a right angle we have ac = dc. Let angle ABC be congruent to angle abc. Without loss of generality we may assume ab = AB and bc = BC. Let D be a point with AB = BD and B(A,B,D). Then BD = bd since congruence of segments is transitive. Then by the ﬁve-segment axiom also dc = DC. Then AC = ac = dc = DC. Hence Line (B, C) is perpendicular to Line (A, B), and ABC is a right angle. That completes the proof. Definition 9.32. Angles amb and cmd are vertical angles if B(a,m,c) and B(b,m,d), or if B(a,m,d) and B(b,m,c). Lemma 9.33. Vertical angles are congruent. Proof. An immediate consequence of Lemma , together with the fact (proved in Lemma 9.24) that angle abc is congruent to angle bca. Lemma 9.34. If angle abc is not not congruent to angle ABC, then angle abc is congruent to angle ABC.

Proof. Suppose angle abc is not not congruent to triangle ABC. Let a′, b′, A′ and B′ be as in Deﬁnition 9.21. Then ba′ = αβ, BA′ = αβ, bc′ = αβ, ¬¬ ¬¬ ¬¬ and BC′ = αβ. By Axiom S1 (stability of segment congruence), the double negations¬¬ can be removed. Then by Deﬁnition 9.21, we have angle abc congruent to angle ABC. That completes the proof. 74 MICHAEL BEESON

9.5. Perpendicular lines. Definition 9.35. Lines L and K, meeting at point m, are perpendicular at m if there are points a and b on L with B(a,m,b) and c on K, with c not on L and a and b not on K, such that triangle amc is congruent to triangle bmc. Lines L and K are perpendicular if they are perpendicular at IntersectLines (L,K). We write ab bc to mean Line (a,b) is perpendicular to Line (b,c), and L K to mean L is⊥ perpendicular to K. ⊥

Figure 15. L and K are perpendicular at m if a, b, and c exist making triangles amc and bmc congruent. K

b c

L b b b am b

Remark. Note that if L and K are perpendicular, they do not coincide, by the clause in the deﬁnition that c is not on L.

Lemma 9.36. If lines L and K meet at m, and a and b are any points on L with B(a,m,b), and c is any point on K different from m, then angles amc and bmc are congruent if and only if L and K are perpendicular. Proof. If L and K are perpendicular, then the congruence of the angles follows immediately from the definition of “congruent angles” and the definition of “perpendicular.” Conversely, suppose the angles are congruent. Then let b′ be a point on L, on the opposite side of m from a, with mb′ = ma. Then triangle cmb′ is congruent to triangle cma by SAS, so K and L are perpendicular. Lemma 9.37. If lines L and K are perpendicular, and lines ℓ and k have the same points as L and K, respectively, then ℓ and k are also perpendicular. Remark. Remember that equality between lines is intensional, so this is not automatic. Proof. The points a, b, and c that witness the perpendicularity of L and K according to the definition, also witness the perpendicularity of ℓ and k. Lemma 9.38. If L and K are perpendicular, then K and L are perpendicular. Proof. Suppose L and K are perpendicular. Let m be their intersection point. By definition of “perpendicular”, there are points a and b on L with B(a,m,b) and am = mb, and c be on K with c = m, such that triangle amc is congruent to triangle bmc. Extend segment cm6 to a point d such that B(c,m,d) and cm = md. Then angle bmd and angle amc are vertical angles, and hence by FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 75

Lemma 9.33 they are congruent. Angle amc is congruent to angle bmc, since they are corresponding angles of congruent triangles. hence by Lemma 9.24, angle bmc is congruent to angle bmd. Then triangle bmc is congruent to triangle bmd, by SAS. Then by deﬁnition of perpendicular, K and L are perpendicular. That completes the proof.

Lemma 9.39 (Stability of perpendicular). If L and K are not not perpendicular, then they are perpendicular

Proof. To prove this we must define perpendicular in a quantifier-free way, getting rid of the existential quantifier in the definition. That we can do so is a consequence of Lemma 9.36. Suppose L and K meet m. Then let C be the circle with center m and radius αβ. Then let p and q be the two intersection points of L and C given by IntersectLineCircle1 (L, C) and IntersectLineCircle2 (L, C), and similarly let a and b be two intersection points of K and C. Then the triangles pma and pmb are congruent if and only if L and K are perpendicular at m. That is, L and K are perpendicular at m if and only if pa = pb. Since p, a, and b are explicitly defined by terms in L and K (putting m = IntersectLines (L,K)), we have an equation between terms of type Point. This is stable by Axiom S1. That completes the proof. 9.6. Existence of midpoints and perpendiculars. Most mathematicians are familiar with the construction in Euclid I.1, spelled out in section 3.4. It is, however, possible to construct midpoints without assuming line-circle continuity, cf. Prop. 4.3, page 167 of [12] or Satz 8.22 of [26]. It is also possible to construct a perpendicular to line L from a point c not on L without using a line-circle continuity or circle-circle continuity. However, at present we are allowing the full use of ruler and compass, so we do not stop to analyze these constructions. Instead, we prove the correctness of the Euclidean constructions. However, the construction Euclid gives for bisecting a segment contains an error (besides not proving the circles intersect), as we will point out below.

Lemma 9.40. Every segment ab with a = b has a midpoint, i.e. a point m between a and b with ma = mb, and a perpendicular6 bisector, i.e. a line perpendicular to Line (a,b) at m. Both m and a perpendicular bisector can be constructed (deﬁned) by terms of ECG.

Remark. The following proof follows Euclid I.1 and I.9 and I.10, but it fills some gaps and corrects an error. Specifically, the circles constructed in I.1 must be shown to actually intersect, the intersection points must be shown not to lie on Line (a,b), and indeed must be shown to lie on opposite sides of ab, all of which is omitted in Euclid; then in I.9, the equilateral triangle that Euclid constructs might have its apex at the vertex of the angle being bisected. One can show that the construction made there will not be continuous in parameters; in a computer animation, one can drag the sides of the angles to make the two points alleged to determine the angle bisector coincide. Of course, one should then take the “other equilateral triangle”, but that cannot be done continuously. In essence the following proof repairs this defect by constructing both equilateral triangles and drawing the angle bisector between their apexes, instead of using the vertex 76 MICHAEL BEESON of the angle being bisected. In other words, there is a non-constructivity in the proof of Euclid I.9, which can be easily removed by a slightly different proof. Proof. Given segment ab, construct circle C with center a and radius ab, and circle K with center b and radius ab. Extend segment ab to point d with B(a,b,d) and bd = ab. Then d lies on K and is outside circle C, because B(a,b,d) and ab is the radius of C. On the other hand point a lies inside C and lies on circle K. Hence by Axiom Cont 6 (circle-circle continuity), both IntersectCircles1 (C,K) and IntersectCircles2 (C,K) are defined. Let these points be p and q respectively. Then pab is an equilateral triangle. Then p does not lie on Line (a,b), by Lemma 9.13. Similarly, abq is equilateral, and hence q does not lie on Line (a,b). Now, by Definition 5, abp is a right turn and abq is a left turn. By Axiom H3 then, p and q are on opposite sides of Line (a,b); that is, Line (p, q) meets Line (a,b) in a point m between p and q. Thus the defects of Euclid’s proof are both succesfully dealt with in ECG. Since ap = ab and bp = ab, and congruence is an equivalence relation, we have ap = bp. Similarly aq = bq. Since pq = pq (every segment is congruent to itself) we have triangle apq congruent to triangle bpq. Then the corresponding angles apq and bpq are congruent. But these are the same angles as apm and bpm (here “same angles” has been precisely defined above). Hence triangle apm is congruent to triangle bpm by SAS. Hence the corresponding sides am and bm are congruent. That is, m is the midpoint of ab. Moreover, the supplementary angles pma and pmb are congruent as corresponding angles of congruent triangles. Hence, by definition of perpendicular, pm is perpendicular to Line (a,b). That completes the proof. Here is a two-dimensional version of Markov’s principle: Lemma 9.41. ECG proves that, if point P does not lie on line L, then some circle with nonzero radius and center P lies entirely on the same side of L as P . Proof. Let point P not lie on line L; we will construct a circle with center P lying on the same side of L as P . Let point K on L be the foot of the perpendicular from P to L. Then K = P . Hence the two circles C1 = Circle (K, P ) and C2 = Circle (P,K) that are used6 to bisect PK have different centers and each contains a point inside the other. Hence the points X = IntersectCircles1 (C1, C2) and Y = IntersectCircles2 (C1, C2) are defined. If X = Y then X is between P and K, contradicting PX = PK. Hence the midpoint of PK is given by M = IntersectLines (Line (X, Y ), Line (P,K)). Hence Circle (P,M) lies on the same side of L as P , because K is nearer to P than any other point of L. That completes the proof of the lemma. 9.7. Right angles. Definition 9.42. abc is a right angle if Line (a,b) is perpendicular to Line (b,c) at b. Lemma 9.43. A right angle is congruent to its supplement. Proof. Let abc be a right angle, and let d be a point with B(a,b,d), so that angle cbd is the supplement of angle abc. Since abc is a right angle, Line (b,c) is perpendicular to Line (a,b), and by definition of “perpendicular”, there are FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 77 points A on Ray (b,a) and C on Ray (b,c) and D on Ray (b, d) such that Ab = eb and AC = eC. That is, triangle AbC is congruent to triangle ebC. Hence by definition of angle congruence, angle abc is congruent to angle dbc. Lemma 9.44. All right angles are congruent. In other words, if abc and ABC are right angles with ab = AB and bc = BC then ac = AC. Remark. Euclid took this as his Postulate 4. Hilbert ([15], p. 20) remarks that this was “unjustified”, and says that the proof of it goes back to Proclus. We note that Hilbert’s proof makes use of a constructively unacceptable argument by cases, but since the conclusion is disjunction-free, we could just double-negate his whole argument. However, since his axioms differ from ours, that still will not produce a proof in our system. We have not yet proved the uniqueness of the perpendicular to a line at a given point, and this lemma will be used to prove it; so the lemma has to be proved without it. Proof. This is Satz 10.12 in [26]. We sketch the proof, with attention to checking that the argument is constructive. We note that the conclusion is stable, i.e. implied by its double negation; hence we can argue classically if we like; in particular a case distinction whether B = b or not is allowed. One begins by proving that reflection in a point, and reflection in a line, both preserve right angles; this is easy since both kinds of reflection preserve congruence of segments (reflections are isometries). Now reflect triangle ABC about the midpoint of segment bB (if b = B, or about b if B = b). Replacing ABC by its reflection, we can assume b 6= B. Again, since the conclusion is stable, we can argue by cases on whether c = C or not. If c = C, then N let y be the midpoint of cC. Reflection in line by takes c into C6, since bc = bC. similarly reflect on the line joining b to the midpoint of cC. This enables us to assume, without loss of generality, that c = C. Then we have two right angles Abc and abc with ab = Ab. If A and a are on opposite sides of Line (b,c), then reflect A in Line (b,c); since the desired conclusion is stable, we may argue by cases on whether A and a are on opposite sides of Line (b,c) or not; hence, without loss of generality we can assume they are not on opposite sides. Then aA does not meet Line (b,c). Again, since the desired conclusion is stable, we can argue by cases; if a = A we are finished, so we can assume a = A. Let z be the midpoint of aA. Then since ba = bA, bza is a right angle and6 A is the reflection in Line (b,z) of a. Let e be the reflection of c in point b. Then since ab bc we have ae = ac. Since Ab bc we have Ac = Ae. Hence Ae = Ac = ac = ae⊥, so Ae = ae. Hence triangle Abe⊥ is congruent to triangle abe. Hence angle Abe is congruent to angle abe. But angle abe is congruent to angle abc and angle Abe is congruent to angle Abc. Hence angle abc is congruent to angle Abc. That completes the proof.

Lemma 9.45. a = b on(c, Line (a,b)) ac = ac′ bc = bc′ c = c′. 6 ∧ ∧ ∧ → Remark. The possible constructive diﬃculty here is that we do not know the order relations of c with a and b. But Lemma 9.10 takes care of this. This is a key lemma needed to prove the uniqueness of perpendiculars. Proof. We apply the six-segment lemma (Lemma 9.10) with (a,b,c,d) replaced by (a,b,c,c), and (A,B,C,D) replaced by a,b,c,c′. The conclusion is cc = cc′. Hence by Axiom B1-ii, we have c = c′. 78 MICHAEL BEESON

Lemma 9.46 (Uniqueness of perpendicular from a point not on a line). Suppose abc and acb are right angles. Then b = c. Proof. (following [26] Satz 8.7) By the stability of equality, we may use proof by contradiction. Suppose b = c. Then extend line segment cb to point c′ with 6 bc = bc′ and B(c,b,c′), and extend ac to a point a′ with B(a,c,a′ and ac = a′c. Then because ab bc, we have ac = ac′, and similarly a′c = a′c′. But since ⊥ B(a,c,a′), we have from Lemma 9.45 that c = c′. Then from B(c,b,c′ we obtain B(c,b,c), and hence by Axiom B1-ii we have b = c. That completes the proof. Lemma 9.47. A supplement of a right angle is a right angle. Proof. Let abc be a right angle and B(a,b,d); we must show abd is a right angle. Let e be a point with B(a,b,e) and ab = be. By the definition of “right angle”, it suffices to show abe is a right angle. We distinguish between “dropped perpendiculars”, dropped from a point not on a line to the line, and “erected perpendiculars”, erected from a given point on the line. We have just proved the uniqueness of dropped perpendiculars; the uniqueness of erected perpendiculars is harder, and will be proved in another section. We now draw some corollaries from the uniqueness of dropped perpendiculars. Lemma 9.48. Suppose lines K and M are both perpendicular to line L and do not coincide. Then K and M do not meet (i.e., they are parallel). Proof. If they do meet in point p, then they meet L in different points, since two points determine a line. But then K and M are two different perpendiculars to L from point p, contradicting Lemma 9.46. That completes the proof. Lemma 9.49. The line from the vertex of an isosceles triangle to the midpoint of the base is perpendicular to the base. Proof. Let a be the vertex and m the midpoint of bc, and ab = ac. Then since am = bm, triangle abm is congruent to triangle acm. Hence am is perpendicular to bc. That completes the proof. Lemma 9.50. Two right triangles with hypotenuse and one leg congruent, are congruent. In symbols: if ac = AC and ab = AB and angles abc and ABC are right triangles, then triangle abc is congruent to triangle ABC.

Proof. Let a′ be the reﬂection of a in Line (b,c), so ab = a′b. Let A′ be the reﬂection of A in Line (B, C), so AB = A′B. Then AA′ = aa′ by Lemma 9.3, and ab = a′b since abc is a right angle, and AB = A′B since ABC is a right angle. Hence triangle aba′ is congruent to triangle ABA′, since all three sides are pairwise congruent. Hence angle baa′ is congruent to angle BAA′. But angle bac is the same angle as angle baa′, and angle BAC is the same angles as angle BAA′, so angle bac is congruent to angle BAC since angle congruence respects the “same angle” relation. Then triangle abc is congruent to triangle ABC by SAS. That completes the proof. Lemma 9.51. A line meets a circle in at most two points. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 79

Remark. This is closely related to Euclid III.2, which is discussed in Section 9.13; but this lemma is easier to prove. Proof. Let a and b lie on the circle with center e. Suppose B(a,c,b) and c lies on the circle. Then ea = ec, so triangle eac is isosceles. Let m be the midpoint of ac; then by Lemma 9.49, em ac. Let q be the midpoint of ab; then since ea = eb, eq ab by the same Lemma.⊥ But a, b, and c are collinear by hypothesis; hence m = ⊥q by the uniqueness of dropped perpendiculars (Lemma 9.46). But then am = aq, and b and c are both extensions of segment am by am, and both b and c lie on Ray (a,b), by the assumption B(a,c,b). Hence by Lemma 9.5, b = c, contrary to the assumption B(a,c,b). That completes the proof. Lemma 9.52. If line L contains a point strictly inside circle C, then L meets C in exactly two points. Remark. See the discussion following Axiom Cont 5. Proof. By Lemma 9.51 there are at most two points of intersection, and by Axiom Cont 5, there are at least two. That completes the proof.

Definition 9.53. Line L is tangent to circle C at x if L meets C at x and only at x. Remark. Some books take the property in the following lemma as the deﬁnition of tangent instead.

Lemma 9.54. (i) If circle C with center e is tangent to L at x then ex L. (ii) Given line L, point x on L, and two points a and e not on L, with ex⊥ L and e lying on a perpendicular bisector of xa, then Circle (e,a) is tangent to⊥ L at x. Remark. We have to say “a perpendicular bisector” rather than “the perpendicular bisector” since the perpendicular bisector cannot be proved unique until we allow the use of the upper dimension axiom. This lemma does not require the upper dimension axiom. Proof. Ad (i). See Fig. 16. Drop a perpendicular K from e to L, and let z be intersection point of K and L. Suppose z = x. Then extend segment xz to a point w such that B(x, z, w) and xz = wz6 . Then since ez L, we have triangle ezx congruent to triangle ezw. Then the corresponding⊥ sides ex and w are equal, so w is a point of intersection of C and L. Then w = x by the deﬁnition of “tangent”, Deﬁnition 9.53. Since B(x, z, w) and w = x we have B(x, z, x). Hence by Axiom B1-ii, we have a contradiction. Hence z = x. By the stability of equality, we have z = x. But by construction ez ¬ L6 ; hence ex L. That completes the proof of (i). ⊥ Ad⊥ (ii). See Fig. 17. Let C = Circle (e,a). Let m be the midpoint of ax. Then em ax, so triangles emx and ema are congruent. Hence their corresponding sides⊥ex and ea are equal. Hence L meets C at x. Now suppose L meets C at a point y = x. Let p be the midpoint of xy. Then triangles epx and epy are congruent,6 since ex = ey = ea and px = py because p is the midpoint of xy, and the third side ep is shared. Then ep is perpendicular to L at p. Then p = x by the uniqueness of the perpendicular from e to L (Lemma 9.46). But then x = y, 80 MICHAEL BEESON

Figure 16. if ez L and xz = zw then ex = ew. So if C is tangent to L, z = w⊥ is impossible. 6

e x b b z w

contradiction. Hence L does not meet C at a point y = x. Hence L is tangent to C at x, by Deﬁnition 9.53. That completes the proof6 of the lemma.

Figure 17. Construction of a circle tangent to L passing through a. If y lies on L and C and px = py then ep L. ⊥

L y b

b p e x b b

b m

b a

Lemma 9.55. There is a point not on a given line. Remark. Classically we could show that one of the three constants α, β, γ is not on L. But constructively we cannot do that, as the choice cannot be made continuously in L. The point that we choose is the one from Euclid Book I Prop. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 81

1, the point E that makes an equilateral triangle with two points A and B on L. Famously Euclid provided no proof that the two relevant circles actually intersect at E; but here we have to prove something equally obvious, which Euclid also overlooked: that E does not lie on L. Constructively or classically, this is equally difficult: why can’t all the points of an equilateral triangle lie on a line? Both betweenness and congruence axioms are needed. Proof. This was proved in the proof of Lemma 9.40; the vertex p of the equilateral triangle constructed on ab does not lie on Line (a,b), by Lemma 9.13. Remark. We note that CircleThree was not used in the above construction; the lemma is thus somewhat remarkable in view of the theorem that any construction defined using only circles constructed from center and point with only one variable must be somewhere undefined. The above construction E(a,b) constructs a point not on Line (A, B), and indeed it is undefined when A = B, so it does not contradict the theorem. It is surprising that we cannot find a construction (without Circle3 ) that always constructs a point different from a given point, but we can find a construction that always constructs a point different from a given line. Lemma 9.56 (Uniform construction of perpendiculars). Let Perp(x, L) be defined as in Definition 5.1. For every point x and line L, Perp(x, L) is defined, and Perp(x, L) is perpendicular to L, and x lies on Perp(x, L). Remark. The important point here is that no case distinction is required as to whether x is or is not on L; that is, there is no distinction in this construction between dropped perpendiculars and erected perpendiculars. That is good, since constructively we cannot decide if a given point x is or is not on a given line, yet we still need to be able to project x onto the line. Proof. Here is the construction script (whose parsing yields the term Perp of the lemma) given in Definition 5.1. Perp(Point x, Line L) { a = pointOn1(L) b = pointOn2(L) Q = Circle3(b,x,a) c = IntersectLineCircleTwo(L,Q) C = Circle3(x,a,c) p = IntersectLineCircle1(L,C) q = IntersectLineCircle2(L,C) K = Circle(p,q) R = Circle(q,p) d = IntersectCircles1(K,R) e = IntersectCircles2(K,R) return Line(d,e) } Proof. We go through the script defining Perp(x, L) and show that the term on the right of each line is defined. Lines 1 and 2 are defined since pointOn1 (L) and pointOn2 are always defined, by Axiom CA1. In line 3, Circle3 (b,x,a) is 82 MICHAEL BEESON defined because Circle3 is everywhere defined, by Axiom CA7. Since the center of that circle is b and b lies on L, b is inside that circle, so by the line-circle continuity axiom, c in line 4 is defined. Circle C in line 5 is defined by Axiom CA7. To show that p and q in lines 5 and 6 are defined and distinct, we must show that L has a point strictly inside C. That point is a, as we now prove. Let e be any point on C. Then ec = ac since C = Circle 3(x,a,c). Since a = b, C is not a degenerate circle; in fact ab xe (equality might hold if x = 6a) and ax < xe. Hence a is strictly inside C.≤ Hence points p and q in lines 6 and 7 of the script are defined, and by Lemma 9.52 they are distinct. Line (d, e) is the perpendicular bisector of segment pq, which exists by Lemma 9.40; the proof of that lemma provides the explict term mentioned in the script for constructing Line (d, e). That completes the proof of the lemma. 9.8. Various forms of the Pasch axiom. The intuition behind the Pasch axiom is this: if a line enters a triangle, it must come out again. In making this precise we must watch out for the case when the line just touches the triangle, either at a vertex or along one side. Perhaps the most natural version is this: “if abc is a triangle (that is the three points are distinct) and line L contains a point x on the closed segment ab (that is T(a,x,b)) then line L contains a point y with T(a,y,c) T(b,y,c).” That version is not constructively valid, as we cannot continuously∨ decide which alternative holds. More precisely, given a fixed triangle abc in the standard model, we can construct a line Lx depending on a real number x such that Lx meets the midpoint of ab, and if Lx meets the closed segment ac then x 0 and if it meets bc then x 0. Hence this version of Pasch would imply the≤ constructively invalid principle≥x 0 x 0. To get a constructively valid principle we could simply take≤the∨ double≥ negation of the conclusion. That is equivalent to saying that it is impossible that L should meet the closed segment ab and not meet the closed segment ac and not meet the closed segment bc. Here is that principle: a = b a = c b = c on (x, L) B(a,x,b) (Negative Pasch) 6 y(on∧(y,L6 ) ∧ 6 (y =∧ a y = b∧ y = c B→(a,y,c) B(b,y,c))) ¬∀ → 6 ∧ 6 ∧ 6 ∧ ¬ ∧ ¬ Negative Pasch has two strikes against it as a candidate axiom: it is not quantifier-free, and it does not actually produce the point of exit of L from the triangle when given the entrance point. Asserting that existence gives us the following principle: a = b a = c b = c on (x, L) B(a,x,b) (Strong Pasch) y6 (on∧(y,L6 ) ∧ (y6 = a∧ y = b y∧ = c B(a,y,c→ ) B(b,y,c)) ∃ ∧ ¬ 6 ∧ 6 ∧ 6 ∧ ¬ ∧ ¬ It turns out that Strong Pasch is provable, but as an axiom it is unappealing, as the construction of the required point y by ruler and compass is not quite evident. Indeed one may suspect at first that it might not be possible to construct y uniformly, since y depends continuously but not differentiably on the line L. But it is not the case that all the functions definable in ECG are differentiable on their domains; for example, IntersectLineCircle2 (Line (a,b), Circle (b,c)) does not depend differentiably on c as c passes through b. (If we did not allow circles zero radius, constructions would be differentiable on their domains.) FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 83

Negative Pasch is too weak; Strong Pasch is too strong; we now give an axiom with a Goldilocks quality: it makes an existence assertion, but one that can be realized with ruler and compass. It says if line L does not meet side ab of triangle abc but does meet side bc then it must meet side ac as well: x(on(x, L) B(a,x,b) x = a x = b) ∀ B(a, IntersectLines→ ¬ (Line (∧a,c)6 ,L),c∧ ) 6 ∧ B(b, IntersectLines (Line (b,c),L),c)→ (Explicit Pasch) Note that that “triangle” abc is allowed to be degenerate, in that b might lie on Line (b,c). Explicit Pasch would be a constructively acceptable axiom, but it is not quantifier-free, and we are aiming for a quantifier-free axiomatization, so we are not willing to take Explicit Pasch as an axiom. We will prove it in Lemma ??. Some of the versions of Pasch considered so far all mention incidence as well as betweenness. That is not essential, as on(y, Line (a,b)) is equivalent to (B(y,a,b) y = a B(a,y,b) y = b B(a,b,y). ¬¬ ∨ ∨ ∨ ∨ To use this way of expressing incidence, we must have a way to get a and b from L. In ECG, that is built-in, but Tarski needed versions of Pasch that specify the line L by using an auxiliary point specified to be on L. Direct translations of the above forms using this idea are too long to be elegant. Several special cases can be succinctly formulated. Three in particular, “Inner Pasch”, “Outer Pasch”, and “Weak Pasch” have been studied by Tarski and his students. We exhibit these formulations in order to examine them constructively. The first one, inner Pasch, is the one taken as an axiom in [26]. Here T is non-strict betweenness. It considers a line Line (p,b) entering a “triangle” caq. We put “triangle” in quotes because it is allowed that some or all of the points be equal, and they can also be collinear. The line is specified by a point b on cq extended. T(a,p,c) T(b,q,c) x(T(p,x,b) T(q,x,a)) (Inner Pasch) ∧ → ∃ ∧ A closely related version is axiom B4 of ECG, which we repeat here for comparison. We will also refer to Axiom B4 as “Explicit Inner Pasch.” The existential quantifier has been replaced by a suitable construction term, so the result is quantifier-free. In addition non-strict betweenness has been replaced by strict betweenness, and the three points are required to form a triangle, i.e. to be distinct and non-collinear. Without the non-collinearity requirement, the term giving the interection point in B4 could possibly be undefined. B(a,p,c) B(b,q,c) p = b q = a on(c, Line (a, q) IntersectLines∧ (∧Line6 (p,b∧ ), 6Line∧(a, ¬ q)) → B(p, IntersectLines (Line (p,b), Line (a,↓ q ∧)),b) B(q, IntersectLines (Line (p,b), Line (a, q)),a)∧ (Pasch’s Axiom B4) If the line Line (p,b) is specified by b on the other side of segment cq we get Outer Pasch26 : T(a,p,c) T(q,c,b) x(T(b,p,x) T(a, x, q)) (Outer Pasch) ∧ → ∃ ∧ 26Note that it is T(a,x,q) rather than T(q,x,a) as in Inner Pasch. That does not matter, after the symmetry axiom has been included (as in ECG) or derived from inner Pasch (as in [26]), but we follow [30] to the letter here. 84 MICHAEL BEESON

As with Inner Pasch, one can replace x by a term of ECG (the same one as in Inner Pasch) to obtain a quantiﬁer-free∃ version in ECG. That version is given here: B(a,p,c) B(q,c,b) p = b a = q on(c, Line (a, q) IntersectLines∧ (∧Line ∧ (6 p,b)∧, Line6 (a,∧ q ¬)) → B(p,b, IntersectLines (Line (p,b), Line↓(a, ∧ q))) B(a, IntersectLines (Line (p,b), Line (a, q)), q)∧ (Explicit Outer Pasch) 9.9. Inner Pasch implies Outer Pasch and plane separation. Outer Pasch was an axiom (instead of, not in addition to, Inner Pasch) in versions of Tarski’s theories until 1965, when it was proved from Inner Pasch in Gupta’s thesis [13].27 Outer Pasch appears as Satz 9.6 in [26], and the plane separation theorem follows it as Satz 9.8. The proofs as given in [26] are constructive, and are valid also for strict betweenness. Lemma 9.57 (Gupta [13]). Explicit Outer Pasch is provable in neutral ECG. Proof. This is Theorem 3.70 in [13], or Satz 9.6 in [26]. Lemma 9.58 (Plane separation, Gupta [13]). SameSide (a,b,L) OppositeSide (a,c,L) OppositeSide (b,c,L). ∧ → Proof. See [26], Satz 9.8. Remark. We note that both theorems are proved using Satz 9.5, which is the special case of the plane separation theorem when Line (a,b) meets L. Neverthe- less there is no argument by cases according as that line does or does not meet L; the proof is completely constructive. Lemma 9.59. SameSide (a,b,L) SameSide (b,c,L) SameSide (a,c,L). ∧ → Proof. Suppose SameSide (a,b,L) SameSide (b,c,L). Then there exists a points x and y such that OppositeSide∧ (a,x,L) and OppositeSide (b,x,L). Since SameSide (b,c,L), by Lemma 9.58 we have OppositeSide (c,x,L). Hence by definition of SameSide , we have SameSide (a,c,L). That completes the proof. Note that OppositeSide (a,b,L) OppositeSide (b,c,L) SameSide (a,c,L) is a trivial consequence of the deﬁnition∧ of SameSide . → Lemma 9.60. If SameSide (a,b,L) then no point of L lies between a and b. Equivalently, no line meets all three (open) sides of a triangle. Proof. The two statements are equivalent since SameSide (a,b,L) means that L meets two sides of some triangle abx. Suppose that B(a,u,x) and B(b,v,x) and B(a,w,b). Then OppositeSide (x,b,L) and OppositeSide (b,a,L), so SameSide (v,a,L). But since OppositeSide (a,v,L), by Lemma 9.58 we have OppositeSide (a,a,L). Hence for some y we have B(a,y,a), contradiction. That completes the proof. The converse will be taken up in section 9.14

27But apparently, judging from footnote 4 on p. 191 of [30], Tarski knew as early as 1956-57 that Outer Pasch implies Inner Pasch; in that footnote Tarski argues against replacing Outer Pasch with Inner Pasch as an axiom, as Szmielew and Schwabh¨auser chose to do. Also on p. 196 of [30], Tarski attributes the idea for the proof of Inner Pasch from Outer Pasch to speciﬁc other people; the history is too long to review here, but he credits only Gupta with the derivation of Outer Pasch from Inner Pasch. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 85

9.10. The crossbar theorem. The following lemma is called the “crossbar theorem” on p. 116 of [12]. It is needed to prove the basic theorems about the ordering of angles. One special case of it occurs as Satz 3.17 in [26], but I do not find the complete theorem there. We prove it here from Outer Pasch and Inner Pasch; but since by Lemma 9.57, Outer Pasch is derivable from Inner Pasch, the crossbar theorem is provable in ECG. Lemma 9.61 (Crossbar Theorem). Suppose given triangle abc, and points A on Ray (b,a) and C on Ray (b,c), with A = B and C = B, and point e with B(a,e,b), and point E the intersection point6 of Line (b,e)6 with Line (A, C). Then B(A, E, C). Proof. If A = E then e is on Line (b,a) and Line (a,c) and hence it is equal to a, since those two lines do not coincide because c is not on Line (b,a). Hence A = E. Similarly C = E. We6 argue by cases6 on the disjunction B(b,c,C) C = c B(b,C,c). We will show that in each case B(A, E, C). That will suffice∨ since∨ the double negation of this disjunction holds, so we will then have B(A, E, C) and by Markov’s principle for betweenness we conclude B(A, E,¬¬ C). Within each case, we will also argue by cases according as B(b,a,A) A = a B(b,A,a), with a similar justification. That gives a total of nine cases;∨ but the∨ problem is symmetrical if variables a, A are changed with variables c, C, so some of the nine cases can be omitted. We first take up the case c = C. If a = A then E = e and we are done. If B(b,a,A) then we apply outer Pasch (Lemma 9.57), with (a,p,c,q,b) replaced by (c,e,a,A,b). The hypotheses of outer Pasch under this substitution become B(c,e,a) and B(A,a,b), both of which we have by the assumptions and the symmetry of betweenness. The conclusion is that the intersection point of Line (e,b) and Line (A, C), which is E, satisfies B(c,E,A). Since c = C we have B(A, E, C) by the symmetry of betweenness, as desired. Now suppose B(b,A,a). We now apply inner Pasch (Axiom B4) with (a,p,c,q,b) replaced by (b,A,a,e,c). The hypotheses become B(b,A,a) and B(c,e,a), which we have. The conclusion is B(A,E,c); since c = C we have B(A, E, C) as desired. That completes the case c = C. The case a = A is symmetrical, so we may drop subcases below when a = A. We next take up the case, B(b,c,C). First assume B(b,a,A). We apply outer Pasch (Lemma 9.57), with (a,p,c,q,b) replaced by (a,e,c,C,b). The hypotheses of outer Pasch under this substitution become B(a,e,c) and B(C,c,b), both of which we have. The conclusion tells us that there is a point x such that B(a, x, C) and B(b,e,x). Now, we apply outer Pasch again, this time with (a,p,c,q,b) replaced by (C,x,a,A,B). Then the hypotheses become B(C,x,a) and B(A,a,B) both of which we have (by the assumptions and the symmetry of betweenness). The conclusion is that there exists a point y with B(b,e,y) and B(A, y, C). But that point y is E, since only one point lies on Line (b,e) and Line (A, C), and that point is E. That completes the subcase B(b,a,A). As noted, we can drop the subcase a = A; we take up the subcase B(b,A,a). We apply inner Pasch with (a,p,c,q,b) replaced by (C,c,b,A,a). The hypotheses are B(C,c,b) and B(a,A,b), which we have by the symmetry of betweenness. 86 MICHAEL BEESON

The conclusion is B(c,E,a) B(A, E, C), the latter of which is our goal. That disposes of the case B(b,c,C∧). The last case is B(b,C,c). The subcase B(b,a,A) is symmetric to the one we just finished, so it is done. The subcase a = A is also done. That leaves as the last subcase, B(b,A,a). This is treated similarly to the first case, except that now we use inner Pasch instead of outer Pasch. Let x be the intersection poit of Line (b,c) and Line (A, c). Then we claim B(b,x,e) as can be proved by one application of inner Pasch. Then we prove B(C,E,A) in one more application of inner Pasch. That completes the proof. 9.11. Ordering of angles. Euclid often mentions a relation < between angles, so in order to claim that ECG is faithful to Euclid, we must define this notion in ECG. Definition 9.62. Point p lies in the interior of angle abc if B(a,e,c) and B(p,b,e), where e = IntersectLines (Line (b,p), Line (a,c). Point p lies in angle¬abc if T(a,e,c) and B(p,b,e) , where T is non-strict betweenness. ¬ Lemma 9.63. If abc and ABC are the same angle and p lies in the interior of abc then p lies in the interior of ABC. Similarly, if p lies in abc then p lies in ABC. Proof. Since ABC and abc are the same angle, B = b, and A lies on Ray (b,a) and C lies on Ray (b,c). Let L = Line (b,p) and let e be the intersection point of L with Line (a,b), and let E be the intersection point of L with Line (A, B). Suppose p lies in the interior of abc; then by definition B(a,e,c) and B(p,b,e). We must show B(A, E, C) and B(p,b,E). Now p is irrelevant and we¬ can apply Lemma 9.61. That completes the¬ proof of the first part. The second part is similar, but we only assume T(a,e,c) instead of B(a,e,c) and must prove T(A, E, C). If e = A then Line (b,p) and Line (b,a) coincide, so E = A and T (A, E, C). Similarly if e = b then E = B and T (A, E, C). But since T(a,e,c), by definition of T we have (e = c e = c B(a,e,c). Since in each of the three cases we have T(a,e,c),¬¬ we have ∨ T(a,e,c∨ ), and hence T(a,e,c). That completes the proof of the lemma. ¬¬ Definition 9.64 (Ordering of angles). abc < ABD if there is a point C in the interior of angle ABD such that angle abc is congruent to angle ABC. Similarly, abc ABD if there is a point C in angle ABD such that angle abc is congruent to angle≤ ABC. Remark. There is no need to use the notation ∠ABC < ∠DEF since ABC < DEF could not have any other meaning. Lemma 9.65 (Angle ordering respects congruence). If abc < def and angles abc and def are congruent to angles ABC and DEF respectively, then ABC < DEF . Similarly with in place of <. ≤ Proof. Since abc < def, there is a point x in the interior of angle def such that angle abc is congruent to angle dex. By Lemma 9.63, we may assume that B(d, x, f). Without loss of generality we can assume ED = ed and EF = ef, since changing the points E and F on Ray (B, E) and Ray (B, F ) makes the same angle. Then by SAS, DF = df. Let X be a point on Ray (D, F ) such that FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 87

DX = dx. Then triangle dex is congruent to triangle DEX. Hence angle dex is congruent to angle DEX. It only remains to show B(D,X,F ). By SAS, triangle def is congruent to triangle DEF , so df = DF . Now we have B(d, x, f) and dx = DX and df = DF , and X is on Ray (D, F ). If X = F then segment df is congruent to its proper subsegment dx, contradiction; hence X = F . If X = D then non-null segment dx is congruent to null segment DX, contradiction.6 Now assume B(D,F,X). Then DF

Lemma 9.66. Let a, b, and x be points on a line L with a = b. Let AB = ab. Then we can ﬁnd a point X on Line (A, B) with AX = ax and6 BX = bx.

Remark. We are not told anything about the order relations of x with a and b.

Proof. By Lemma 9.14, construct a point e on Line (b,a) such that B(e,a,b) and B(e,x,b). Then extend segment BA by ea to construct a point E with B(E,A,B) and EA = ea. Then ﬁnd a point X on Ray (E, A) with EX = ex. We claim that X is the desired point, i.e. AX = ax and BX = bx. Since equality is stable, we can prove this by cases on the possible order relations of a, b, and x. (The point is that the construction of X did not demand such case distinctions; a single term with parameters (a,b,x,A,B) gives X.) Case 1, B(a,x,b). Then since ex = EX and ea = EA we have by Lemma 9.6 that ax = AX. Since ab = AB, applying Lemma 9.6 again we have xb = XB (and hence BX = bx). Case 2, x = a. Then since ex = ea we have X = A by Lemma 9.4. Hence ax = aa = AA since all null segments are congruent; and bx = ba = BA = BX, so bx = BX. Case 3, B(x,a,b). Then since ea = EA and ex = EX we have xa = XA by Lemma 9.6; so ax = AX. Then since ab = AB we have xb = XB by Lemma 9.6, so bx = BX. The other cases reduce to these cases by interchanging a with b and A with B. That completes the proof of the lemma.

Lemma 9.67 (Triangle construction theorem). Let abc be a triangle (so the vertices are distinct and not collinear). Let AB = ab and let z be a point not on Line (A, B). Then there is a point C such that triangle ABC is congruent to 88 MICHAEL BEESON triangle abc and zC does not meet L. (In fact point C exists if we only assume a, b, and c are distinct, without assuming the are non-collinear.)

Remark. The conclusion says zC does not meet L, rather than SameSide (z,C,L). We will prove the equivalance of these two statements, but only later, using the upper dimension axiom.

Proof. Let x be a point on Line (a,b) such that cx ab. The existence of such an x was proved in Lemma 9.56. By Lemma 9.66,⊥ we can find a point X on L = Line (A, B) such that ax = AX and bx = bX. By Lemma 9.56, we can find a line K perpendicular to Line (A, B) at X. Now we want to take C to be a point on K with CX = cx, and C on the same side of L as z. To construct C, drop a perpendicular from z to K (which we can do without knowing whether z is on K or not, by Lemma 9.56). Let w be the foot of that perpendicular, so w is on K. Then wz does not meet L, by Lemma 9.48, since wz K and K L. In particular w = X, since X lies on L, so Line (X, w) is defined,⊥ and it makes⊥ sense to speak of6 a point lying on Ray (X, w). Then we can construct C as a point on Ray (X, w) with XC = xc. Note that we have not yet used that a b, and c are non-collinear; so the parenthetical remark in the lemma is established. But now, since c does not lie on Line (a,b), axc is a triangle, and X does not lie on Line (A, B), so AXC is a triangle. Then since all right angles are congruent, we have triangle axc congruent to triangle AXC, and triangle BXC congruent to triangle bxc. Hence AC = ac and BC = bc. Since we are given AB = ab, triangle ABC is congruent to triangle abc by SSS (the definition of triangle congruence. We must still show that Cz does not meet L. Suppose that Cz meets L in point p. Then C = w since zw does not meet L. Since C is on Ray (X, w), we have (B(X,C,w6 ) B(X,w,C)). If B(X,w,C) then we apply Inner Pasch, with (a,p,c,b,q¬¬ ) replaced∨ by (z,p,C,X,w). The conclusion of Inner Pasch gives as point r with B(p,r,X) and B(w,r,z). But p and X are on L, so r is on L; but then since B(w,r,z), L meets wz, contradiction. Hence B(X,w,C). If B(X,C,w) we instead apply Outer Pasch, with (a,p,c,q,b) replaced¬ by (z,p,C,w,X). The conclusion gives us a point r with B(z, r, w) and B(r,p,X). Again since r and p are on L, B(r,p,X) implies r is on L, and then B(z, r, w) shows that zw meets L, contradiction. Hence B(X,C,w). But now we have contradicted (B(X,C,w) B(X,w,C); hence¬ Cz does not meet L after all. That completes¬¬ the proof. ∨ Note that intuitively, C is the intersection point of two circles, circle C with center A and radius ac, and circle K with center B and radius bc. The problem is to prove those circles intersect. If there is a shorter way to show the circles do intersect than the above proof, I do not know it. The uniqueness of the triangle, which is the next lemma, is more difficult than the existence. Even if we knew those circles intersected, we have not proved that two circles intersect in only one point on a given side of the line joining their centers; indeed, this lemma will be used to prove that.

Lemma 9.68 (Triangle uniqueness theorem). In the previous lemma the point C is unique in the sense that if ABC and ABD are both congruent to abc and C and D are on the same side of Line (A, B) as x, then C = D. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 89

Remark. We follow the general idea of Satz 10.16 of [26], but appealing to our Lemma 9.56 for the existence of perpendiculars. Note that uniqueness in the sense that Cx does not meet Line (A, B) will not hold in three dimensions and we have not yet used the upper dimension axiom; but uniqueness in the same-side sense does hold in higher dimensions. Proof. Suppose C and D are two points on the same side of AB as x, and triangles ABC and ABD are both congruent to triangle abc. Then by Lemma 9.1, triangle ABC is congruent to triangle ABD. Let s be a point on Line (A, B) such that Cs AB; such a point exists by Lemma 9.56. Extend segment Ds (not Cs) to ⊥ point C∗ such that B(D,s,C∗) and Ds = sC∗. Now C∗ and D lie on opposite sides of Line (A, B), as witnessed by point s, and D and C lie on the same side of Line (A, B), by hypothesis. By Lemma 9.38, Line (A, B) is perpendicular to Line (D, C∗), and hence, if s = B, triangles DsB and C∗sB are congruent, and so 6 DB = C∗B. If s = B then DB = Ds = sC∗ = BC∗. Since (s = B s = B), ¬¬ ∨ 6 we have DB = C∗B, and hence DB = C∗B. Since triangles ABD and ABC are congruent,¬¬ we also have DB = CB, so by Lemma 9.1 we have DB = CB. By the plane separation principle, C and C∗ lie on opposite sides of Line (A, B). Hence there is a point t on Line (A, B) between C and C∗. Since triangles ABC and ABD are both congruent to triangle ABC, we have AC = AD and BC = BD. Angle BsC∗ is congruent to angle BsC by Lemma 9.43, so by SAS, triangle BsC∗ is congruent to triangle BsC. Hence BC∗ = BC. Similarly, exchanging A and B in the argument, we have AC∗ = AC. We claim tC = tC∗. To prove this, we have to argue by cases on the possible orderings of A, t, and B on Line (A, B). If we show in every case tC = tC∗ then since not all the cases can fail, we conclude tC = tC∗ and hence tc = cC∗ by the stability of equality. First assume B(A,t,B¬¬ ) or B(A,B,t). By Lemma 9.14, let E be a point on Line (A, B) such that B(A, t, E) and B(A, B, E). Apply Lemma 9.10 with (a,b,c,d,e) replaced by (A,t,B,C,E) and (A,B,C,D,E) replaced by (A,t,B,C∗, E). The hypotheses are satisfied since AC = AC∗ and BC = BC∗. The conclusion is tC = tC∗. The next case is t = B; then tC = BC = BC∗ = tC∗. The next case is t = A; then tC = AC = AC∗ = tC∗. The remaining cases reduce to these cases by interchanging A and B. Hence tC = tC∗. Now we have AC = AC∗ and BC = BC∗ and tC = tC∗. Assume A = t, so 6 AtC and AtC∗ are triangles. Then triangle AtC is congruent to triangle AtC , by definition of triangle congruence (SSS). Then angle AtC is congruent to angle∗ AtC∗, as corresponding angles of congruent triangles. Then AB CC∗, by s ⊥ definition of perpendicular. Then C and C∗t are two perpendiculars from C∗ to Line (A, B). Hence by Lemma 9.46 we have s = t. Then both C and D lie on Line (C∗t), and both extend segment C∗t by tC, since Dt = Ds = sC∗ by construction of C∗, and Ct = tC∗ so since s = t we have Dt = sC∗ = tC∗ = Ct. But then by Lemma 9.4, C = C∗. That is the desired conclusion, but it was reached under the assumption A = t. Under the assumption B = t, interchanging 6 6 A and B in the argument we also reach the desired conclusion C = C∗. But since A = B, we have (t = A t = B) (which is equivalent to (t = A t = B)). 6 ¬¬ 6 ∨ 6 ¬ ∧ Hence C = C∗; hence C = C∗. That completes the proof. ¬¬ 90 MICHAEL BEESON

Lemma 9.69. [SAA] If angles abc and acb are congruent respectively to angles ABC and ACB, and ab = AB, then triangles abc and ABC are congruent. Proof. By Lemma 9.67 there exists a point E on the same side of Line (A, B) as C such that triangle ABC is congruent to triangle abc. Then angle ABE is congruent to angle ABC since both are congruent to angle abc. Let x be a point on Ray (A, C) such that ac = AX. Then since angle ABC is congruent to angle ABE, we have triangle ABC congruent to triangle ABE. But since E and C are on the same side of Line (A, B), by Lemma 9.68, E = C. Hence triangle ABC is congruent to triangle abc, since triangle ABE is congruent to triangle abc. That completes the proof. Lemma 9.70. Suppose angle abc is congruent to angle abd. Then d lies on Ray (b,c). Proof. Let A lie on Ray (b,a) with AB = αβ, and let C lie on Ray (b,c) with BC = αβ, and D on Ray (b, d) with bd = αβ. By the definition of angle congruence, triangle AbC is congruent to triangle AbD. Hence, by Lemma 9.68, C = D. Then Ray (b,c) coincides with Ray (b, d), so d lies on Ray (b,c) as claimed. That completes the proof. Lemma 9.71. Let z be a point not on line L. Two circles with distinct centers on L intersect in at most one point C such that Cz does not meet L. Proof. Let C and D be two intersection points of circles C and K with centers A and B respectively. Triangle ABC is congruent to ABD, so by Lemma 9.68, not both Cz and Dz can fail to meet L. That completes the proof. Lemma 9.72 (Stability of angle ordering). abc < def abc < def ¬¬ → Remark. The proof is nontrivial; after the definitions are unwound, one sees the key idea, but it still requires several of the difficult theorems already proved to complete the proof. The proof is illustrated in Fig. 18. Note also that this is the first time we have used IntersectCirclesSame .

Figure 18. Stability of angle ordering. Given angle dep congruent to angle abc, we must have R = r.

b d a A b b

r b b b e b b c b b q R b p b f FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 91

Proof. By the deﬁnition of angle ordering, abc < def means that there exists a point p in the interior of angle def such that angle abc is congruent to angle dep. By Deﬁnition 9.62, p is in the interior of angle def if and only if B(d,q,f) B(p,e,q), where ∧ ¬ q = IntersectLines (Line (e,p), Line (d, f)).

That is, p lies on Ray (e, q). The key observation is that there is a canonical choice of point p, namely the one produced by copying triangle abc onto segment de with its vertex at b. (That observation is key, because it means that we do not have to pass a double negation through an existential quantifier p ; the quantifier can be replaced by a term defining p, as we shall see in detail.)∃ Let C = Circle (e,a,b) and let A be a point on Ray (e, d) such that eA = ba; then A = IntersectLineCircle2 (Line (e, d), C), so A is given by a term. Let K = Circle (A,a,c), and let

r = IntersectCirclesSame (C,K,f).

By the triangle construction theorem (Lemma 9.67) we see that r is defined. We claim that abc < def if and only if r is in the interior of angle def. That is, if there is any p in the interior of angle def such that abc is congruent to dep, then this r is such a p. To prove this, we argue as follows. Suppose that abc is congruent to dep and p lies in the interior of angle dep. Triangle Aer is congruent to triangle abc by definition of triangle congruence, since bc = er and ac = Ar by the definition of r, and ab = Ae by construction of A. Hence angle abc is congruent to angle Aer. Hence angle Aer is congruent to angle dep, since both are congruent to angle abc. Let point R on Ray (e,p) have eR = er. Then triangle Aer is congruent to triangle AeR by SAS. We claim r = R. Let L = Line (e, A). By the triangle uniqueness theorem, it suffices to prove that r and R are on the same side of L. By construction, r is on the same side of Line (e, A) as f, since L joins the centers of C and K. Since p is in the interior of angle def, the point q defined above lies between d and f, so by Lemma 9.83, q and f are on the same side of L, since segment qf does not meet L. But R lies on Ray (e,p). Hence Rq does not meet L, since Line (e,p) meets L in e and q does not lie on L. Hence by Lemma 9.83, R and q are on the same side of L. Since we already proved q and f are on the same side of L, by the plane separation theorem, r and R lie on the same side of L. That completes the proof that if any p in the interior of angle def has abc congruent to dep, then r is such a p. In other words,

abc < def B(d,q,f) B(r, e, q) ↔ ∧ ¬ where q = IntersectLines (Line (e, r), Line (d, f)) and where r is given by a speciﬁc term with variables a,b,c,d, and e, as deﬁned above. Hence 92 MICHAEL BEESON

abc < def (B(d,q,f) B(r, e, q)) ¬¬ ↔ ¬¬ ∧ ¬ B(d,q,f) B(r, e, q) ↔ ¬¬ ∧ ¬¬¬ B(d,q,f) B(r, e, q) ↔ ¬¬ ∧ ¬ B(d,q,f) B(r, e, q) ↔ ∧ ¬ abc < def by the stability of betweenness ↔ That completes the proof of the lemma. 9.13. Euclid Books I, II, and III. We are now in a position to prove the theorems in the first three books of Euclid. We mention a few of these that we will need later. Since we are still working in neutral geometry, we will explicitly mention which parallel axiom is needed. Euclid’s first use of the parallel postulate is in Prop. I.29, and it is also needed for theorems about circles in Book III. These propositions, even the ones in Book I, could not be proved earlier, since many of them mention the ordering of angles, and some of them require us to lay off an angle on a given side of a given segment; our ability to do that requires the triangle construction lemma; indeed Euclid I.23 is the triangle construction lemma. For example, a corollary of the congruence of vertical angles and the definition of angle ordering is the exterior angle theorem, Euclid 1.16. (Without the parallel axiom, one only gets an inequality, not the stronger theorem that an exterior angle can be divided into two angles, each congruent to one of the interior and opposite angles.) Moreover, in order to support arguments by cases on whether one angle is larger than another or not, we need the stability of angle ordering (Lemma 9.72), which as we have seen was not trivial to prove. Lemma 9.73 (Exterior angle theorem). Suppose B(B,C,D) and A does not lie on Line (B, C). Then angle ACD is greater than angle BAC. That is, there is a point E such that B(A, E, C) and and point F such that B(B,E,F ) such that angle BAC equals angle ACF . Proof. This is Euclid I.16. We note that according to our precise definition of angle ordering, the meaning of the proposition is as stated here. Compare Euclid’s proof to the proof in [12], Theorem 4.2, page 165, which requires the much deeper crossbar theorem (Lemma 9.61); Euclid’s proof does not require it. In particular it is not required to prove that F and G in Euclid’s diagram lie on opposite sides of Line (B, C). Lemma 9.74. A leg of a right triangle is less than the hypotenuse. Proof. Euclid does not state this explicitly, but by the exterior angle theorem (Euclid I.16), or Lemma 9.73 above, the right angle is greater than each of the other two angles, so the lemma follows from Euclid I.19, which in turn follows from Euclid I.18. Euclid’s proofs make use of an argument by contradiction on angle ordering, but since angle ordering is stable by Lemma 9.72, Euclid’s proofs are essentially constructive. That is as much as we will say about this proof. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 93

Lemma 9.75 (Euclid III.2). Let line L meet circle C in two points a and b, and let point x on L be different from both a and b. Then x is inside c if (and only if) x is between a and b, and otherwise x is outside c. Remark. Euclid’s proof has been criticized (see e.g. Heath’s commentary on the proposition in [8]), since III.2 uses III.1 and vice-versa. The alleged flaw is that Euclid’s proof of III.2 starts by using III.1 to find the center of the given circle, but III.1 implicitly uses III.2. That criticism does not apply in our theory, since circles always “come equipped” with their centers. That is, instead of needing III.1, we can just take the center to be center(C). To put the matter another way: the issue is that if one views Euclid III.1 as an attempt to show that center(C) is definable and hence redundant, the attempt is circular, and hence fails; but that is a different matter from the correctness of Euclid III.2. Proof. Euclid’s proof proceeds by contradiction, but that is constructively acceptable due to the stability of segment ordering, proved in Lemma 9.15. His proof appeals to the equality of the base angles of an isosceles triangle (Lemma 9.28) and to euclid I.16 and I.19, whose proofs have been discussed above. Lemma 9.76. If line L meets chord pq of circle C, then the two intersection points of L with C lie on opposite sides of Line (p, q).

Figure 19. Lemma 9.76. B(p, x, q) implies B(a,x,b).

p b

b e b b

b x

q ab b

Remark. This innocent-looking lemma requires Euclid III.2, which as we have seen requires the exterior angle theorem and the triangle construction and uniqueness theorems. See Fig. 19. Proof. Let L meet pq in point x, so B(p, x, q), and let a and b be the intersection points of L with C. We have to show B(a,x,b). Let e be the center of C. Then by Lemma 9.75, since x lies on the chord pq, x is strictly inside C, i.e. ex < ep. Suppose B(a,b,x). Then by Lemma 9.75, x is strictly outside C, i.e. ex > ep. 94 MICHAEL BEESON

This is a contradiction; hence B(a,b,x). Similarly B(x,a,b). Also x = a and x = b since x is strictly inside¬ C. Hence by axiom¬ B3, B(a,x,b) as required.6 That6 completes the proof.

Lemma 9.77 (Euclid I.29). The Playfair parallel axiom implies that the alternate interior angles formed by a transversal to parallel lines are congruent. Proof. By Lemma 9.34, if the alternate interior angles are not not congruent, they are congruent. Hence we can prove the proposition by contradiction. Then Euclid’s proof is valid. That completes the proof. Lemma 9.78 (Euclid III.20). The Playfair parallel axiom implies that an exterior angle of a triangle is equal to the sum of the alternate interior angles. Remark. The use of the word “sum” does not imply addition. It merely means that the exterior angle can be decomposed into two angles, each congruent to one of the opposite interior angles. Addition of angles has not been defined (either here or in Euclid). Proof. Let abc be the triangle and led d be a point on Line (a,c) with B(a,c,d), and consider the exterior angle bcd. The line mentioned in the remark is Para(c, Line (a,b)). Then angle abc is congruent to the alternate interior angle bce (by Lemma 9.77), and angle bca is congruent to angle ecd, as can be proved classically from the parallel postulate. Double negating this proof, and using the stability of angle congruence (Lemma 9.34), we find that these congruence assertions are derivable constructively from Playfair. That completes the proof. Lemma 9.79 (Euclid III.20). The Playfair parallel axiom implies: An angle inscribed in a semicircle is a right angle. That is, if ea = eb = ec and B(a,e,b), then angle acb is a right angle. More generally, if a and b are any points on the circle, and c is any point on the circumference, angle aeb is double angle aec. Remark. Note that, although we allow degenerate circles, for this lemma the semicircle must be non-degenerate, as otherwise the notion of “inscribed angle” makes no sense. Proof. The proof in Euclid is constructive; see Fig. ??. Euclid’s proof depends on Euclid I.32, that an exterior angle of a triangle equals the sum of the opposite interior angles; we have proved I.32 above from the Playfair axiom. That completes the proof. 9.14. Consequences of the upper dimension axiom. So far, we have not used the upper dimension axiom. Now we will use it. Lemma 9.80. Suppose K and M are lines perpendicular to L at m. Then K and M coincide (have the same points). Remark. This result depends on the upper dimension axiom, since in 3-space there are many perpendiculars to L at m. Proof. Suppose K and M are perpendicular to L at m. Let a be a point on K not on L. Extend segment am to b such that am = mb and B(a,m,b). Let c be a point on M with mc = ma. Let p and q be two points on L with FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 95 pm = mq (to be specific take pm = mq = ma although the distance pm is not essential). Then since K M we have triangle amp congruent to amq and triangle bmp congruent to ⊥bmq, and triangle cmp congruent to cmq. Let Ca = Circle (a, q), let Cb = Circle (b, q), and Cc = Circle (c, q). Then p and q lie on all three circles. By the upper dimension axiom, their centers a, b, and c are collinear. By Lemma 9.11 we do not have B(m,c,a) or B(m,a,c) or B(m,b,c) or B(m,c,b). Since we also have c = m, by Axiom B3 we have (c = b c = a). If c = b c = a then K and M coincide,6 so in either case c is¬¬ on K; therefore∨ we have∨ on(c,K). By stability axiom S3, we can drop the double negation, so on(c,K¬¬). Hence K and M coincide. That completes the proof. Lemma 9.81. Suppose K and M are lines perpendicular to L and x lies on both K and M. Then K and M coincide. Remark. The point is that we do not require a case distinction as to whether x is or is not on L. Proof. Let u be on K; we must show u is on M. If x is on L, then Lemma 9.80 applies, so on (u,M) in that case. If x is not on L, then Lemma 9.46 applies, so on (u,M) in that case. But we have (on (x, L) on (x, L). Hence on(u,M). But then by Axiom S3 we have¬¬on (u,M).∨ That ¬ completes the proof.¬¬ Lemma 9.82. Let ab be a chord of circle C. Then the center of C lies on the perpendicular bisector of ab. Proof. Let m be the midpoint of ab and p the center of C. Then pa = pb and ma = mb and pm = pm, so triangle pam is congruent to triangle pbm. Hence mp is perpendicular to ab at m. But also the perpendicular bisector of ab is perpendicular to ab at m; so by Lemma 9.80, p is on the perpendicular bisector. That completes the proof.

Figure 20. If ab does not meet L then a and b are on the same side of L because bc meets L. (Lemma 9.83).

w b b b a b

b m

b b r x z L

c b

Lemma 9.83. If a and b are not on L and no point of L is between a and b then SameSide (a,b,L). Indeed a speciﬁc point c, namely the reﬂection of a in 96 MICHAEL BEESON

L, is constructible (i.e. given by a term t(a,L)) such that c is on the opposite side of L from both a and b. Remark. Recall that SameSide (a,b,L) means a and b are both on the opposite side of L from some point c. This lemma obviously fails in three-space, since if a and b are on the same side of L then they are coplanar with L, but it may be that segment ab does not meet L even though they are not coplanar. Proof. Drop a perpendicular from a to L and let x be its foot, so x is on L and ax L. By Lemma 9.56, construct a line K perpendicular to ax; let w be a point⊥ on both K and Line (a, x). Then w = x since then, by Lemma 9.80, L and K would coincide, so b would lie on L, contrary6 to hypothesis. Let m be the midpoint of segment wx. Let rm = mb with B(r,m,b). Assume w = b. Then angles rmx and bmw are vertical angles, so they are congruent by Lemma6 9.33. Then triangles rmx and bmw are congruent, by SAS. On the other hand (still assuming w = b), let s be a point on L with sx = wb. Since w = b, we have s = x. Since these6 are both right angles, by Lemma 9.44 they are6 congruent. Then6 triangles sxm and bwm are congruent, by SAS. Then rx and sx are both perpendicular to K at x. By Lemma 9.80, Line (x, r) and Line (x, s) coincide. Therefore r lies on L. (For all we know s might be r or it might be the reflection of r in x, so we cannot say s = r, but we are finished with s now that we know r is on L.) Let c be the reflection of a in x, that is, the point on K with B(c,x,a) and ax = cx. Assume B(m,x,c). We wish to apply Outer Pasch with (a,p,c,q,b) replaced by c,x,m,b,r). We check the hypotheses of Outer Pasch under this substitution. One is B(b,m,r), which we have by construction of r. The other is B(m,x,c), which we have assumed. Hence Outer Pasch can be applied. The result is a point z with B(c,z,b) and B(z, x, r). Since B(z, x, r), z is on L, and hence bc meets L. Then b = c since B(c,z,b), so z = IntersectLines (L, Line (b,c)). We have thus proved 6 w = b on(IntersectLines (L, Line (b,c)) 6 → under the assumptions w = b and B(m,x,c). Now keep the assumption6 w = b, but replace B(m,x,c) with the assumption B(m,x,a). Interchanging a and6 c in the above argument we find by Outer Pasch a point z on L with B(a,z,b), contradicting the assumption that ab does not meet L. Hence B(m,x,a). We also have m = x, since if m = x then wx is a null segment,¬ so bw and L are both perpendicular6 to ax at x, contradicting Lemma 9.80. Hence by Axiom B3, (B(x,m,a) m = a B(x,a,m). ¬¬ ∨ ∨ We claim that in each of the three cases we have B(m,x,c). We have B(a,x,c) by construction of c, so if m = a we have B(m,x,c) immediately. If B(x,m,a) then B(a,m,x); together with B(a,x,c) we get B(m,x,c) by Lemma 9.9. If B(x,a,m) then from B(c,x,a) we obtain B(c,x,m) by outer transitivity (Lemma 9.8) and hence B(m,x,c). Hence all three cases imply B(m,x,c). Hence B(m,x,c). By Markov’s principle for betweenness we have B(m,x,c). But above¬¬ we have shown that this assumption leads to the conclusion that bc meets L. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 97

Now drop all betweenness assumptions, and assume w = b. Then b = c, since ba does not meet L, but bc meets L in x. Hence Line (b,c) is defined and6 equals K. Hence IntersectLines (L, Line (b,c)) = x and hence IntersectLines (L, Line (b,c)) lies on L. We have thus proved w = b w = b on(IntersectLines (L, Line (b,c)) ∨ 6 → But we have (w = b w = b). Hence ¬¬ ∨ 6 on(IntersectLines (L, Line (b,c)). ¬¬ By stability axiom S3, we can drop the double negation. Then b and c are on opposite sides of L, by definition of OppositeSide . Since B(c,x,a), a and c are on opposite sides of L. Then a and b are on the same side of L, by definition of SameSide . That completes the proof of the first part of the lemma. For the second part, we use the fact that there is a term representing every ruler and compass construction, and we showed how to construct c, which we now summarize. p = pointOn1(L) q = pointOn2(L) C = Circle(p,a) K = Circle(q,a) u = IntersectCircles1(C,K) v = IntersectCircles2(C,K) M = Line(u,v); x = IntersectLines(L,M); c = IntersectLineCircle2(Line(a,x),Circle(x,a))

There is, of course, a term t(a,L) representing this construction. That completes the proof. Lemma 9.84. Not both SameSide (a,b,L) and OppositeSide (a,b,L). Proof. Suppose SameSide (a,b,L) and OppositeSide (a,b,L). Then some c is on the opposite side of L from both a and b. Since b is on the opposite side from c and a is on the opposite side from b, a is on the same side as c. Since a and c are on opposite sides, and a and c are on the same side, by the plane separation lemma (Lemma 9.58), a and a are on opposite sides. Hence there is a point u with B(a,u,a), contradicting Axiom B1-ii. That completes the proof. Lemma 9.85 (Stability of SameSide ). SameSide (a,b,L) SameSide (a,b,L). ¬¬ → Proof. Suppose a is not on L and SameSide (a,b,L). We claim that no point of L is between a and b. Assume that¬¬ B(a,z,b). Then a and b are on opposite sides of L. Then SameSide (a,b,L), by Lemma 9.84. But that contradicts SameSide (a,b,L¬). Hence there does not exist any z between a and b. Then by¬¬ Lemma 9.83, a and b are on the same side of L. That completes the proof. 98 MICHAEL BEESON

Remark. It is not hard to show that the stability of OppositeSide is equivalent to the special case of the principle t t , when t is IntersectLines (L,K). This principle, which we have called¬¬ S6,↓ →is not↓ taken as an axiom of ECG, but we will show below that it follows from the strong parallel axiom, so in ECG OppositeSide is also stable. We do not prove that here since we wish to put the consequences of the parallel axiom in a separate section. Lemma 9.86. Explicit Pasch is provable in ECG. Remark. Clearly ExplicitPasch can fail in three-space, so it should require the upper dimension axiom to prove it. By contrast, Inner Pasch and Outer Pasch do not require the upper dimension axiom. Proof. Suppose B(a, IntersectLines (Line (a,c),L),c) and suppose L does not meet ab and that a and b do not lie on L. Then by Lemma 9.83, a and b are on the same side of L. By definition of opposite sides, a and c are on opposite sides of L. Hence, by the plane separation theorem (Lemma 9.58), b and c are on opposite sides of L. By definition of opposite sides, there is a point y on L with B(b,y,c). Since point b is not on L, IntersectLines (Line (b,c),L) is defined, by Axiom I12. Then by Axiom I13, y = IntersectLines (Line (b,c),L). Since B(b,y,c), the conclusion of Explicit Pasch is verified. That completes the proof of the lemma. Lemma 9.87. Explicit Outer Pasch implies Negative Pasch. Proof. Suppose abc is a triangle, i.e. the points are distinct and non-collinear, and suppose line L meets ab in a point p between a and b. Suppose L meets Line (b,c); let x be the point of intersection (there is only one since a, b, and c are not collinear). We have to prove not all of the following fail: a, b, or c is on L, or L meets ac, or L meets bc. Suppose then that they all do fail. Suppose L meets Line (b,c). Let x be the point of intersection. Since L does not meet bc and is not equal to b or c, then (B(x,b,c) B(b,c,x)). If B(x,b,c), then by Axiom B4 (explicit inner Pasch),¬¬ L meets ac∨, contradiction. If B(b,c,x), then by Explict Outer Pasch, L meets bc, contradiction. Hence L does not meet Line (b,c). Then b and c are on the same side of L. Since L meets segment ab, a and b are on opposite sides of L. By the plane separation theorem, a and c are on opposite sides of L. Hence L meets segment ac, contradiction. That completes the proof of the lemma. Tarski also considered the following variant of Pasch’s axiom: B(a,t,d) B(b,d,c) x y((a,x,b) B(a,y,c) B(y,t,x)) (Weak Pasch) ∧ → ∃ ∃ ¯ ∧ ∧ Of course Tarski used T instead of B, but in case a, b, and c are collinear, or t = a, or t = d, one can take x = y = t, so weak Pasch with T is classically equivalent to weak Pasch with B. Weak Pasch differs from the others in that the points x and y are neither unique nor in any way specified by the axiom. Nevertheless they can be constructed: Lemma 9.88. Explicit Pasch implies weak Pasch. Proof. Construct a parallel K to L = Line (a, d) through t and apply Explicit Pasch to triangle abd. Line K meets side ad at t and does not meet side bd since FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 99 it is parallel to L. Hence by Explicit Pasch, it meets side ab in x. Similarly, using triangle acd, K meets side ac in point y. That completes the proof. 9.15. Right and left turns and sides of lines. Recall that Right and Left are define in ECG in terms of IntersectCircles1 and IntersectCircles2 . Of course, these axioms can also be viewed as defining which intersection point of circles is which, in terms of the fundamental and intuitive notions of right and left turns. These notions are in turn axiomatized by other axioms of ECG that connect Right and Left to the notions of SameSide and OppositeSide . These connections can be completed with a few lemmas, which we now give. Lemma 9.89. B(a,b,e) Right (a,b,c) Right (a,e,c) ∧ → B(a,e,b) Right (a,b,c) Right (a,e,c) ∧ → B(e,a,b) Right (a,b,c) Right (e,b,c) ∧ → and similarly with Left instead of Right . Proof. Suppose B(a,b,e) Right (a,b,c). Then by Axiom H8, Right (c,a,b). Then e and b are on the same∧ side of Line (a,c), as witnessed for example by the reflection x of b in point a, i.e. the point x such that B(x,a,b) and bx = ab. Then B(x,a,e) by Lemma 9.7. Hence by Axiom H4, we have Right (a,e,c). That proves the first assertion of the lemma; the other five assertions are proved similarly. We omit the proofs. The following rather basic lemma seems to require Lemma 9.89: Lemma 9.90. If c is on Ray (a,b) and c = a then 6 IntersectLines (Line (a,b),K) ∼= IntersectLines (Line (a,c),K) IntersectLineCircle1 (Line (a,b), C) ∼= IntersectLineCircle1 (Line (a,c), C) IntersectLineCircle2 (Line (a,b), C) ∼= IntersectLineCircle2 (Line (a,c), C) Proof. To prove the first assertion, suppose IntersectLines (Line (a,b),K) is defined. Then Line (a,b) and K do not coincide, by Axiom I21, and there is a point m on both Line (a,b) and K, by Axiom I4. Since c and a are on both Line (a,b) and Line (a,c), by Axiom I3, those lines coincide; in particular m is on Line (a,c). Since m is also on K, and since K and Line (a,c) do not coincide, (or else K and Line (a,b) would coincide, which they do not), IntersectLines (Line (a,c),K) is defined, by Axiom I12, and is equal to m, by Axiom I13. That completes the proof of the first assertion. Let p = IntersectLineCircle1 (Line (a,b), C) q = IntersectLineCircle2 (Line (a,b), C)

p′ = IntersectLineCircle1 (Line (a,c), C)

q′ = IntersectLineCircle1 (Line (a,b), C) Then for points x not on Line (a,b), Right (x, p, q) Right (x,a,b) byAxiomCont4 ↔ Right (x,a,c) by Lemma 9.89 ↔ Right (x, p′, q′) by Axioms Cont 4 ↔ 100 MICHAEL BEESON

But the points p′ and q′ are the two intersection points of the line and circle, so

((p′ = p q′ = q) (p′ = q q′ = p)). ¬¬ ∧ ∨ ∧ But since x is not on L, the second alternative would violate Axiom H10 and Lemma ??. Hence, by the stability of equality, the first alternative holds. That completes the proof of the lemma. Lemma 9.91. Let a and b lie on line L, and x and y not lie on line L, and suppose Right (a,b,x). Then x and y are on the same side of L if and only if Right (a,b,y), and x and y are on opposite sides of L if and only if Left (a,b,y). Similarly with Right and Left interchanged. Proof. The left-to-right implication in the first claim is Axiom H4 for Right and Axiom H5 for Left . Now suppose Right (a,b,x) and Right (a,b,y). We must show x and y are on the same side of L. By Lemma 9.83, it suffices to show that no point of L is between x and y. Suppose B(x,z,y) and z is on L. Then x and y are on opposite sides of L. Then by Axiom H6, Left (a,b,y). Taking p = q = y in Axiom H3, the hypotheses of H3 hold, and we get B(y, IntersectLines (Line (y,y), Line (a,b),y), which false since Line (y,y) is undefined, and even if it were it would violate Axiom B1-ii. That completes the proof of both directions of the first claim. The second claim is proved similarly, with Right and Left interchanged, appealing to Axiom H7 instead of H6. That completes the proof. Lemma 9.92. Left (a,b,p) Right (a,b,p) on(p, Line (a,b)). ∧ → Proof. Let C = Circle (a,p) and K = Circle (b,p). By definition of Left and Right , if Left (a,b,p) or Right (a,b,p), then both intersection points of C and K are defined. Hence a = b, by Axiom I17 or Axiom I18. Then L = Line (a,b) is defined. Assume, for proof6 by contradiction , that p does not lie on L. Let M be the perpendicular from p to L, let x be its intersection point with L, and let q be the reflection of p in L, i.e. the point with qx = px and B(p, x, q). Then p and q are on opposite sides of L so by Axiom H6, we have Left (a,b,q). Now triangle axp is congruent to triagle axq by SAS, since L is perpendicular to M by Lemma 9.38 (or by SAS since all right angles are congruent). Hence ap = aq. Similarly bp = bq. Then C = Circle (a, q) and K = Circle (b, q). Then by definition of Left and Right , since Left (a,b,q) we have q = IntersectCircles2 (C,K). But then by hypothesis, q = p. Since B(p, x, q) we now have B(p,x,p), contradicting Axiom B1-ii. This contradiction shows on(p,L). By Axiom S3 we have on(p,L). That completes the proof of the lemma.¬¬ Lemma 9.93. Suppose p = IntersectCircles1 (C,K)= IntersectCircles2 (C,K). Then p is collinear with the centers of C and K. Proof. Let a and b be the centers of circles C and K, and let L = Line (a,b). By the definitions of Right and Left we have Right (a,b,p) and Left (a,b,p). By Lemma 9.92, p lies on Line (a,b). That completes the proof. Lemma 9.94. Let C and K be two distinct circles, and suppose x lies on both C and K. Then x is not different from both IntersectCircles1 (C,K) and IntersectCircles2 (C,K). FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 101

Proof. Let p = IntersectCircles1 (C,K) and q = IntersectCircles2 (C,K). These points exist by circle-circle continuity, since the point c on C is both non-strictly inside and non-strictly outside of K. Let a and b be the centers of C and K respectively. Since the circles are distinct and have a point in common, a = b. Then L = Line (a,b) is defined. If p = q then Line (p, q) is perpedicular to6 a. Let e on L be the intersection of L with6 Line(p, q). Since x lies on both circles, we have ax = ap and bx = bp, so triangle abx is congruent to triangle abp. By Lemma 9.68, if x is on the same side of L as p then x = p, and if it is on the same side of L as q then x = q. Since p and q are on opposite sides of L, by the plane separation theorem if x is on the opposite side of L from q then it is on the same side as p, so x = p. On the other hand if x is not on the opposite side of L from q then by Lemma 9.83, it is on the same side as q, so x = q. But OppositeSide (x,q,L) OppositeSide (x,q,L)) ¬¬ ∨ ¬ is logically valid; hence (x = p x = q). That completes the proof. ¬¬ ∨ Lemma 9.95. Suppose a, b, and c are not collinear. Then (Right (a,b,c) Left (a,b,c)). ¬¬ ∨ Remark. We cannot remove the double negation, since ECG cannot prove any disjunctive theorem, as we will show in a subsequent section. Proof. Suppose a, b, and c are not colliear, and suppose Right (a,b,c) and Left (a,b,c). We must derive a contradiction. Let C = Circle¬ (a,c) and K = ¬Circle (b,c). Then by definition of Right and Left , c = IntersectCircles1 (C,K) and c = IntersectCircles2 (C,K). By Lemma 9.94,6 this is impossible. That completes6 the proof. 9.16. Definability of IntersectCirclesSame and IntersectCirclesOpp. We have included a function symbol IntersectCirclesSame to produce the intersection point of two circles on the same side of the line joining their centers as a given point x. In this section we show that it was superfluous to have a primitive symbol for that purpose, as a compound term already plays that role. The term, however, is rather complicated, so in the interest of supporting Euclid more directly, it may be handy to have a primitive symbol. Up until now, we have mentioned IntersectCirclesSame only once, when we needed it to prove the stability of angle ordering in Lemma 9.72. That lemma in turn is needed for the propositions in the latter part of Euclid Book I, but has not been used anywhere in this paper; it was proved only to lay the foundation for Euclid. In particular none of the constructions we have defined so far used IntersectCirclesSame, so we may use them in the proof that it is definable. Theorem 9.96 (Definability of IntersectCirclesSame ). There is a term t(R,K,x) of ECG not containing IntersectCirclesSame or IntersectCirclesOpp , such that ECG proves t(R,K,x)= IntersectCirclesSame (R,K,x). More precisely, the following is provable in ECG without the axioms for IntersectCirclesSame and IntersectCirclesOpp : If circles R and K are not concentric, and L is the line joining their centers, and x is a point not on L, then 102 MICHAEL BEESON

Figure 21. Deﬁnability of IntersectCirclesSame

r b w b b x

q b

b b b a e b L

pb c

t(R,K,x) if and only if R and K have a point in common, and in that case p = t(R,K,x↓ ) is a point of intersection of R and K, and if there are two distinct points of intersection of R and K, then p is on the same side of L as x. Moreover if t(R,K,x) then the centers of R and K are different and x is not on L. ↓ Remark. If there is only one point of intersection, then by Lemma 9.93 that point is on L. Proof. Here is the construction script defining the term t: t(Circle R, Circle K, Point x) { a = center(R) b = center(K) L = Line(a,b) c = IntersectCircles1(R,K) J = Perp(c,L) e = IntersectLines(J,L) M = Perp(x,J) w = IntersectLines(J,M) p = IntersectLineCircle2(Line(e,w),Circle(e,c) return p } Here is the correctness proof of the script. Since R and K are not concentric, L in line 3 is defined. If R and K have no point in common, then c in line 4 is not defined, so t(R,K,x) is not defined, as required. We may therefore suppose that R and K have a point in common. Then c in line 4 of the script is defined. (The point c might lie on L if the circles are tangent.) Since Perp is always defined, J in line 4 is defined, whether or not c is on L. By definition of perpendicular, J meets L, so e is defined. We note that the definition of Perp in Lemma 9.56 involves IntersectCircles1 and IntersectCircles2 , but not IntersectCirclesSame or IntersectCirclesOpp . Since Perp is always defined, M is defined, and by definition of perpendicular, w is defined. We claim w is not FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 103 on L. Since wz and L are both perpendicular to J, if w were on L, then by the uniqueness of perpendiculars at a point (Lemma 9.80), Line (w, x) and L would coincide, but x is on Line (w, x) but not on L. Hence w is not on L. Hence e = w; hence J coincides with Line (e, w). Since J passes through the center of6 Circle (e,c), J meets Circle (e,c), so z in the penultimate line of the script is defined, and hence t(R,K,x) is defined. Observe that t(R,K,x) then the two circles do intersect, as the subterm of t(R,K,x) defining c must be↓ defined, according to the laws of the logic of partial terms. Also if t(R,K,x) is defined, then Line (e, w) is defined, so e = w, which implies x is not on L (since if x is on L then both L and ew are perpendiculars6 to J from x). That proves the first claim of the lemma, that t(R,K,x) is defined if and only if R and K have a point in common. Note that although J coincides with Line (e, w), they may possibly have opposite orientations, so it is important to put Line (e, w) in the penultimate line of the script and not J. Now suppose x is not on L and t(R,K,z) is defined, and let p = t(R,K,x). We must show that p is on the same side of L as x. ( Fig. 21 shows p on the opposite side of L from x, because it illustrates the situation in a proof by contradiction; in reality p and q are switched from what is shown in Fig. 21.) Since wx does not meet L, by Lemma 9.83 x and w are on the same side of L. Therefore it suffices to show that w and p are on the same side of L. Suppose w and p are not on the same side of L. Then by Lemma 9.83, wp meets L, so B(w,e,p). Let q be the other intersection point of the two circles, namely q = IntersectLineCircle1 (Line (e, w), Circle (e,c)) Now suppose p = q. Observe that triangles abp and abq are congruent. By Lemma 9.68 p and6 q are not on the same side of L. By the plane separation theorem p and q are opposite sides of L. Then B(p,e,q). Let r be any point not on J. By Axioms Cont 4 and Cont4, we have Right (r,q,p) if and only if Right (r, e, w), and similarly with Left in place of Right. Suppose Right (r,q,p). Then Right (r, e, w). We have Right (q,p,r) by Axiom H8 since Right (r,q,p) Right (e,w,r) by Axiom H8 since Right (r, e, w) Right (e,p,r) by Lemma 9.89 since Right (r,q,p) and B(q,e,p) Now since B(w,e,p) and B(q,e,p), we have by Axiom B3 (B(e,q,w) B(e,w,q) q = w). ¬¬ ∨ ∨ If q = w we have Right (w,p,r) since Right (q,p,r). In the other two cases we have Right (w,p,r) since Right (q,p,r), by Lemma 9.89. Hence Right (w,p,r). But Right is stable (since it is defined by an equality); hence Right¬¬ (w,p,r). We then have Right (w,e,r) by Lemma 9.89 since Right (w,p,r) and B(w,e,p) Left (e,w,r) by Axioms H8 and H10 104 MICHAEL BEESON

Now we have Right (e,w,r) and Left (e,w,r). But then r lies on Line (e, w), by Lemma 9.92. But since J coincides with Line (e, w), that contradicts our choice of r. This contradiction shows that the assumption SameSide (w,p,L) cannot hold; that is, we have proved SameSide (w,p,L).¬ But SameSide is stable, by Lemma 9.85. Hence SameSide¬¬(w,p,L). That completes the proof. Corollary 9.97. There is a term s(R,K,x) of ECG not containing IntersectCirclesSame or IntersectCirclesOpp such that ECG proves s(R,K,x)= IntersectCirclesOpp (R,K,x). Remarks. This theorem, like the previous one, is to be understood as meaning that the defining property of s can be proved without using the axioms for IntersectCirclesSame or IntersectCirclesOpp . Proof. If x is not on L then we can reflect x in L, by dropping a perpendicular from x to L , with foot u, and extending the non-null segment xu by segment xu to point z. Then B(x,u,z) with z on L, so z is on the opposite side of L from x. Now let s(R,K,x) be t(R,K,z). Then (provided R and K are not concentric and do meet) s(R,K,x) is an intersection point of R and K on the same side of L as z. But then by the plane separation theorem, since z is on the opposite side of L from x, s(R,K,x) is on the opposite side of L from x. That completes the proof. 9.17. Rotation and Uniform Reflection. In this section, we define a construction that can rotate a given point through a given angle (without a case distinction whether the point lies on the vertex of the angle or not), and apply it to define a construction that reflects a point in a line, without a case distinction whether the point is on the line or not. These “uniform” constructions, along with the uniform perpendicular construction Perp of Lemma 9.56, are important in constructive geometry, and their definitions are far from obvious. Lemma 9.98 (Construction of angle bisector). There is a term Bisect(a,b,c) such that if a, b, c are distinct points, and e = Bisect(a,b,c), then e = b and angle abe is congruent to angle cbe, and e is between two points p and6 q on Ray (b,a) and Ray (b,c) respectively. Proof sketch. Let Midpoint be the construction given in the proof of Lemma 9.40. Here is a construction script for the bisector: Bisect(Point a, Point b, Point c) { C = Circle(a,b) e = IntersectLineCircle2(Line(b,c),C) d = Midpoint(a,e) return d } Then as in the proof of Lemma 9.40, triangle abd is congruent to triangle ebd, so angle abd is congruent to angle ebd as required. That completes the proof. Lemma 9.99 (Uniqueness of angle bisector). Given three non-collinear points a, b, and c, and points d and e such that angles abd and cbd are congruent, and angles abe and cbe are congruent. Then b, e, and d are collinear. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 105

Proof. Let C be a point on Ray (b,c) with aC = ab. Since a, b, and c are non- collinear, L = Line (a, C) is defined. By the crossbar theorem (Lemma 9.61), L meets Line (b, d) in a point D, and L meets Line (b,e) in a point E, with B(a,D,C) and B(a,E,C). Then by SAS, triangles abD and DbC are congruent. Hence L is perpendicular to Line (b,D). Similarly L is perpendicular to Line (b, E). Suppose Line (b,D) and Line (b, E) do not coincide. Then they are parallel by Lemma 9.48. But b lies on both those lines, contradiction. Hence those lines cannot fail to coincide. Then b, E, and D cannot fail to be collinear. But collinearity is defined by a negation, so it is stable. Hence b, E, and D are collinear. Hence b, e, and d are collinear, as claimed. That completes the proof. Definition 9.100. Let Rotate be the term represented by the following construction script: Rotate(P,O,Q,A) { E = Bisect(P,O,Q) K = Perp(A,Line(O,P)) B = IntersectLines(K,Line(O,E)) Z = project(B,Line(O,Q)) return Z } Lemma 9.101. There are terms of ECG which we denote Project, Para, and Rotate , such that the properties of these terms used in Section 5 are provable in ECG. Specifically, (i) Project(x, L) is a (and in fact the unique) point on L such that a line perpendicular to L at x passes through x, and (ii) If P , O, and Q are not collinear, and A lies on Line (O, P ) (but not necessarily on Ray (O, P ), then the value Z = Rotate (P,O,Q,A) is defined and lies on Line (O,Q). (iii) OA = OZ, and if A = O, then POQ is a right turn if and only if AOZ is a right turn, and POQ is6 a left turn if and only if AOZ is a left turn; and if A = O then Z = O. Proof. The uniform construction of the perpendicular through point x to line L, which is Perp, was exhibited and proved correct in Lemma 9.56. The projection Project(x, L) of x on L is the foot of that perpendicular, i.e. Project(x, L)= IntersectLines (L, Perp(x, L)). The uniform parallel construction, called Para, is defined by Para(x, L)= Perp(x, Perp(x, L)). It produces a line parallel to L through x if x is not on L, and a line that coincides with L if x is on L, but without requiring a case distinction. Now we turn to (ii). Let Bisect be as in Lemma 9.98. Then we can define Rotate as in Section 5, namely, by a term corresponding to the script in Def- inition 9.100 (see Fig. 2). We now argue for (ii). Suppose P , O, and Q are not collinear, and A lies on Line (O, P ). Then e = Bisect (P,O,Q) is defined, and K = Perp(A, Line (O, P ) is defined. Since P , O, and Q are not collinear, Ray(b,e) is the angle bisector of angle POQ, which means that angle POB 106 MICHAEL BEESON is congruent to QOE. According to Lemma 9.98, E is between two points on Ray (O,Q) and Ray (O, P ) and not equal to O. Since these rays do not line on the same line (because O, P , and Q are not collinear), E does not lie on Line (O, P ). Then A does not lie on Line (O, E). Hence K and Line (O, E) do not coincide. Hence B, the intersection point of these two lines, is defined. Since O = Q, Line (O,Q) is defined and hence Z, the projection of B on Line (O,Q), is defined.6 Note that in spite of the appearance of Fig. 1, we did not assume anything about the position of A on Line (O, P ). That is, Z is defined, no matter whether A lies on Ray (O, P ) or the opposite ray. That completes the proof of (ii). Turning to (iii), first suppose A = O. Then B = O and Z = 0, as claimed, by the properties of Project. In particular OA = OZ as both segments are OO. Now suppose A is on Ray (O, P ). Then Z is on Ray (O,Q). But angle AOB is the same angle as angle P OE, and angle QOE is the same angle as angle ZOB; hence angle ZOB is congruent to angle AOB. Angles BAO and BZO are right angles, by the property of Perp, which was used to construct points B and Z. Since all right angles are congruent, angle BAO is congruent to angle BZO. Side OB is congruent to itself. Hence triangles BAO and BZO are congruent, by SAA (Lemma 9.69). Hence OA = OZ as corresponding sides of congruent triangles. On the other hand, if A is not on Ray (O, P ), then since A is on Line (O, P ) we have B(A, O, P ). Let P ′ and Q′ be the reflections of P and Q in point O. Let E′ be the reflection of E in O. By the congruence of vertical angles, OE′ bisects angle P ′OQ′. Let E′′ = Bisect (P ′,O,Q′). Then both OE′′ and OE ′ bisect angle P ′OQ′, and OE = OE′ = OE′′, so E = E′′, by Lemma 9.99. Thus the situation is exactly the reflection in O of Fig. 2;′ arguing as above with primes on P , E, and Q, conclude OA = OZ. Now we have proved OA = OZ in three cases that are classically mutually exclusive. Hence OA = OZ. Then by the stability of equality, OA = OZ. Now suppose A¬¬= O and A lies on Ray (O, P ). We will show in this case POQ is a right turn if and6 only if AOZ is a right turn. We claim Z lies on Ray (O,Q). Since lying on a ray is defined by a negation, it is stable, so we may prove this by contradiction. Suppose Z does not lie on Ray (O,Q). Then B(Z,O,Q). Then Z is on the opposite side of L = Line (O, E) from Q. But A is on the same side of L as P , by Lemma 9.83, since AP meets L only at O, and B(O, A, P ). Since P and Q are on opposite sides of L, then A is on the opposite side of L from Q, by the plane separation property. Now A and Z are both on the opposite side of L from Q; hence they are on the same side of L. But then triangles OBA and OBZ are two congruent triangles on segment OB with third vertex on the same side of L, so by Lemma 9.68 we have Z = A. But then Q and P lie on Line (O,Z), contradicting the assumption that P , Q, and O are not collinear. Hence Z does lie on Ray (O,Q). We have Z = O since if Z = O then B = O and then A = O, contradiction. Now by Lemma6 9.89, POQ is a right turn, if and only if AOQ is a right turn, and since Z lies on Ray (O,Q) by Lemma 9.89 again, if and only if AOZ is a right turn. Similarly, assuming A lies on Ray (O, P ) and A = O, POQ is a left turn if and only if AOZ is a left turn; and similarly, we reach6 the desired conclusions FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 107 if B(A, O, P . (Alternately we could prove as a lemma that Right and Left are preserved under reflection in a point, and apply that lemma.) Since these cases are classically mutually exclusive, we have proved

A =0 (Right (P,O,Q) Right (A,O,Z)) (Left (P,O,Q) Left (A,O,Z)) 6 → ¬¬ ↔ ∧ ↔ Right and Left are deﬁned by the equality of certain terms (see Deﬁnition 8.8), so they are stable (implied by their double negations) by the stability of equality, Axiom S1. Hence we can push the double negation inwards, and drop it in front of Left and Right , obtaining the desired conclusion: A =0 (Right (P,O,Q) Right (A,O,Z)) (Left (P,O,Q) Left (A,O,Z)) 6 → ↔ ∧ ↔ That completes the proof of the lemma. Lemma 9.102 (Inverse rotation). If Z = Rotate (P,O,C,A) then Z = Rotate (C,O,P,Z). Proof. The bisector of angle QOP is the same as the bisector of angle POQ, by Lemma 9.99. Then when, in the second rotation, we erect the perpendicular to OQ at Z, it intersects OL at B because of the uniqueness of the perpendicular at Z (Lemma 9.80). Then when we drop the perpendicular to OP , it meets Line (O, P ) at A because of the uniqueness of the projected point, proved in Lemma 9.101. That completes the proof.

Definition 9.103. The term I is given by the following construction script. (The script takes no arguments, which means that the term I has no variables.) I() { L = Line(0,1) J = Perp(0,L) K = Perp(gamma,J) p = IntersectLines(J,K) return IntersectLineCircle2(Line(0,p),Circle(0,1)) } The following lemma states an “obvious fact” whose proof is surprisingly in- tricate. Lemma 9.104. Right (1, 0, I). Proof. Since 0 is another name for α and 1 is another name for β, and αβγ is a left turn by Axiom H12, 10γ is a right turn by Axiom H9. Then we deﬁne p = IntersectLines (J, Perp(γ, J)). Then p = 0 since if p = 0, K and L coincide, since both are perpendicular to J at 0, but6 then α, β, and γ would be collinear, which they are not. Since J and Perp(γ, J) meet only at p, we conclude that 10p is a right turn, by Lemma 9.89. Let q = IntersectLineCircle2 (Line (0,p), Circle (0, 1)). (This is the point you might call I, on the y-axis J opposite I.) By Axiom Cont 4, for all points x not on Line (0,p−) (which coincides with J, but might have the same or opposite orientation), we have Right (x, q, I) if and only if Right (x, 0,p). 108 MICHAEL BEESON

Taking x to be 1, we have Right (1,q,I) if and only if Right (1, 0,p). But we have proved Right (1, 0,p). Hence Right (1,q,I). By Lemma 9.12 we have B(q, 0, I). Hence by Lemma 9.89, we have Right (1, 0, I). That completes the proof of the lemma. We now turn to uniform reflection; by that we mean a construction to reflect point x in line L without a case distinction as to whether x is on L or not. By the uniform perpendicular construction we can get a line through x perpendicular to L, meeting L at a point e, but since only non-null segments can be extended, we cannot extend segment xe as we desire to do, unless x is not on L. The key to avoiding this case distinction is to rotate x twice by ninety degrees! The construction is illlustrated in Fig. 22. Point x is rotated twice, first to point y and then to the desired answer, point z. The dashed lines show the construction used to perform the rotation, according to the Rotate construction, using two angle bisectors and Perp to construct points c and a. Here a, b, and c are points equidistant from e and definitely different than e, while we do not assume that x = e. 6

Figure 22. Uniform Reﬂection

y b

b c

b b

b b b b b L z a e b x

Lemma 9.105 (Uniform reflection). There is a term Reflect (x, L) that produces the reflection of x in line L, without a case distinction whether x is on L or not. That is, if z = Reflect (x, L) and K perpendicular to L contains x and meets L at e, then z lies on K with ez = ex, and unless x lies on L, B(x,e,z). Proof. Here is the construction script illustrated in Fig. 22. Reflect(Point x, Line L) { K = Perp(x,L) e = IntersectLines(K,L) C = Circle(e, FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 109 alpha, beta ) a = IntersectLineCircle1(L,C) b = IntersectLineCircle1(K,C) c = IntersectLineCircle2(L,C) y = Rotate(a,e,b,x) z = Rotate(b,e,c,y) Return z; }

Since Perp is always defined, K is defined, and since K is perpendicular to L, e in line 2 is defined. The constants α and β are used in line 3 to ensure that C has a nonzero radius. Since L and K both pass through the center of C, the points a, b and c are defined. Since these three points are different from e (because C is not a degenerate circle), y and z are defined by Lemma 9.101. Hence Reflect (x, L) is defined (regardless of whether x is or is not on L). By the properties of Rotate , we have ex = ey and ey = ez. Hence ex = ez. It remains to show that if x does not lie on L (i.e. x = e) then B(x,e,z). Suppose x = e. Then x and z both lie on circle C and line6 K. Since K is a diameter of 6 C, its center lies between its endpoints, by Lemma 9.12. That completes the proof of the lemma. Lemma 9.106 (Reflection in a diameter). Let line L pass through the center of circle C and let p lie on C. Then Reflect (p) lies on C. Proof. Let L pass through the center e of C and let p lie on L. Let M = Perp(p,L). Let x be the intersection point of M and L. Let q = Reflect (p,L). Then q lies on M and xp = xq, by Lemma 9.105. We must show q lies on C, which will follow from qe = pe. By Axiom S4, the stability of On, we can argue by cases. Case 1, e = p. Then circle C is degenerate and so q = p and we are finished. Case 2, p does not lie on L and e does not lie on M. Then e, p, q, and x are all distinct, and since px = qx and all right angles are congruent, we have triangle pxe congruent to triangle qxe by SAS. Hence ep = eq, completing Case 2. Case 3, p lies on L. Then p = q by Lemma 9.105, so pe = qe, completing Case 3. Case 4, e lies on M. Then Line (e,p) coincides with M, so q lies on Line (e,p). Then x = e by the uniqueness of dropped perpendiculars. Since px = qx by Lemma 9.105, we have pe = qe. That completes Case 4, and with it, the proof of the lemma, since these cases are clasically mutually exclusive. Lemma 9.107 (Reflection in a point). There is a construction Reflect (a,x,L) such that if a and x are points on line L then Reflect (a,x,L) is defined and if p = Reflect (a,x,L) then pa = ax and T(p,a,x), where T is non-strict betweenness. Remarks. This trivial-sounding lemma is not trivial to prove, because we cannot make a case distinction whether x = a or not, so we cannot form Line (x, a) or extend the segment xa. Once constructed, we say “p is the reflection of x in a” if we know that x = a; in that case it is not necessary to mention L since L 6 110 MICHAEL BEESON necessarily coincides with Line (a, x); but being given L is necessary in order to construct the reflection without a case distinction. Proof. Here is the construction script. It calls on the construction of Lemma 9.105; we use Reflect for both terms; since one has more arguments than the other, they can be distinguished. Reflect(point a, point x, Line L) { K = Perp(a,L); return Reflect(a,K) } The claimed properties of this term follow immediately from Lemma 9.105. That completes the proof of the lemma. 9.18. The other intersection point. Many Euclidean constructions involve constructing one intersection point p of a line L = Line (a,b) and a circle C, and then we say “Let q be the other intersection point of L and C”. The question is, whether in ECG we can give a uniform method of constructing Q from p, L, and C. Yes, we can, as we will now show.

Figure 23. Deﬁnability of “the other intersection point”.

b a

p x q b b b L

Theorem 9.108. There is a term Other(p,L,C) such that, if point p lies on line L and circle C, then Other(p,L,C) is deﬁned and lies on both L and C, and if z = p and z lies on both L and C then Other(p,L,C)= z. 6 Proof. Here is the construction script for Other: Other(Point P, Line L, Circle C) { a = center(C) K = Perp(a,L) q = Reflect(p,K); FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 111

return q; }

Since Perp is always defined, K in line 2 is defined. Let x = IntersectLines(K,L). By Lemma 9.105, qx = px and q lies on L. We must prove q lies on C. If a does not lie on L then since K and L are perpendicular at x we have triangle axp congruent to triangle axq. Hence corresponding sides ap and aq are equal. If a does lie on L then since a also lies on K we have a = x, hence ap = aq. But we have (on(a,L) on(a,L)). Hence ap = aq. Hence ap = aq by the stability¬¬ of equality. Hence∨ ¬ q lies on C. ¬¬ Now suppose z is any point different from p that lies on both L and C. We must show z = q. By Lemma 9.52 there are exactly two points on both L and C. But p and q are two such points, and z = p. Hence z = q. By the stability of equality then z = q. That completes the6 proof. ¬¬ Remark. In case one thinks, “this is not what Euclid had in mind!” (when he speaks of the other intersection point), that is of course true; but Euclid never tried to give uniform constructions; and in general it takes much more care to produce uniform constructions to avoid case distinctions, as we have seen already with uniform perpendiculars and uniform midpoints. We next want to show that similarly, one can construct the “other intersection point” of two circles. Theorem 9.109 (The other intersection point of two circles). There is a term Other2(p,C,K) such that, if point p lies on circle C and on circle K, and C and K are not coincident, then Other2(p,C,K) is defined and lies on both C and K, and if z = p lies on both C and K then Other2(p,C,K)= z. 6 Proof. Here is the construction script:

Other2(Point p, Circle C, Circle K) { a = center(C); b = center(K); L = Line(a,b); q = Reflect(p,L); return q; }

Here is the correctness proof of the script. Then C and K are not concentric, since if a = b then C and K coincide (because then ap = bp, so for all z, On(z, C) if and only if za = ap if and only if ab = bp if and only if On(z,K)). But by hypothesis, C and K do not coincide. Hence a = b. Therefore L in line 3 is defined. Since Reflect is always defined, by Lemma6 9.105, q is defined, and by Lemma 9.106, q lies on circles C and K. That completes the proof. 9.19. Defining the order of points on a line. We have earlier (see section 9.3) defined an ordering on segments. That corresponds to arithmetic on positive numbers. We now define an ordering on the points of a line, corresonding to positive and negative numbers. 112 MICHAEL BEESON

Let 0 be another name for β and let 1 be another name for α, and let L = Line (α, β) be called the x-axis. Let K be perpendicular to L at 0. By Lemma 9.56, there exists a line M containing γ and perpendicular to K (we do not need to know whether γ is on K or not). Let z on K be the foot of this perpendicular. By Lemma 9.48, M is parallel to L. On Ray (0,z) construct a point I such that 0I = 01. Then I is on the same side of L as γ, since M is parallel to L by Lemma 9.48. (Note, this does not use the parallel axiom; that lemma was derived in neutral geometry.) Since αβγ is a right turn (by Axioms H8, H10 and H12), and I is on the same side of L as γ, then αβγ, which can also be written 10I, is also a right turn, by Axiom H4. Definition 9.110. For points x and y on L: xy := Right (x, y, I) x y := y < x ≤ ¬ x y := y x ≥ ≤ Note that Left (x, y, I) already implies x = y, so it is not necessary to stipulate x = y. 6 6 Lemma 9.111. With notation as above, 0 < x x < 0 x =0. 6 ∧ 6 → Proof. Suppose 0 < x and x < 0. Then 0xI is not a left turn and x0I is not a left turn. But that6 contradicts6 Lemma 9.95. That completes the proof. Note that ECG cannot prove a 0 on L. Then for points x on L, x> 0 if and only if x is on the same side of 0 as 1; and 0 < x if and only if B(x, 0, 1). ¬ Remark. This lemma connects the deﬁnition of order in terms of Right and Left with the primitive notion of betweenness. Proof. Let K be the line perpendicular to L at 0. Then we must show that x> 0 if and only if x is on the same side of K as 1. Suppose x> 0; we must prove x and 1 are on the same side of K. Since x > 0, we have x = 0 and 0xI is a left turn. Similarly 01I is a left turn. 6 By Axiom H9 (applied twice) I0x is a left turn. Since 0 < 1, we have by the deﬁnition of < that 01I is a left turn. Hence by Axiom H9, I01 is a left turn. Hence by Axiom H2, x and 0 are on the same side of Line (0, I), which has the same points (and hence the same sides) as line K. Hence x and 0 are on the same side of K, which was what was to be proved. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 113

Conversely, suppose x and 1 are on the same side of K. Since 0 < 1, as above I01 is a left turn. By Axiom H5 then, I0x is a left turn. By Axiom H10, Ix0isa right turn, and then by Axiom H8, x0I is a right turn; then x> 0 by Definition 9.110. That completes the proof of the first claim of the lemma. The second claim can be derived from the first, but it can also be proved directly, since (for points x on L), if x is on the same side of 0 as 1 then by the plane separation theorem B(x, 0, 1), so by Lemma 9.89, and the fact that Right (1, 0, I), we have Right (x,¬0, I), hence Left (0, x, I), which is the definition of 0 < x. Conversely, to prove 0 < x B(x, 0, 1), suppose 0 < x and B(x, 0, 1). Let 1 be the extension of segment→ ¬ 10 by 10. Then B( 1, 0, 1), so by definition of−SameSide , 1 and x are on the same side of K. By− definition of 0 < x we have Left (0, x,− I); by Lemma 9.89 we have Left (0, 1, I). Hence Right ( 1, 0, I). By Lemma 9.89 we have Right ( 1, 1, I) and then−Right (0, 1, I). But since− 0 < 1 we have Left (0, 1, I), contradicting− Lemma 9.92. That completes the proof of the lemma.

§10. The parallel postulate. We will introduce three versions of the parallel postulate: Euclid’s own version (Euclid 5), which says that under certain conditions, two lines will meet in a point; a version we call the strong parallel postulate, which weakens Euclid’s conditions for the two lines to meet; and Play- fair’s version, which says two parallels to the same line through the same point must coincide (and makes no assertion at all about the existence of an intersection point). We then start to consider the relations of implication between these versions (relative to neutral ECG), and finally draw some consequences of the strong parallel axiom. 10.1. Alternate Interior Angles. In connection with parallel lines, the following terminology is traditional: If L and K are two lines, and p lies on L and q lies on K but not on L, then pq is a transversal of L and K. Then Line (p, q) makes four angles with L and four angles with K; certain pairs of them are called alternate interior angles or adjacent interior angles. In ECG, there is no primitive concept of “angle”; angles are treated as triples of points, and angle congruence is a defined concept (it is defined in Definition 9.21). In the next section we will discuss Euclid’s parallel postulate, and we will need to express it without mentioning angles directly. We therefore discuss how to do that now. Angle congruence is defined in terms of triangle congruence, which is in turned defined by the SSS criterion: the three sides are pairwise congruent. We will now unwind these definitions to express the concept of alternate interior angles being equal, without mentioning angles. The following definition is illustrated in Fig. 24.

Definition 10.1. “pq makes alternate interior angles equal with K and L”, written AI(p,q,L,K), means on (p,L) on (q,K) on (q,L) p = q ∧ ∧ ¬ ∧ 6 r,s,t (on(r, K) on (s,L) B(r,t,s) B(p,t,q) r = p q = s ∧ ∃ ∧ ∧ ∧ ∧ 6 ∧ 6 pt = qt rt = st) ∧ ∧ 114 MICHAEL BEESON

Figure 24. Transversal pq makes alternate interior angles equal with L and K, if pt = tq and rt = st.

p r K b b

b t

L b b s q

This can also be expressed as “K and L make interior angles equal with pq”.

The reason for requiring B(r,s,t) is that r and t need to be on opposite sides of Line (p, q) in order that the angles rpt and qst are alternate interior angles. At the cost of introducing two more points into the definition, we could have made it easier to verify; any two congruent triangles with those angles as a vertex could replace prt and qst, as illustrated in Fig. 25. As the lemmas given after angle congruence is defined in Definition 9.21 show, if we can produce x and y so that Fig. 25 is correct, then AI(p,q,L,K) will hold. A careful reader might have noticed that we omitted the congruence rp = qs in the definition, which would be required to express the congruence of the shaded triangles according to the definition of triangle congruence. This was not an accident. In case the lines K and L are as shown in the diagram, the triangles are congruent anyway by SAS, since the angles at t are vertical angles, and (as we prove in Lemma 9.33), vertical angles are congruent in neutral ECG. But not only is rp = qs not necessary, there is a good reason to omit it: it allows the definition to make sense even when K and L coincide! This is useful because it enables us to avoid making case distinctions according as K and L do or do not coincide.

Figure 25. Another way to say that transversal pq makes alternate interior angles equal with L and K. The shaded triangles are congruent.

p r K b b x b b t b y L b b s q FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 115

On the other hand, instead of introducing more points using , we can get rid of entirely by constructing specific points r, s, and t: ∃ ∃ r = IntersectLineCircle2 (K, Circle (p,α,β)) t = Midpoint (p, q) s = IntersectLineCircle2 (Line (r, t)Circle (t, r)) These points will do for r, t, and s in AI(p,q,L,K) if any points will do. In this way AI(p,q,L,K) can be expressed by a quantifier-free formula. Similarly x and y can be calculated in Fig. 25, by requiring xp = rp and yq = sq. That is actually what we would get by directly expanding the definition of angle congruence. 10.2. Euclid’s parallel postulate. Euclid’s postulate 5 is If a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles. We consider the formal expression of Euclid’s parallel axiom. Let L and M be two straight lines, and let pq be the “straight line falling on” L and M, with p on M and q on L. We think that what Euclid meant by “makes the interior angles on the same side less than two right angles” was that, if K is another line K through p, making the interior angles with pq equal to two right angles, then M would lie in the interior of one of those angles (see Fig. 26). For K to make the interior angles on the same side of pq equal to two right angles, is the same as for K to make the alternate interior angles equal. In the previous section we discussed the formal expression of that concept. Once we use this criterion to express that L and K make interior angles on the same side equal to two right angles, we can then use three more points to “witness” that one ray of line M emanating from p lies in the interior of one of the interior angles made by M. Fig. 26 illustrates the situation.

Figure 26. Euclid 5: M and L must meet on the right side, provided B(q,a,r) and pq makes alternate interior angles equal with K and L.

M p r K b b

b a

L b q

Here is a formal version of Euclid’s parallel axiom, using the formula AIE from the previous section to express that pq makes alternate interior angles equal with K and L: 116 MICHAEL BEESON on(p,K) on(p,M) on(a,M) on(r, K) on(q,L) B(q,a,r) AIE∧ (p,q,K,L∧) B(p,a,∧ IntersectLines∧ (L,M∧)) ∧ → The logical axioms of LPT make it superfluous to state in the conclusion that IntersectLines (L,M) is defined. That follows automatically. We now write out Euclid 5 in the primitive syntax of ECG, eliminating the defined concept AIE. The result, illustrated in Fig. 27, is the official version of Euclid 5; that is, when we refer to Euclid 5 in subsequent sections, this is what we mean.

Figure 27. Euclid 5: M and L must meet on the right side, provided B(q,a,r) and pt = qt and rt = st.

M p r K b b

b b a t L b b s q

on(p,K) on(p,M) on(a,M) on(r, K) on(q,L) B(q,a,r) (Euclid 5) on (s,L∧) B(p,t,q∧) pr = qs∧ pt = qt ∧rt = st ∧ on (p,L) ∧ B(∧p,a, IntersectLines∧ (L,M∧ )) ∧ ∧ ¬ → We consider strengthening this axiom by removing the hypothesis on (p,L). Classically it makes no difference, since if p is on L then the conclusion¬ that M meets L is already satisfied; but the relation between the two versions is not so obvious with constructive logic. If on (p,L), then because B(p,t,q) and on (q,L), we have on (t,L); hence also on (r, L). But then we have on (q,L) and on (r, L) and B(r, a, q), so on (a,L). Similarly if on (a,L), there is a similar “collapse”, and one can show on (p,L). So the hypothesis on (p,L) could be replaced by on (a,L). We do not know if the axiom really¬ becomes stronger if we remove ¬ on(p,L), but we leave that hypothesis there, because it seems to be what ¬Euclid intended. 10.3. The strong parallel postulate. Although we have finally arrived at a satisfactory formulation of Euclid 5, that formulation is satisfactory only in the sense that it accurately expresses what Euclid said. It turns out that this axiom is not satisfactory as a parallel postulate for ECG. The main reason is that it is inadequate to define division geometrically. The details will be made clear when we define a geometric construction Reciprocal geometrically that constructs a segment of length 1/x from a segment of length x. For now, it is enough to explain that as x gets nearer and nearer to 0, the number 1/x requires a line of smaller and smaller slope to meet a certain horizontal line. If x passes through zero, this intersection point “goes to infinity”, then is undefined when x = 0, but then “reappears on the other side”, coming in from minus infinity. Without FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 117 knowing the sign of x, we will not know on which side of the transversal pq the two adjacent interior angles will make less than two right angles. In other words, with Euclid 5, we will only be able to divide by a number whose sign we know; and the principle x = 0 x < 0 x > 0 is not an axiom (or theorem) of ECG. The conclusion6 is that→ if we want∨ to divide by nonzero numbers, we need to strengthen Euclid’s parallel axiom. We make three changes in Euclid 5 to get what we call the strong parallel postulate: (i) We change the hypothesis B(q,a,r) in Euclid’s axiom to on(a,K). In other words, we require that the two adjacent interior angles do not¬ make exactly two right angles, instead of requiring that they make less than two right angles.

(ii) We change the conclusion to state only that M meets L, without specifying on which side of the transversal pq the intersection lies. (iii) We drop the hypothesis on (p,L). ¬ Here is the strong parallel postulate, which we take as an axiom of ECG: on(p,K) on(p,M) on(r, K) on(q,L) (Strong Parallel Postulate, SPP) on(a,M∧) on (a,K∧ ) ∧ ∧ on (s,L) ∧B ¬(p,t,q) pr = qs pt = qt rt = st ∧ IntersectLines∧ ∧(L,M) ∧ ∧ → ↓

Figure 28. Strong Parallel Postulate: M and L must meet (somewhere) provided a is not on K and pt = qt and rt = st.

M p r K b b

b b a t L b b s q

The strong parallel axiom differs from Euclid’s version in that we are not required to know in what direction M passes through P ; but also the conclusion is weaker, in that it does not specify where M must meet L. In other words, the betweenness hypothesis of Euclid 5 is removed, and so is the betweenness conclusion. Since both the hypothesis and conclusion have been changed, it is not immediate whether this new postulate is stronger the Euclid 5, or equivalent, or possibly even weaker, but it turns out to be stronger–hence the name. Also we call attention to the fact that nothing in the postulate prevents p from lying on L. In that case the “triangles” (shaded in the diagram) are no longer triangles, but the statement that their sides are pairwise congruent still makes sense. The condition B(t,p,q) will force L and K to coincide if p lies on L. We 118 MICHAEL BEESON show below, in Corollary 10.13, that it does not matter if we add the hypotheses on (p,L) and on (a,L) to the strong parallel postulate–the modified axiom is equivalent¬ to the¬ one given above. In particular, change (iii) that we made in passing from Euclid 5 to the SPP is insignificant. 10.4. Playfair’s axiom. The astute reader may have observed that we have now discussed two “parallel postulates” in detail, yet have never defined or even mentioned the word “parallel”. Euclid did not give his Postulate 5 the name “parallel postulate” (or any other name). A case can be made that it is more of a “triangle construction postulate.” Be that as it may, we now define Parallel (L,K) for lines L and K to mean that the lines do not meet: Definition 10.2. Parallel (L,K) x (on(x, L) on(x, K)). ↔∀ ¬ ∧ This requires a universal quantifier. Compare it to the quantifier-free formula IntersectLines (K,L) . ¬ ↓ If L and K are parallel, then IntersectLines (K,L) is undefined (by Axiom I4); but it will also be undefined if L and K coincide, by Axiom I21. The relation defined by IntersectLines (L,K) can be referred to informally as “weakly parallel”; classically¬ it means “parallel↓ or coincident”, but constructively, we generally cannot make the decision of two weakly parallel lines whether they are coincident or parallel. Instead, weakly parallel means “not not parallel or coincident.” (Consider, for example, the x-axis and the line y = a; these lines are parallel if a = 0 and coincident if y = 0, so if we could decide whether two weakly parallel lines6 are coincident or parallel, we could decide y =0 y = 0.) That this concept does have something to do with Euclid’s parallel∨ postulate6 is hinted at by the following lemma: Lemma 10.3. Let K and L be two lines, with point p on K but not on L, and let traversal pq make equal alternate interior angles with K and L. Then K and L are parallel. Proof. Refer to Fig. 24 for an illustration; let points p, r, s, q, and t be as shown in the figure with triangle prt congruent to triangle qst. Suppose, for proof by contradiction, that K and L meet in point x. (Since Parallel is defined as not meeting, a constructive proof of it consists in deriving a contradiction.) Consider triangle pqx. If x is on Ray (p, r), then the angles of triangle pqx are pxq, qpx, and pqx. The two adjacent interior angles on pq make two right angles together, but in neutral geometry one can prove that two angles of a triangle are less than two right angles. (This is Euclid I.17, proved in two lines from Euclid I.16, which we prove constructively in Lemma 9.73 below.) This contradiction shows that K and L cannot meet in a point x on Ray (p, r). Similarly, they cannot meet in a point x on Ray (r, p). But, if they meet in a point x at all, then by Axiom B3, (B(x, p, r) x = p B(p, x, r) x = r B(p, r, x)). ¬¬ ∨ ∨ ∨ ∨ In each of those cases, x lies on Ray (r, p) or on Ray (p, r); so we have a contradiction, and K cannot meet L at all. Hence K is parallel to L as claimed. That completes the proof of the lemma. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 119

Most modern treatments of geometry formulate the parallel axiom in a diﬀer- ent way: If two lines K and M are parallel to L through point p, then K = M. We call this the “Playfair’s postulate”, or for short just “Playfair”, after John Playfair, who published it in 1795, although (according to Greenberg [12], p. 19) it was referred to by Proclus. Since we can always construct one parallel, it would be equivalent to assert that, given a line L and a point p not on L, there exists exactly one parallel to L through p.

Figure 29. Playfair: M and L can’t fail to meet.

M p K b

The formal expression of Playfair’s postulate says that if there are two lines through point p that are weakly parallel to L then those two lines coincide. Classically, “weakly parallel” means “parallel or coincident”, but constructively, it just means that IntersectLines (K,L) is not defined (without specifying why not). IntersectLines (K,L) IntersectLines (M,L) ( Playfair’s Postulate) ¬ on(p,K) on(p,M↓ ∧¬) on(q,K) on(q,M)↓ ∧ ∧ → ↔ We do not say L = M since equality between lines is intensional; instead we assert that the same points are on L as are on M. Playfair’s postulate is classically equivalent to the more usual formulation, in which the hypothesis mentions Parallel instead of “weakly parallel”: Parallel (K,L) Parallel (M,L) (weak Playfair) on(p,K)∧ on(p,M) on(q,K) on(q,M) ∧ ∧ → ↔ Neither version of Playfair’s axiom makes any existential assertion at all; to see this just recall that the definition of Parallel only uses a universal quantifier. They are therefore the weakest versions of the parallel postulate that we will consider. It is an empirical fact that Playfair’s axiom suffices to prove all the consequences of the parallel postulate that do not themselves make existential assertions. “Weakly parallel” means we do not make a case distinction whether the lines coincide or not. Of course if they do coincide the conclusion is trivial. Since we need to deal with weakly parallel lines, in order to support coordinatization without case distinction, Playfair’s axiom is more useful than weak Playfair. Another relevant fact is that Playfair is equivalent to its double negation, which one cannot say about weak Playfair (since Parallel involves a universal quantifier and , is not the same as ). ¬¬ ∀ ∀ ¬¬ 120 MICHAEL BEESON

Lemma 10.4. Playfair’s axiom implies (in neutral ECG) that weakly parallel lines make equal alternate interior angles with any transversal. Similarly, weak Playfair’s axiom implies that parallel lines make equal alternate interior angles with any transversal. Proof. Let L and K be weakly parallel lines, and let pq be a transversal, with p on K and q on L and p = q. Let s be any point on L not equal to q. Let t be the midpoint of pq, and extend6 segment st to point r with B(s,t,r) and tr = ts. Let M = Line (p, r). Then by Definition ??, pq makes equal alternate interior angles with L and M. If p does not lie on L, then by Lemma 10.3, M and L are parallel. But then M and K are two weakly parallel lines to L passing through p. Hence by Playfair’s axiom, they coincide. Hence pq makes alternate interior angles with L and K. On the other hand, if p does lie on L, then M and K coincide and we are finished in that case too. But by Axiom S3, the stability of on , we have (on (p,L) on (p,L)), so it is constructively legal to argue by these two ¬¬cases. That completes∨ ¬ the proof. The next lemma will be needed below. Lemma 10.5. Given a point p and a line L, we can construct a point on L different from p. Remark. Of course the lemma is classically trivial. But notice that the similar- sounding lemma, given lines K and L that do not coincide, there is a point on L that is not on K, is not constructively correct, since such a point cannot be found continuously in K and L. Proof. Let a = pointOn1 (L) and b = pointOn2 (L). Then a = b. Extend segment ab by ap to point z; then extend segment az by ab to point6 t. Then az > ap (by Definition 9.64). Let C = Circle (a,t). Since ap = az < ap, point p does not lie on C (by Lemma 9.65 and Axiom C5). Hence p = t, so point t meets the conditions of the lemma. That completes the proof of6 the lemma. 10.5. Another version of the strong parallel postulate. We now consider yet another expression of the parallel postulate, which we call SPPE for “strong parallel postulate equivalent.” We show that this version is closely related to the stability principle t↓ t . ¬¬ → ↓ IntersectLines (K,L) on(p,K) on(p,M) (SPPE) ¬ on(a,M) on(a,K↓) ∧ IntersectLines∧ (L,M) ∧ ∧ ¬ → ↓ In other words, if K is weakly parallel to L through p, then any other line M through p must meet L. Lemma 10.6. Playfair’s postulate proves that SPPE is equivalent to the strong parallel postulate SPP. Proof. We show first that the strong parallel postulate implies SPPE (with Playfair). Let K and L be weakly parallel, i.e. IntersectLines (K,L) is not defined, and let p be on K, and let M be another line through p. By Lemma 10.5, we can find a point q on L different from p. By Playfair’s axiom and Lemma 10.4, pq makes alternate interior angles equal with K and L. Then by the strong parallel axiom, M meets L, but that is the conclusion of SPPE. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 121

Figure 30. SPPE: K and L are parallel, so M and L must meet, but we don’t say where, and we assume nothing about a except that it is not on K.

M p K b

b a

Conversely, suppose SPPE, and let K and L be lines, and pq a transversal making equal alternate interior angles with K and L, and M another line through p, with p on K, and a a point of M that is not on K. By Lemma 10.3, K and L are parallel, so by SPPE, M meets L. That completes the proof of the lemma. 10.6. Coordinates. In this section, we show that the strong parallel axiom is sufficient to introduce the usual Cartesian coordinates. The “coordinates” of a point b are given by terms X(b) and Y (b). These terms are defined by the following construction scripts. The term I in the second script is defined in Definition 9.103.

Definition 10.7. X(Point b) { L = Line(0,1) J = Perp(0,L) H = Perp(b,L) return IntersectLines(H,L) }

Y(Point b) { L = Line(0,1) J = Perp(0,L) M = Perp(b,J) P = IntersectLines(M,J) return Rotate(I,0,1,P) }

Note that Y (b) returns a point on the x-axis L, not a point on the Y -axis J. Coordinatization can be done without any parallel axiom. To “uncoordinatize” we define a term MakePoint (x, y). This term takes two points on Line (0, 1) and makes a point whose coordinates are the given points. A related term MakePoint2 starts with a point on the x-axis and another point on the y-axis. For technical reasons we define MakePoint slightly more generally, so that it works with any points x and y and first projects them on Line (0, 1); this step does nothing if the points are on Line (0, 1) to begin with. 122 MICHAEL BEESON

Definition 10.8. The term MakePoint (x, y) is defined by the following construction script, which is illustrated in Fig. 31. MakePoint(Point x, Point y) { L = Line(0,1) J = Perp(0,L) U = Perp(x,L) W = Perp(y,L) z = Rotate(1,0,I,\IntersectLines(W,L)) V = Perp(z,J) return IntersectLines(U,V) } We already proved in Lemma 9.104 that Right (1, 0, I). A similar fact, which could not be stated until coordinates were defined, is the following: Lemma 10.9. Right (1, 0, Y (γ)). Proof. By definition of Y , Y (γ)= Rotate (I, 0, 1,p). By Lemma 9.101, Y (γ) is on Ray (0, 1). Hence by Lemma 9.89, Right (Y (γ), 0, I) if and only if Right (1, 0, I), which we proved in Lemma 9.104. Hence Right (Y (γ), 0, I). That completes the proof of the lemma. Next we prove the basic properties of MakePoint , and the coordinate functions X and Y .

Figure 31. MakePoint constructs a point with coordinates x and y, shown by the small open circle. J U W

z b

b b b L 0 x y

Lemma 10.10 (Uncoordinatization). If b = MakePoint (x, y) is defined, then x = X(b) and y = Y (b). The strong parallel postulate implies that for every pair of points (x, y) on L, MakePoint (x, y) is defined. Proof. Let U = Perp(x, L) and V = Perp(z, J) as in the script for MakePoint (x, y). If b = IntersectLines (U, V ) is defined then X(b)= x and Y (b)= y, by Lemma 9.80 FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 123

(the uniqueness of perpendiculars at a point). To prove the second claim, assume the strong parallel postulate. We only have to prove that U and V do intersect. Since V and L are both perpendicular to J, they are weakly parallel, that is, IntersectLines (V,L) is not defined, since if it were defined then by Axiom I14, p = IntersectLines (V,L) would be on both V and L, and hence by Lemma 9.81, V and L would coincide, contradicting Axiom I21. (Note that we needed the uniform version of uniqueness of perpendiculars, since we do not know whether p is on J or not.) U does not coincide with L, and in fact we can construct a point a on U that is not on L, for example a = IntersectLineCircle1 (U, Circle (x, 0, 1)). That point is not on L since it is different from x, and if it were on L, then L and U would coincide, but perpendicular lines cannot coincide, as remarked after Definition 9.35. Now L is weakly parallel to V , and U and L have a common point x and point a lies on U but not on L, so SPPE (which is equivalent to the strong parallel postulate) applies, with (L,K,M,p,a) in SPPE taken to be (V,L,U,x,a), and the conclusion is that IntersectLines (V,U) . That completes the proof. ↓ Remark. If SPPE is weakened by changing the hypothesis IntersectLines (K,L) ¬ ↓ to Parallel (K,L), then we need a case distinction y =0 y = 0. In either case we can conclude IntersectLines (U, V ) , but since the stability∨ 6 of t is not an axiom of ECG, we cannot complete the↓ argument by double-negating↓ the case distinction. 10.7. The strong parallel postulate and stability of IntersectLines. Re- call that the stability axioms omitted the axiom t t , because it was provable except in the case when t involves IntersectLines¬¬ ↓ →. We↓ now show that in that case, it is relatively strong! If we assume that stability axiom, then we can no longer distinguish the various versions of the parallel postulate. In a sense, that simply argues for accepting both SPP and the stability postulate, which is what we actually do in ECG, but we also wish to investigate the constructive relations between the different forms of the parallel postulate relative to some weaker theory, which we take to be Playfair plus neutral ECG. Lemma 10.11. Let L and K be lines that do not coincide and are not parallel. Then there exists a point on K that is not on L. Remark. Of course classically the lemma is trivial and does not even need the hypothesis that the lines are not parallel. Constructively that hypothesis is crucial, since without it, a point on M that is not on L cannot be found continuously as M varies (ever so slightly from a parallel). Proof. By making a rotation and translation, we can assume that L is Line (0, 1). Let J = Perp(0,L). Then from each point b of K, we can drop a perpendicular M(b) to J. We claim that M(b) meets K only in b. To prove this, suppose that M(b) meets K at another point; then M(b) and K coincide, so K is perpendicular to J; then K is parallel to L by Lemma 9.48), but that contradicts the hypothesis that K is not parallel to L. Hence, as claimed, M(b) meets K only in b. Let Y (b) be the intersection point of M(b) with J, which exists since M(b) is perpendicular 124 MICHAEL BEESON to J. We only need to show that Y (b) takes on a nonzero value for some b, since if b is on L, then H(b) coincides with L, so H(b) = 0. As usual let I be a point on J with 0I = 01. By Lemma 10.10, using the strong parallel axiom there is a point b such that X(b)= x and Y (b)= I. That b lies on M but not on L. That completes the proof.

Lemma 10.12. SPP (and SPPE) imply the stability principle:

IntersectLines (L,M) IntersectLines (L,M) . ¬¬ ↓ → ↓ Conversely, that stability principle plus Playfair implies SPP (and SPPE).

Proof. First we show that SPPE implies the stability principle for IntersectLines . IntersectLines (L,M) is the definition of “weakly parallel.” Suppose then ¬that L and M are not weakly↓ parallel. Then they do not coincide and are not parallel. Let p be a point on M (by Lemma 10.11 we can even choose p so that p is not on L). Let K be weakly parallel to L through p (for example K = Perp(p, Perp(p,L)) will do). Let a be a point on M different from p (for example IntersectLineCircle1 (M, Circle (a,α,β)) will do). Then a is not on K since if it were, M and K would coincide, but M is not weakly parallel to L and K is weakly parallel to L. Then by SPPE, L and M meet. That completes the proof that SPPE implies the stability principle for IntersectLines . Conversely, suppose that stability principle holds. We will prove SPPE. Let K be weakly parallel to L through p, and let M be another line through p, containing a point a that is not on K. By S6 it suffices to prove IntersectLines (L,M) . Suppose then that IntersectLines (L,M) . Since L¬¬and M do not coincide,↓ they are parallel. But¬ that contradicts strong↓ Playfair. That completes the proof of the lemma.

Corollary 10.13. If, in the strong parallel postulate or in SPPE, we add

on (a,L) on (p,L) ¬ ∧ ¬ to the hypothesis (so the lines L and K are required to be parallel in SPPE, not just weakly parallel), and the case of L coinciding with M in the strong parallel postulate is ruled out, then the resulting principles are equivalent to the principles as formulated, and to the stability principle for IntersectLines .

Proof. The principles with one or two additions to the hypothesis become prima facie weaker, so we only have to show that the modified principles still imply the stability principle for IntersectLines . The proof of Lemma 10.12, by appealing to Lemma 10.11, works just as well for the modified SPPE; hence the modified SPPE implies the stability principle. On the other hand the modified strong parallel postulate implies the modified SPP, by the proof of Lemma 10.6, since the additional hypotheses in question are the same in both principles. That completes the proof. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 125

10.8. Stability of definedness and ∼=. When listing the stability axioms of ECG, we discussed the principle S6, namely t t , which we did not take as an axiom. In this section we will prove¬¬ ↓ that → princi↓ ple in ECG, and show that the strong parallel postulate is needed only in case t contains IntersectLines . The key to this theorem is to establish geometric conditions for t to be defined. To that end we begin with lemmas. The first one is used to give geometric conditions for a line and circle to intersect. Lemma 10.14. Let C be a circle with center a and L a line. Let K be the perpendicular to L through a. Let p be the intersection point of K and L. If any point of L is inside C then p is inside C; if any point of L is strictly inside C then p is strictly inside C. Proof. If any point of L is inside C, then by the line-circle continuity axiom, L meets C at one or two points u, v. Case 1: If u = v then triangles upa and vpa are congruent by Lemma 9.50, since they are right6 angles sharing a leg and with equal hypotenuses ua = va. Hence up = vp. Hence by Lemma 9.5, u and v do not lie on the same side of p. Hence by Markov’s principle for betweenness, B(u,p,v). Hence by Lemma 9.75, ap < ab, since ab is the radius of C. Then p is inside C, by definition of “inside.” Case 2: If u = v then by the definition of “tangent”, L is tangent to C at v, so by Lemma 9.54, Line (a,u) is perpendicular to L; then K and Line (a,u) are both perpendicular to L from a, so by the uniqueness of perpendiculars, p = u = v. Then ap = au = ab, so p is inside C in that case too. Now we justify an argument by cases. Since (u = v u = v), we have ap ab; hence by the stability of segment ordering¬¬ (Lemma∨ 6 9.15) we have ¬¬ap ab≤, so p is inside C. That completes the proof of the lemma. ≤ The following lemma is used to give geometric conditions for two circles to intersect. Lemma 10.15. Let C and K be circles with centers a and b, respectively, and suppose a = b. Let L = Line (a,b) and let p = IntersectLineCircle1 (L,K) and q = IntersectLineCircle26 (L,K). Then if any point of K is inside C, p is inside C; if any point of K is strictly inside C, p is strictly inside C; if any point of K is outside C, q is outside C, and if any point of K is strictly outside C, q is strictly outside C. Proof. Let u = IntersectLineCircle1 (L, C) and v = IntersectLineCircle2 (L, C). According to the definitions of “inside” and “outside” (Definition 8.11), what we have to prove is that if x is any point on K, then xa < au pa < au → xa au pa au ≤ → ≤ xa > au qa > au → xa au qz au ≥ → ≥ It will suffice to show pu au with equality only when x = p, and ua qu with equality only when x =≤ q. By Lemma 9.72 (the stability of angle ordering),≤ these statements are stable; so we can argue by cases on the possible betweenness 126 MICHAEL BEESON relations of the six points in question. In each case we assume the previous cases do not hold. Case 1, p = q. Then K is a degenerate circle and x = p. Then the conclusions are all immediate, completing this case. Then p = q, so by Lemma 9.12, we have B(p,b,q), and by Lemma 9.91, p is on Ray (b,a)6 and q is on the opposite ray. We have q = a, since q is on the opposite ray to Ray (b,a), but a is not on that ray since a =6 b. Case 2, C is a degenerate circle, i.e. u = v = a6 . Then if any point x of K is inside C, we have x = a, and so x lies on L. Then by Lemma 9.52, x cannot be different from both p and q. But x = q since x = a and a = q. Hence x = p. Hence x = p by the stability of equality.6 Hence a = p. Hence6 p is inside¬¬x. Since q = a, q is certainly outside a. That completes this case. 6 Now neither K nor C is a degenerate circle; hence by Lemma 9.12 we have B(u,a,v). Assume q is strictly inside C. Then B(p,b,q) by Lemma 9.12 B(a,b,q) B(u,a,b) by Lemma 9.91 ∧ B(p,a,b) by inner transitivity of betweenness B(u,q,v) since q is strictly inside C We also have B(u,a,b) derived above B(a,b,q) derived above B(u,a,q) inner transitivity B(u,q,v) derived above B(a,q,v) inner transitivity We will now show that p is inside C. Since B(p,a,b) we have ap< pb by definition of < = bq since p and q are on K with center b < aq since B(a,b,q) < av since B(a,q,v) = au since u and v are on C with center a Hence ap < au, so p is strictly inside C. We have now proved that if q is strictly inside C, so is p. Now suppose x is any point of K that is inside C. If p is strictly inside C, then the first inequality is satisfied. Hence if q or p is strictly inside C then the first inequality is satisfied. Then, by the stability of congruence, we may assume that both q and p are outside C. Then B(v,p,b). Let x be any point on K, and let m be the foot of the perpendicular from x to L. Then we have xa > am since the leg of a right triangle is less than the hypotenuse by Lemma 9.74; and B(p,m,b); so since B(a,v,p) and B(v,p,m) we have B(a,v,m) and hence av < am. Hence au = av < am < ax, so x is strictly outside K. Hence the first inequality holds. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 127

The second inequality is proved the same way, using “inside” instead of “strictly inside”; we need to first prove that if q is inside C, so is p, which is done the same way as for “strictly inside.” Now for the third inequality. Assume x is strictly outside C. Assume q is inside C. Then, as shown above, p is inside C also. Then bx = bq since x and q lie on K < aq since B(a,b,q) < av since B(a,q,v) since q is inside C = au since u and v both lie on C Hence x does not lie outside C, contradiction. Hence q does not lie inside C. Hence, by the stability of congruence, q lies outside C, which establishes the third inequality. The fourth inequality is proved the same way, using “outside” instead of “strictly outside.” That completes the proof of the lemma. Theorem 10.16. (i) Let t be a term not containing IntersectLines . Then t t ¬¬ ↓ → ↓ is provable in neutral ECG, i.e. without any parallel axiom. (ii) The strong parallel axiom proves every instance of the schema t t . ¬¬ ↓ → ↓ Remark. This theorem shows that the principle S6, namely t t , that we did not take as an axiom of ECG, turns out to be the¬¬ strong↓ → parallel↓ postulate in disguise. Principle S6 seems quite natural in connection with the other stability axioms, because it looks like another form of Markov’s principle– if two lines cannot fail to meet then they meet. This might be considered as an argument in favor of the strong parallel axiom. At the very least it shows that if we want to study the logical relationships of various forms of the parallel axiom, the base theory had better not include S6. Proof. It suffices to prove the lemma for terms t that do not contain IntersectCirclesSame or IntersectCirclesOpp , by Theorem 9.96. We proceed by induction on the complexity of such terms t. The basis case is when t is a term consisting of one function symbol with variables for arguments. We have to consider each of the function symbols of ECG separately. Several of the function symbols are total, i.e. f(x) is provable, immediately from the axioms. These include pointOn1 , pointOn2↓, pointOnCircle , and center. For example, given circle C, we have by Axiom CA6, Circle (center(C), pointOnCircle (C)) = C, from which it follows that center(C) and pointOnCircle (C) . Similarly from Axiom CA1 we have pointOn1 (L) and↓ pointOn2 (L) . ↓ When f is IntersectLineCircle1 or↓ IntersectLineCircle2↓ , we appeal to the line- circle continuity axiom Cont 1. Suppose IntersectLineCircle1 (L, C) , and let z = IntersectLineCircle1 (L, C). Then by Axiom I6, z lies on both ↓L and C. Then with b = a = x = p, the hypothesis of Axiom Cont 1 is satisfied. That is, IntersectLineCircle1 (L, Circle (a,b)) if and only if ↓ p, x(on(p,L) ax = ap T(a,x,b). ∃ ∧ ∧ 128 MICHAEL BEESON

We now claim that this last condition is stable. That is,

p, x(on(p,L) ax = ap T(a,x,b)) p, x(on(p,L) ax = ap T(a,x,b)) ¬¬∃ ∧ ∧ → ∃ ∧ ∧ Suppose p, x (on(p,L) ax = ap T(a,x,b)). The hypothesis is the double negation¬¬∃ of “some point on∧L is inside∧C”. We will construct p and x as required. Let K = (a,L) and let p be the intersection point of K and L. By Lemma 10.14, if any point⊥ of L is inside C, then this point p is inside C. Let x be a point on Ray (a,b) such that ax = ap. Then on(p,L) by construction of p and ax = ap by construction of x. Then by the definition of “inside”, if any point of L is inside C, then T(a,x,b). But we have assumed that not not some point of L is inside C; hence T(a,x,b). Since T is defined by a negated formula, and triple negation is the¬¬ same as single negation, we have T(a,x,b). That completes the proof of the stability of IntersectLineCircle1 (L, C) . The case of IntersectLineCircle2 is treated the same way. ↓ We next consider the case when f is IntersectCircles1 or IntersectCircles2 . By Axiom I18, if C and K are concentric, then IntersectCircles1 (C,K) and IntersectCircles2 (C,K) are not defined. Let C and K be two circles that are not concentric. By the circle-circle continuity axiom, C and K intersect if there is a point of K inside C and a point of K outside C. In fact they intersect if and only if this condition holds, since if they do intersect, the intersection points are both inside and outside C. Let a = center(C) and b = center(K). Let L = Line (b,a), which is defined since C and K are not concentric. Let p = IntersectLineCircle1 (L,K) and q = IntersectLineCircle2 (L,K). By Lemma 10.15, if K has a point inside C and another point outside C, then p is inside C and q is outside C. Hence K and C intersect if and only if p is inside C and q is outside C. Let c = pointOnCircle (C). Then K and C intersect if and only if ap ac and ac aq. But since segment ordering is stable by Lemma 9.15, this≤ last condition≤ is stable. Hence “non-concentric circles C and K intersect” is stable. Now suppose IntersectCircles1 (C,K) . Then by Axiom I17 and the stability of equality,¬¬C and K are not concentric,↓ and by Axiom I8, not not there is a point on both circles. But as just shown, this condition is stable, so there is indeed a point on both circles. Hence by Axiom I16, IntersectCircles1 (C,K) . That completes the proof in the case f = IntersectCircles1 (C,K). The case of↓ IntersectCircles2 (C,K) is treated similarly, appealing to Axiom I19, I9, and I17 instead of I18, I8, and I16. That completes the basis case of the induction on the complexity of t for claim (i), since we have considered all the function symbols except the two excluded in the first sentence of the proof. For claim (ii), we must also treat the case when t = IntersectLines (r, s). By Lemma 10.12, with the strong parallel postulate we have IntersectLines (L,M) IntersectLines (L,M) . ¬¬ ↓ → ↓ Suppose t . Then by the strictness axiom of LPT, r and s . Then by the induction¬¬ ↓ hypothesis r and s . Then we can substitute¬¬ ↓ r and¬¬s for↓ L and M in the stability principle for↓ IntersectLines↓ , obtaining IntersectLines (r, s) ; but that is t . That completes the proof of the lemma. ↓ ↓ FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 129

Here is the induction step: Consider a term t = f(s1,... ,sn), where f is a function symbol and the sk are terms. Suppose t . Then for each k, sk . ¬¬ ↓ ¬¬ ↓ By the induction hypothesis, there is a proof of sk (in neutral ECG unless t contains IntersectLines , and in ECG if it does contain↓ IntersectLines ). By the basis case we have

f(x1,... ,xn) f(x1,... ,xn) . ¬¬ ↓ → ↓ Since we have proved sk for each k = 1, 2,...,n, we can substitute sk for xk and conclude ↓ f(s1,... ,sn) f(s1,... ,sn) . ¬¬ ↓ → ↓ That is the desired conclusion t t . That completes the proof. ¬¬ ↓ → ↓ We now ask whether t ∼= q is stable. The answer is, that the strong parallel axiom implies that t ∼= q is stable, but we really need the strong parallel axiom to prove that.

Lemma 10.17. The strong parallel axiom implies t = q t = q. ¬¬ ∼ → ∼ Proof. We have t = s ((t s t = s) s t t = s) ¬¬ ∼ ↔ ¬¬ ↓ → ↓ ∧ ∧ ↓ → ↓ ∧ ( t s t = s) ( s t t = s) ↔ ¬¬ ↓ → ¬¬ ↓ ∧¬¬ ∧ ¬¬ ↓ → ¬¬ ↓ ∧ ¬¬ ( t s t = s) ( t s t = s) ↔ ¬¬ ↓ → ¬¬ ↓ ∧ ∧ ¬¬ ↓ → ¬¬ ↓ ∧ We will be able to complete the proof if we know that t and s are stable. In ECG, by the stability axioms, this is true as long¬¬ as↓ t and¬¬s do↓ not contain the function symbol IntersectLines ; to prove the stability of definedness of terms involving IntersectLines , we need the strong parallel axiom. However, with the aid of the strong parallel axiom we do have that stability, by Theorem 10.16. That completes the proof. 10.9. ECG proves Euclid’s parallel postulate. In this section, we give a direct proof of Euclid 5 in ECG. Later we give a more informative proof, by characterizing the models of each theory as different types of Euclidean fields. But it is still instructive to have a direct proof. The strong parallel postulate SPP has a weaker conclusion than Euclid 5, because it does not specify on which side of P the intersection point will lie. On the other hand, SPP also has a weaker hypothesis than Euclid 5, so its exact relationship to Euclid 5 is not immediately clear. One direction is settled by the following theorem: Theorem 10.18. ECG proves Euclid’s Postulate 5. Proof. Since we proved SPPE is equivalent to SPP using Playfair, we can use SPPE. Let L be a line, p a point not on L, K parallel to L through P , M another line through P , q a point on L, q a point on M not on pq, and r the intersection of QA with K. Suppose that the interior angles made by L, M, and PQ make less than two right angles, which formally means that a is between q and r. See Fig. 26 for an illustration. By SPPE, since M and K do not coincide, M does meet L at some point e (indicated by the open circle in the diagram). It remains to show that a is between p and e. By Markov’s principle for betweenness, it 130 MICHAEL BEESON suffices to prove that p is not between e and a, and e is not between p and a. Suppose first that p is between e and a. Then e is on the opposite side of line K from a. Segment eq lies on line L, and line L does not meet K, so segment eq does not meet K. Hence by Lemma 9.83, e and q are on the same side of line K. Since e is on the opposite side of K from a, it follows that a and q are on opposite sides of line K. Hence point r, the intersection of aq with K, must be between a and q. But that contradicts the fact that a is between q and r. Hence the assumption that p is between e and a has led to a contradiction. Now suppose instead that e is between p and a. Then a and p are on opposite sides of L. But rp does not meet L, since rp lies on K, and K does not meet L. Hence by Lemma 9.83, r and p are on the same side of line L. Hence a and r are on opposite sides of L. But then the intersection point of ar and L, which is q, lies between a and e, contradicting the fact that a lies between q and r. Hence the assumption that e is between p and a has also led to a contradiction. Hence, as noted already, by Markov’s principle for betweenness, a is between p and e. That completes the proof of Euclid’s Postulate 5 from the strong parallel postulate. Now we consider the converse problem, of deriving Axiom 58 from Euclid’s Postulate 5. The obvious proof attempt works only if we assume an additional axiom. A convenient formulation of what is required is “of two unequal segments one is longer”: ab = cd ab < cd cd < ab (Segment Comparison Principle) 6 → ∨ Note that, unlike any axioms of ECG this principle involves a disjunction. It is equivalent to the ordering principle x =0 x> 0 x< 0 6 → ∨ We shall discuss the constructive acceptability (or non-acceptability) of this principle far below. For now, we just note that, unlike the axioms of ECG, it involves a disjunction. Lemma 10.19. Let T be the theory ECG without any parallel postulate. Then in T, the Segment Comparison Principle plus Euclid 5 implies the strong parallel postulate. Proof. Assume Euclid’s postulate and let p be a point not on line L, and K parallel to L through p, and line M another line through p as in Axiom 58. Let q be the point on L at the foot of the perpendicular to L from p. Let A on line M and B on line L be on the same side of pq. Then K makes the interior angles Apq and pqB equal to two right angles. Hence M does not, or it would coincide with K. If angle Apq is less than a right angle, then by Euclid’s Postulate 5, M meets L. Similarly, if angle Apq is more than a right angle, we can show by Euclid’s Postulate that M meets L (on the other side of q from B). By Axiom 75, one of these alternatives must hold. More formally, let S be the intersection point of Line (A, B) with K. Then angle Apq is less than a right angle if A is between S and B, and more than a right angle if S is between A and B. By the Segment Comparison Principle, one of these alternatives holds. Hence by Euclid’s Proposition 5, M meets L. That completes the proof of the lemma. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 131

10.10. Various forms of the parallel axiom. Within neutral geometry, we can consider the logical relations between various forms of the parallel axiom. In the introduction we have already listed the Playfair axiom, Euclid 5, and the strong parallel axiom of ECG. There are also a number of other propositions that are classically equivalent to Euclid 5. Lemmas in this section continue to be proved in neutral geometry, i.e. in ECG without any parallel postulate. The first of these is the “triangle circumcription principle”, which says that any three non-collinear points lie on a circle. This makes sense constructively, but it also has a stronger version, the “strong triangle circumscription principle”, which says that there is a construction (term of ECG) t(a,b,c,L) such that if a and b lie on L, and c does not lie on L, then t(a,b,c,L) is a circle passing through a, b, and c, and if a = b then t(a,b,c,L) is tangent to L. We have proved that a line intersects a circle in at most two points, and from that it is not difficult to prove Euclid III.1, that a chord of a circle lies inside the circle. Hence the wording, “triangle circumscription” is appropriate; once a circle passes through the vertices of a triangle, the whole triangle lies inside the circle. The triangle circumscription principle is a convenient form for a theory with only point variables, since it can be naturally stated without mentioning circles or lines: given three non-collinear points, there is a fourth point equidistant from all three. We show below that it is constructively equivalent to the strong parallel postulate, so we could have taken it for the parallel postulate in ECG. Some may find it a more natural formulation. According to [30], p. 190, Szmielew took it for her parallel axiom in her influential manuscript that formed the basis for [26], but Schwabhäuser adopted another form of the parallel axiom for publication.

Figure 32. Parallel axiom implies triangle circumscription. The center is where non-parallel lines M and L meet.

M b x

c b

a b b b L 132 MICHAEL BEESON

Lemma 10.20. The strong parallel axiom implies the strong triangle circumscription principle.

Proof. Suppose a and b lie on line L, and c does not lie on L. Then by Lemma 12.4, we can construct a line K perpendicular to L that is the perpendicular bisector of ab if a = b, and contains a if a = b. Let M be the perpendicular bisector of ac; since6 c is not on L, a = c, so this perpendicular bisector exists. Let x be the intersection point of L and6 K (so x is the midpoint of ab if a = b, and is a if a = b. Now if K and M meet, that intersection point will be6 the center of the desired circle, as is easily proved, at least when a = b a = b; but once the intersection point e of K and M exists, we can argue by6 cases∨ that ae = be = ce, since equality is stable. The crux of the matter is to prove that K and M do meet. By Lemma 9.56 there is a line N perpendicular to M containing x. Let y on M be the intersection point of N with M. Let J be perpendicular to N at x. Then N is perpendicular to both J and M; so by Lemma 9.48, J and M are parallel. Now assume that K and M are parallel, and both contain x. Hence by Playfair (which is a consequence of the strong parallel axiom), K and M coincide. Now K is perpendicular to N and to L at x; by the uniqueness of the perpendicular, L and N coincide. But then c lies on L, contrary to hypothesis. Hence K and M are not parallel. Then by the strong parallel axiom, they meet. That completes the proof.

Lemma 10.21. The triangle circumscription principle implies the strong parallel axiom.

Proof. Suppose L is a line and p is a point not on L; let pw be perpendicular to L at point w on L and let K be perpendicular to Line (p, w) at p. Then K is parallel to L. Let M be a line through p parallel to L that does not coincide with K. Let J be perpendicular to M at p and let x be a point on J not on M. We can choose x so that x is not on L (For example, by dropping a perpendicular from p to L, finding its midpoint m, and choosing px = pm.) By Lemma 9.56, there exists a line N perpendicular to L and containing x. Let y be the reflection of x in M (so x = y since x is not on M, and x is on J and J is the perpendicular bisector of xy.6 Let z be the reflection of x in L; since x is not on L, z = x. If x, y, and z are all collinear, then lines J and N coincide, since x and y 6 lie on J, and x and z lie on N. Then J is perpendicular to both L and M. Hence M and K are both perpedicular to J at p. By the uniqueness of the perpendicular, M and K coincide, contradicting out assumption that M and K do not coincide. That proves that x, y, and z are not all collinear. Then by the triangle circumscription principle, there is a circle C through those three points. Let its center be e. Then since xy is a chord of C, its perpendicular bisector M passes through e. Since xz is a chord of C, its perpendicular bisector L passes through C. Hence L and M meet in point e. This contradicts the assumption that M is parallel to L. Hence we have proved that every line M parallel to p through L coincides with K. In particular any two such lines coincide with each other. That completes the proof. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 133

Corollary 10.22. The triangle circumscription principle implies the strong triangle circumscription principle. Proof. By the previous two lemmas. Remark. The “uncertainty” in the strong parallel axiom about which side of the transversal has angles making less than two right angles corresponds to a similar uncertainty about which side of Line (a,b) the noncollinear point c might lie on. If we weaken the triangle construction principle by strengthening its hypothesis to say that c lies on the same side of Line (a,b) as a given point x, then the resulting “weak triangle circumscription principle” is equivalent to Euclid 5. Lemma 10.23. Playfair’s axiom implies the alternate interior angle theorem, that any line traversing a pair of parallel lines makes alternate interior angles equal. Proof. Since ordering of angles is stable, we can argue by contradiction. Hence the usual classical proof of the theorem applies. Definition 10.24. A rectangle is a quadrilateral with four right angles. Lemma 10.25. Playfair’s axiom implies that if a quadrilateral has congruent diagonals, then it is a rectangle. Proof. “Angle ABC is a right angle” is stable by Lemma 9.39 and Definition 9.42. Hence the classical proof applies. Lemma 10.26. The opposite sides of a rectangle are equal. Proof. In quadrilateral ABCD let AC = BD. Then triangle ABC is congruent to triangle ABD since they share side AB and AC = BD and the two right angles are congruent by Lemma 9.44. Hence AD = BC. Similarly the other two sides are congruent. That completes the proof. Lemma 10.27. The existence of any rectangle implies Playfair’s axiom. Proof. FINISH THIS (why did I need this anyway?) As mentioned above, Tarski [29] and later [26] took a different form of the parallel postulate, illustrated in Fig. 33.28 The following axiom is similar to Tarki’s axiom, and we give it his name, but his axiom used non-strict betweenness and did not include the hypothesis that a, b, and c are not collinear. The degenerate cases are trivial: if a, b, and c are collinear, then we can (classically, or with more work also constructively) find x and y without any parallel axiom, and if (say) d = b then we can take x = t and y = c, etc. Hence the following axiom is classically equivalent in neutral geometry to the one used by Tarski: B(a,d,t) B(b,d,c) a = d (Tarski parallel axiom) ( B(a,b,c∧ ) B(∧b,c,a6 ) B(c,a,b)) ∧ ¬ x y∧(B ¬(a,b,x) B∧( ¬a,c,y) B(x,t,y)) → ∃ ∃ ∧ ∧ 28Technically, according to [30], the axioms taken op. cit differed in the order of arguments to the last betweenness statement, but that is of no consequence. 134 MICHAEL BEESON

Figure 33. Tarski’s parallel axiom a b

b d

b b

t x b y

Something like this axiom was ﬁrst considered by Legendre (see [12], p. 223), but he required angle xay to be acute, so Legendre’s axiom is not quite the same as Tarski’s parallel axiom. Theorem 10.28. Euclid 5 implies Tarski’s parallel axiom in neutral ECG.

Figure 34. Constructive proof of Tarski’s parallel axiom from Euclid 5. M is constructed parallel to Line (b,c) and cd = ce and bd = bg. Then x and y exist by Euclid 5.

b e c b

b b f d b b M y b g b h b t

Proof. Assume the hypothesis of Tarski’s parallel axiom. Construct line M parallel to Line (b,c) through t. Construct point e by extending segment dc by dc; then ec = dc and B(d,c,e), as illustrated in Fig. 34. Let L be Line (a,c). Then Line (d, t) meets L at a, and does not coincide with L since, if it did coincide with L, then points d and c would be on L, and hence point b, which is on Line (b,c), would lie on L by Axiom I3; but that would contradict the hypothesis that a, b, and c are not collinear. Hence Line (d, t) meets L only at a, by Axiom I3. Hence segment dt does not meet L. Hence, by Lemma 9.83, d and t are on the same side of L. Since B(d,c,e), d and e are on opposite sides of L. Hence, by the plane separation theorem, e and t are on opposite FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 135 sides of L. Hence, there is a point f on L with B(e,f,t). Now we apply Euclid 5; the two parallel lines are Line (b,c) and M, and the conclusion is that L meets M in some point, which we call y. Speciﬁcally we match the variables (L,K,M,p,r,a,q) in Euclid 5 to the following terms in the present situation: (M, Line (b,c), Line (a,c),c,e,f,t). Then all the hypotheses have been proved, except that we have B(e,f,t) while what is required is B(t,f,e); but those are equivalent by the symmetry of betweenness. Hence Euclid 5 is indeed applicable and we have proved the existence of point y on M and L. Now, we do the same thing on the other side of angle bac, extending segment db to point g with db = bg and B(g,d,b), and using the plane separation property to show that gt meets Line (a,b) in a point h with B(g,h,t). Then Euclid 5 applies to give us a point x on M and Line (a,b). It only remains to prove B(x,t,y). By the crossbar theorem (Lemma 9.61), Ray (a, d) meets Line (x, y) in a point between x and y; but since a is not on Line (x, y) (because a, b, and c are not collinear), there is only one point of intersection of Ray (a, d) and Line (x, y), and that point is t; hence B(x,t,y). That completes the proof of the theorem. We next show that Tarski’s parallel axiom implies Euclid 5 in neutral ECG. Tarski proved29 that it implies Playfair’s axiom (see [26], Satz 12.11, p. 123), and we constructivize his proof here to prove Euclid 5. The proof is short and beautiful (and the reference is in German and out of print), and we need to check its constructivity, so we repeat it here.

Figure 35. Tarski’s parallel axiom implies Euclid 5. C is parallel to L. The points x and y produced by Tarski’s axiom show that B is not parallel to L.

C be

bc b b b d B y

b L x t

Theorem 10.29. Tarski’s parallel axiom implies Euclid 5 in neutral ECG. Proof. Let L be a line, a a point not on L, and suppose C is parallels to L through a. Let B be another line through a, and points e, b, and t satisfy B(e,b,t) with e on C and t on L, and b not on L or Line (a,t), as in the hypothesis of Euclid

29The cited proof is in a book with two co-authors, but Tarski used this axiom from the beginning of his work in geometry, and it seems certain that he had this proof before Szmielew and Schwabh¨auser were involved. 136 MICHAEL BEESON

5. Specifically at is a transversal of the parallel lines C and L, with which the adjacent interior angles of line B make less than two right angles on the side where b is, as witnessed by B(e,b,t). Formally, the variables (L,K,p,r,a,q) of Euclid 5 are matched to (L,C,a,e,b,t) in this application. We must show that B meets L. The situation is illustrated in Fig. 35. Since b does not lie on Line (a,t), but does lie on Line (e,t), we have e = a. Hence Line (e,a) is defined. Define 6 c = IntersectLineCircle2 (Line (e,a), Circle (a,e)). Then c is on C and ae = ac and B(e,a,c). Then let b be the point of intersection of segment et with B. Now let d be the intersection point of Line (b,c) and Line (a,t). By Outer Pasch (Lemma 9.57), applied to triangle eat and segment bc, we have B(a,d,t) and B(b,d,c). Now we are in a position to apply Tarski’s parallel axiom to angle bac and point t, which is in the interior of angle bac as witnessed by crossbar bc and point d. According to Tarski’s parallel axiom, there exist points x and y on Ray (a,b) and Ray (a,c) respectively such that B(x,t,y). Then x and y are on opposite sides of L, since point t is on L between x and y. But ya does not meet L, since ya lies on C, which is parallel to L, so y is on the same side of L as a, by Lemma 9.83. Then by the plane separation theorem, x and a are on opposite sides of L. Hence segment ax meets L. But then B is not parallel to L. That completes the proof.

§11. Comparisons of ECG to other formalizations. 11.1. Comparison of ECG to Hilbert-style theories. By a Hilbert-style theory, we mean one with variables for lines and whose axioms follow those of Hilbert. For example, the textbook [12] that we used in [3] to compare the ﬁrst formulation of ECG with classical logic ues a Hilbert-style theory. The present version of ECG does have variables for lines, and for circles too, but its axioms follow Tarski more closely than Hilbert. Just as in classical geometry, one may choose formalisms with more sorts of variables and more axioms, in which case the formal development looks a bit more like Euclid, or one may choose points- only, and seek for a minimal axiom set, in which case some infrastructure must be developed. Here we have taken some steps towards reducing the number of axioms, in order to see how this infrastructure is can be constructively developed. We have retained variables of type Line and Circle, but they are easily eliminated if one so desires. One should then show (directly or indirectly) that there are translations in both directions, i.e. the Tarski-style axioms are provable in Hilbert systems, and the Hilbert sytems are interpretable in the Tarski-style systems. 11.2. Comparison of ECG to Tarski’s theories. There are several differences: Tarski’s language is smaller, with only variables for points and the relations of non-strict betweenness and equidistance. He uses non-strict betweenness, while ECG uses strict betweenness. His logic is classical instead of intuitionistic. His theory has no function symbols. His axioms are (consequently) not quantiﬁer-free, though they are all in AE (or ) form. Since he has no terms for the results of constructions, the question o∀∃f the continuity of those constructions does not arise. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 137

The use of a larger language is a matter of convenience, and of the attempt to make ECG faithful to Euclid. One can eliminate the variables for lines and circles, introducing new versions of the function symbols that work only with points. One can also eliminate the function symbols in favor of quantifiers, and (in view of Markov’s principle for beweenness and stable equality) one can translate back and forth between strict and non-strict betweenness. There are two remaining innovations in ECG: the use of intuitionistic logic, and the effort to ensure the continuity of the terms for constructions, making use of the concepts of right and left turns. These are independent innovations, in the sense that either could be considered without the other. We next demonstrate that one can consider intutionistic logic without construction terms. It is some interest, both foundationally and technically, to formulate a geometrical theory in Tarski’s language with intuitionistic logic, that is constructively acceptable. One cannot just adopt Tarski’s axioms because he has (purposely) not excluded degenerate cases, and some of those cases (for example in his “inner Pasch” axiom) make the axiom as formulated constructively invalid. We modify the axiom of segment construction (T(q,a,x) ax = bc) to this version: ∃ ∧ q = a x( ( T(q,a,x) x = a) (Ax. 4I) 6 → ∃ ¬ ¬ ∧ 6 This avoids extending a null segment but allows extending a segment by a null segment. The case of extending a null segment, i.e. Euclid I.2, is dealt with by constructing a circle of given radius. But in the circle axiom of Tarski’s theory, one cannot get the intersection point of a line and circle without already having some point on the circle (to specify it). Therefore one either has to modify the circle axiom, or explicitly assume a version of Euclid I.2: yxy = uv (Axiom for Euclid I.2) ∃ It is interesting to see how the necessity to assume Euclid I.2 in constructive geometry, which we saw already in ECG, comes up in Tarski’s theories. It turns out that the inner Pasch axiom is the only other one that needs a modification. Here is the modified axiom: B(a,p,c) B(b,q,c) p = b q = a B(c,a,q∧) B(a,q,c∧ ) 6 (q,c,a∧ 6 ) ∧ ¬x[B(p,x,b∧) ¬ B(q,x,a)]∧ ¬ → (Ax. 7I-T) ∃ ∧ One can also formulate a similar theory using B instead of T. If one replaces T with B throughout (including in the modified Ax. 7I-T, which is then called Ax. 7I-B), then one has to also add back a few axioms that were present in Tarski’s original theories but were derivable using T in degenerate cases. Specifically, one has to add back the symmetry of betweenness B(a,b,c) B(c,b,a) and the null segment axiom ab = cc a = b and the congruence→ axiom ab = ba. We also take B(a,b,c) a = b→and B(a,b,c) b = c as axioms. The identity axiom for betweenness→ becomes6 B(a,b,a) instead→ 6 of T(a,b,a) a = b. To either of these theories we should¬ add stability of atomic formulas.→ We call these theories “Intuitionistic Tarski-T” and “Intuitionistic Tarski-B.” Theorem 11.1. Tarski’s geometry with intuitionistic logic, either with B or with T, is satisfied in the recursive model. With classical logic, these theories are 138 MICHAEL BEESON equivalent to Tarski’s theory. The two versions are equivalent when T is defined in terms of B or vice-versa, without using disjunction. Proof. All the axioms of intuitionistic Tarski-B are evidently provable in ECG. Hence this theory holds in the recursive model. We turn to the equivalence of the two versions. First we define B by B(a,b,c) := T(a,b,c) a = b b = c ∧ 6 ∧ 6 Then we have to check that the interpretations of all the axioms of Tarsi-B are provable in Tarski-T. First we note that the extra axioms added to Tarksi- B that are not in Tarski-T are provable (that was why they are not already included). See [26] for details. Since B(a,b,c) implies T(a,b,c), hypotheses become stronger; we only need to check that the stronger conclusions are provable. For example, we check the Pasch axiom. Under the hypotheses of Ax 7I(B) one gets from Ax. 7I(T) the existence of x such that T(p,x,b) T(q,x,a), and one has to prove B(p,x,b) B(q,x,a). So one has to rule out∧ the degenerate cases and use Markov’s principle∧ for betweenness. The degenerate case are p = x, x = b, q = x, and x = a. If p = x then a, b, and c are collinear, which is ruled out by hypothesis. If x = b then q = b, contradicting B(c,q,b). If q = x then (since x = b), p lies on Line (c,b) and hence p = c, contradicting B(a,p,c). If x = a then6 p = a, contradicting B(a,p,c). The other axioms are very easy and we omit the verifications. That completes the proof that the two intuitionistic versions of Tarski’s theory are equivalent. To check that with classical logic, these theories are equivalent to Tarski’s theory, we just need to check that the extra hypotheses imposed in axioms Ax. 4-I and Ax. 7-I do not classically restrict the power of the axioms. With regard to the segment extension axiom Ax. 4, we need to prove that a null segment can be extended; but we added the “Axiom for Euclid 1.2”, which takes care of this. With regard to Ax. 7, we need to check the case when a, c, and q are collinear, which is excluded by hypothesis in the intuitionistic theories, as well as the cases p = b and q = a. The latter two cases are easy: take x = p or x = q, respectively. Suppose a, c, and q are collinear. Then we have B(a,p,c) and B(c,q,b), and all the points collinear, from which it follows that B(p,c,b) by the inner transitivity of betweenness. So technically, we have to check that this theorem is provable in Tarski’s intuitionistic theory. We do check below that it is provable in ECG, and the same proof works in Tarksi’s intuitionistic theory. That completes the proof of the theorem. In subsequent sections, we will show how to interpret ECG in intuitionistic Tarski-B, and we will also prove that Tarski’s geometry with intuitionistic logic, either with B or with T, proves every AE theorem that is provable with classical logic in Tarski’s theories, and that these are the same AE theorems provable in ECG.

§12. Connections between geometry and Euclidean fields. We now turn to the characterization of models of ECG as planes over Euclidean fields. Classically, the models of Euclidean geometry (for example Tarski’s theory) are all planes over a Euclidean field, that is, an ordered field in which every positive element has a square root. The main point of this section is to prove a similar FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 139 theorem for the models of ECG. There are two interesting parts of this work: First, it is not trivial to defined signed arithmetic without case distinctions, i.e. to give a single construction for the sum of two points on Line (0, 1), regardless of the positions of those points relative to the fixed points 0 and 1. Second, the different forms of the parallel axiom turn out to correspond to different axioms for Euclidean fields. In Section 6, we have already discussed the constructive version(s) of the axioms for Euclidean fields. Recall that there were four versions, which we called “Playfair”, “weakly Euclidean”, “Euclidean”, and “strongly Eu- clidean”, according to what axiom about the existence of reciprocals is assumed. Euclidean means nonzero elements have reciprocals; weakly Euclidean means positive elements have reciprocals; Playfair means that elements without reciprocals are zero; and strongly Euclidean means that nonzero elements are either positive or negative (in addition to weakly Euclidean or Euclidean, which are then equivalent). We will show that these different versions of the Euclidean field axioms correspond directly to the different versions of the parallel postulate we have considered. Euclid 5 corresponds to positive elements having reciprocals; the strong parallel axiom corresponds to nonzero elements having reciprocals; and Playfair’s axiom corresponds to “elements without reciprocals are zero.” 12.1. Signed addition. In order to establish the connection between ECG and the theory of Euclidean fields, it will be necessary to formalize the definitions of Perp, Project, and Rotate that are required to define Add and Multiply as in Section 5. Definition 12.1. Minus (x)= Reflect (x, 0, Line (0, 1)). Equivalently, Minus (x)= Reflect (x, Line (0, I)). Next we need a way to construct a strict lower bound for given points x and y on Line (0, 1). We also define the absolute value of x. Definition 12.2. Abs (x)= Extend (Minus (1), 0, 0, x) UpperBound (x, y)= Extend (0, 1, 0, Extend (0, 1, 0, Extend (0, 1, 0, x))) LowerBound (x, y)= Extend (0,Minus(1), 0, Extend (0, 1, 0, Extend (0, 1, 0,y))) Lemma 12.3. If p = LowerBound (x, y) and q = UpperBound (x, y), we have B(p, x, q) and B(p,y,q), and p < x x < q and p

Figure 36. Additive inverse: Addition when OA = OB. A is rotated to U, then projected to V , then rotated to W = O.

C H

b b V U

b bb b R B 0 A tated onto the line H =0I. Then it is projected onto the line perpendicular to Line (0, 1) at B; that projection is V . Then it is rotated back to Line (0, 1) at point W . We have WB = BV = OU = OA. Hence WB = 0A. But 0B = 0A since B = Minus (A) (or A = Minus (B)). Since A and B are reflections in 0, we have B(B, 0, A), so 0 is on Ray (B, A). Then by Lemma 9.5 we have B = 0. That completes the proof of Add (A,Minus(A)) = 0 and Add (Minus (B),B)=0. Evidently Minus (0) = 0, since 0 is its own reflection in Line (0, I). The constructively difficult part of this proof was getting Add (A, B) to be defined (without a case distinction). Since equality is stable, we may argue for the equations in the lemma by cases, according to the signs of the arguments. Recall that x is positive (x > 0) if x lies on Ray (0, 1) and x = 0, and similarly for “negative”; hence if x is positive, Minus (x) is negative and6 vice-versa. The law Add (Minus (x), Minus (y)) = Minus (Add (x, y)) follows from the observation that just as Minus (x) is the reflection in 0 of x, Add (Minus (x), Minus (y)) is the reflection in 0 of Add (x, y); in fact the whole diagram for the definition of addition is reflected in 0 when A and B are reflected in 0. Next we take up the commutative law Add (A, B) = Add (B, A). The case when either of A or B is zero has already been proved; we now take up the case when A and B are both positive, which is illustrated in Fig. 37. Starting with A, we rotate to U, project to V , rotate to W = B + A. Starting with B, we rotate to X, project to Y , and rotate to Z = B + A. We must prove W = Z. As mentioned, we will argue by cases. First assume A and B are both positive, i.e. they lie on Ray (O, 1). By the properties of Rotate in Lemma 9.101, the points Z and W lie on Ray (0, 1), as illustrated in Fig. 37. We have B(0, A, W ), by construction of W , and B(0,B,Z), by construction of Z. We also have OA = BW and AZ = AY = OX = OB, so AZ = OB. Thus segment OW is composed of OB and BW , and segment OZ is composed of segment OA and AZ, and we have OB = AZ and BW = BV = OU = OA, so BW = OA. Then by Lemma 9.3, we have OW = ZO = OZ. Hence W = Z by Lemma 9.5, since both W and Z lie on Ray (0, 1). That completes the proof of commutativity of addition in case the arguments are positive. If A is negative and B is positive, then Fig. 38 illustrates the situation. A is rotated to U, projected to B, and rotated back to F = Add (A, B), labeled 142 MICHAEL BEESON

Figure 37. Commutativity of addition with A and B positive. C

b b X Y

b b U V

b b b b R 0 A B W = A + B = B + A

Figure 38. Commutativity of addition, one argument negative.

b b Y X

A b b 0 b A + B b B

b b U V

A + B in the diagram. B is rotated to X, projected to Y , then rotated back to G. We have to prove F = G. Since A is negative and B is positive, we have B(A,O,B). We have OB = OX = AY = AG, so AG = OB. Therefore by Lemma ?? we have GB = AO, since AB is composed of AO and OB, but also of AG and GB = AO. We have AO = OU = BV = FB. Hence AO = FB. Hence GB = FB. But both F and G lie on Ray (A, B), since B(A,O,B), according to the properties of Rotate . Hence by Lemma 9.5, we have F = G as desired. That completes the proof of commutativity in case A is negative and B is positive. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 143

The other two cases are similar to the two we have treated, and we omit the details. Now we turn to the associative law, Add (Add (A, B), C)= Add (A, Add (B, C)). Again, since the conclusion is an equality and equality is stable, we can argue by cases. Since we have already proved Add (0, x)= Add (x, 0) = x, if any one of A, B, or C is zero we are done. Since Add (Minus (A),B)) = Add (A, Minus (B)) = Minus (Add (A, B)), general associativity can be reduced to the case when A, B, and C are all positive. That case is illustrated in Fig. 39.

Figure 39. Associativity of addition, positive arguments.

b b X Y

b b Z E

b b b P U V

b b b b b b 0 A B A + BC G = B + CA + B + C The reader should follow the computation of A+B, then (A+B)+C, arriving at the point labeled A+B +C. Then follow the computation of B +C, and then A + (B + C), arriving again at the point labeled A + B + C. We have to prove that you really arrive at the same point. The proof is a diagram-chase producing a chain of conguences, using the properties of projection and rotation. Here are the details: Let W = Add (A, Add (B, C)) and let W ′ = Add (Add (A, B), C), and F = Add (A, B), with other points as in Fig. 39. We have

OA = OU = VB = BF where F is labeled A + B OB = OZ = CE = CG

OF = OX = Y C = CW ′ OA = OU = GP = GW

Then OW is composed of OG and GW , and GW = OA. In turn OG is composed of OC and CG, and CG = OB as shown above. So OW is composed of OC, CG, and GW , which are respectively congruent to OC, OB, and OA. Similarly, OW is composed of OC and CW ′. By Lemma 9.6, we have CW = CW ′ ′ (cutting segment OC of both OW and OW ′). Now CW ′ is composed of CG and GW ′, and CW is composed of CG and GW ; applying Lemma 9.6 again we have GW = GW ′. Then by Lemme 9.5, which applies since B(O, G, W ) and B(O, G, W ′), we have W = W ′. That completes the proof of the associative law, and also the proof of the lemma. 144 MICHAEL BEESON

There are, of course, several other cases according to the signs of A, B, and C; but we omit all the other cases, on the grounds that they are very similar to this case. That completes the proof of the lemma. 12.2. Signed multiplication. We will now show how to adapt Descartes’s deﬁnition of multiplication to allow signed arguments. We specify that B in Descartes’s diagram is 0, and A is 1, and the angle at B is a right angle (which does not correspond to Descartes’s diagram, but is convenient). Descartes’s diagram (Fig. 5) is for multiplying BD by BC, but we want both points to start out on Line (0, 1), so the ﬁrst step is to obtain C in the diagram by rotating the given point from Line (0, 1) to Line (0, I). Fig. 40 illustrates the following construction script: Multiply(Point a, Point b) { c = Rotate(1,0,I,a) J = Line(1,c) K = para(b,J) e = IntersectLines(K,\Line(0,I)) d = Rotate(I,0,1,E) return d }

Figure 40. A variation of Descartes’s multiplication; here d = Multiply(a,b).

K e b

J c b

b b b L 0 a 1 d b

Lemma 12.7. The strong parallel axiom implies Multiply (x, y) is deﬁned for all x and y on Line (0, 1). Remark. The script works also in the problematic cases a = b, a = 0, b =, and even a = b = 0. Proof. We work through the script line by line, checking that each point constructed is provably deﬁned. Suppose given points D and Q to multiply. Since FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 145

Rotate is everywhere defined, C in the first line of the script is defined. Since C is on Ray (0, I), and Line (0, I) meets Line (0, 1) only in 0, C = 1, so L in line 2 of the script is defined. Since Para is everywhere defined, K in6 line 3 of the script is defined, and is parallel to Line (1, C) through D, if D = C, and coincides with Line (1, C) if D = C. Now we wish to prove that K meets6 Line (0, I). Assume K fails to meet Line (0, I). Then Q is not zero, since if Q = 0 also C = 0 so C lies on K and Line (0, I). Also D is not 1, since if D = 1 then the two lines both pass through C. Then K is parallel to Line (1, C) through D. If it does not meet Line (0, I) then by Lemma ??, Line (1, C) is parallel to Line (0, I), which is false since they both contain C. Hence the assumption that K fails to meet Line (0, I) has led to a contradiction. That is, we have proved

(IntersectLines (K, Line (0, I)) ) ¬¬ ↓ Then by the strong parallel axiom, IntersectLines (K, Line (0, I)) ). Then E in line 4 of the script is defined, and then N, the script’s result, is defined↓ in line 5, since the points I, 0, and 1 are distinct. That completes the proof of the lemma. Next we exhibit a construction script for Hilbert’s multiplication method, described in Section 5.4. It uses the construction Reflect from Lemma 9.105, which reflects a point e in line L, without a case distinction as to whether e is on L or not. Note the use of LowerBound to ensure that we are extending a non-null segment in the next line.

HilbertMultiply(Point a, Point b) { A = LowerBound(a,b) B = UpperBound(a,b) m = Midpoint(A,B) J = Perp(m,L) x = Midpoint(I,a) K = Perp(x,Line(I,a)) e = IntersectLines(J,K) C = Circle(e,a) M = Perp(e,Line(0,1)) Q = Perp(e,M) w = Reflect(I,Q); return w }

Lemma 12.8. The strong parallel axiom implies that HilbertMultiply (a,b) is deﬁned for all points a and b on Line (0, 1), and returns the result of Hilbert’s multiplication construction; that is, the point w produced by the script lies on the circle C and is diﬀerent from I if there is another intersection point of C and Line (0, 1), and is equal to I if not.

Remark. This script, like Multiply , works in the degenerate cases, including a = b = 0, thanks to the use of LowerBound and UpperBound , and the uniform perpendicular Perp. It is instructive to trace through what happens when a = b = 0. 146 MICHAEL BEESON

Proof. By the properties of LowerBound and UpperBound , we have B(A,a,B) and B(A,b,B), so A = B. Hence m is defined, and is the midpoint of AB, and if a = b m is the midpoint6 of ab, and if a = b then m = a. Since Perp is always defined,6 J is defined; since I is not on L = Line (0, 1) but a is, x is defined; since Perp is always defined, K is defined. Then since I does not lie on Line (a,b), by the strong triangle circumscription principle (Lemma 10.20), there is a circle C′ through I, a, and b. Let E be the center of that circle. Then aI is a chord of C′, so its perpendicular bisector K passes through E. If a = b then J is the perpendicular bisector of chord ab, so J passes through E. If6 a = b then L is tangent to C′ at x, by the strong triangle circumscription principle. Then J also passes through E. But (a = b a = b); hence on(E, J). By Axiom S3, on(E, J). Hence E lies on¬¬J and on∨ K6. Then the e¬¬defined in the construction script is defined and equal to E, the center C′. Hence C′ coincides with C, since they have the same center e and both pass through a. Since Perp is always defined, M and Q are defined; and since Reflect is always defined, so is w. Now w lies on circle C since, if I is not on C and z = IntersectLines (Q, Line (0, I)), then triangle Iez is congruent to triangle wez; and if I is on C then w = I is on C; since (On(I, C) On(I, C)), we have On(w, C), and hence by Axiom S5 we have¬¬On(w, C).∨¬ That completes the proof.¬¬ Lemma 12.9. The strong parallel axiom implies that Multiply satisfies the commutative and associative laws and the law Multiply (1, x)= Multiply (x, 1) = x, and HilbertMultiply (a,b)= Multiply (a,b). Proof. The verification of the commutative and associative laws for Multiply is carried out, for positive arguments, in Hilbert [15], pp. 46–55. (Hilbert is multiplying segments, and he starts by laying them off to the right of 0 on Line (0, 1), so only multiplication of positive arguments is defined.) Each case where the arguments have known signs can be argued similarly; for example if a < 0 and b < 0, then the picture in Fig. 6 is reflected in the center point e; if only one argument is negative, then the picture is reflected in one of the axes, and Hilbert’s arguments for the equivalence and correctness of the two definitions carry over easily. There are also the cases in which one or more arguments are zero. One can check that Multiply (0, x) and Multiply (x, 0) are 0, and the same for HilbertMultiply . Hence our conclusion holds in those cases also. But the double negation of the disjunction of those cases holds, since they cannot all fail. Pushing the double negation inwards, and using the stability of equality, the desired equations hold without requiring a case distinction. It is true that Hilbert’s axiom system is different than that of ECG, which is more closely related to Tarski’s axioms. But Chapter 12 of [26] is primarily devoted to a very rigorous proof of the theorem of Pappus-Pascal (which Hilbert calls Pascal’s theorem), which is the key to the correctness of Multiply. We therefore do not repeat either proof here. That completes the proof of the lemma. We now check the validity of Descartes’s construction of square roots. Let SquareRoot (x) be the term defined by the construction script in Section 5.5. Then we have FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 147

Lemma 12.10. Playfair’s parallel axiom implies that iff x lies on Ray (0, 1), and z = SquareRoot (x), then HilbertMultiply (z,z)= x. Remark. We only need Playfair’s axiom, not the strong parallel axiom. The assumption implies SquareRoot (x) is defined, and the fact that HilbertMultiply (z,z) is defined is part of the conclusion. Proof. We first give a “proof by diagram.” In the case of squaring, Hilbert multiplication has the circle tangent to the horizontal axis, since a and b coincide, as illustrated in Fig. 41. Replacing a by √x and reflecting the figure in a diagonal line from upper left to lower right, we obtain the first diagram in Fig. 42. Relabeling that diagram, and omitting the bottom half of the circle, we obtain Descartes’s method of calculating the square root of x, shown in the second diagram in Fig. 42. We will now give a rigorous proof. Since what is to be proved is an equation, by the stability of equality we can argue by cases. The first case is when x = 0. Then SquareRoot (x) = 0 and HilbertMultiply(0, 0)= 0, disposing of that case. So now we can assume x = 0. Let b and d be constructed by Descartes’s method from input b, where by assumption6 bp is congruent to 01, a segment whose length is taken as unity. The horizontal line at the bottom is a diameter of the circle, whose endpoints are p and s. Erect the perpendicular to this diameter at p and let cp = bd. Then d = c since if d = c then b = p, by the uniqueness of dropped perpendiculars, but b6 = 0 since 0 = 1 and bp 01. Hence L = Line (c, d) is defined; it meets the circle at 6d. Let r be6 the other intersection point of the circle with Line (c, d) (which is defined without a case distinction as to whether the circle has two or only one intersections with the line). Now triangle dbp is congruent to triangle cpb by SAS, since db = cp and bp = pb and all right angles are congruent, so angle dbp is congruent to angle cpb. Then dp = bc as corresponding sides of congruent triangles. Now dcpb is a quadrilateral whose diagonals are equal. By Playfair’s axiom and Lemma 10.25, dcpb is a rectangle. Hence by Lemma 10.26, its opposite sides are equal. Hence dc = bp. The segment that SquareRoot constructs (before the last rotation in the SquareRoot script) is db, which is labeled √x in the figure. The actual result of SquareRoot is a point z such that 0z = db. Since db = cp, the result of HilbertMultiply (z,z) is a point w such that 0w = rc. What we must prove then is that rc = sb. Point q is defined as the foot of the perpendicular from r to the diameter. (Point t has not yet been defined.) Define m to be the midpoint of rd and e to be the center of the circle. Then em L since triangle rme is congruent to triangle dme, by the definition of triangle⊥ congruence (SSS). Then L em by Lemma 9.38, so r is the reflection of d in Line (e,m). Define t = Reflect⊥ (c, Line (e,m)). Then quadrilateral trqs is the reflection of quadrilateral cdbp, and hence it is also a rectangle. Hence tr = dc = bp = sq = 01. Now all the angles of quadrilateral rdbq are right angles. Therefore its opposite sides are equal, by Lemma 10.26. Hence rd = qb. Hence by Lemma 9.3, together with dc = sq, which was proved above, we have rc = sb. That completes the proof of the lemma. 148 MICHAEL BEESON

Definition 12.11. Point HilbertSquare(Point x) { return Multiply(x,x) }

Figure 41. Hilbert multiplication in the special case of squaring.

2 a b

I b

a b

Figure 42. If z = SquareRoot (x), then HilbertMultiply (z,z)= x. The left circle is Fig. 41 reﬂected in the diagonal line. The right circle is a relabeling that is recognized as Descartes’s square root.

1 b b b

a √x

b b x = a2 1 FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 149

Figure 43. Rigorous proof that if z = SquareRoot (x), then HilbertMultiply (z,z)= x.

t r m d b b b b b c

√x

b b b s q e b p x = a2 1

Figure 44. Proof of the distributive law. The shaded triangles are congruent and the opposite sides of the parallelogram are equal.

a(b + c) b

ac b

a b ab b

b b b b b 1 c b + c

Lemma 12.12. on (x, Line (0, 1) HilbertSquare (x) 0. → ≥ Proof. Let L = Line (0, 1). Suppose on (x, L). Let C be the circle tangent to L at x, and passing through I. Then the other intersection point p of C and the y-axis J = Perp(0,L) (which might be I itself) is on the same side of L as I, since if not then p = I and the y-axis meets C in a third point, contradicting Lemma 9.52. By definition6 z = HilbertSquare (x) = Rotate (I, 0, 1,p). Then by Lemma 9.101, z lies on Ray (0, 1). Hence z 0. That completes the proof. ≥ 12.3. The distributive law. I do not know how to prove the distributive law directly from the definitions of Add and HilbertMultiply . On the other hand, from Descartes’s definition of Multiply , it is relatively easy. Hilbert [15], p. 55, gives the diagram in Fig. 44. 150 MICHAEL BEESON

Lemma 12.13. Multiply (and hence HilbertMultiply ) satisfy the distributive law with Add .

Proof. See Fig. 44. The point on the y-axis labeled a is actually Rotate (1, 0,I,a). labeled ab is the point which, when rotated clockwise ninety degrees, will be Multiply (a,b). Similarly for ac. Assuming b and c and a are all positive, the addition shown is that produced by Add . Then the shaded triangles are congruent, since they are right triangles each with horizontal leg congruent to 0b. Hence the vertical sides of the shaded triangles are congruent (to ab). Hence the upper left side of the parallelogram is also congruent to ab. Hence the point labeled a(b + c) is also ac + ab. Since addition is commutative, that proves the distributive law in case a, b, and c are all positive. If a< 0 we can use the laws ( a) b = (a b) and ( a) + ( b) = (a + b), which we have already proved for− HilbertMultiply· − · and Add− , to− reduce− the problem to the case a > 0. There are then three more cases to treat, when one or both of b and c is negative. If both are negative, the diagram is just reflected in the y-axis and the argument is essentially unchanged. To treat the case when b is negative, just replace b + c in the diagram by c, and c by c b, and change the labels on the y-axis accordingly. Then the same argument applies.− And if any of the three points is zero, the law simplifies to an identity using x+0 = x and x 0 = 0. Now comes the point where we must consider constructive logic. Since classically· these cases are exhaustive, constructively not not one of them holds; hence we have proved a(b + c)= a b + a c ¬¬ · · (where the normal notation abbreviates Multiply and Add ). But equality is stable; hence we can drop the double negation. That completes the proof of the distributive law for Multiply . But Hilbert proved that HilbertMultiply (a,b) = Multiply (a,b), so the distributive law also holds for HilbertMultiply . That completes the proof of the lemma. 12.4. Reciprocals. According to the definition of HilbertMultiply , a reciprocal of a nonzero point a on L = Line (0, 1) is a point b on L such that there is a circle C through a, b, and I, and C is tangent to K = Line (0, I) at I. The reciprocal of x can be constructed by the following script. It uses the script Other for the other intersection point of a line and circle, which was defined and proved correct in Theorem 9.108. The script is illustrated in Fig. 45. The dashed line N is used in the proof, but not in the construction.

Point Reciprocal(Point a) { K = Perp(L,0); H = Perp(I,K); M = Line(a,I); m = Midpoint(a,I); J = Perp(m,M); e = IntersectLines(H,J); C = Circle(e,a); b = Other(a,L,C); } FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 151

Figure 45. Construction of the reciprocal of a K C

M N J e b b H I

b m

b b L a b

The following lemma proves the correctness of this script. The hard part is to show that Reciprocal (x) is defined for x = 0. 6 Lemma 12.14 (Reciprocals). Let Reciprocal be defined as in the construction script above. The strong parallel postulate implies that for a =0, Reciprocal (a) is defined, and if b = Reciprocal (a), then HilbertMultiply (a,b)=16 .

Proof. Assume a = 0. We will show Reciprocal (x) is defined, by going through the script line by6 line. Since Perp is everywhere defined, K and H are defined; since I does not lie on L, a = I, so M is defined. Since a = I, m is defined; since Perp is everywhere defined,6 J is defined. Now we have6 to prove that lines H and J meet in some point e. That is where we use the strong parallel axiom. Let N = Perp(I,M). Then by Lemma 9.48, N and J are parallel, since both are perpendicular to M. Both N and J pass through I. M is not perpendicular to H at I, since if it were, then Ia and I0 would be two perpendiculars to H at I, so they would coincide, and it follows that a = 0, contrary to hypothesis. Then let x be any point on H other than I, for example, x = IntersectLineCircle1 (H, Circle (I, 0, 1). Then x is not on N, since if it were then H and N would coincide. Hence we can apply the strong parallel axiom to conclude that H meets J. Then e in the script is defined. Since Circle is always defined, C is defined. Since a lies on both L and C, b in the penultimate line is defined. Hence Reciprocal (a) is defined. By Lemma 9.54, C is tangent to K at I. Then by definition of HilbertMultiply , we have HilbertMultiply (a,b) = 1, since circle C passes through a and b and meets K in I, and the “fourth point of intersection” is the point I, since C is tangent to K at I. That completes the proof. 12.5. Interpreting ECG in field theory. For readers who are not expert in logic, we begin with a short discussion of reducing one theory T (for example a theory of ordered fields) to another theory S (for example ECG or some other 152 MICHAEL BEESON geometric theory). Classically there are two approaches: we try to characterize the models of T in terms of models of S (for example, models of a geometry are planes over a field); or more technically we try to interpret T in S, which means to give a formal translation of sentences φ in the language of T into sentences φˆ in the language of S, such that if T proves φ then S proves φˆ. The two approaches are connected by Gödel’s completeness theorem, since if S does not prove φˆ then there is a model F of S in which φˆ fails, and if T does not prove φ there is a model of T where φ fails, and usually a connection between the models is established by what amounts to an interpretation. When we use intuitionistic logic, we do not have Gödel’s completeness theorem, so when we speak of a model-theoretic characterization of a geometric theory, this could be interpreted as (i) meaning just what it says, using a constructive logic at the meta-level, or (ii) as a shorthand for the claim that a formal interpretation exists. In the case of geometry and field theory, to interpret field theory in geometry we have to define addition and multiplication and order on the points of a line, by formulas in the language of geometry, and verify the field axioms. That corresponds to showing how to construct a field within each model of the geometry. Logicians write T φ to mean that theory T proves formula φ. An interpre- ⊢ tation is sound if T φ implies S φˆ. It is faithful if S φˆ implies T φ. Here is a model-theoretic⊢ description⊢ of the results in⊢ this section. Assume⊢ F is a Euclidean field. We will show how to turn F 2 into a model of ECG (or, to describe the construction more formally, we will show how to interpret geometry in field theory). As usual in the corresponding classical theories, we take the points to be elements of F 2, and let lines, circles, arcs, and segments have their usual analytic definitions; in particular we define circles so that circles of zero radius are allowed. Hence Circle3 can be interpreted. Markov’s principle in F allows us to verify that axiom of ECG. The intersection points of circles and lines, and the intersection points of circles and circles, can be defined by the solution of quadratic equations. Before beginning the more technical part of the proof, we make a general remark about mathematics in EF. The axioms permit one to do ordinary analytic geometry in EF; since we can take square roots of nonnegative elements, and sums of squares are nonnegative, we can define x ; dot product and cross product can also be defined; vectors are simply treated| as| pairs (or, in principle, n-tuples) of field elements. Linear transformations can be represented as two-by-two matrices, which in turn can be represented in EF by mentioning their entries. When one says “there exists a linear transformation. . . ” one means “ a,b,c,d... ”. One can prove, for example, that given points (that is, pairs of field∃ elements) a,b with a = b, and points A, B with A = B, there is a linear transformation that takes (a,b,c6 ) to (A, B, C). Namely, translate6 by A a, then rotate and dilate to bring b to B; of course for a formal proof one has− to write out the transformation explicitly in terms of the coordinates of the given points and verify by algebraic calculation that it works. We turn now to the technical problem of making the definitions sketched above into a formal interpretation. The idea of the interpretation is that each point x should be represented by a pair of field elements (x1, x2). Circles will be represented by their centers, and by one point on the circle, which requires four field elements; similarly, lines are represented by two points, which requires four FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 153

field elements. To manipulate these pairs and quadruples we make use of lists of points, which we write as (x1, x2,...xn); usually n will be 2,4, or 6. It will be necessary that one variable for a point, circle, or line corresponds in the interpretation to two, three, or four (respectively) variables for field elements. Officially, there is a fixed list of variables p0,p1,... of type Point, a list ℓ0,ℓ1,... of variables of type Line, and a list c0,c1,... of variables of type Circle, and in field theory EF there is similarly a fixed list of variables v0, v1,... ,vn. (Technically when logicians write “x is a variable of type Point”, they mean that x is one of the pn; thus “x” is a metavariable ranging over the true variables.) We associate to each variable of ECG, two, four, or three variables of EF in a unique way. Specifically, pn is associated to v9n, v9n+1, and ℓn is associated to v9n+k for k = 2, 3, 4, 5, and cn is associated to v9n+k for k = 6, 7, 8. If x is a variable of type Point, we denote the variables of EF associated to it by x1 and x2; if x has type Line we use x1, x2, x3, and x4, and if x has type Circle, we use x1, x2, and x3. Here the subscript notation is used in a special, technical way. Usually x1 would just be another metavariable ranging over official variables. Now, the subscript is used to associate a variable of EF to a variable of ECG. This will mean that, for the duration of this proof, we will abstain from using subscripts in the usual way, to avoid confusion. For example, if the variable x has type Point, and is really u5, then x1 is v50 and x2 is v51. As a typographical convenience, if ℓ is a list of terms, we will write ℓn to mean the n-th member of the list. For example, when x is a variable of type Point, we will define below x◦ = (x1, x2); then x1◦ = x1, so there is no ambiguity with the use of subscripts on variables. When working in or with the theory EF or related theories, we use some common abbreviations: xy means x y for field elements. If x and y are lists of the same length, x y means the dot product· defined using the multiplication sign 2 · 2 2 of EF. x means x x. If t is a list of two terms (a 2-vector) then t = t1 + t2. · 2 2 | | Hence, if t is a term of type Point, t◦ means (t ) + (t ) . If t is ap term of | | 1◦ 2◦ EF+ then t means √t2, which means √t t. Ifpt and s are two terms of type | | · Point, t◦ + s◦ is the list (t1 + s1,t2 + s2); binary minus x y means x + ( y); − − t◦ s◦ means (t1 s1,t2 s2). Cross product x y is definable in EF on lists of − − − × length 2 as x1y2 x2y1 (so the result is a scalar). If t is a list (t1,... ,tn), then − t means the conjunction of the tk for k =1,... ,n. ↓We want to interpret ECG in the↓ field theory EF defined in Definition 8.2, and vice-versa; but there is a technical detail concerning the treatment of the constants α, β, and γ that must be dealt with without delay. In geometry these constants stand for “unspecified” noncollinear points. When interpreting geometry in field theory, we intend to interpret α as (0, 0) and β as (1, 0). It may seem natural to pick a third specific point, such as (0, 1), for the interpretation of γ. The problem with this is that it introduces “extra” or “accidental” theorems, since any specific point definable in EF will be definable in geometry too. If we want our interpretations to be faithful (the interpretation of φ is provable if and only if φ is provable), then we cannot pick a specific point as the interpretation of γ. The way around this difficulty is to augment EF with two new constants. At the same time we will solve another problem, namely that ECG does not 154 MICHAEL BEESON specify which point on a circle is given by pointOnCircle , while we must pick a particular point in the interpretation in EF. We can pick one arbitrarily, such as always picking the “northernmost” point, and the interpretation will sound, but then it will not be faithful, as when we go from ECG to EF and then back, we may not get back the point we started with. We fix this by introducing another arbitrary constant to serve as the x-coordinate of pointOnCircle (Circle (0, 1)). + Definition 12.15. The theory EF is EF with three additional constants γ1, 2 + + γ2, and γ3, and the axioms 0 = γ2 and γ3 1. “Weak EF ” is EF with Axiom EF1 replaced by Axiom EF7,6 positive elements≤ have reciprocals. “Playfair EF+” is EF+ with Axiom EF1 replaced by Axioms EF9 and EF10. Using EF+ instead of EF does not introduce any new theorems: Lemma 12.16. EF+ is conservative over EF. That is, every theorem of EF+ that does not contain the constants γ1 and γ2 is provable in EF.

Proof. Let φ be any formula and let π be a formal proof of φ. Replace γ1 and γ2 throughout π by 0 and 1, and γ3 by 0. The axiom γ2 = 0 becomes 0 = 1, 2 6 6 which is Axiom EF0. The axiom γ3 1 is an abbreviation for P (γ3 γ3 1); this becomes P (0 0 1), which is≤ provably equivalent to P¬( 1),· which− is provable in EF¬. Hence· − the result of the substitution is still a¬ formal− proof. If φ does not contain γ1 and γ2, then the last line of the proof after substitution is still φ. That completes the proof. In the interpretation we are about to define, points will be represented as pairs of field elements; we treat these pairs as lists of length two, (a,b). Lines will be represented as lists of four field elements, two for each of two points on the line. We write these lists as (a,b,c,d). Circles similarly will be represented as lists of three field elements, two for the coordinates of the center, and one for the radius.

Definition 12.17. The deﬁnition of t◦ for terms t of ECG will now be given.

x◦ is (x1, x2) for x a variable of type Point

L◦ is (L1,L2,L3,L4) for L a variable of type Line

C◦ is (C1, C2, C3 ) for C a variable of type Circle | | α◦ is (0, 0) a list with two elements

β◦ is (1, 0)

γ◦ is (γ0,γ1)

pointOn1 (L)◦ is (L1◦,L2◦)

pointOn2 (L)◦ is (L3◦,L4◦) Line (t,s)◦ is Append(t◦,s◦) In the last line, Append is list concatenation, and multiplication of a list by a field element means to multiply each member of the list by the element. The point of using Append and Nonzero is to ensure that Line (a,b) is undefined when a = b, or more generally when a b is not invertible, when working with weaker reciprocal axioms than EF1. With− these definitions, the direction of a line is recoverable from the order of the two associated points. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 155

A circle is represented by its center and radius:

Circle (a,b,c)◦ is (a◦,a◦, b◦ c◦ ) 1 2 | − | center (C)◦ is (C1◦, C2◦) The interpretation of pointOnCircle (C) is a point on C such that the radius vector from the center to that point makes an angle with the horizontal whose + cosine is γ3 (the third new constant introduced in EF ). This is formally expressed by

2 pointOnCircle (C)◦ is (C1◦ + γ3 C3◦ , C2◦ + 1 γ3 C3◦ ) · | | q − · | | + 2 Since EF includes the axiom γ3 1, we have pointOnCircle (C)◦ provable in EF+. ≤ ↓ Next we want to define the interpretation of IntersectLines (L,K). The idea is that from L◦ and K◦ we can get two point on each line, then compute the coordinates of the point of intersection, and the formulas giving those two coordinates are the members of the list IntersectLines (L,K)◦. Note that for this to work, there must be a single formula, i.e. no definitions by cases are allowed, e.g. according as the lines are vertical or not. This does work out, as the formula for the intersection turns out to have a denominator that is non zero exactly when the two lines are neither parallel nor coincident. To work this out, suppose that a = (L1◦,L2◦), b = (L3◦,L4◦), c = (K1◦,K2◦), and d = (K3◦,K4◦). Then solve for the intersection point x of the line through a and b with the line through b and c. One way to do that is to note that on (x, Line (a,b)) means (x a) (x b)=0, so we have that equation and (x c) (x d) = 0. Writing those− equations× − out by components and using Cramer’s− rule× to− solve them we find

a2 b2 b1 a1 D = − − c2 d2 a1 c1 − −

1 a2b1 a1b2 b1 a1 x1 = − − D c2d1 c1d2 d1 c1 − −

1 a2 b2 a2b1 a1b2 x2 = − − D c2 d2 c2d1 c1d2 − −

The denominator D of the solution is zero just when its rows are linearly depen- dent; but the rows are the vectors determining the lines K and L respectively, so the determinant is nonzero just when L and K are lines in different directions; hence, the terms of EF that gives the intersection points are defined exactly when lines L and K intersect. Hence, using the axiom EF1, lines that are neither parallel nor coincident will meet, verifying the strong parallel axiom; if in field theory we only know positive elements have reciprocals then we must be able to determine the sign of D; and in Playfair field theory, we must show that D is invertible before two lines can be proved to intersect. Some of the algebra needed to define IntersectLineCircle1 (L, C)◦ and IntersectLineCircle2 correctly has been done in Section 6.2, but we still need to pay attention to whether one can get a single term of EF that interprets IntersectLineCircle1 (L, C). 156 MICHAEL BEESON

To define IntersectLineCircle1 (L, C) and IntersectLineCircle2 (L, C), let a = (L1◦,L2◦) and b = (L3◦,L4◦), and let e = (C1◦, C2◦) be the center of C and r = C3◦ be the radius of C. Then the equations for the intersection points x are (x a) (x b)=0 (where is cross product) − × − × (x e)2 = r2 (squaring a vector means dot product) − We need to prove that these equations can be solved, and the solutions given by two terms defined without a case distinction. We look for the intersection points x in the form x = a + λ(b a) (so they will automatically lie on Line (a,b), and then the second equation− becomes a quadratic equation for λ: (λ(b a)+ a e)2 = r2 − − λ2(b a)2 + 2(a e) (b a)λ + (a e)2 r2 =0 − − · − − − This equation can be solved by the quadratic formula: (a e) (b a) ((a e) (b a))2 (b a)2((a e)2 r2) (1) λ = − · − ± − · − − − − − − p (b a)2 − and the denominator is zero just when b = a; hence just when the interpretation of Line (a,b) is undefined. We define the interpretation of IntersectLineCircle2 (L, C) to be the term given by taking the positive sign for the square root and IntersectLineCircle2 (L, C) to be the term given by taking the negative sign. Here we understand that in the above expressions for λ, we will also make the following substitutions:

e := (C1◦, C2◦)

c := (C3◦, C4◦)

a := (L1◦,L2◦)

b := (L3◦,L4◦) It remains to check that if p and q are the two points that interpret IntersectLineCircle1 (L, C) and IntersectLineCircle2 (L, C), respectively, then they occur on L in the same order as (a,b). We will come to that point after we have finished defining the interpretation. Our next task is to define the interpretations of IntersectCircles1 and IntersectCircles2 . Much of the required work has already been done in Sec- tion 6.2, where we saw that the equation for the intersection points is quadratic, so it can be solved using the quadratic formula (which is both expressible and provable in EF); that gives us two terms for the two intersection points, according as the plus or the minus sign is used in the quadratic formula. But do these two terms correspond exactly to IntersectLineCircle1 (C,K) and IntersectLineCircle2 (C,K)? And which solution of the equations serves as the interpretation of which term? The latter question is easy to answer for the circles C and K considered in Section 6.2, which have their centers at (0, 0) and (1, 0) respectively. In that case IntersectCircles1 (C,K) is the one with y-coordinate 0, i.e. where we take the negative square root to solve the last equation in that≤ section, and IntersectCircles2 (C,K) is the other solution, where we take the positive square root. To correctly define the interpretations of IntersectCircles1 FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 157 and IntersectCircles2 , we cannot assume the circles have their centers at (0, 0) and (1, 0). Let λ = b a , where a and b are the centers of C and K. Let r = (1/λ) times the radius| − of| C, and s = (1/λ) times the radius of b. Let e be a unit vector (b a)/ b a , which is defined since a = b. Let f be a unit vector orthogonal to e−such| that− | the cross product e f =6 1. Then solve the equations in Section 6.2 as is done there; if (x, y) solves those× equations then a+λxe+λyf is a point on both circles. The (term defining the) solution with y 0 is the interpretation of IntersectCircles2 (C,K), and the (term defining the)≥ solution with y 0 is the interpretation of IntersectCircles2 (C,K). Since≤ we have proved that IntersectCirclesSame and IntersectCirclesOpp are redundant, we do not need to interpret them; or putting the matter another way, their interpretations can be defined using their definitions in terms of the rest of ECG. Now we have completed the definition of a term t◦ of EF for every term t of ECG. Next we define a formula φ◦ of EF for each formula φ of ECG. If a and b are terms of type Point, then a◦ = b◦ means a◦ = b◦ a◦ = b◦, and 0 0 ∧ 1 1 a◦ = b◦ means (a◦ = b◦). Similarly, if t is a list of terms of EF, then t is the 6 ¬ ↓ conjunction of the tk for tk in the list t. The following predicate will be used in defining the interpretation↓ of lines. Definition 12.18.

(x) := (x1 = x3 x2 = x4) L ¬ ∧ For terms t of type Point or Circle, we deﬁne

(t )◦ is t◦ ↓ ↓ and for terms L of type Line, we deﬁne

(t )◦ is t (t) ↓ ↓ ∧L

(a = b)◦ is a◦ = b◦

B(a,b,c)◦ is a◦ = b◦ b◦ = c◦ c a = b a + c b 6 2∧ 6 ∧2 | − | | − | | − | E(a,b,c,d)◦ is (a b) = (c d) − − on(p,L)◦ is (p◦ (L1◦,L2◦)) ((L3◦,L4◦) (L1◦,L2◦))=0 − 2× − 2 On(p, C)◦ is (p◦ (C◦, C◦)) = ((C◦, C◦) (C◦, C◦)) − 1 2 3 4 − 1 2 We deﬁne for x a variable of type Point or Circle,

( x A)◦ is x◦A◦ ∀ ∀ ( x A)◦ is x◦A◦ ∃ ∃ where when we quantify over a list of variables, as in x◦, we mean to quantify over each of the variables separately. The interpretat∃ ion of quantification over variables of type Line has a twist. We need to ensure that Line (a,a) is undefined (as that is Axiom CA4); but as it stands, there is nothing to prevent (a0,a1,a0,a1) from being L◦ for a line L. We could try to rule that out by defining Nonzero (x) = x / x , which in EF would be defined and equal to 1 when | | | | x = 0, and then just multiply each element in the list defining Line (a,b)◦ by 6 158 MICHAEL BEESON

Nonzero (a b). That works in EF but it causes trouble when we work in field theories with− weaker reciprocal axioms, as Nonzero (x) is only defined when x is invertible, which may not be the same as x = 0. The solution to this problem 6 is to simply interpret variables of type Line to range over lists (a1,a2,b1,b2) in which we do not have a1 = b1 a2 = b2. To accomplish this we recall from Definition 12.18 that ∧

(x) := (x1 = x3 x2 = x4) L ¬ ∧ the condition just mentioned. Then we deﬁne for L a variable of type Line,

( L A)◦ is L◦( (L◦) A◦ ∀ ∀ L → ( L A)◦ is L◦( (L◦) A◦) ∃ ∃ L ∧ We write φ[x := t] for the result of substituting t for x in φ. If x is a list of variables and t is a list of terms, we use the same notation for simultaneous substitution. Lemma 12.19 (Substitution commutes with the interpretation). For each for- + mula A of ECG, with free variables x = x1,... ,xn, EF proves A(x)◦[x◦ : t◦] is equivalent to A[x := t]◦. Proof. By induction on the complexity of the formula A. We have deﬁned the interpretation so that the lemma is true for atomic formulas. For example, B(p,q,r)◦ is B(x,y,z)◦[(x,y,z) := (p◦, q◦, r◦)] for terms p, q, and r, and similarly for the other atomic sentences. Since both the interpretation and substitution commute with the propositional connectives, the induction steps corresponding to propositional operations are immediate. The two quantiﬁers are treated the same way; we write out the proof for . In the following calculation, each line is provably equivalent to the next in EF.∀ Here x is a variable of type Point or Circle and x◦ is a list of two or three variables as explained above.

( x A(x, y))◦[y◦ := t◦] ∀ x◦ (A(x, y))◦[y◦ := t◦] by deﬁnition of the interpretation ∀ x◦ (A(x, y)[y := t])◦ by the induction hypothesis ∀ ( x A(x, y))[y = t]◦ by deﬁnition of substitution ∀ If x is a variable of type Line, then we have to account for the use of . We have L

( x A(x, y))◦[y◦ := t◦] ∀ x◦ ( (x◦) (A(x, y))◦[y◦ := t◦]) by definition of the interpretation ∀ L → x◦ ( (x◦)[y◦ := t◦] (A(x, y)◦[y◦ := t◦]) since y◦ is not in the list x◦ ∀ L → x◦ ( (x◦) (A(x, y)◦[y◦ := t◦]) by definition of the interpretation ∀ L → x◦ ( (x◦) (A(x, y)[y := t])◦ by the induction hypothesis ∀ L → ( x A(x, y))[y = t]◦ by definition of substitution ∀ That completes the proof of the lemma. The following lemma will be needed below, and it also may help the reader understand the purpose of : L FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 159

Lemma 12.20. Let L be a variable of type Line. Then ECG proves (L ). L ◦ Proof (L ) (L1 = L3 L2 = L4) L ◦ ↔ ¬ ∧ (L = L L = L ) ↔ ¬ 1◦ 3◦ ∧ 2◦ 3◦

Now, for the k such that L is ℓk (the k-th variable of type Line), we have Lj◦ = v9k+j , for j = 1 to 4, by the definition of L◦. Then, by the definition of v9k+j (see Definition 12.47), we have

L1◦ = X(pointOn1 (L))

L2◦ = Y (pointOn1 (L))

L3◦ = X(pointOn2 (L))

L4◦ = Y (pointOn2 (L)) Putting this result into the previous calculation we have (L ) (X(pointOn1 (L)) = X(pointOn2 (L)) L ◦ ↔ ¬ Y (pointOn1 (L)) = Y (pointOn2 (L))) But ECG proves y = z X(y)= X(z) Y (y)= Y (z). Hence ↔ ∧ (L ) pointOn1 (L) = pointOn2 (L) L ◦ ↔ 6 But pointOn1 (L) = pointOn2 (L) is provable from Axioms CA1 and CA4, as 6 pointed out just after the statement of axiom CA7 on page 47. Hence (L◦) is provable in ECG. That completes the proof of the lemma. L

Theorem 12.21 (Interpretation of ECG in EF). (i) Suppose ECG proves φ. + Then EF proves φ◦. That is, ECG is interpreted by the theory of Euclidean fields (where nonzero elements have reciprocals). (ii) If the strong parallel axiom is replaced by Euclid 5, then EF1 can be replaced by axiom that positive elements have reciprocals (EF7). That is, Euclid 5 is interpreted by the theory of weakly Euclidean fields. (iii) If the strong parallel axiom is replaced by Playfair’s axiom, then the axiom that nonzero elements have reciprocals (EF1) can be replaced by axioms EF9 and EF10 (elements without reciprocals are zero, and elements greater than a positive invertible are invertible). That is, Playfair geometry is interpreted by the theory of Playfair fields. Remark. The issues about multiplicative inverse come up only in verifying that IntersectLines (L,K)◦ is defined when it needs to be. There are only two axioms requiring the intersection points of lines to be defined; those are Pasch’s axiom and the parallel postulates. We need Axioms EF9 and EF10 to verify Pasch’s axiom, and EF9 is needed only for Pasch’s axiom. Proof. By induction on the lengths of proofs in ECG. We first check the axioms of the logic of partial terms. There are the axioms c for c = α,β,γ. Since α◦ is (0, 0), α becomes 0 0 , which is provable in EF↓ by that same axiom of ↓ ↓ ∧ ↓ 160 MICHAEL BEESON

LPT. Now β◦ is (1, 0), so β becomes 1 0 , which is provable. The axiom ↓ ↓ ∧ ↓ + + γ becomes γ1 γ2 , which are axioms of EF since EF uses the logic of ↓ ↓ ∧ ↓ partia terms. Similarly, the interpretation of the axiom x is x◦ ; since x is a ↓ ↓ list of variables, this is a conjunction of formulas of the form xk , each of which is provable by this same axiom of LPT. ↓ We turn to the strictness axioms φ(p) p for atomic φ. Using abbreviations introduced above, this notation includes→ the↓ case where p is a list of several variables. Arguing in EF, suppose φ(p)◦. By Lemma 12.19, that is equivalent to φ(x)◦[x := p◦]. So what has to be proved is φ◦(t◦) t◦ . In other words → ↓ we must show that A[x := t] t whenever A is φ◦ for some φ. That leads us to ask, for what A we have A→[x :=↓ t] t . That property holds for atomic A that contain x, and holds for conjunctions→ if↓ it holds for at least one conjunct that actually contains x, as one easily checks. (Of course it does not hold in general, as the example A(x) := x shows, since we certainly do not have ¬ ↓ t t .) One can check by referring to the deﬁnitions of B◦, E◦, on ◦, and ¬ ↓ → ↓ On ◦ above that in each case, the interpretation of an atomic formula is either an equation containing all the variables, or a conjunction, one of whose conjuncts is an equation containing all the variables. That completes the veriﬁcation of the strictness axioms We consider the axiom x(t A(x) A(t); more formally we should write A[x := t] instead of A(t).∀ Consider↓ ∧ the case→ when x is a variable of type Point. The interpretation of the formula is

x(t◦ t◦ A(x)◦ A(t)◦). ∀ 1 ↓ ∧ 2 ↓ ∧ → Argue in EF as follows: suppose t◦ t◦ A(x)◦. Then by this same axiom, 1 ↓ ∧ 2 ↓ ∧ we have (A(x)◦)[x◦ := t◦]. Then by Lemma 12.19 we have A[x := t]◦, which is the desired conclusion. Now consider the axiom t A[x := t] x A. The interpretation of this is ↓ ∧ → ∃ t◦ A[x := t]◦ x◦A◦. Argue in EF as follows: Suppose t◦ A[x := t]◦. ↓ ∧ → ∃ ↓ ∧ By Lemma 12.19 we have A◦[x◦ := t◦]. Since t◦ , we can use this same axiom to ↓ conclude x◦ A◦, which is the desired conclusion. That completes the veriﬁcation that the logical∃ axioms of LPT are correctly interpreted. We now turn to the nonlogical axioms of ECG. First we take up the stability axioms S0 to S5. Each of these has the form φ φ, where φ is atomic, with main symbol =, E, on , On , or B. The¬¬ interpretation→ preserves logical operations, so we have to check that the interpretations of these are stable in EF. For t = s, E(a,b,c,d), on (p,L) and On (p, C), the interpretations are polynomial equations in t◦, s◦, a◦, etc. The interpretation of B(a,b,c) is a conjunction of polynomial equations and inequations. All these are stable. That completes the veriﬁcation of the stability axioms. Next we turn to the incidence axioms. Consider Axiom I1, on (a, Line (a,b)). The interpretation of the term Line (a,b) is the list (a1◦,a2◦,b1◦,b2◦). So the interpretation of on (a, Line (a,b)) is

(a◦ (a◦,a◦)) ((b◦,b◦) (a◦,a◦))=0 − 1 2 × 1 2 − 1 2 Since a◦ = (a1◦,a2◦), the ﬁrst factor in the cross product is zero; since polynomial identities can be veriﬁed in EF, this is provable in EF. Axiom I2, which is FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 161 on (b, Line (a,b)) is treated similarly–this time the second factor in the cross product is zero. Axiom I3 is on (p,L) on (q,L) on (r, Line (p, q)) on (r, L). For notational ∧ ∧ → convenience let a = (L1◦,L2◦) and b = L3◦,L4◦, so that a and b are two distinct points on L◦ (since (L◦)). Then, assuming the of the hypothesis of Axiom L ◦ I3, we have p = q (since Line ◦(p, q) ), and 6 ↓ 0 = (q◦ a) (q◦ b) since on (q,L)◦ − × − 0 = (p◦ a) (p◦ b) since on (p,L)◦ − × − 0 = (r◦ p◦) (r◦ q◦) since on (r, Line (p, q))◦ − × − and we must prove (r◦ a) (r◦ b) = 0. The above equations are invariant under translation, and cross− product× − is invariant under rotations about the origin. Therefore we can assume without loss of generality that a = (0, 0) and b = (b1, 0). Then we have (dropping the superscript ◦ for simplicity) 0 = (q a) (q b) − × − = q (q b) × − = q1q2 q2(q1 b1) − − = q2b1

But b1 = 0 since a = b. Hence q2 = 0, i.e q lies on the x-axis. Similarly from 6 6 the second equation in the hypothesis we obtain p2 = 0. Now we have from the third equation, 0 = (r p) (r q) − × − = (r1 p1)r2 r2(r1 q1) − − − = (q1 p1)r2 − But q1 = p1 since p = q and both p and q lie on the x-axis. Hence r2 = 0; that is r also6 lies on the x6 -axis. Now we calculate (r a) (r b)= r (r b) − × − × − = r1(r2 b2) r2(r1 b1) − − − =0 since r2 = 0, b2 = 0, and r2 =0 That completes the verification of Axiom I3. Axiom I4 is p = IntersectLines (L,K) on (p,L) on (p,K). We have sketched the proof that it is correctly interpreted→ when∧ we explained how to calculate IntersectLines (L,K)◦; the complete calculation will not be given here. Axiom I5 is IntersectLines (L,K) ∼= IntersectLines (K,L). The interpretations of the two sides are the solution of the same two equations, given in the opposite order; so they agree. Axiom I6 says that if p = IntersectLineCircle1 (L, C), then p is on both L and C; and Axiom I7 says the same for IntersectLineCircle2 . Since the interpretations of IntersectLineCircle1 and IntersectLineCircle2 are defined by solving the the equations for L◦ and C◦, verifying this axiom amounts to proving the quadratic formula in EF, which is straightforward. Axiom I8 and Axiom I9 similarly treat the intersection points of two circles; again the interpretations of the terms give solutions of equations that say the 162 MICHAEL BEESON points are on the circles, so by the quadratic formula in EF, the interpretations of these axioms can be verified. Axiom I10 and Axiom I11 have been shown to be redundant in Theorem 9.96. Therefore we do not have to verify them. Axiom I12 says that if L and K do not coincide (because p is on both of them) then their intersection point is defined. As remarked in the definition of IntersectLines (L,K)◦, the term giving the solution has a denominator that is nonzero just when L and K do not coincide and are not parallel. Since terms with zero denominator are undefined in EF, since if a/0 , then 0 a/0 would be both 1 and 0, but 1 = 0 by Axiom EF0. ↓ · Axiom I13 says that6 if IntersectLines (L,K) , then the only intersection point of L and K is the one given by that term. As just↓ remarked, if the interpretation of that term is defined, then the lines (specified by) L◦ and K◦ do not coincide and are not parallel, so there is only that solution. Axiom I14 through I17 say that if lines and circles intersect, or circles and circles intersect, then the terms giving the intersection points are defined. Under the interpretation, that means that the equations for the intersection points have solutions given by the terms we have defined to be the interpretations of IntersectLineCircle1 (L, C), etc. Again, this is just verifying in EF that the solution terms do indeed solve the equations. Axiom I17 and Axiom I18 specify that if the terms for the intersection points of circles are defined, then the circles are not concentric. In Theorem 6.1 we derived a formula for the intersection points, but we began by assuming the circles were not concentric and making a linear transformation to bring the centers to (0, 0) and (1, 0). The formula for that transformation, which will be part of the interpretation of IntersectCircles1 (C,K)◦ and IntersectCircles2 (C,K)◦, will have a denominator that is zero when the circles are concentric. That will result in IntersectCircles1 (C,K)◦ and IntersectCircles2 (C,K)◦ being undefined when (C1◦, C2◦) = (K1◦,K2◦). That completes the verification of Axioms I17 and I18. Axioms I19 and I20 concern IntersectCirclesSame , which has been shown redundant; so we do not have to verify these axioms. Axiom I21 says that if IntersectLines (L,K) , then K and L do not coincide. ↓ We have already remarked that the denominator will be zero if K◦ and L◦ coincide, which verifies Axiom I21. We now turn to the constructor and accessor axioms. The verifications of these are just a matter of checking that the bookkeeping is correct; lines and circles are represented by lists of four field elements in a consistent way. Consider Axiom CA1,

Line (pointOn1 (L), pointOn2 (L)) = L.

Here L◦ is a list of four elements (L1◦,L2◦,L3◦,L4◦), of which the ﬁrst two represent one point on L, and the second two represent another point. The interpretation of pointOn1 (L) will be pointOn1 (L)◦ = (L1◦,L2◦), and the interpretation of pointOn2 (L) will be pointOn2 (L)◦ = (L3◦,L4◦). Then the interpretation of the left hand side Line (pointOn1 (L), pointOn2 (L)) concatenates those two lists, producing L◦. That veriﬁes Axiom CA1. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 163

Axiom CA2 says that if a = b then pointOn1 (Line (a,b)) = a. If a◦ = (a◦,a◦) 6 1 2 and b◦ = (b1◦,b2◦) then Line (a,b)◦ = (a1◦,a2◦,b1◦,b2◦) and

pointOn1 (Line (a,b))◦ = (a1◦,a2◦)= a◦.

The hypothesis a = b tells us that (Line (a,b)◦). Axiom CA3 is verified similarly, with pointOn26 . L Axiom CA4 says that Line (a,a) is not defined. According to the definition of (t↓)◦, the interpretation of (Line (a,a) )◦ is (equivalent to) ↓ (Line (a,a)◦) a◦ . L ∧ ↓ But Line (a,a)◦ is the list (a1◦,a2◦,a1◦,a2◦), which does not satisfy (see Definition 12.18). That completes the verification of Axiom CA4. L Axiom CA5 is center (Circle3 (a,b,c)) = a. We have

Circle3 (a,b,c)◦ = (a1◦,a2◦,d,e) where the terms d and e do not matter here. Then

center (Circle3 (a,b,c))◦ = (a1◦,a2◦)= a◦. That completes the veriﬁcation of Axiom CA5. Axiom CA6 is Circle (center (C), pointOnCircle (C)) = C. By deﬁnition,

2 C◦ = (C1◦, C2◦, (C3◦) ) q 2 pointOnCircle (C)◦ = (C1◦ + γ3 C3◦ , C2◦ + 1 γ3 C3◦ ) · | | q − · | | center (C) = (C1◦, C2◦) Circle (center (C), pointOnCircle (C)) is an abbreviation for Circle3 (center (C), center (C), pointOnCircle (C)).

So its interpretation is (C1◦, C2◦, d), where

2 2 2 d = (C1◦ + γ3 C3◦ C1◦) + (C2◦ + 1 γ3 C3◦ C2◦) r · | |− q − | |−

2 2 2 2 = γ3 C3◦ + ( 1 γ3 C3◦ ) r | | q − | | = C◦ | 3 | 2 = (C3◦) officially, since x is just an abbreviation q | | Hence Circle (center (C), pointOnCircle (C))◦ = C◦ as desired. That completes the verification of Axiom CA6. Consider Axiom CA7, which says Circle3 (A, B, C) . Note that the term for ↓ Circle3 ◦ includes a square root (to calculate the radius); but the argument of the square root is a sum of squares, so it can be proved non-negative in EF. To deal with circles of zero radius, we do need to prove closure of the non-negative elements (those whose additive inverses are not positive) under sum and product, but that is easy. Now we come to the betweenness axioms. We have defined the interpretation of B(a,b,c) to be an algebraic formula expressing that the distance from a◦ to c◦ is the sum of the distances from a◦ to b◦ and from b◦ to c◦, and that a◦ = b◦ and 6 164 MICHAEL BEESON b◦ = c◦. (Without these last two conditions we would get non-strict betweenness.)6 These two conditions make Axioms B1-ii and B1-iii automatically valid, and Axiom B1-iv (symmetry of B) reduces to the commutativity of addition. Axiom B1-i requires us to check that this condition on the distances can only be satisfied when b is on the line connecting a and c. We can check that algebraically directly, or we can first derive the triangle inequality in EF and use that. The details are of approximately equal length, so we do the direct verification; for simplicity we drop the superscripts, writing a instead of a . We will prove that for a, b, and c distinct pairs of field elements (thought of as◦ 2-vectors), a c a b + b c | − | ≤ | − | | − | with equality only when b is on the line connecting a and c. It suffices by a linear change of variables to prove it when a = 0. Then it becomes c b + b c . We have the following reversible calculation (in which b2 means| | the ≤ |dot| product| − | of vector b with itself, etc.): c b + b c | | ≤ | | | − | c2 b2 +2b (b c)+ b c 2 ≤ · − | − | = b2 +2b2 2b c + c2 + b2 2b c + c2 − · − · = c2 +4b2 4b c + c2 − · = c2 + (2b c)2 − 0 (2b c)2 ≤ − This calculation can be checked in EF from the bottom up. That verifies Axiom B1-i. Axiom B2 says that the point calculated by the Midpoint (a,b) script, when interpreted, lies between a◦ and b◦. Let p be that midpoint; then to verify Axiom B1 we need to check that

a◦ p◦ + b◦ p◦ = a◦ b◦ . | − | | − | | − | That follows if p◦ is indeed the midpoint of segment a◦b◦. To verify this we make the calculation one would make in ordinary analytic geometry, and check that we used only the axioms of EF to do it. In particular, the square roots one needs to solve for the intersection points exist, because what is under them is nonnegative. We omit the details. We turn to Axiom B3, which says that if a, b, and c are distinct points on a line, and two of the three possible betweenness relations among them fail, then the third one cannot fail, i.e. it not not holds. Since equality is stable in EF we might as well say that the third one holds. Suppose then (dropping the superscripts for convenience) we have B◦(a,b,c) and B◦(b,c,a). Then we ¬ ¬ must show that B◦(c,a,b). That means that we have c a = b a + c b | − | 6 | − | | − | b a = c b + a c | − | 6 | − | | − | and we must show c b = c a + a b | − | | − | | − | FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 165

Since one can prove in EF that linear transformations preserve distance, we can make a linear transformation that brings the line to the x-axis, and point a to zero. Then we can assume that b, and c are scalars rather than vectors. Since equality is stable we can argue by cases on the possible order relations of b and c. Our hypotheses are now c = b + c b and b = c b + c , and our desired conclusion is c b = c +| b|. 6 That| | | will− be| true| | if 6 b|and− |c have| | opposite signs, i.e. not both are| − nonnegative| | | | | and not both are nonpositive. But the hypotheses tell us that b and c both do not have opposite signs to the same expression; hence they do have opposite signs to each other. The other clauses in Axiom B3 are treated similarly. That completes the veriﬁcation of Axiom B3. Before going on to the next axiom, we note that the interpretation of B(a,b,c), which is a c = a b + b c , implies in EF that a, b, and c are collinear, in the usual| − algebraic| | − sense| | of− satisfying| a linear equation, which can be written without a case distinction (as to to whether the line is vertical or not) by saying that the cross product of (a b) (b c) is zero. If we know that the direction of the line is not vertical then− the× usual− form of a linear equation applies. Then one can prove in EF that the interpretation of B(a,b,c) is equivalent to the conjunction of the two coordinate inequalties (for i =1, 2)

((a◦

(2) z◦ := IntersectLines (Line (p,b), Line (a, q))◦. We must show that z is deﬁned, i.e. that using the axioms of EF (or the ◦ restricted axioms for reciprocals) we can calculate the coordinates of z◦. We may simplify the algebra by ﬁrst applying a linear transformation to bring the points to some standard position, since linear transformations preserve B◦ (even though they may not preserve distance). By a translation, we can bring a◦ to (0, 0); then by the transformation with the matrix

p1 p2 p2 p1 − 2 2 we can bring p◦ to the point (p1 + p2, 0) on the positive x-axis. Note that no division was required so far. Then c lies on the x-axis. We continue to use only the axioms of Playfair ﬁeld theory, i.e., we cannot just divide by any nonzero element, but only by one known to be invertible. Since p is the intersection point of Line (a,c) and Line (b,p), and these lines do not coincide since b does not lie on Line (a,c), the cross product (b p) (a c) is − × − invertible. Since p2 = 0 and a = (0, 0) and c2 = 0 we have (b p) (a c)= b2c1, − × − so b2c1 is invertible. Generally if xy is invertible then x and y are invertible, since if (xy)z = 1 then x(yz) = 1 and y(xz) = 1. Hence b2 is invertible. Then the linear transformation with matrix

b2 b1 + p1/b2 − 0 1 166 MICHAEL BEESON is deﬁned; it ﬁxes the x-axis and brings the point b directly over point p, as illustrated in Fig. 46.

Figure 46. Veriﬁcation of Axiom B4 (Explicit Inner Pasch)

b b

b q

b b b a = (0, 0) p c

Now recall the formula for for

z = IntersectLines (Line (p,b), Line (a, q))◦. Its denominator in this case is D = (p b) (a q) − × − = b2 q1 since p2 = a2 = 0 and b1 = p1 · The desired intersection point z◦ will be defined if and only D is invertible. Since b2 is invertible and D = b2 q1, it suffices to show that q1 is invertible. Note that (b p) (a ·p) is invertible, since it is the denominator of − × − IntersectLines (Line (b,p), Line (a,p))◦ and this is term is equal to p, so it is defined. But (b p) (a p)= b2p1, and − × − − b2 is invertible. Therefore p1 is a quotient of invertibles, hence it is invertible. But q1 >b1 since q is between b and c and c1 >p1 = b1. Hence q1 >p1 = p1 . | | By Axiom EF10, anything greater than a positive invertible is invertible, so q1 is invertible. That completes the proof that z◦ is defined. It remains to verify the conclusion of the axiom, namely B(p,z,b) B(q,z,a). ∧ Since a◦ = (0, 0) the collinearity of a◦, z◦, and q◦ is expressed by z◦ q◦ = 0. × That is, z◦q◦ z◦q◦ = 0. But z◦ = 1, so q◦ = z◦q◦. Then (dropping the super- 2 1 − 1 2 1 2 2 1 script ◦ for readability), we have by the above remarks that the interpretation of B(q,z,a) is q1 > 1. (That can also be derived by a six-line calculation directly from the definition.) The interpretation of the hypothesis B(a,p,c) tells us that FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 167 c1 > 1, since a1 = 0 and b1 = 1; and then since b1 = 1 and c1 > 1, the hypothesis B(b,q,c) tells us 1 < q1 < c1. Hence the interpretation of the conclusion B(p,z,q) is verified. The interpretation of the conclusion B(p,z,b) is equivalent, by the above remarks, to 0 < z2 < b2 . We have z2 < q2 since B(a,z,q) has | | | | | | | | been verified, and q2 < b2 since B(b,q,c) is a hypothesis. Hence z1 < b2 as required to verify B| (p,z,b| | ).| That completes the verification of Axiom| | B4.| | We turn now to the handedness axioms. Note that Right (a,b,c) implies a = b, as does Left (a,b,c), since it is defined using the intersection points of circles with6 centers a and b, and Axiom I17 and A18 guarantee that these intersection points are not defined for concentric circles; this observation will used below without repetition. Although Right and Left are not official predicate symbols of ECG, it will still be convenient to give them interpretations. The concept “ABC is a left turn” can be defined as usual in computer graphics, by the cross product. Specifically the cross product is (c1 b1)(a2 b2) (c2 − − − − b2)(a1 b1), where a = (a1,a2) and b = (b1,b2). Note that no division is required. The cross− product is defined even when a = b = c, while Right (a,b,c) should be undefined when a = b. We therefore define Right ◦(a,b,c) to mean that a = b and the cross product (a b) (c b) is non-negative. We have already discussed6 this idea when giving the− interpretation× − of IntersectCircles1 and IntersectCircles2 . With this definition of Right ◦ and Left ◦, we claim that the interpretation of Definition 8.8 is provable. (That definition would be taken as an axiom, if Right and Left were taken as an official part of the theory.) Arguing in EF (i.e., using analytic geometry), let C be a circle with center a, and K be a circle with center b = a. Let 6 p = IntersectCircles1 ◦(C,K)

q = IntersectCircles2 ◦(C,K)

We must show, using the axioms of EF, that bap is a right turn (as defined using cross product) and baq is a left turn. To do this we recall the definition of IntersectCircles1 (C,K). That definition says that with λ = b a , and e and f a pair of orthogonal unit vectors with e = (b a)/ b |a −and| e f = 1, we can put p and q in the form a + λxeλyf,− and the| − solution| × with y 0 is IntersectCircles2 ◦(C,K), that is q; and the one with y 0 is ≥ ≤ IntersectCircles1 ◦(C,K), that is q. Thus

q = a + λxe + λyf with y 0 ≥

p = a + λxe + λyf with y 0. ≤ Now to verify the interpretation of Definition 8.8, we must check that abp is a right turn, that is, (a b) (p b) 0; and also that abq is a left turn, that is, (a b) (q b)− 0.× If we− make≥ a linear transformation of positive determinant,− then× the− signs≤ of cross products are preserved; so we can make a linear transformation with positive determinant that brings a to (0, 0) and b to (1, 0); that makes e = (1, 0), f = (0, 1), and λ = 1. Then we have q = (x, y) 168 MICHAEL BEESON with y 0 and p = (x, y) with y 0. We calculate ≥ ≤ (a b) (p b) = ((0, 0) (1, 0)) ((x, y) (1, 0)) − × − − × − = ( 1, 0) (x 1,y) − × − = y − 0 since y 0 ≥ ≤ Hence abp is a right turn. Similarly (a b) (q b)= y 0 since for q, y 0; so abq is a left turn. That completes the− verification× − that− Definitio≤ n 8.8 is correctly≥ interpreted; or to put it another way, that the definitions of IntersectCircles1 ◦ and IntersectCircles2 ◦ work as intended. Having verified Definition 8.8, we may use that result in verifying the handedness axioms; in other words, we can proceed as if Right and Left were symbols of the language. By definition, the interpretation of on (x, Line (a,b)) is just the usual analytic-geometry condition for x to lie on the line connecting a and b (formulated in terms of cross product, but equivalent to the formulation with linear equations if we know the line is not vertical). Linear transformations with positive determinant preserve incidence, betweenness, and the signs of cross products, as well as collinearity; and the handedness axioms H1-H12 mention only those predicates (i.e., there are no occurrences of the equidistance relation E). Hence, in verifying them, we may first apply a linear transformation of positive determinant to bring the points into a more convenient position. For example, consider Axiom H1. Applying a linear transformation, we may assume a and b lie on the X-axis, indeed we may assume a = (0, 0) and b = (1, 0); and the further hypotheses are that p and q do not lie on the x-axis, and abp and abq are both right turns. We must show that no point on the x-axis is between p and q. Let p = (x, y). We have (a b) (p b)= y as above, so since abp is a right turn, we have y 0. Similarly− if×q =− (u, v),− we have (a b) (q b)= v, so ≤ − × − − v 0. If z is a point between p and q, then we must have z2 < 0, so z does not lie≤ on the x-axis. That verifies Axiom H1. Axiom H2 is similarly, but with Left instead of Right , so we have y 0 and v 0, and again z cannot lie on the x-axis. Axiom H3 has Right (a,b,p) Left≥ (a,b,q≥), so when we make the same calculation, we find y 0 and v 0. Hence∧ the line between p and q does cross the x-axis, so the intersection≤ of≥ the x-axis (which is the interpretation of Line (a,b)) meets the interpretation of Line (p, q). That verifies Axiom H3. Turning to Axiom H4, again we apply a linear transformation of positive determinant to bring a to (0, 0) and b to (1, 0). As before, the hyptothesis that Right (a,b,p) is equivalent to p2 0, and the desired conclusion Right (a,b,q) ≤ is equivalent to q2 0. By hypothesis there is a point on the x-axis between ≤ p and y; hence y2 > 0 and p2 < 0. From the interpretation of B(q,z,y) ∧ on (z, Line (a,b)) we have z on the x-axis and since y2 > 0, we have q2 < 0; hence Right (a,b,q), which is the desired conclusion. That verifies Axiom H4. Axiom H5 is similar with Left instead of Right ; the signs are reversed, so we find Left (a,b,q) in the end. Turning to Axiom H6, again we apply the same linear transformation; so the hypothesis Right (a,b,p) tells us that p2 0; the hypothesis B(p,q,x) ≤ ∧ FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 169 on (x, Line (a,b)) tells us that q2 > 0, which we have seen above implies Left (a,b,q). That verifies Axiom H6. H7 is similar with Right and Left interchanged. Axioms H8 through H11 concern turns in a triangle. By means of a linear transformation with positive determinant, we may assume that a = (0, 0), b = (1, 0), and c lies on the y-axis; but we do not know anything about the sign of c2 (it may even be zero). In this situation, as we have proved above, Right (a,b,c) is equivalent to c2 0, and Left (a,b,c) to c2 0. Let us verify Axiom H8. The ≤ ≥ hypothesis tells us c2 0 and b = c. To check that Right (b,c,a), we compute the cross product ≤ 6

(b c) (a c)=(1 c1, c2) ( c1, c2) − × − − − × − − = (1 c1)( c2) ( c2)( c1) − − − − − = c2 + c1c2 c2c2 − − = c2 − 0 since c2 0 by hypothesis ≥ ≤ Hence Right (b,c,a) as required. That verifies Axiom H8. Axioms H9, H10, and H11 are verified similarly. Axiom H12 is Left (α,β,γ). We have specified α◦ = (0, 0), β◦ = (1, 0), and γ◦ = (γ1,γ2). We compute

((α β) (γ β))◦ = ((0, 0) (1, 0)) ((γ1,γ2) (1, 0)) − × − − × − = ( 1, 0) (γ1 1,γ2) − × − = γ2 − 0 since γ2 > 0 is an axiom ≤ Hence Left (α,β,γ) as desired. That completes the verification Axiom H12, which is the last handedness axiom. The lower dimension axioms D1 through D4 are immediate, since we have chosen α◦,β◦ to lie on the x-axis and be distinct, γ◦ does not lie on the x-axis since γ2 > 0. Next we verify the upper dimension Axiom D5, which says that if three distinct circles have a common chord, their centers are collinear. Let three circles with distinct centers p, q, and r have the common chord ab. Make a linear transformation that brings the chord to the y axis with its midpoint at the origin; we may also assume the endpoints are a = (0, 1) and b = (0, 1). Points equidistant from a and b must lie on the x-axis, and hence the centers− of the circles are collinear. Note that linear transformations take vectors with zero cross product into vectors with zero cross product, so collinearity in the cross product sense used to define on ◦ is preserved. That completes the verification of Axiom D5. We now turn to the congruence axioms. Axiom C1 says that we can lay off segment cd along segment ba, by taking the point q = IntersectLineCircle2 (Line (a,b), Circle3 (a,c,d)). The possibility c = d is allowed. We have to prove that the conclusions aq = cd and B(x,a,b) are correctly interpreted. Of course the interpretation of aq = cd ¬ is immediate, since q◦ lies on the circle of radius c◦ d◦ about a. We have to | − | 170 MICHAEL BEESON prove (dropping the superscript ◦) that (x = a a = b x b = x a + b a ) ¬ 6 ∧ 6 ∧ | − | | − | | − | Suppose x = a and a = b and x b = x a + b a ; we must derive a 6 6 | − | | − | | − | contradiction. By definition of IntersectLineCircle2 ◦, we know that for z not on Line (a,b), zab is a right turn exactly when zpq is a right turn, where p = IntersectLineCircle1 (Line (a,b), Circle3 (a,c,d)). That is, the signs of (z a) (z b) and (z p) (z q) are the same. Again we make a linear transformation− × − that brings−a to× (0,−0) and b to (1, 0). With i = (0, 1) we have Right ◦( i,a,b). Hence Right ( i,p,q). The two intersection points of the circle with Line− (a,b) are ( c d ,−0); therefore p = ( c d , 0) ±| − | −| − | and q = ( c d , 0). Therefore q1 0. But B◦(x,a,b) means x2 < 0, as is | − | ≥ easily checked. Hence B◦(q,a,b) as desired. That completes the verification of Axiom C1. ¬ Axioms C2, C3, and C4 mention only the equidistance relation, and their interpretations are immediate consequences of the definition of x; for example consider C4, which says ab = ba. Its interpretation is a b =| b a , which is immediate from the definition and the ring laws. We| − say| no| mo− re| about these three axioms. Consider Axiom C5, which defines incidence for circles. The interpretation of On (x, Circle3 (a,b,c)) is x◦ a◦ = b◦ c◦, which is exactly the interpretation of ax = bc. That verifies| Axiom− − C5. | | − Axiom C6 is Tarski’s 5-segment axiom. Here we have four points a, b, c, d in the configuration illustrated in Fig. 8, and four more points A, B, C, D, in a similar configuration, with the segments shown as solid lines pairwise congruent, i.e. ab = AB, ad = AD, bd = BD, and bc = BC. The conclusion is that dc = DC. To verify this, we make a linear transformation T that takes (a,b) onto (A, B). Since ab = AB, this can be done by taking T as the composition of a translation and a rotation, so T has determinant 1 and preserves distances. Since T preserves distances, we have A Tc = a c = A C and B Tc = b c = B C. Therefore Tc lies on| the− circle| | of− radius| | AC− about| |A −and the| |circle− | of radius| − BC about B. Since a, b, and c are collinear, and T is linear, A, B, and C are collinear; two non-concentric circles with an intersection point on the line joining the centers have only one intersection point (as is easily proved in analytic geometry, i.e. in EF); therefore Tc = C. Similarly, T d lies on the circle of radius AD about A (since AD = ad) and on the circle of radius BD about B (since BD = bd). Let D′ be the reflection of D in the line joining A and B. Then (T d = D T d = D′). Since we are trying to prove an equality, namely D C¬¬= d c ,∨ and equality is stable, we can argue by cases. If T d = D then | − | | − | we are finished, as we have D C T d C = d c . If T d = D′ then we have | − || − | | − | D C = D′ C since C lies on the axis of the reflection that takes D to D′, | − | − and D C = D′ C = T d C = d c . That completes the verification of Axiom| C6.− | | − | | − | | − | For readers who are not trained in formal logic, we offer the following remarks. The “proof” we have just given that Axiom C6 is verified, like the verifications of the other axioms above, should be regarded as a high-level description of an algebraic calculation explicitly representing the components of the points and the FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 171 transformation T . In turn that algebraic calculation would technically need to be formalized in EF, producing a long list of formulas constituting a formal proof. If we wrote out explicitly either the calculation or the formal proof, it would be difficult for a human to follow, so we prefer to give the high-level description. Now we turn to the line-circle continuity axioms. In Definition 8.11, we have defined the concept“inside” in ECG. That definition makes it evident that the interpretation of “x is inside Circle (a,b)” is x◦ a◦ a◦ b◦. Please refer to the definition of the interpretation, where| − we defined| ≤ | terms− IntersectLineCircle1 ◦ and IntersectLineCircle2 ◦ of EF; see especially equation (1). We proved that if line L◦ meets C◦, then IntersectLineCircle1 ◦(L◦, C◦) and IntersectLineCircle2 ◦(L◦, C◦) are defined and lie on line L◦ and circle C◦. But that is just what the interpretation of Axiom Cont 1 says; so we have already verified Axiom Cont 1, at least using the strong parallel axiom. Now we discuss the verification of that axiom using only Playfair field theory. The question is whether the denominator in (1) is invertible. It is (a b) (a b) where the line that should intersect the circle is Line (a,b). We only− know· that− a = b, not that a b 2 is invertible. This problem is easily remedied. By Axiom6 EF10, it will | − | suffice if we replace a and b by two points a′ and b′ such that a′b′ > 01, i.e. the distance between a′ and b′ is more than 1. That can be done, for example, by taking a′ = a and

b′ = IntersectLineCircle2 (Line (a,b), Circle3 (b, 0, 1)).

Then B(a,b,b′) and bb′ = 01. Since Line (a,b′) coincides with Line (a,b), it is enough to verify that IntersectLineCircle1 (Line (a,b′), C)◦ is deﬁned, since by Lemma 9.90 we have

IntersectLineCircle1 (Line (a,b′), C)= IntersectLineCircle1 (Line (a,b), C).

But (b′)◦ a◦ is invertible, by Axiom EF10, since it is > 1 and 1 is invertible. |IntersectLineCircle2− | is treated similarly. That completes the verification of Axiom Cont 1. Axiom Cont 2 says that if L has a point strictly inside C, then the two intersection points are distinct. To prove this, suppose that the two intersection points are not distinct. Refer to the definition of IntersectLineCircle1 ◦ and IntersectLineCircle2 ◦; then the expression under the square root in the solution for λ is zero: (3) ((a e) (b a))2 (b a)2((a e)2 (e c)2)=0 − · − − − − − − In this formula, e is the center of the circle, a and b are the points determining line L, and c is a point on C (given by the third and fourth coordinates of the interpretation of the circle). Technically, all these letters should have superscript ◦, since we are working in EF, but we omit the superscripts. We have to show, using analytic geometry, that this formula implies that line L is tangent to circle C. One can do this in a natural and direct way using the expression for a dot product in terms of the cosine of the included angle; but trigonometry is not available in EF, at least not directly, so that proof is not legal here. It is also not necessary. In EF, the quadratic formula tells us that there are at most two points of intersection; and since we have already verified Cont 1, both the terms giving points of intersection are defined, so there is at least one. Then 172 MICHAEL BEESON all we have to do is verify in EF that if point (t,s) on L lies inside C, then the two intersection points of L and C are distinct. Linear transformations preserve circles and collinearity, so we may assume the center of the circle is at the origin and the line is horizontal. The equation of C then is x2 + y2 = r2 for some r, and the equation of L is y = c for some c. Since (t,s) is on L, we have s c. Since (t,s) is inside C we have t2 + c2 < r2. The intersection points of the− line with C are the solutions of x2 + c2 = r2, i.e. x = √r2 c2. If these are equal then r2 = c2. Then from t2 + c2 < r2 we have t2 <± 0; but− squares are positive, so this is a contradiction. Hence the two intersection points are not equal, which was what we had to prove. That completes the verification of Axiom Cont 2. Axiom Cont 3 says that IntersectLineCircle1 (L, C) and IntersectLineCircle2 (L, C) depend extensionally on the circle C, i.e. if two circles C and K have the same points on them, then IntersectLineCircle1 (L, C)= IntersectLineCircle1 (L,K) and IntersectLineCircle1 (L, C)= IntersectLineCircle1 (L,K) Suppose the interpretation of “C and K have the same points on them” holds. Then C◦ and K◦ have the same points on them, i.e. the same coordinate pairs satisfy the equation of C◦ and the equation of K◦. The equation of C◦ is 2 2 2 2 (x C◦) + (y C◦) = (C◦ C◦) + (C◦ C◦) − 1 − 2 3 − 1 4 − 2 according to the definition of On ◦. But according to the definition of C◦, we will have C◦ = K◦, since the centers will be the same, and the third and fourth coordinates are determined by the center and radius (they specify a point on the circle due “north” of the center). Hence the equations of C◦ and K◦ are the same. That completes the verification of Axiom Cont 3. We turn to Axioms Cont 4 and Cont 5, which say that the order of the two intersection points on line L = Line (a,b) is the same as the order of a and b, in the sense that, if p = IntersectLineCircle1 (L, C) and q = IntersectLineCircle2 (L, C), then for each point z not on L, zpq is a right (or left) turn if zab is a right (or left, respectively) turn. Suppose that z is not on L, and zab is a right turn, i.e. Right ◦(a,z,b). We must show that zpq is a right turn. Since Right is stable (it is defined by an equality), we can argue by cases. If p = q then zpq is automatically a right (and a left) turn, so there is nothing to prove. Therefore we can assume p = q. Hence C is not a degenerate circle. By the handedness axioms, which we have6 already verified in EF, we can change z to any convenient point on the same side of L as the original z. A convenient point is one of the two intersections of the diameter M of C that is perpendicular to L. Also by the handedness axioms we can change a and b to any other points a′ and b′ on the line in the same order as the original a and b, in the sense that the points are moved one at a time along L without crossing. Let us use “west” and “east” to refer to directions on L, with a originally west of b. Then we can slide a west of C, and b east of C, and then slide a and b back towards the center of e until they intersect C, thus defining a′ and b′; and we have Right ◦(z,a′,b′) by the handedness axioms. Now a′ and b′ coincide, in some order, with p and q. But looking at the definition of IntersectLineCircle1 (L, C)◦, we find that p = a λ(b a) and q = a + λ(b a), − − − FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 173 where λ 0 because it is given by a square root. Hence p is the intersection ≥ point to the “west” and q is the intersection point closest to the “east”. Hence a′ coincides with p and b′ coincides with q. Hence Right ◦(z,p,q) as desired. That completes the verification of Axioms Cont 4. Axiom Cont 5 is verified in the same way, replacing Right ◦ by Left ◦. Next we turn to the circle-circle continuity axioms. Axiom Cont 6, illustrated in Fig. 12, says that if circle K has a point p inside circle C and a point q outside circle C, then both intersection points IntersectCircles1 (C,K) and IntersectCircles2 (C,K) are defined. (An incidence axiom that we have already verified says that they lie on both C and K.) We have already given a detailed verification of this axiom in Theorem 6.1. Axiom Cont 7 concerns IntersectCirclesSame , and does not need to be verified in view of Theorem 9.96. Axiom Cont 8 specifies that the intersection points of circles depend extensionally on the circles. That is evident from our definition of On ◦. Axiom Cont 9 (illustrated in Fig. 13) says that if a and b are the centers of C and K, and p = pointOnCircle (C) and q = pointOnCircle (K), and t is non- 2 strictly between a and b with bt = ap, then pt = ab. Let us write γ4 = 1 γ , − 3 and r = C3◦, to simplify the formulas. Then by definition, p

p◦ = (a◦ + γ3 r, a◦ + γ4 r) 1 · 2 · q◦ = (b◦ + γ3 r, b◦ + γ4 r) 1 · 2 · The hypothesis of Axiom Cont 9 mentions point t, non-strictly between b and q, such that bt = ap. Then we have

t◦ = b◦ + λ(q◦ b◦) − where λ is chosen to make t b = p a ; that requires λ = p◦ a◦ / q◦ b◦ . The denominator is not zero,| since− | K |is− a| non-degenerate circle,| − by| the| hypothesis− | B(u,b,q). Now we calculate

t◦ b◦ = λ q◦ b◦ | − | | − | = p◦ a◦ | − | as required. We also have to verify the second part of the axiom, which says that segment pt does not meet segment ab. That is true in the ◦ interpretation since p◦t◦ is parallel to a◦b◦. That completes the verification of Axiom Cont 9. We now take up the strong parallel axiom. Let L be a line, and let K be a line such that IntersectLines (L,K) is not defined; that is, K cannot fail both to be parallel to L and to be coincident with L. Let p be on K and let a be a point not on K, and let M = Line (p,a). We have to show that IntersectLines (L,M) is defined, i.e. M meets L. Let J = Perp(a,L), so J is perpendicular to L; let m be the intersection point of J and L. By applying a linear transformation, we may assume that L = Line (0, 1) and m = 0. Now J does not coincide with K, since it contains point a, which is not on K. Then if J does not meet K it is parallel to K, and since then m is on J but not on K, K is parallel to L. Then J and L are two parallels to K through m, contradicting even Playfair’s axiom; hence J does meet K. Call the point of intersection z. Then z = a since a is not on K. We 6 174 MICHAEL BEESON have z◦ = (0,z2◦) since z is on J; similarly a◦ = (0,a2◦). Now we can write the equation of line M ◦ and solve for its meeting point with the x-axis.

a1◦ z2◦ 0 = y = a◦ + − x 1 a p 1◦ − 1◦ (p◦ a◦) a◦ x = 1 − 1 · 2 a z 2◦ − 2◦ and since a = z, the denominator is not zero, and hence using the strong reciprocal axiom,6 it is invertible, so this equation does provably define x in EF. That completes the verification of the strong parallel postulate in EF. Next we check that Euclid 5 can be verified using only P (x) 1/x . The algebra for this is the same as above; we must use the extra hypotheses→ of↓ Euclid 5 to control the sign of the denominator. The extra hypotheses are that K does not meet L, and there are points r on K and q on L with B(r, a, q). Then we have r2◦ = q2◦ and a2◦ < r2◦, and hence a2◦ < q2◦, controlling the sign of the denominator as required. That completes the verification of Euclid 5 from the axiom P (x) 1/x . Finally we→ check that↓ Playfair’s axiom can be verified using only weak Eu- clidean field theory, i.e. Axioms EF9 and EF10 instead of the strong reciprocal axiom EF1. Suppose Playfair’s axiom. Then with the hypotheses as for the strong parallel axiom, but also assuming K and L are parallel (rather than coincident). Let a be any point on M. We have to verify that M ◦ and L◦ cannot fail to meet. That is, x in the above equation cannot fail to be defined. Suppose it does fail to be defined; then the denominator of the expression for x is not invertible. Hence, by Axiom EF9, it is zero. Then a2 = q2, so a lies on K. Since a was any point on M, we have proved M and K coincide, contradicting the hypothesis. That completes the proof of the theorem. 12.6. Area in Euclid, Hilbert, and ECG. In the time of Euclid, area was not considered as a real-valued function. In fact, “area” is never mentioned in Euclid; nor is “distance” mentioned. When Euclid states theorems that would now be considered to be about area, he speaks of “equality” between (specific) polygons. Hilbert [15], Chapter IV, has a chapter on Theory of Plane Area in which he proceeds as follows. He first makes Euclid’s notion of “equality” of polygons precise: two polygons are equidecomposable if they can be decomposed into a finite number of triangles that are congruent in pairs. Two polygons (or unions of polygons) A and B are equicomplementable if one can find a finite number of pairs of equidecomposable polygons (Pi,Qi) and polygons A′, B′, such that A′ is composed of A and the Pi and B′ is composed of B and the Qi. Euclid’s “equal” in the latter part of Book I should be read as “equicomplementable”, according to Hilbert. He gives an argument for reading it as “equicomplementable” rather than “equidcomposable”. Of course neither of these notions is a first-order (geometric) notion, because of the mention of “finite number”; but to formalize Euclid Book I (including its culmination, the Pythagorean theorem) a fixed (and relatively small) number of triangles in the decompositions is required, and we do not have a theorem with “equal polygons” in the hypothesis, so Book I remains first-order. We take it FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 175 then that Euclid’s I.47, the Pythagorean theorem, can be proved in ECG. Euclid I.47 says that the square on the side of the hypotenuse is equicomplementable (with, say, less than 100 triangles, to stay first-order) to the two squares on the sides. To define area as a function from polygons to points on Line (0, 1), we make use of the fact that we already know how to introduce coordinates (X(a), Y (a)) geometrically. We can then introduce the usual apparatus of vector algebra, but using geometric definitions for the underlying arithmetic.

Definition 12.22.

Point VectorSubtract(Point a, Point b) { u = Add(X(a),Minus(X(b))) v = Add(Y(a),Minus(Y(b))) return MakePoint(u,v) }

Point CrossProduct(a,b) { p = Multiply(X(a),Y(b)) q = Multiply(X(b),Y(a)) return Add(p,Minus(q)) } Point SignedArea(Point a, Point b, Point c) { u = VectorSubtract(c,b) v = VectorSubtract(a,b) w = CrossProduct(u,v) 1/2 = Divide(1,Add(1,1)) return Multiply(1/2, w) } Point Area(Point a, Point b, Point c) { return SquareRoot(HilbertSquare(SignedArea(a,b,c))) }

Then the area of a triangle can be deﬁned in the way that is now customary in computer graphics:

Lemma 12.23. Let y lie on segment xz. Then

SignedArea (0,z,x)= SignedArea (0,y,x)+ SignedArea (z,y, 0)

Proof. This is a routine computation in vector algebra, but it has to be done using Add and HilbertMultiply and CrossProduct . Nevertheless for readability we write it in the ordinary notation; so for instance X(y) becomes y1 and CrossProduct (x, y) becomes x y. Since y lies on segment xz we can write, for × 176 MICHAEL BEESON some t, y = tz + (1 t)x. Then − y x = y1x2 y2x1 × − = (tz1 + (1 t)x1)x2 (tz2 + (1 t)x2)x1 − − − = t(z1x2 x1x2 z2x1 + x2x1) − − = t(z1x2 z2x1) − z y = z1(tz2 + (1 t)x2) z2(tz1 + (1 t)x1) × − − − = t(z1z2 z1x2 z2z1 + z2x1)+ z1x2 z2x1 − − − = t( z1x2 + z2x1)+ z1x2 z2x1 − − y x + z y = z1x2 z2x1 × × − = z x × That completes the proof. Lemma 12.24. SignedArea (a,b,c)= SignedArea (b,c,a) and SignedArea (a,b,c)= Minus (SignedArea (b,a,c). Proof. These are also proved by routine computation. For readability we write b a instead of VectorSubtract (b,a), and so on: − (b a) (a c) = (b a) + (a c) (b c) − × − × × − × (c b) (b a) = (c b) (c a) + (b a) − × − × − × × and the two right sides are equal. That completes the proof. We now introduce several more standard concepts in ECG: Definition 12.25. Norm(Point a) { xsq = HilbertSquare(X(a)) ysq = HilbertSquare(Y(b)) return SquareRoot(Add(xsq,ysq)) } VectorAdd(Point a, Point b) { u = Add(X(a),X(b)) v = Add(Y(a),Y(b)) return MakePoint(u,v) } Distance(Point a, Point b) { u = VectorSubtract(a,b) return Norm(u) } Radius(Circle C) { e = center(C) b = pointOnCircle(C) return Distance(e,b) } ScalarMultiply(Point c, Point u) FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 177

{ x = HilbertMultiply(c,X(u)) y = HilbertMultiply(c,Y(u)) return MakePoint(x,y) } We develop in ECG the theory of two-by-two matrices and matrix multiplication Av, where A is a matrix whose entries are points on Line (0, 1) and v is any point (thought of as a 2-vector (X(v), Y (v))). Point MatrixMultiply( Point a, Point b, Point c, Point d, Point v) { u = Add(HilbertMultiply(a,X(v)),HilbertMultiply(b,Y(v))) v = Add(HilbertMultiply(c,X(v)),HilbertMultiply(d,Y(v))) return MakePoint(u,v) } Point Determinant(Point a, Point b, Point c, Point d) { return Add(Multiply(a,d),Minus(Multiply(b,c))) } We can define linear transformations (using six points); we define translations, rotations, and reflections as usual: translations are maps u VectorAdd (u,b), and rotations about the origin are maps with matrices 7→ a b where a2 + b2 =1 b a − where for readability we have written b instead of Minus b and a2 instead of HilbertSquare (a), etc. − Lemma 12.26. Translations preserve both congruence and distance. That is, if A = VectorAdd (a,t) and B = VectorAdd (b,t) then AB = ab and Distance (A, B)= Distance (a,b). Proof. First we show translations preserve distance. A = VectorAdd (a,t) B = VectorAdd (b,t) VectorSubtract (A, B)= VectorSubtract (a,b)

Hence Distance (A, B)= Distance (a,b) by definition of Distance . Now we show translations preserve congruence. Let a and t lie on Line (0, 1) and let p = Add (a,t); then ap =0t, by Lemma 9.6, since segment 0p is composed on the one hand of 0a and ap and on the other of 0t and tp, but tp = 0a by the definition of Add (a,b), according to which a is first rotated to the y-axis, then projected onto the vertical line through t, then rotated back to the x-axis to construct p. Those two rotations have the same radius 0a; hence tp =0a. Now consider a vertical translation A = VectorAdd (a, MakePoint (0,t)) B = VectorAdd (b, MakePoint (0,t)). Then quadrilateral abBA has four right angles (for t = 0 and a = b) and hence is a rectangle; hence ab = AB. But any translation6 is the composition6 of a 178 MICHAEL BEESON horizontal and a vertical translation; hence if ab lies on Line (0, 1) and AB is the image of ab under any translation, then AB = ab. Thus any two horizontal segments that can be transformed into each other by a translation are congruent. Since rotating through ninety degrees by Rotate also preserves congruence, the same is true of vertical segments, i.e. any translation preserves congruence of vertical segments. Note, rotation in the sense of Rotate is not (so far proved to be) the same as rotation by a matrix of the form specified as a rotation. But now consider an arbitrary segment ab, transformed into A and B by the formulas in the lemma. Then let

a′ = VectorAdd (a, MakePoint (X(t), 0)

A′ = VectorAdd (A, MakePoint (X(t), 0).

Then aa′ = AA′ since aa′ is a horizontal segment and AA′ is its translation by t. Similarly a′b is a vertical segment, so a′b = A′B. Then right triangle aa′b is congruent to right triangle AA′B; hence their corresponding sides ab and AB are congruent. That completes the proof of the lemma. Lemma 12.27. Rotations preserve congruence and distance. That is, if T is a matrix of the form specified above as a rotation, and A = Ta and B = Tb, then Distance (A, B)= Distance (a,b) and ab = AB. Proof. First we show rotations preserve distance. By a routine computation, one proves that if A is a matrix and v is a point then Norm (Av)= HilbertMultiply (Determinant (A), Norm (v))). (But the routine computation takes place in ECG, using the geometric definitions of arithmetic.) Hence matrices of determinant one preserve distance. These include rotations (defined as above), as one easily computes that the determinant of a rotation is 1. Next we prove that rotations preserve congruence. Let A be a rotation. Then for some u and v with Norm (u, v) = 1, we have u v A = . Minus (v) u Let p = MakePoint (u, v). (See Fig. 47.) Then we claim A(MakePoint (x, 0)) = Rotate (1, 0,p,x). This is proved by a calculation: u Minus (v) x A(MakePoint (x, 0)) = v u 0 HilbertMultiply (u, x) = HilbertMultiply (v, x) = MakePoint (HilbertMultiply (u, x), HilbertMultiply (v, x)) Hence the point q = A(MakePoint (x, 0)) lies on Line (0,p), since it satisfies the equation of that line (as shown by Hilbert, using similar triangles, see [15], p. 57), and X(q)= HilbertMultiply (u, x) FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 179

Figure 47. Rotations preserve congruence. Point q is deﬁned by matrix multiplication, but it is the same as the point computed by Rotate (1, 0,p,x).

q b

p b

b u b b b 0 X(p) X(q) x

Y (q)= HilbertMultiply (v, x). On the other hand Rotate (1, 0,p,x) is the point s on Ray (0,p) with 0s =0x, at least for x 0. We are trying to prove s = q. We know now that they lie on the same ray and≥ that they have the same distance x from 0. That is tantalizingly close, but still some argument is required to show s = q. It will suﬃce to prove that there is only one point on Ray (0,p) at a given distance from 0. Any point on that ray has the form w = MakePoint (HilbertMultiply (c,u), HilbertMultiply (c, v)) for some c > 0, so its distance from 0 is c, since MakePoint (u, v) has norm 1. Then c = Divide (X(w),u) if u = 0 and c = Divide (Y (w), v) if u = 0. In either case c is determined by w so there6 cannot be two points on the ray at the same distance from 0. Since equality is stable, we are allowed to argue by cases here. That completes the proof of the lemma. Lemma 12.28. If ab = AB then Distance (a,b)= Distance (A, B). Conversely if Distance (a,b)= Distance (A, B) then ab = AB. Proof. By a translation and rotation we can bring a to the origin and b to lie on Ray (0, 1). By the previous lemmas, ab is congruent to this new segment and distance is preserved. Without loss of generality, then, we may assume a = A = 0 and both b and B lie on Ray (0, 1). Now assume ab = AB. Then by Lemma 9.5, b = B. Hence Distance (a,b) = Distance (a,B) = Distance (A, B). 180 MICHAEL BEESON

Conversely, assume Distance (a,b) = Distance (A, B). But Distance (a,b) = X(b), and Distance (A, B) = X(B). Since b and B lie on Ray (0, 1), we have Y (b) = Y (B) = 0. Hence b = MakePoint (X(b), 0) = B. Hence 0b = 0B; but 0= a = A so ab = AB. That completes the proof. Lemma 12.29. Let triangle abc be congruent to triangle ABC. Then Area (a,b,c)= Area (A, B, C). Proof. By a routine computation, one proves that if A is a matrix and v is a point then Norm (Av)= HilbertMultiply (Determinant (A), Norm (v))). (But the routine computation takes place in ECG, using the geometric definitions of arithmetic.) Hence matrices of determinant one preserve area. These include rotations (defined as above) and reflections. We can bring a and A to 0 by translations, and then by a rotation we can bring b and B to lie on Ray (0, 1). By the previous lemmas, these translations and rotations preserve both congruence and distance. Hence, without loss of generality we can assume a = A = 0 and b = B, with b lying on Ray (0, 1). By the uniqueness of triangle construction (Lemma 9.68), then not not C = c or C is the reflection of c in Line (0, 1). In the latter case, we can apply a reflection (which also has determinant one) to make c = C. Therefore not not Area (a,b,c)= Area (A, B, C); but by the stability of equality, we can drop the double negation, so Area (a,b,c) = Area (A, B, C) as claimed. Hilbert [15], defines the area of a triangle more traditionally, using the formula “half the base times the altitude”, but our way is perhaps easier, as it reduces everything to mechanical computations with the cross product. Once the area of a triangle is defined, however, we follow Hilbert: Lemma 12.30 (Hilbert’s theorem 49). Let point o lie outside triangle abc. Then SignedArea (a,b,c)= SignedArea (o,a,b)+SignedArea (o,b,c)+SignedArea (o,c,a). Proof. Again, what is to be proved is an equality, so we may argue by cases as to the position of o, as Hilbert does. Hilbert’s argument depends only on the previous two lemmas, so it goes through without change. After this theorem, one can define the area of a positively oriented simple polygon as the sum of the areas of triangles in a triangular decomposition of that polygon, and prove that it is independent of the decomposition. The culmination of Hilbert’s theory of area is his Theorem 51 (see [15], p. 69), which says Two equicomplementable polygons have the same area, and two polygons with the same area are equicomplementable. Since “equicomplementable” is not a first-order concept, Hilbert’s Theorem 51 is not a first-order theorem as it stands (even if we restrict the number of sides of the polygon to 4 or some other fixed number). The first half could be taken as a first-order schema, one theorem for each fixed number of triangles allowed for the decompositions. The second half raises the question whether a fixed number of triangles suffices. Hilbert’s proof shows that the required number of triangles is bounded by a small constant times the number of sides of the polygon; hence, for a fixed number of sides of the polygon, this is a first-order theorem. Since the conclusion is an equality, and equality is FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 181 stable, we are free to prove the theorem by contradiction, so we do not have to worry about whether Hilbert’s proof is constructive or not. From Hilbert’s Theorem 51 and Euclid I.47, we can draw as a corollary the following version of the Pythagorean theorem: Lemma 12.31. If p is on Circle(0, r), then Add (HilbertSquare (X(p)), HilbertSquare (Y (p))) = HilbertSquare (r). Proof. Consider the triangle with vertices 0, MakePoint(X(p), 0), and p. The square on the hypotenuse has area HilbertSquare (r), as one can compute using the cross product and a decomposition of the square into two triangles. The square on the horizontal side has area HilbertSquare (X(p)). The square on the vertical side has area HilbertSquare (Y (p)). By Euclid I.47, the two smaller squares are equicomplementable to the large square. The lemma then follows from Hilbert’s Theorem 51. That completes the proof. Actually, Lemma 12.31 has a much simpler proof: Since p and the point b = MakePoint (r, 0) both lie on Circle (0, r), then by Axiom C5, segment 0p is congruent to segment 0b. Hence, by Lemma 12.28, Distance (0,p)= Distance (0,b). But Distance (0,p)= Add (HilbertSquare (X(p)), HilbertSquare (Y (p))), and Distance (0,b) = HilbertSquare (r). Hence Distance (0,p) = Distance (0,b), proving Lemma 12.31 the easy way. We can then use Hilbert’s Theorem 51 to prove Euclid’s version of the Pythagorean theorem, but without the specific decompositions given by Euclid’s proof to establish that the two small squares and the large square are equicomplementable. Nevertheless, the fact that proof is constructive shows that some decomposition can be found. It would be possible to trace through the proof and extract the Euclid-style proof in ECG of the Pythagorean theorem that is implicit in the short proof of Lemma 12.31 plus Hilbert’s Theorem 51. 12.7. Handedness, cross product, and linear transformations. In this section we show that Right (a,b,c) is related to the cross product (a b) (c b) as one would expect. This turns out to be somewhat more difficu− lt× than− expected. In this section we also formalize some other lemmas in vector algebra; we need these results because in the course of studying the faithfulness of the interpretation of ECG in field theory, we need to formalize in ECG a number of algebraic arguments. We start with the theory of similar triangles. Definition 12.32. Triangle abc is similar to triangle ABC if the three corresponding angles are pairwise congruent, i.e. angles abc, bca, and cab are congruent to ABC, BCA, and CAB respectively. One easily checks that similarity is an equivalence relation. Definition 12.33. The proportion a : b = c : d means HilbertMultiply (a, d)= HilbertMultiply (b,c). The proportion ab : AB = bc : BC means Distance (a,b) : Distance (A, B)= Distance (b,c) : Distance (B, C). 182 MICHAEL BEESON

Lemma 12.34. Triangles abc and ABC are similar if and only if corresponding sides are proportional. A : a = B : b and B : b = B : c and A : a = C : c. Proof. See [15], pp. 55-57. Since the conclusions are stable, we do not have to check line by line whether the proofs are constructive. The proof goes back to Descartes’s definition of Multiply , so as we have stated the lemma, we also need to appeal to the equivalence of Multiply and HilbertMultiply . Lemma 12.35. . Let ABC and abc be right triangles with right angles at b and B and legs ab and AB parallel to Line (0, 1) and legs bc and BC perpendicular to Line (0, 1), and suppose BC : bc = AB : ab. Then triangle ABC is similar to triangle abc. Remark. Of course the lemma is true without the restriction that the legs be parallel to the coordinate axes; but to prove that we would naturally use the Pythagorean theorem in the form A C 2= A B 2 + C B 2, and to prove that we need a chain of lemmas reaching| − back| | to− this| lemma;| − so| at this point we prove only the special case where the legs are parallel to the axes. In that special case, the Pythagorean theorem is immediate from the definitions of Norm and Distance . Proof. Using normal notation (to abbreviate geometric arithmetic) let λ = (B − A)1/(b a)1. Then by hypothesis also λ = (C B)2/(c a)2, and since the legs − − − are parallel to the axes we have C1 = B1 and A2 = B2, and c1 = b1 and a2 = b2. Then C A 2 = (C A)2 + (C A)2 | − | − 1 − 2 = (A B)2 + (C B)2 − 1 − 2 = λ2(b a)2 + λ2(b c)2 − 1 − 2 = λ2((b a)2 + (b c)2) − 1 − 2 = λ2((c a)2 + (a c)2 − 1 − 2 = λ2 a c 2 | − | and taking the square root of both sides we have C A = λ c a . It follows that the third pair of sides is also proportional, and| − then| by Lemma| − | 12.34, the triangles are similar. That completes the proof of the lemma. Next we show that the points on a line are exactly the points satisfying the equation of the line, in either of two senses: Lemma 12.36. The following three conditions are all equivalent: (i) Point p is on Line (a,b) (ii) for some λ we have p = VectorAdd (a, ScalarMultiply (λ, VectorSubtract (b,a))). (iii) CrossProduct (VectorSubtract (p,a), VectorSubtract (b,a))=0. Proof. We first show that (ii) implies (i). By Axiom S3, on (p,L) on (p,L), so we can argue by cases. Suppose p is given by the formula¬¬ in (ii).→ Case 1, X(a)= X(b) (i.e. the line is vertical). Then X(p)= X(a), from the formula for p. Then Line (b,a) is perpendicular to Line (0, 1) at X(a), by Lemma FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 183

9.81. Then p is on Line (a,b) because, by definition of X(p), the perpendicular from p to Line (0, 1) meets Line (0, 1) at the same point as Line (a,b) does, namely at X(a), so by Lemma9.81 that perpendicular coincides with Line (b,a). Case 2, Y (p)= Y (a) (i.e. the line is horizontal). This is analogous to Case 1, with Y and X interchanged. Case 3, p = a. Then there is nothing to prove as p is already on Line (a,b). Case 4, X(b) = X(a) and Y (b) = Y (a) and p = a. Let c = MakePoint6 (X(b), Y (a)).6 Then a, b,6 and c are not collinear, since c lies on a horizontal line through a, so if they were collinear, then b would also lie on that line, so Y (b) would be equal to Y (a), contradiction. Let d = MakePoint (X(p), Y (a)). Then (since p = a) similarly a, d, and p are not collinear. We wish to show triangles acb and6 adp are similar. We have VectorSubtract (p,a)= HilbertMultiply (λ, VectorSubtract (b,a)). Taking the norm of both sides we have Distance (a,p)= HilbertMultiply (λ, Distance (b,a)). Taking X instead of norm, we have Distance (a, d)= HilbertMultiply (λ(Distance (a,c))) and taking Y instead of norm, we have Distance (p, d)= HilbertMultiply (λ, Distance (b,c)). Therefore the corresponding sides of triangles acb and adb are proportional. Hence by Lemma 12.34, the triangles are similar. Hence angle pad is congruent to angle bac. Hence p lies on Line (a,b), as one can prove from the definition of angle congruence. That completes the proof that (ii) implies (i). Now we prove (i) implies (iii). Again we can argue by cases, since equality is stable. For readability we use normal notation for components and cross products, but it only abbreviates geometric arithmetic. Let points d and c be as defined above. Then We have

(p a) (b a) = (p a)1 (b a)2 (p a)2(b a)1 − × − − · − − − − In the cases of a vertical or horizontal Line (a,b), the right side is zero. Hence we may assume the line is not horizontal or vertical. Then by Deﬁnition 12.33, the cross product is zero if and only pd : bc = ad : ac Since triangles pda and bca are right triangles with legs parallel to the coordinate axes, by Lemma ?? we conclude that the cross product is zero if and only if triangle pda is similar to triangle bca. But triangle pda is similar to triangle bca if and only if point p lies on Line (a,b). Hence the cross product is zero if and only if p lies on Line (a,b). That completes the proof that (i) implies (iii). Now we prove (iii) implies (ii). Suppose (p a) (b a) = 0. Then as shown above − × −

0 = (p a)1 (b a)2 (p a)2(b a)1 − · − − − − Note that we are not allowed to argue by cases for a conclusion that starts with λ. We must deﬁne λ without a case split, which is slightly tricky since ∃ 184 MICHAEL BEESON we must avoid a zero denominator that might arise in some cases. But we do know a = b, so the denominator should be Distance (b,a), which is equal to Norm (VectorSubtract6 (b,a)). At the same time we have to take care about the sign of λ, which could be positive or negative. The correct value of λ is then found by taking the component of p a in the direction b a, and dividing by Distance (b,a): − −

(p a) (b a) λ := − · − HilbertSquare (Distance (b,a))

The numerator abbreviates a DotProduct term. Now we claim that with this value of λ, (ii) holds. (Once we have speciﬁed λ, what has to be proved is an equality, so now we could argue by cases, but it turns out not to be necessary to do so.) In normal notation, what has to be proved is p a = λ(b a). Putting in the value of λ, what has to be proved (in normal notation)− is −

(p a) (b a) p a = − · − − b a 2 | − | This can be obtained by setting x = p a and y = b a in the formula − − x y x = · y for y =0 y 2 6 | | or equivalently

y 2x = (x y)y | | · which is easily veriﬁed by a componentwise calculation:

2 2 2 y x = (y1 + y2)(x1, x2) | | 2 2 2 2 = (x1(y1 + y2), x2(y1 + y2) = (x1y1 + x2y2)(y1,y2)

= (x y)(y1,y2) ·

That completes the proof of the lemma.

Lemma 12.37. The sign of CrossProduct (a,b) is invariant under linear transformations with positive determinant. In fact if A is a linear transformation then

CrossProduct (A(x), A(y)) = HilbertMultiply (Determinant (A), CrossProduct (x, y)).

Proof. A straightforward computation, but it has to be performed using geometric arithmetic inside ECG. Nevertheless for readability we write it in normal FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 185 notation, writing for example x1 instead of X(x1) and x2 instead of Y (x). a b x A(x)= 1 c d x2 ax + bx = 1 2 cx1 + dx2 ay + by A(y)= 1 2 cy1 + dy2 ax + bx ay + by A(x) A(y)= 1 2 1 2 × cx1 + dx2 × cy1 + dy2

= (ax1 + bx2)(cy1 + dy2) (cx1 + dx2)(ay1 + by2) − = bcx2y1 adx2y1 + adx1y2 bcx1y2 − − = (ad bc)(x1y2 x2y1) − − = Determinant (A)(x y) × That completes the proof of the lemma. The following lemma formalizes in ECG a well-known test for parallelism or coincidence. Note how the use of linear transformations simplifies its proof. Lemma 12.38. L = Line (a,b) and K = Line (c, d) are weakly parallel (that means IntersectLines (L,K) is not defined) if and only if CrossProduct (VectorSubtract (b,a), VectorSubtract (d, c))=0. Proof. We note that if all four points are transformed by the same linear transformation, since lines are carried to lines, the transformed lines intersect if and only if the original lines do, provided the determinant of the transformation is nonzero. By Lemma 12.37, the cross product is nonzero after such a transformation if and only if it is nonzero before. Hence we may assume that a = 0 and b lies on Ray (0, 1). Then the cross product in the lemma is (in normal notation) b1(d2 c2). Since b1 = 0, the cross product vanishes if and only if d2 = c2; but that− is exactly the6 condition IntersectLines (K, Line (0, 1)) to be undefined. That completes the proof of the lemma. Lemma 12.39. Right (a,b,c) is invariant under rotations about b. Proof. To study this issue we make use of “local coordinates” for the first time. We treat Line (b,a) as we formerly treated Line (0, 1), and temporarily consider 0 as another name for b and 1 as another name for the point on Ray (b,a) such that αβ = 01. Let J = Perp(0, Line (b,a)) be the new y-axis, and let X and Y be projections on Line (b,a) and on J. In order to define addition in these new coordinates we need to determine an orientation of J; that is, we need to determine a point on J to temporarily play the role of I. That should be a point at distance 1 from b on the same side of Line (b,a) as c. It is not quite trivial to construct such a point, but the following script does the job: Point NewI(Point a, Point b, Point c) { r = Y(c) C = Circle(b,alpha,beta) 186 MICHAEL BEESON

L = Line(b,a) K = Perp(a,L) s = IntersectLineCircle1(K,C) K = Circle(r,s) I = IntersectCirclesSame(C,K,c) return I } Then I and c are on the same side of Line (b,a)= Line (0, 1), so by Axiom H4, Right (1, 0, I) if and only if Right (a,b,c). Now consider a rotation T , whose matrix in these coordinates is

u Minus (v) T = v u 

Consider ﬁrst the case v> 0, i.e., a non-trivial rotation counterclockwise by less than 180 degrees. Let A = T (a) and C = Tc(c). We have to prove Right (a,b,c) if and only if Right (A, b, C). Suppose Right (a,b,c). We will prove Right (A, b, C). We need to construct an auxiliary point d. We need d to be in the upper half plane (the side of Line (0, 1) containing c) and also to be on the same side of Line (b, A) as C. See Fig. 48 for an illustration.

Figure 48. Right (a,b,c) is invariant under rotation. abc changes to abd, then to Abd, then to AbC.

A b d b

b b b q b a

C b

Since we are trying to prove Right (A, b, C), and Right is deﬁned by an equality, we may legally argue by cases. Case 1 is when C lies in the lower half plane. Then segment AC meets the x-axis in a point q. We have A = q since A lies in the upper half plane. Let d be the midpoint of Aq. Then d is6 in the upper half plane. We claim d is on the same side of Line (b, A) as C. To prove that, suppose dC meets Line (b, A). Then Line (b, A) and Line (b, C) coincide, so A C = 0; but we have proved that rotations preserve cross products, so A C = a× c = 0. Hence dc does not meet Line (b, A). Then by Lemma 9.83, d and× C are× on6 the same side of Line (b, A). That completes the construction of the desired point d in Case 1. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 187

Case 2, C lies in the closed upper half plane. Then let d be the midpoint of AC. Then d lies in the upper half plane, since its y-coordinate is the average of a positive and a non-negative number. As in Case 1, d lies on the same side of Line (b, A) as C. Thus in either case we have the desired point d. Now we can argue as follows: Right (a,b,c) by assumption Right (a,b,d) ↔ by Axiom H4, since d and c are on the same side of Line (b,a) Right (A, b, d) ↔ by Axiom H4, since a and A are on the same side of Line (b, d) Right (A, b, C) ↔ by Axiom H4, since d and C are on the same side of Line (b, A) Thus in either case we reach the conclusion Right (A, b, C). Since not not (Case 1 or Case 2), we have Right (A, b, C), and hence by the stability of equality and the definition of Right¬¬as an equality, we have Right (A, b, C). Since we assumed Right (a,b,c), we have proved without an assumption that Right (a,b,c) Right (A, b, C). → Similarly, appealing to Axiom H5, we can prove Left (a,b,c) Left (A, b, C). Hence Left (A, b, C) Left (a,b,c). We wish to prove → ¬ → ¬ Right (a,b,c) Right (A, b, C). ↔ Since Right is stable, we are allowed to prove this by cases. If a, b and c are collinear, then also A, b and C are collinear, so both sides are automatically true. Otherwise Right (a,b,c) is equivalent to Left (a,b,c) and we are done. That completes the proof in case v> 0, i.e., rotation¬ T is a rotation by less than 180 degrees. To finish the proof, we decompose any rotation into a product of two rotations, each of which is by less than 180 degrees. Then we can apply the above argument to each rotation separately. We need to prove formally that any rotation can be so decomposed. Given rotation u v T = − v u with u2 + v2 = 1 and v< 0 we define 1 u V = − r 2 1+ u U = −r 2 (These formulas are based on the usual half-angle formulas for sine and cosine, but of course no trigonometry is involved here. How we thought of these formulas is not relevant for the proof.) Now define U V S = − V U 188 MICHAEL BEESON

Then routine algebraic computation (done using geometric arithmetic, here indicated with “normal” notation), shows that S2 = T . That completes the proof of the lemma. Lemma 12.40. Right (a,b,c) is invariant under translations along Line (a,b). That is, if A(x)= VectorAdd (x, HilbertMultiply (t, VectorSubtract (a,b))) then Right (A(a), A(b), A(c)) Right (a,b,c). ↔ Proof. For any t, the point A(x) lies on Line (a,b) if x lies on Line (a,b), because it satisfies the equation of that line. We claim that Right (a,b,c) is equivalent to Right (a, A(b), A(c). By the stability of equality, we can argue by cases on the sign of t. If t = 0 there is nothing to prove. If t> 0 then A(a) lies on Ray (b,a), so by Lemma 9.89, Right (a,b,c) is equivalent to Right (A(a),b,c). Then A(b) lies on Ray (A(a),b), so a second application of Lemma 9.89 makes Right (A(a),b,c) equivalent to Right (A(a), A(b),c). Combining the two we have Right (a,b,c) equivalent to Right (A(a), A(b),c). Now we claim A(c) is on the same side of Line (a,b) as c. By Lemma 9.83 it suffices to show that line segment cA(c) does not meet Line (a,b). For that it suffices to show that Line (c, A(c)) is parallel to Line (b,a). That follows from Lemma 12.38, since one easily computes that CrossProduct (VectorSubtract (A(c),c), VectorSubtract (b,a))=0. Hence A(c) is on the same side of Line (a,b) as c. Hence by Axiom H4, Right (A(a), A(b), A(c)) is equivalent to Right (A(a), A(b),c), which we have shown is equivalent to Right (a,b,c). That completes the case t> 0. The case t< 0 is treated similarly. That completes the proof of the lemma. The following lemma is the computer-graphic definition of Right , but now “internalized” using geometric arithmetic: Lemma 12.41. For a = b, we have Right (a,b,c) if and only if 6 CrossProduct (VectorSubtract (a,b), VectorSubtract (c,b)) 0. ≥ Proof Since both Right and are stable, we can argue by cases. If a, b, and c are collinear then Right (a,b,c≥ ), and the cross product is zero, so the claimed result is valid in that case. Therefore we can assume that a, b, and c are not collinear. We also argue by cases according as b = 0 or b = 0. Assume b = 0. By Lemma 12.39, we can rotate abc about b until a lies on6 Line (0,b). Then,6 by Lemma 12.40, we can translate abc along Line (0,b) until b = 0. By the cited lemmas, Right (a,b,c) still has the same value as originally. Moreover, by Lemma 12.37, the sign of the cross product in the lemma is also not changed. Hence, we can assume b = 0. Now, by Lemma 12.39 and Lemma 12.37, we can rotate again, bringing a onto Ray (0, 1). Finally we can compute. We know Right (1, 0, I) by Lemma 9.104. Hence Right (a,b,c) is equivalent to Y (c) > 0, by Axiom H4, since Y (c) > 0 if and only if c is on the same side of Line (0, 1) as I. Since b = 0, the cross product in the lemma is CrossProduct (a,c), which by definition, and since Y (a) = 0, is HilbertMultiply (a, Y (c)). Since a lies on Ray (0, 1), we have a> 0, so the cross product has the same sign as Y (c). Hence Right (a,b,c) and CrossProduct (a,c) > 0 are equivalent. Since we have assumed FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 189 a, b, and c are not collinear, the cross product is not zero, so we can replace > by . That completes the proof. ≥ Corollary 12.42. Right (a,b,c) is invariant under arbitrary linear transformations with positive determinant. Proof. By Lemma 12.41, Right (a,b,c) is equivalent to a (a b) (c b) 0 (written in normal instead of geometric notation). By Lemm− a× 12.37,− the≥ latter is invariant under linear transformations with positive determinant. Hence Right (a,b,c) is also invariant. That completes the proof. Lemma 12.43 (Triangle inequality). Norm (a+b) Norm (a)+Norm (b), with equality if and only if 0, a, and b are collinear, if and≤ only if a b =0. × Proof. For readability we write x 2 instead of HilbertSquare (Norm (x)), a + b instead of Add (a,b), etc. k k a + b 2 = a 2 + b 2 +2DotProduct (a,b) k k k k k k a 2 + b 2 +2 DotProduct (a,b) ≤k k k k k k a 2 + b 2 +2 a b ≤k k k k k kk k ( a + b )2 ≤ k k k k Taking SquareRoot of both sides yields the inequality. We have equality only if DotProduct (a,b) = a b . A short computation shows this is equivalent to k kk k a b = 0. We write a1 and a2 for X(a) and Y (a), etc.: × 2 2 (DotProduct (a,b)) = (a1b1 + a2b2) 2 2 2 2 = a1b1 + a2b2 +2a1b1a2b2 2 2 2 2 a b = (a1 + a2)(b1 + b2) k kk k 2 2 2 2 1 2 2 2 = a1b1 + a2b2 + a2b1 + a1b2 Subtracting, we find equality in the lemma if and only if 2 2 2 2 2a1a2b1b2 = a2b1 + a1b2 2 0 = (a2b1 a1b2) − 0= a b × It only remains to show that a b = 0 if and only if 0, a, and b are collinear. × But b is on the line containing 0 and a if and only if b2a1 = a2b1 (that is, b satisfies the equation of the line, which is x2a1 = a2x1). But that is exactly the condition a b = 0. That completes the proof of the lemma. × Lemma 12.44. B(a,b,c) Add (Distance (a,b), Distance (b,c)) = Distance (a,c). ↔ Proof. Recall Distance (a,b) = Norm (VectorSubtract (b,a)). We have (in more conventional notation) c a = (c b) + (b a) − − − c a = (c b) + (b a) k − k k − − k c b + b a by the triangle inequality ≤k − k k − k 190 MICHAEL BEESON

When the conventional notation is replaced by the geometrically defined concepts here abbreviated, we have the statement of the lemma. That completes the proof. Lemma 12.45. If center (Q) = center (K) and Radius (Q) = Radius (K) then Q = K. Proof. Suppose we can prove pointOnCircle (Q)= pointOnCircle (K). Then we have Q = Circle (center (Q), pointOnCircle (Q)) by Axiom CA6 = Circle (center (K), pointOnCircle (K)) = K by Axiom CA6 again Therefore it suffices to prove pointOnCircle (Q) = pointOnCircle (K). The idea is that Axiom Cont 9 forces pointOnCircle (K) to lie in the same direction from center (K), for any circle K. Consider the following construction script: f(Circle K) { C = Circle(0,1) p = pointOnCircle(C) b = CircleCenter(K) r = Radius(K) s = MakePoint(HilbertMultiply(X(p),r),HilbertMultiply(Y(p),r)) z = VectorAdd(b,s) return z; } Then we claim that f(K) = pointOnCircle (K). To prove this we will use Axiom Cont 9. Refer to Fig. 13. In the statement of Axiom Cont 9, take C to be Circle (0, 1), and let p = pointOnCircle (C) and q = pointOnCircle (K). Let b = center (K). Lay off 0p on Ray (b, q), to produce point t on Ray (b, q) with bt = 0p. Then according to Axiom Cont 9, we have pt = 0b. Then 0ptb is a parallelogram; then b and t are respectively the images of 0 and p under the translation by b. But translations take lines to lines; so Line (0,p) goes to Line (b,t) under translation by b. Indeed, translations take rays to rays, so Ray (0,p) goes to Ray (b,t). The point s in the script for f lies on Ray (0,p) (since it satisfies the equation of Line (0,p) and r 0), so its image z = VectorAdd (b,s) under that translation lies on Ray (b,t). Since≥ translations preserve distance, and Distance (0,s)= r, we have Distance (b,z)= r. By Lemma 12.28, bz = bq. Hence by Axiom C5, On (z,K). But Line (b,t) intersects K in at most two points, by Lemma 9.51, and by Lemma 9.12, only one of those two points lies on Ray (b,t). Hence z = q as claimed. That completes the proof of the lemma. The following lemma shows that Axiom Cont 9 really does specify what pointOnCircle (C) is, relative to the arbitrary choice of pointOnCircle (Circle (0, 1)). Lemma 12.46. Let b be the center of circle K, and let p = pointOnCircle (Circle (0, 1)). Then pointOnCircle (K)= VectorAdd (b, ScalarMultiply (Radius (K),p))). Proof. Refer to Fig. 13. Let q = pointOnCircle (K). Since what has to be proved is an equality, by the stability of equality we can argue by cases. First consider FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 191 the case when b = 0 and Radius (K) 1. Then let point t lie on Ray (b, q) with bt = 1; that point t is inside K because≥ the radius of K is at least 1, so T(b,t,q) as in the hypothesis of Axiom Cont 9 (where T is non-strict betweenness). Then by Axiom Cont 9, p and q are collinear; so X(p) is to X(q) as Y (p) is to Y (q). That is, HilbertMultiply (X(q), Y (p)) = HilbertMultiply (X(p), Y (q)). Since e = 0, we have

q′ := VectorAdd (b, ScalarMultiply (Radius (K),p)) = ScalarMultiply (Radius (K),p))

We are trying to show that q = q′. We will ﬁrst show the sum of the squares of the coordinates of both sides are the same. For readability we write this in normal notation, rather than in geometric arithmetic notation, using q1 = X(q) and q2 = Y (q). Here is the computation: 2 q′ = HilbertSquare (Norm (ScalarMultiply (Radius (K),p))) | | 2 2 2 2 2 2 = p1 q1 + q2 + p2 q1 + q2 q q 2 2 2 2 2 = q1 + q2 (p1 + p2) q 2 2 2 2 = (q1 + q2)(p1 + p2) 2 2 = (q1 + q2) since p lies on the unit circle At the last step we have used Lemma 12.31. The previous steps are valid because we have checked that the laws of EF work in geometric arithmetic. But the last expression is HilbertSquare (Radius (K)). We have now shown that q′ = q ; and | | | q′ is a scalar multiple of p, it lies on Line (0,p); But q lies on Line (0,p) too. According to Axiom Cont 9, pt = ab; but in this case ab is a null segment, so p = t. Then B(b,t,q) means B(0,p,q). Since Radius (K) 0, q′ is a positive scalar multiple of p, so it lies on the same side of 0 as p does;≥ but that is the same side as q. Hence q′ and q are two points on Ray (0, q) with the same distance from 0. By Lemma 12.28, 0q = 0q′. Then by Lemma 9.5, q = q′ as desired. That completes the case when e = 0 and Radius (K) 1. The case when e =0 and Radius (K) < 1 is handled similarly, changing the≥ roles of C = Circle (0, 1) and K. We now take up the case when b = 0 and Radius (K) 1. Then we can ﬁnd t as in the hypothesis of Axiom Cont6 9 (note that the case≥ of radius 1 is covered because Axiom Cont 9 only requires t to be non-strictly between b and q). Then Axiom Cont 9 tells us that ptba is a parallelogram. Note that the part about pt not meeting ab is essential here. Hence t = VectorAdd (b,p). Then since tb =0p, by Lemma 12.28 we have Distance (t,b) = 1. Since q lies on Ray (b,t) we have (in normal notation for readability, but it abbreviates geometric vector arithmetic) q b = r(t b), where r = q b since t b = 1. But since t = p and b = 0, we− have q −b = rp. But that| − is (translated| | − into| normal notation) what had to be proved.− That completes the proof in the case b = 0 and Radius (K) > 1. 6 192 MICHAEL BEESON

Now consider the case when Radius (K) < 1. Then we interchange the roles of C = Circle (0, 1) and K in Axiom Cont 9, and a similar calculation proves the theorem in that case. These cases are classically exhaustive; and in each case we have reached the desired conclusion; hence we have proved not not the desired conclusion. But since the conclusion is an equality, and equality is stable, we then finally reach the conclusion itself. That completes the proof of the lemma. 12.8. Interpreting field theory in ECG. We now define an interpretation from field theory to the language of ECG. That involves defining multiplication, addition, and square root in geometry. The definition of multiplication requires constructing a circle through three points, for which we use the strong parallel axiom, so we do not obtain interpretations from the theory of weak Euclidean fields or Playfair fields into the corresponding geometries. We return to this issue in a subsequent section; in this section, we concentrate on interpreting EF+ in ECG. We begin by associating a term t¯of ECG to each term t of EF+. Of course we wantx ¯, when x is a variable, to be a variable of ECG. One may be tempted to just sayx ¯ is x, since after all one can consider that the list of variables v1, v2,... is the same for any first-order theory; but we want to define our interpretation so that it will be the inverse of the interpretation φ¯ defined in the previous section. That interpretation takes the k-th point variable pk of ECG onto (p9k,p9k+1), etc. We therefore definex ¯ in a more complicated way. The reasons for arranging the definition in this way will become apparent in Lemma 12.59. Recall that X(b) is a term giving the intersection point of Perp(b, Line (0, 1)) with Line (0, 1); in other words the “x-coordinate of b”. Similarly Y (b) gives the intersection point of Perp(b,K) with K = Perp(0, Line (0, 1)), that is, the “y-coordinate of b”. Here is the complete definition ofx ¯ for variables x of EF+.

Definition 12.47.

v9k := X(pk)

v9k+1 := Y (pk)

v9k+2 := X(pointOn1 (ℓk))

v9k+3 := Y (pointOn1 (ℓk))

v9k+4 := X(pointOn2 (ℓk))

v9k+5 := Y (pointOn2 (ℓk))

v9k+6 := X(center (ck))

v9k+7 := Y (center (ck))

v9k+8 := Radius (ck)

Thus when x is a variable of EF+,x ¯ is generally not a variable, but a term of ECG. That term contains a variable, which we call x∗. Speciﬁcally

+ Definition 12.48. Let vk be the variables of EF , and pk, ℓk, and ck the Point, Line, and Circle variables of ECG. To each variable x of EF+ we associate a variable x∗ of ECG as follows: FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 193

If x is v9k+j then x∗ is pk if j = 0, 1, or ℓk if j = 2, 3, 4, 5, or ck if j = 6, 7, 8. If x is a list of variables then x∗ is the list of u∗ for u in list x. A “group of variables” is either v9k, v9k+1 , or v9k+2, v9k+3, v9k+4, v9k+5 , or { } { } v9k+6, v9k+7, v9k+8 . { } Then x and y are in the same group if and only if x∗ is the same variable as y∗. The point of the deﬁnition is that the groups will be used to represent coordinates specifying points, lines, and circles, respectively. It takes two coordinates for a point, four for a line, and three for a circle (center and radius). Here is the deﬁnition of t¯ for any term t of EF+, or any list of two, three or four terms: Definition 12.49. 0¯ is 0 (another name for α) 1¯ is 1 (another name for β)

γ1 is X(γ) where X is as deﬁned just above

γ2 is Y (γ) where Y is as deﬁned just above

γ3 is X(pointOnCircle (Circle (0, 1))) x¯ is as deﬁned above, when x is a variable t + s is Add (t¯+¯s) t s is HilbertMultiply (t,¯ s¯) · √t is SquareRoot (t¯) 1/t is Reciprocal (t¯) (a,b) := MakePoint (¯a, ¯b) (a,b,c,d) := Line (MakePoint (¯a, ¯b), MakePoint (¯c, d¯)) (a,b,c) := Circle (MakePoint (¯a, ¯b),

MakePoint (Add (¯a, HilbertMultiply (γ3, c¯)), Add (¯b, HilbertMultiply (SquareRoot (Add (1,

Minus (HilbertSquare (γ3)))), c¯)))) In the last line, c is the radius of the circle, so to get a point on the circle, we take a point one radius away from the center, at an angle whose cosine is γ¯3. Since the term given there is a bit long, we give an equivalent construction script (but with bars omitted): f(Point a, Point b, Point c) { e = MakePoint(a,b) // the center of C x = Add(a,HilbertMultiply(gamma3,c)) // x-coordinate of pointOnCircle(C) s = Sqrt(Add(1,Minus(HilbertSquare(gamma3)))) // sine of the angle y = Add(b,HilbertMultiply(s,c)) // y-coordinate of pointOnCircle(C) p = MakePoint(x,y) // pointOnCircle(C) return Circle(e,p) } 194 MICHAEL BEESON

We next want to define φ¯ for every formula φ of EF+. The definition of x φ(x) is somewhat complicated. There is only one type of variable in EF, for∀ field elements; but there are three types of variables in ECG, and the correspondence we have already set up uses the variables of EF in different ways, according to their index (subscript) in the official list of variables. Recall that of the nine variables starting with v9k, the first two represent (under the ◦ interpretation) coordinates of a point pk, the next four represent coordinates of two points of the line ℓk, and the last three represent the center and radius of a circle ck. The following definition is designed to make φ φ¯ the inverse of that interpretation. 7→ Definition 12.50.

P (t) is B(t,¯ 0, 1) t¯= 0) ¬ ∧ 6 t = s is t¯=¯s t is t¯ ↓ ↓ φ ψ is φ¯ ψ¯ ∧ ∧ φ ψ is φ¯ ψ¯ ∨ ∨ φ ψ is φ¯ ψ¯ → → φ is φ¯ ¬ ¬ xφ is x∗ φ¯ ∀ ∀ xφ is x∗ φ¯ ∃ ∃ The strange thing about this definition is that different variables x can correspond to the same x∗. For example, v9k and 9k+1 both correspond to pk. Hence if a formula has the form v9k v9k+1φ, its interpretation is going to get ∀ ∀ two quantifiers over pk. That does make sense: when we introduce an arbitrary point to represent an arbitrary field element, the other coordinate of the point is not yet used. These double, triple, or quadruple quantifiers can of course be erased (up to provable equivalence), so we have

¯ v9k v9k +1φ pkφ ∀ ∀ ↔ ∀ ¯ v9k+2 v9k +3φ v9k +4φ v9k +5φ ℓkφ ∀ ∀ ∀ ∀ ↔ ∀ ¯ v9k+6 v9k +7 v9k +8φ ckφ ∀ ∀ ∀ ↔ ∀ ¯ v9k v9k +1φ pkφ ∃ ∃ ↔ ∃ ¯ v9k+2 v9k +3φ v9k +4φ v9k +5φ ℓkφ ∃ ∃ ∃ ∃ ↔ ∃ ¯ v9k+6 v9k +7 v9k +8φ ckφ ∃ ∃ ∃ ↔ ∃

It might seem natural to take this as the definition, but then one has to worry about defining the interpretation when not all the variables in the same group are quantified; in essence the definition given above says, when some of the variables in a group are not quantified over, the right hand side is still the same as it would be if they all were quantified over. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 195

Example. Suppose we start with the formula φ equal to x(on (x, L)). Let us ∃ suppose L is actually the variable ℓ3 and x is the variable p2. We compute

φ◦ ( ℓ3(on (p2,ℓ3))◦ ↔ ∃ v9 3+2, v9 3+3, v9 3+4, v9 3+5(on(p2,ℓ3)◦ ↔ ∃ · · · · v29, v30, v31, v32((p◦ ((ℓ◦)1, (ℓ◦)2)) (p◦ ((ℓ◦)3, (ℓ◦)4)) = 0) ↔ ∃ 2 − 3 3 × 2 − 3 3 (Recall that the ◦ interpretation uses a cross product to interpret on .) Now p2◦ = (v18, v19), and ℓ3◦ = (v29, v30, v31, v32). Then

φ◦ v29, v30, v31, v32(((v18, v19) (v29, v30)) ((v18, v19) (v31, v32)) = 0) ↔ ∃ − × − Now we interpret back into ECG. We want to compute φ◦ and get back the original φ. The sequence of quantiﬁers v29, v30, v31, v32 will be converted back ∃ to ℓ3, so we get ∃ φ ℓ3ψ ◦ ↔ ∃ where ψ is the equation

((v18, v19) (v29, v30)) ((v18, v19) (v31, v32))=0. − × − Let us take this piece by piece. We have

(v18, v19)= MakePoint (v18, v19)

= MakePoint (X(p2), Y (p2))

= p2 by Lemma 10.10 Similarly

(v29, v30)= MakePoint (v29, v30)

= MakePoint (X(pointOn1 (ℓ2)), Y (pointOn1 (ℓ2)))

= pointOn1 (ℓ2) byLemma10.10 Thus the variables come out right in ECG. That was our point for now; to explain why the quantifiers are handled as they are in Definition 12.50. But since we started this example, we remark on how it will finish: the subtraction and cross product will be converted to arithmetic between points using Add , Minus , and HilbertMultiply . What we get back is thus not the original φ, but a formula in geometry expressing that the cross product used to interpret on is zero (not a formula in EF, a formula in ECG); and to prove that it is equivalent to the original φ, we will need to give geometric proofs of some formulas of algebra. That will be done below; for now we are just motivating the definition of φ¯. That completes the example. Next we have to investigate how substitution commutes with this interpretation. First we note that if x is a variable, generallyx ¯ is a term of the form X(MakePoint (a,b)) or Y (MakePoint (a,b)), where a and b are variables of ECG; hence by Lemma 10.10,x ¯ is provably equal to a variable; moreover, that variable is a subterm ofx ¯, which could be specified by a long definition by cases. We therefore define A[¯x := s] to mean the result of substituting s for the variable equal tox ¯ in A. In other words, we can substitute forx ¯ even though technically x¯ is not a variable. 196 MICHAEL BEESON

Lemma 12.51. For terms r of EF+, we have r[x := t]=¯r[¯x := t¯]. Proof. If x and y are diﬀerent variables of EF, thenr ¯ andx ¯ may contain more than one variable (each), but these variables will not clash: suppose x is v9J and y is v9K+1, with 0 j < 9 and 0 k < 9. Then the variables ofx ¯ are p9J and ≤ ≤ p9J+1, and the variables ofy ¯ are p9K and p9K+1, so they do not clash. Similarly for every other possible combination of indices j and k where x is v9J+j and y is v9K+k; it would take a long time to write out all the combinations, but the deﬁnition has been concocted to make this true. We proceed by induction on the complexity of r. If r is a constant or a variable other than x, the left side is justr ¯, and the right side isr ¯ too, sincer ¯ does not containx ¯. If r is the variable x, then r[x := t] = t¯ andr ¯[¯x := t¯] is also t¯. That takes care of the base case of the induction. Since substitution commutes with function symbols, there is nothing to prove for the induction step. That completes the proof. Example. We give an example to illustrate the previous lemma. Suppose r is v18 and x is also v18 and t is v20. Then

r¯ = v18 = X(p2)

x¯ =¯r = X(p2)

t¯= v20 = X(pointOn1 (ℓ2))

r[x := t]i = v20

r[x := t]= v20 = X(pointOn1 (ℓ2)) On the other hand

r¯[¯x := t¯]= X(p2)[X(p2) : X(pointOn1 (ℓ2))]

= X(pointOn1 (ℓ2)) = r[x := t] as calculated above The example is meant to illustrate whyr ¯ does not containx ¯: namely, all the variables mentioned inx ¯, where x is vk, have subscripts between 9k and 9k + 8, so they do not clash with variables arising fromy ¯ where y is another variable of EF.

Lemma 12.52. Let φ be a formula of EF. Then ECG proves φ[x := t] is equivalent to φ¯[¯x := t¯]. Proof. By induction on the complexity of φ. When φ is atomic, say r = s, we have φ[x := t] r = s[x := t] ↔ r[x := t]= s[x := t] ↔ r[x := t]= s[x := t] ↔ r¯[¯x := t¯]=¯s[¯x := t¯] byLemma12.51 ↔ (¯r =s ¯)[¯x := t¯] ↔ φ¯[¯x := t¯] ↔ FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 197 as required. When φ is atomic, of the form r , we have ↓ φ[x := t] r [x := t] ↔ ↓ r[x := t] ↔ ↓ r¯[¯x := t¯] by Lemma 12.51 ↔ ↓ r¯ [¯x := t¯] ↔ ↓ φ¯[¯x := t¯] ↔ as required. When φ is atomic, of the form P (r), we have φ[x := t] P (r)[x := t] ↔ P (r[x := t]) ↔ 0 < r[x := t] ↔ 0 < r¯[¯x := t¯] ↔ (0 < r¯)[¯x := t¯] ↔ P (r)[¯x := t¯] ↔ φ¯[¯x := t¯] ↔ as required. For the induction step, we consider the case of φ ψ; the other propositional connectives are treated in the same way. ∧ (φ ψ)[x := t] φ[x := t] ψ[x := t] ∧ ↔ ∧ φ[x := t] ψ[x := t] ↔ ∧ φ¯[¯x := t¯] ψ¯[¯x := t¯] by the induction hypothesis ↔ ∧ (φ¯ ψ¯)[¯x := t¯] ↔ ∧ φ ψ[¯x := t¯] ↔ ∧ as required. Now consider the case of a formula that begins with a quantifier. Refer to Definition 12.50 for the definition of φ¯ in this case; it is complicated. In particular we do not have yφ y¯φ¯ ∀ ↔ ∀ since in generaly ¯ is not even a variable; it is a term like X(pk). Instead, we have

yφ y∗φ,¯ ∀ ↔ ∀ where y∗ is given in Deﬁnition 12.48. Then ( yφ)[x := t] y(φ[x := t]) ∀ ↔ ∀ y∗φ[x := t] ↔ ∀ y∗φ¯[¯x := t¯] by the induction hypothesis ↔ ∀ yφ[¯x := t¯] ↔ ∀ as required. 198 MICHAEL BEESON

The case of an existential quantifier is treated exactly the same, just changing to . That completes the proof of the lemma. ∀ ∃ Theorem 12.53. Euclidean field theory is interpretable in ECG. That is, if EF+ proves φ then ECG proves φ¯. Proof. We first show that on (t,¯ Line (0, 1)) is provable for every term t of EF. This is proved by induction on the complexity of the term t. When t is 0 or 1, t¯ is α or β, and these both lie on Line (0, 1), since α = 0 and β = 1. When t is γ or a variable, t¯ is produced by the coordinate functions X and Y , whose values by definition lie on Line (0, 1). For all other terms, t¯ is either produced directly by X and Y , or is produced by Add , HilbertMultiply , SquareRoot , or Reciprocal , all of which are defined so that if their arguments lie on Line (0, 1), then so do their values. That completes the proof that t¯ lies on Line (0, 1) for all terms t. In case t is a list of two or three terms, then t¯ is (provably) a point or circle, respectively, according to the definition of t¯, since Circle3 is everywhere defined, and MakePoint is everywhere defined since we are assuming the strong parallel axiom. In case t is a list of four terms such that (t), then t¯ is a line, because the two points specified by the first two and last twoL members of the list will be distinct, by definition of . We have already checkedL that the interpretations of the commutative ring axioms are provable, using the strong parallel axiom of ECG. Technically, we ought to exhibit formal proofs in ECG of the (interpretations of) the ring axioms, and it may well be possible to produce such proofs using a theorem-prover or proof-checker, but we rely on the reader to be convinced that such proofs exist based on an examination of the detailed constructions given above. We turn to the verification of the (interpretations of the) axioms involving P (x). Let L = Line (0, 1). We note that by Lemma 9.91, P¯(x) is equivalent (for x on L) to Right (x, 0, I), where as above I is a point on the line K perpendicular to L at 0 such that Right (1, 0, I) and 0I = 01. Recall also that we defined xy> 0. Suppose then that v does not lie on Ray (0,D); then v and D are on opposite sides of L, since y is between them. ID does not meet L, since by Lemma 9.48 and the fact that ID and L are both perpendicular to K. Hence by Lemma 9.83, D and I are on the same side of L. Since v and D are on opposite sides of L, by Lemma 9.58, v and I are on opposite sides of L. Since FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 199

I and u are on the same side of L, u and v are on opposite sides of L. But uv and L are both perpendicular to K, and they do not coincide since x> 0. Hence by Lemma 9.48 they are parallel, so uv cannot meet L, which is a contradiction, since u and v are on opposite sides of L. That shows that v cannot fail to lie on Ray (0,D). But since lying on a ray is defined by a negation, v lies on Ray (0,D). That completes the verification that Add(x, y) is positive if x and y are positive. Next we check that HilbertMultiply (a,b) is positive if a and b are positive. Refer to Fig. 6. Let C be the circle passing through a, b, and I. We must show that if a and b lie on Ray (0, 1) and are not zero, then the fourth point of intersection z lies on Ray (0, I) and is not 0. Since lying on a ray is defined by a negation, we can prove this by cases. The first case is z = I, i.e., a and b are reciprocals and C is tangent to K at I. Then z = I lies on Ray (0, I). The second case is z = 0; then there are three points of C on L, contradicting Lemma 9.52. Now suppose z = 0 and z = I. If z and I are on opposite sides of L, then L meets chord zI of circle6 C. By6 Lemma 9.76, L meets C in two points on opposite sides of K, which coincides with Line (z, I). But it meets C in a and b, which are on the same side of K, and by Lemma 9.52, those are the only two intersection points. This is a contradiction. Hence z and I are not on opposite sides of L; that is, B(z, 0, I). Hence z lies on Ray (0, I), which was what we had to show. That completes¬ the verification that the product of positive elements is positive. Next we check axiom EF3, which is x + y = 0 (P (x) P (y)). Let L = Line (0, 1). The interpretation of this axiom is → ¬ ∧ on(x, L) on(y,L) (Add (x, y)=0 (Pˆ(x) Pˆ(y)) ∧ → → ¬ ∧ Assume x and y are on L and Add (x, y) = 0. Then y = Minus (x), as can be proved as follows using the laws of ring theory that we have already checked in Lemma 12.9: Add (x, y)=0 Add (Minus(x), Add(x, y)) = Add (Minus(x), 0) Add (Add (Minus(x), x),y)= Minus (x) Add (0,y)= Minus (x) y = Minus (x) Thus we have to prove Pˆ(x) Pˆ(Minus (x)). Suppose Pˆ(x) Pˆ(Minus (x)); we must derive a contradiction.¬ Then∧ B(x, 0, 1) and x = 0 and ∧B(Minus (x), 0, 1). Then by definition of SameSide , x ¬and Minus (x) areonthe6 same¬ side of0, i.e. on the same side of the line perpendicular to L at 0. But by definition of Minus (x), if x = 0 then B(x, 0, Minus (x)), so x and Minus (x) are on opposite sides of 0. Then6 by the plane separation theorem (Lemma 9.58), x and x are on opposite sides of L; that is, for some y we have B(x, y, x), contradicting Axiom B1-ii. That completes the verification of axiom EF3. Next we check EF4, which is x + y =0 P (x) P (y) x = 0. As above, Add(x, y) = 0 implies y = Minus (x), so what∧ ¬ has∧ to ¬ be proved→ is that, for x on L, (x > 0) and (Minus (x) >)0 implies x = 0. Since equality is stable, we can¬ prove x = 0 by¬ contradiction, so assume x = 0 and (x> 0) and (y > 0). 6 ¬ ¬ 200 MICHAEL BEESON

Then with y = Minus (x) we have B(y, 0, x) and B(x, 0, 1) and B(y, 0, 1). Since B(x, 0, 1) and B(x, 0,y), 1 and y are on the same side of 0. Hence by Lemma 9.58, B(1, 0,y). But this contradicts B(y, 0, 1). That completes the verification of ¬axiom EF4. Now we turn to axiom EF5, which is x + y =0 P (y) z(z z = x). The interpretation is that, for x, y on L, if Add (x, y)=0∧ ¬ P¯(→y) then ∃ for· some z on L, HilbertMultiply (z,z)= x. Suppose x and y are on∧L ¬and Add (x, y) = 0; then y = Minus (x). Suppose P¯(y); then y > 0, that is, Minus (x) 0; that is x lies on Ray (0, 1). Let SquareRoot¬ be defined¬ by the construction≤ script in Section 5.5. Let z = SquareRoot (x). Then by Lemma 12.10, HilbertMultiply (z,z)= x. That completes the verification of axiom EF5. There are axioms about the special constants γ2 and γ3. The axiom γ2 0, ≥ or officially P ( γ2), becomes 0 γ2, which is 0 Y (γ); according to the definition of¬ in−ECG, this means≤ Right (Y (γ), 0, I≤). By Lemma 10.9 this is ≤ 2 provable. The axiom γ 1 becomes HilbertSquare (γ3) 1. By definition, 3 ≤ ≤ γ3 = X(pointOnCircle (C)), where C = Circle (0, 1). Let p = pointOnCircle (C); then we have to prove HilbertSquare (X(p)) 1. By Lemma 12.31, we have ≤ Add (HilbertSquare (X(p)), HilbertSquare (Y (p))) = Add (HilbertSquare (X(1)), HilbertSquare (Y (1))).

But X(1) = 1 and Y (1) = 0, so the right-hand side is equal to Add (1, 0), which is 1, because we have already checked the commutative ring axioms. By Lemma 12.12, HilbertSquare (Y (p)) 0. Adding HilbertSquare (X(p)) to both sides, which is legitimate since a b≥is equivalent to Add (b, Minus (a)) 0, and we have already veriﬁed the ﬁeld≤ laws, we obtain ≥

Add (HilbertSquare (X(p), HilbertSquare (Y (p)) HilbertSquare (X(p)). ≥ But the left side is 1, so we have 1 HilbertSquare (X(p)) as desired. Next we verify Markov’s principle≥ P (x) P (x). The interpretation of this is ¬¬ →

on(x, Line (0, 1)) ( ( B(x, 0, 1) x = 0) ( B(x, 0, 1) x = 0)). → ¬¬ ¬ ∧ 6 → ¬ ∧ 6 The double negation can be pushed inside the conjunction and dropped, making the statement equivalent to one of the form A (B B), which is provable. → → We now turn to the veriﬁcation of Axiom EF7′, which is

x =0 (x (1/x)=1), 6 → · i.e. nonzero elements have reciprocals. The interpretation of this is that if a is a point on L different from 0, then there is a point b on L such that HilbertMultiply (a,b) = 1. We take b to be the point Reciprocal (a), where Reciprocal is defined by the construction script given just before Lemma 12.14. Then Lemma 12.14 gives the desired conclusion. We have now verified the interpretation of every non-logical axiom of EF+. Let us check the logical axioms of LPT. The first one is xA t A[x := t]. ∀ ∧ ↓ → FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 201

We have xA t A[x := t] ∀ ∧ ↓ → xA t¯ A[x := t] ↔ ∀ ∧ ↓ → xA t¯ A¯[¯x := t¯] byLemma12.52 ↔ ∀ ∧ ↓ →

Now in the formula xA, A might begin with other universal quantifiers over variables in the same∀ group as x, where “same group” is defined just after Defi- nition 12.47. Throwing these variables into a list y we can write xA in the form x yB. Then Definition 12.50 tells us ∀ ∀ ∀ x,yB x∗B¯ ∀ ↔ ∀ Then we can continue the argument: xA t A[x := t] ∀ ∧ ↓ → x,yB t¯ A¯[¯x := t¯] ↔ ∀ ∧ ↓ → x∗B¯ t¯ A¯[¯x := t¯] ↔ ∀ ∧ ↓ → x∗B¯ t¯ yB[¯x := t¯] ↔ ∀ ∧ ↓ → ∀ x∗B¯ t¯ ( y∗ B¯)[¯x := t¯] ↔ ∀ ∧ ↓ → ∀ But y∗ and x∗ are the same variable (which was why we did not write y∗ on the left); so the last formula is another instance of the axiom schema whose intepretation we are verifying. That completes the verification of that axiom. The axiom A[x := t] t x A is treated similarly; the only differences are that we have instead∧ of↓ →, and ∃ the quantifier appears on the right instead of the left of the implication∃ sign.∀ That completes the base case of an induction on the length of proofs in EF+(except for any logical axioms there may be). In addition to checking the logical axioms, for the induction step we have to check that the rules of inference preserve the interpretation. For example if modus ponens is a rule of inference, we have to check that if φ¯ and φ ψ are provable, then so is ψ. That is evident since φ ψ is just φ¯ ψ¯. Actually,→ we did not specify the exact rules of inference→ of any of our theories,→ having been content to mention “multi-sorted intuitionistic predicate calculus with the logic of partial terms.” But since the definition of φ¯ commutes with the propositional connectives, and since it almost commutes with quantifiers (except for changing variables x to x⋆, it does not matter whether we choose, for example, a Gentzen-style formulation or a Hilbert-style formulation; in either case the induction step is just commuting the interpretation with the connectives or quantifiers to get another instance of the same inference rule or logical axiom. That completes the proof of the theorem. 12.9. Faithfulness of the interpretations. We have now proved that the interpretation φ¯ from EF to ECG is sound, and the interpretation φ◦ from ECG to EF+ is sound; that is, if φ is provable then its interpretation is provable. An interpretation is called faithful if the interpretation of a formula is provable if and only if the formula is provable. In this section, we prove that both the 202 MICHAEL BEESON interpretations we have introduced are faithful. The reason is simple: they are inverses. Before embarking on the details, we comment on the significance of this result. It shows that geometry is completely equivalent to algebra: each is reduced to the other, and nothing is added or left out by the reductions. With classical theories, the issue of faithfulness is handled by model theory and an appeal to Gödel’s completeness theorem. With intuitionistic theories, we do not have that option. Besides, even for classical theories, it is of interest to have an algorithm for translating, say, an algebraic proof into an axiomatic geometric proof. For example, there are various computer programs that “prove” geometry theorems by translating them to algebra, and then using computer algebra techniques. Using the work in this section, it would be possible to write computer programs to convert those algebraic “proofs” (which are really calculations) first to formal proofs in field theory, and then to formal proofs in ECG (or ECG plus classical logic). That possibility, however, is not the main reason for studying the faithfulness of the interpretations. It is simply to check that these things work out as expected; any failure would be a surprise and would be either a “bug” in the formal systems or the definitions of the interpretations, or would represent a new and scientifically interesting phenomenon. The latter is illustrated by the fact that the interpretations are not faithful for Euclid 5 or Playfair, but only for the strong parallel axiom. The essence of what is required to prove the interpretations faithful is that we have to formalize analytic geometry in ECG. We had to define arithmetic just to define the interpretation from field theory to geometry, but now we will have to show, for example, that the points on a line are exactly those satisfying the equation of the line. Euclid proved the Pythagorean theorem, but in a different sense of “area” than we now use; we will need to derive the modern form, which uses arithmetic to calculate the areas. We now begin the technical preparations. We start with some lemmas that help us compute t◦ for various terms t.

Lemma 12.54. Perp(p,L)◦ = (q1, q2, r1, r2), where (q1, q2) and (r1, r2) are the coordinates of the two intersection points of circles C and K, whose centers are given by the intersection points of L and the circle centered at p with radius r = a p◦ + a b where a = (L1◦,L2◦) and b = (L3◦,L4◦); the circles C and K pass| through− | each| − other’s| centers. Proof. Perp(p,L) is the line deﬁned by the two intersection points of two circles centered on L and passing through p. Since the ◦ interpretation preserves incidence the lemma follows. A detailed proof would require computing the ◦ interpretation of each line of the script deﬁning Perp; we omit those details. That completes the proof. Lemma 12.55. The interpretation of being perpendicular is that the interpreted line vectors have dot product zero. Formally,

(L K)◦ = ((L◦ L◦), (L◦ L◦)) ((K◦ K◦), (K◦ K◦))=0 ⊥ 1 − 3 2 − 4 · 1 − 3 2 − 4 Proof. By definition of perpendicular, L and K are perpendicular if there exist points a, m, and b on L and c on K such that m is also on K and triangle amc is FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 203 congruent to bmc. Under the ◦ interpretation we have a◦ m◦ = b◦ m◦ , etc. The proof then reduces to an exercise in the computation| − of| the| dot− product,| which we omit. That completes the proof. Lemma 12.56. EF+ proves a = b c = b on (d, Line (b,c)) 6 ∧ 6 ∧ d1◦ b1◦ b1◦ Rotate (a,b,c,d)◦ = A − + → d◦ b◦ b◦ 2 − 2 2 where A is the matrix of the rotation; specifically A A A = 11 12 A21 A22 where (a◦ b◦) (c◦ d◦) A = A = − · − 11 22 a b c d | ◦ − ◦|| ◦ − ◦| (a◦ b◦) (c◦ d◦) A12 = A21 = − × − − a b c d | ◦ − ◦|| ◦ − ◦| Remark. We cannot mention cos θ and sin θ in EF+, but intuitively, the entries of the rotation matrix involve those quantities, which are here computed in terms of dot product and cross product. Note that no case distinctions are needed. Proof. The denominators are nonzero by the hypotheses a = b and c = b; so at least the matrix A is defined. We claim that the determinant6 of the matrix6 A is 1. If we define (a◦ b◦) (c◦ d◦) cos θ = − · − a b c d | ◦ − ◦|| ◦ − ◦| (a◦ b◦) (c◦ d◦) sin θ = − × − a b c d | ◦ − ◦|| ◦ − ◦| then the claim boils down to sin2 θ + cos2 θ = 1. Since we are working in EF+, this argument is only motivation, not proof. The actual proof consists in writing the identity out in terms of the components of a, b, c, and d and simplifying. What has to be verified is an instance of the following proposition: if u and v are unit vectors, then (u v)2 + (u v)2 = 1. Writing this out by components we have · × (u v)2 + (u v)2 · × 2 2 = (u1v1 + u2v2) + (u1v2 u2v1) 2 2 2−2 2 2 2 2 = (u1v1 +2u1v1u2v2 + u2v2 ) + (u1v2 2u1v2u2v1 + u2v1) 2 2 2 2 2 2 2 2 − = u1v1 + u2v2 + u1v2 + u2v1 2 2 2 2 = (u1 + u2)(v1 + v2) =1 since u and v are unit vectors Hence the determinant of A is indeed 1. According to Lemma 9.101, the value x of Rotate (a,b,c,d) is a point on Line (b,a) such that bx = bd and Right (d, b, x) if and only if Right (c,b,a). There is only one such point x, so if we show the point defined in the lemma 204 MICHAEL BEESON and asserted to be Rotate (a,b,c,d)◦ has those two properties, then indeed it is Rotate (a,b,d,c)◦. Let us write Ax◦ for the result of multiplying matrix A by the column vector with entries x1◦ and x2◦. Then the point defined in the lemma is z = A(d◦ b◦)+b◦. We have −

z b◦ = (A(d◦ b◦)+ b◦) b◦ | − | | − − | = A(d◦ b◦) | − | = d◦ b◦ since A has determinant 1 | − | Since A is a linear transformation, it takes d to a point on the line determined by b◦ and a◦. There are only two points on that line at the distance d◦ b◦ | − | from b◦. One of them is z and the other is on the other side of b◦, so just one of them satisfies the second condition of the lemma. Hence if we show that z satisfies that condition, we will be finished. Right (c,b,a)◦ is equivalent to (c◦ b◦) (a◦ b◦) > 0, as we verified in − × − the proof of Theorem 12.21; and that condition is equivalent to A21 > 0. Now Right (d,b,z) is equivalent to (d◦ b◦) (z◦ b◦) > 0, so we calculate that cross − × − product. We have z◦ = A(d◦ b◦)+ b◦. For simplicity we set u = d◦ b◦; then − − z◦ b◦ = Au◦. Then − (d◦ b◦) (z◦ b◦)= u Au − × − × = u1(Au)2 u2(Au)1 − = u1(A21u1 + A22u2) u2(A11u1 + A12u2) 2 2 − = A21u1 A12u2 + u1u2(A22 A11) 2 − 2 − = A21u1 A12u2 since A11 = A22 −2 2 =2A21(u + u ) since A21 = A12 1 2 − 0 if and only if A21 > 0 ≥ Hence Right (c,b,a)◦ if and only if Right (d,b,z)◦. Hence z is the correct point, i.e. z = Rotate (a,b,c,d)◦ as claimed. That completes the proof of the lemma. Lemma 12.57. EF+ proves the following:

I◦ = (0, 1)

MakePoint (x, y)◦ = (x◦,y◦) Proof. We follow the script given in Deﬁnition 9.103 to deﬁne I. For convenience we repeat that script here; you may wish to refer to Fig. 31, which we do not repeat. I() { L = Line(0,1) J = Perp(0,L) K = Perp(gamma,J) p = IntersectLines(J,K) return IntersectLineCircle2(Line(0,p),Circle(0,1)) } FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 205

We have

L◦ = (0, 0, 1, 0)

We now calculate J ◦ according to Lemma 12.54. The circles C and K mentioned in that lemma have radius 1, and are centered at ( 1, 0) and (1, 0). In order to pass through each other’s centers they have radius− 2. Therefore they meet the y-axis at (0, √3). Hence ± J ◦ = (0, √3, 0, +√3) − The order of the two points represented in J ◦ is determined by the script for Perp, although it is not stated explicitly in Lemma 12.54 and does not matter. + Now we calculate K◦. Recall γ◦ = (γ1,γ2), and γ2 > 0 is an axiom of EF . This time, the radius of the circle used to ﬁnd the centers of the new C and K is

2 2 r = (γ2 + √3) + γ1 +2√3 q so the coordinates of the centers are

2 2 0,γ2 r γ1 ± q − and the intersection points of the two circles are ( c,γ2) for a certain c given by a solving a quadratic equation. The value c is thrown± away anyway when we calculate p◦ in line 4 of the script, and we find p◦ = (0,γ2). Since γ2 > 0 then in the last line of the script the return value is (0, 1). That is the value of I◦. That proves the first claim of the lemma. (It may seem like a lot of work for an obvious result, but the point is that proving these things formally shows that the machinery is set up correctly.) Consider MakePoint (a,b)◦; recall from Definition 10.8 that MakePoint is defined by the following script: MakePoint(Point x, Point y) { L = Line(0,1) J = Perp(0,L) U = Perp(x,L) z = Rotate(1,0,I,y) V = Perp(z,J) return IntersectLines(U,V) }

We will go through the script line by line and calculate the ◦ interpretation of each line. In line 1 we have L◦ = (0, 0, 1, 0). In line 2 as above we have J ◦ = (0, √3, 0, √3). In line 3 we have U ◦ = (x, c,x,c) where again c could be explicitly− calculated but does not matter. According− to Lemma 9.101 we have 2 1 z◦ = ( y ,y ). According to Lemma 9.56, J L and U L. Let us abuse − ⊥ ⊥ the dot-product notation by writing L◦ K◦ for the dot product mentioned in · Lemma 12.55. Then by that lemma we have J ◦ L◦ = 0 and U ◦ L◦ = 0. Hence U and J are in the same direction. Since z lies on· V by Lemma· 9.56, all points on V have the same second coordinate as z◦, which is y1◦, and since U contains 206 MICHAEL BEESON x, all points on U have the same ﬁrst coordinate as x◦, namely x1◦. Hence the intersection point has coordinates (x1◦,y1◦). That completes the proof. The next lemma says that that EF proves the correctness of the algorithms HilbertMultiply , Add , and SquareRoot .

+ Lemma 12.58. EF proves the following, assuming a2◦ = b2◦ =0:

HilbertMultiply (a,b)◦ = (a◦ b◦, 0) 1 · 1 Add (a,b)◦ = (a1◦ + b1◦, 0) Minus (a)◦ = ( a◦, 0) − 1 Reciprocal (a)◦ = (1/a◦, 0)

P ( x◦) SquareRoot (x)◦ SquareRoot (x)◦ = (x◦, 0) ¬ − 1 → · 1 Proof. First we consider the term Reﬂect (a,L)◦. That is a term in EF that gives the reﬂection of a◦ in line L◦. In particular

Reﬂect (a, Line (0, I))◦ = ( a◦,a◦). − 1 2 We start with the theorem about Minus .

Minus (a)◦ = Minus (MakePoint (X(a), Y (a)) = Reﬂect (MakePoint (X(a), Y (a)), Line (0, I))

Now apply ◦ to both sides.

Minus (¯a)◦ = Reﬂect (¯a, Line (0, I))◦

Minus (¯a)◦ = Reﬂect ◦(MakePoint ◦(X(a)◦, Y (a)◦), Line (0, I)◦)

= Reﬂect ◦(MakePoint ◦(a0◦,a1◦), (0, 0, 0, 1))

= Reflect ◦((a0◦,a1◦), (0, 0, 0, 1)) = ( a◦,a◦) as discussed above − 0 1 = a◦ − as claimed. HilbertMultiply (a,b)◦ is a term of EF that gives the fourth intersection point of circle C with the y-axis, where the first three intersection points are (a◦, 0), + (b◦, 0), and (0, 1). We have to prove in EF that that fourth intersection point is (0,a b). The way the term HilbertMultiply is defined, the formula will first · construct the perpendicular bisector of the line segment from (a◦, 0) to (0, 1), and the line x = (a◦ + b◦)/2, and then find the center of the circle at the intersection point of those lines, which will be

e◦ = ((a b + 1)/2, (a◦ + b◦)/2). · The radius of the circle is e 1 . Then the fourth intersection point is (0,y) where y solves | − | e (0, 1) 2 = (0,y) e 2. | − | | − | That is, ab +1 2 ab+ 2 y = 1 − 2 − 2 FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 207

It is an easy piece of algebra to check that the solutions are just y = 1 and y = ab; that algebra can be done in EF, since only the ring laws are involved. (Denominators of 2 can be cleared). Similarly, Reciprocal (a)◦ is the second point on the x-axis lying on the circle that is tangent to the y-axis at (0, 1), and passes through (a1◦, 0). That circle’s center is at (a + (1/a))/2, 1) The term Add involves Rotate , so we first check that Rotate ◦ really performs a rotation. We only need rotations by plus or minus ninety degrees to define Add , using angles with one side on the x-axis. Rotate is defined by bisecting an angle and dropping two perpendiculars, so it has to be checked that the construction we gave for bisecting an angle, at least, a ninety-degree angle with its sides on the x and y axis, is analytically correct. That algorithm involved constructing the points at distance one from the vertex (that would be ( 1, 0) and (0, I)), and then drawing the circles with those centers passing through± the other point,± and then connecting their intersection points, which will be in the four cases at hand will be points on the lines y = x, so the angle bisectors will be those lines; then when we drop perpendiculars± from points on the x-axis or y-axis to those lines, and then again to the target axis, the effect will be the same as a rotation. For example, Rotate (1, 0,I,x)◦ will be ( x2◦, x1◦). To describe this analytically we can represent rotations as two-by-two− matrices and show that

0 1 x1◦ Rotate (1, 0,I,x)◦ = − . 1 0 x2◦ 

0 1 x1◦ Rotate (I, 0, 1, x)◦ = . 1 0 x◦ − 2 More generally to rotate clockwise about the vertex (b1◦, 0), the result is

0 1 x◦ b◦ b◦ 1 − 1 + 1 . 1 0 x◦ 0 − 2 Now, the script (or term) for Add (a,b)◦ calls for a rotation of (a0◦, 0) by the ﬁrst matrix, followed by a projection onto the vertical line at (b1◦, 0), followed by a rotation by the second matrix. In order to verify the correctness of Add , the result should be (a1◦ + b1◦, 0). Let us check it in EF. We start with (a1◦, 0) (where a1◦ is not assumed to have any particular sign). Rotating by the ﬁrst matrix, we obtain (0,a1◦). Projecting we have (b1◦,a1◦). Using the equation given above for the second rotation, we have

0 1 b◦ b◦ b◦ b◦ 1 − 1 1 1 1 0 a◦ 0 0 − 1 a + b = 1◦ 1◦ 0 as desired. That completes the correctness proof of Add . Next we turn to the correctness of SquareRoot . According to the algorithm, we ﬁnd SquareRoot (x)◦ = (y1◦, 0) by ﬁrst constructing the circle with center at (x1◦ +1)/2, 0 and passing through 0, and then computing the length of the vertical chord over (x1◦, 0). So the radius of the circle is r = (x◦ + 1)/2, and if y is the 208 MICHAEL BEESON height of the chord we have

2 2 x◦ +1 2 x◦ +1 x◦ + (y◦) = − 2 2 

2 which is easily solved to derive (y◦) = x◦. That completes the proof of the lemma.

Lemma 12.59. For each term t of EF+, or list of two or four terms, + ¯ EF proves t ∼= t◦. When t is a list of three terms t = (a,b,c), we have EF+ proves t = (a,b, c ). | | Remark. Lists of three terms represent circles given by center and radius. We have arranged it so a negative value r in the third position is converted byr ¯ to a positive radius r . So when we taker ¯◦ we get back r instead of the original r. | | | |

Proof. We proceed by induction on the complexity of t. First consider the basis case when t is a constant. If t is a constant, it is 0, 1, γ1, or γ2. If t is 0, then t¯ = α and t¯◦ = α◦ = 0. If t is 1, then t¯ = β and t¯◦ = β◦ =1. If t = γ1, then t¯= X(γ1), and

t¯◦ = X(γ)◦

Here X(γ) is by definition a term of ECG giving the foot of the perpendicular from γ to Line (0, 1). Therefore it (provably in ECG) satisfies the geometric formula A(γ,X(γ)) expressing that it is the foot of that perpendicular, where A(p, q) is on(q, Line (0, 1)) qp Line (0, 1). By the soundness of the ◦ interpre- ∧ ⊥ tation, X(γ)◦ is the field element q◦ satisfying A◦(γ◦, q◦). Since γ◦ = (γ1,γ2), that field element q is γ1. Hence γ1◦ = γ1 as claimed. Similarly when t is γ2, but using Y instead of X. Now consider the basis case when t is a variable x. Then x occurs as the j-th variable vj for some j, andx ¯ was defined according to the value of j mod 9. There are 9 cases to consider. If x is v9k thenx ¯ = X(pk), where pk is the k-th Point variable of ECG. Thenx ¯◦ = (X(pk)◦) = (pk)0◦. By definition, pk◦ is the list (v9k, v9k+1), so (pk)0◦ = v9k = x as required. If x is v9k+1 thenx ¯ = Y (pk), where pk is the k-th Point variable of ECG. Thenx ¯◦ = (Y (pk)◦) = (pk)1◦. By definition, pk◦ is the list (v9k, v9k+1), so (pk)1◦ = v9k+1 = x as required. If x is v9k+2 thenx ¯ = X(pointOn1 (ℓk)), where ℓk is the k-th Line variable. Thenx ¯◦ = (ℓk)1◦. Now (ℓk)◦ is (v9k+2, v9k+3, v9k+4, v9k+5), as one can verify by checking the definition on page 153. Hence (ℓk)1◦ = v9k+2 = x as required. Sim- ilarly when x = v9k+3, v9k+4, v9k+5, using Y instead of x and/or pointOn2 (ℓk) instead of pointOn1 (ℓk). FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 209

If x is v9k+6, thenx ¯ = X(center (ck)), where ck is the k-th Circle variable. Then

x¯◦ = (ck)1◦

= (v9k+6, v9k+7)1

= v9k+6 = x as required. Similarly if x = v9k+7, with Y instead of x and (Ck)2◦ instead of (Ck)1◦. If x is v9k+8 then

x¯ = X(SquareRoot (HilbertSquare (Y (pointOnCircle (ck)),

Minus (Y (center (ck))))))) by deﬁnition ofx ¯, Deﬁnition 12.47

Applying ◦ to both sides we have

x¯◦ = X(SquareRoot (HilbertSquare (Y (pointOnCircle (ck)),

Minus (Y (center (ck)))))))◦

By Lemma 12.58, we can convert the geometrized arithmetic of SquareRoot and HilbertSquare to ﬁeld operations in EF:

2 x¯◦ = (Y (pointOnCircle (ck)) Y (center (ck)) ) ◦ − ◦ p 2 = (pointOnCircle (ck)2◦ center (ck)2) q − By deﬁnition

pointOnCircle (ck)◦ = (ck◦)1, (ck◦)2 + (ck◦)3 = (v9k+6, v9k+7 + v9k+8) by Deﬁnition 12.17 | center (ck) = (v9k+6, v9k+7)

center (ck)2 = v9k+7 so we have

2 x¯ = ((v9k+7 + v9k+8) v9k+7) ◦ | − p 2 = v9k+7 q = √x2 = x | | = x which is always deﬁned, sox ¯ = x as claimed. (That was the reason for using x | | instead of just x in the deﬁnition of v9◦k+6.) That completes the base case of the induction. 210 MICHAEL BEESON

Consider the case when t is r s for terms r and s. Then t¯= HilbertMultiply (¯r, s¯). · When we apply ◦ to both sides we get

(t¯)◦ = HilbertMultiply (¯r, s¯)◦

=¯r◦ s¯◦ by Lemma 12.58 1 · 1 = r s by the induction hypothesis ∼ · = t Similarly the cases when t = r + s or r are treated using Add or Minus in place of HilbertMultiply , and again appealing− to Lemma 12.58. Suppose t is √r. Then t¯= SquareRoot (¯r). Applying ◦ to both sides we have

(t¯)◦ = (√r)◦

= SquareRoot (¯r)◦

= √r¯◦ by Lemma 12.58 √ ∼= r by the induction hypothesis = t Now consider the case when t is 1/r. Then t¯= Reciprocal (¯r), so

t¯◦ = Reciprocal (¯r)◦ ∼= 1/r¯◦ by Lemma 12.58 ∼= 1/r by the induction hypothesis ∼= t Next consider the case when t is a list of two terms, t = (a,b). Then by deﬁnition t¯= MakePoint (¯a, ¯b), so

t¯◦ = MakePoint (¯a, ¯b)◦

= (¯a◦, ¯b◦) byLemma12.57 = (a,b) by the induction hypothesis = t as desired Next consider the case when t is a list of four terms, t = (a,b,c,d). The “is” of deﬁnition can be replaced formally by ∼=. (We cannot use = since perhaps a¯ = ¯b.) Then by deﬁnition t¯ is Line (MakePoint (¯a, ¯b), MakePoint (¯c, d¯)).

The “is” of deﬁnition can be replaced formally by ∼=. (We cannot use = since perhapsa ¯ = ¯b.) Applying ◦ to both sides we have ¯ ¯ ¯ t◦ ∼= Line (MakePoint (¯a, b), MakePoint (¯c, d))◦ = (X(MakePoint (¯a, ¯b))◦, Y (MakePoint (¯a, ¯b))◦), ¯ ¯ ∼= X(MakePoint (¯c, d))◦, Y (MakePoint (¯c, d))◦) ¯ ¯ ∼= (¯a◦, b◦, c¯◦, d◦) ∼= (a,b,c,d) by the induction hypothesis ∼= t as desired FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 211

Finally consider the case when t is a list of three terms, t = (a,b,c). Then by deﬁnition t¯= Circle (MakePoint (¯a, ¯b), MakePoint (¯a, Add (¯b, c¯))

Applying ◦ to both sides we have

t¯◦ = Circle (MakePoint (¯a, ¯b), MakePoint (¯a, Add (¯b, c¯))◦

= (X(MakePoint (¯a, ¯b))◦, Y (MakePoint (¯a, ¯b))◦,

MakePoint (¯a, ¯b)◦ MakePoint (¯a, Add (¯b, c¯))◦ ) | − | = (¯a◦, ¯b◦, (¯a◦, ¯b◦) (¯a◦, Add (¯b, c¯)◦) ) | − | = (a,b, (a,b) (a, Add (¯b, c¯)◦) ) by the induction hypothesis | − | = (a,b, (a,b) (a, ¯b◦ +¯c◦) ) byLemma12.58 | − | = (a,b, (a,b) (a,b + c)) by the induction hypothesis | − = (a,b, (0,c) ) | | = (a,b, c ) | | That completes the proof of the lemma.

Lemma 12.60. For each term t of ECG, ECG proves t = t◦. Moreover the following are also provable: (i) If p is a term of type Point, then

t = MakePoint (t1◦, t2◦)

X(t) = t1◦

Y (t) = t2◦ (ii) If C is a term of type Circle, then

X(center (C)) = C1◦

Y (center (C)) = C2◦

Radius (C) = C3◦ (iii) if L is a term of type Line, then

X(pointOn1 (L)) = L1◦

Y (pointOn1 (L)) = L2◦

X(pointOn2 (L)) = L3◦

Y (pointOn2 (L)) = L4◦

L = Line (MakePoint (L1◦, L2◦), MakePoint (L3◦, L4◦)) Proof. We proceed by induction on the complexity of the term. The basis case is when t is a variable or constant. Consider the case of a Point variable t = pk. Then t◦ is (v9k, v9k+1), so t1◦ is v9k, and t1◦ is X(pk), that is, X(t). t2◦ is v9k+1, and t2◦ is Y (pk), that is, Y (t). Then by Lemma 10.10 we have t = MakePoint (X(t), Y (t)) = MakePoint (t1◦, t2◦). 212 MICHAEL BEESON

But by deﬁnition,

t◦ = (v9k, v9k+1)

= MakePoint (v9k, v9k+1)

= MakePoint (t1◦, t2◦) = t as shown just above

Then X(t◦) = t1◦ and Y (t◦) = t2◦ by Lemma 10.10. That completes the case when t is a Point variable. Now consider the case of a line variable t = ℓk. Then t◦ is a list of length 4, giving the coordinates of two specified points on the line; specifically t◦ = (v9k+2, v9k+3, v9k+4, v9k+5). Then according to the definition ofx ¯ for variables x, we have

t1◦ = X(pointOn1 (t))

t1◦ = Y (pointOn1 (t))

t3◦ = X(pointOn2 (t))

t4◦ = Y (pointOn2 (t)) Applying Lemma 10.10, we have

pointOn1 (t)= MakePoint (t1◦, t1◦)

pointOn2 (t)= MakePoint (t3◦, t4◦) By Axiom CA1 we have t = Line (pointOn1 (t), pointOn2 (t)), and substituting the values we calculated for pointOn1 (t) and pointOn2 (t), we have

t = Line (MakePoint (t1◦, t1◦), MakePoint (t3◦, t4◦))

= t◦ by the definition ofu ¯ when u is a list of length 4 (see Definition 12.49). That completes the case of a Line variable. Now consider the case of a Circle variable t = ck. Then t◦ is a list of length 3, giving the coordinates of the center and the radius; specifically t◦ = (v9k+6, v9k+7, v9k+8), so t1◦ is v9k+6, t2◦ is v9k+3, and t3◦ is v9k+4. Then according to the definition ofx ¯ for variables x (Definition 12.47), we have

t1◦ = X(center (t)

t2◦ = Y (center (t)

t3◦ = Radius (t) Applying Lemma 10.10, we have

center (t)= MakePoint (t1◦, t1◦). That completes the case of a Circle variable. Now consider the case when t is constant. Then t is α, β, or γ. Suppose t = α. ¯ ¯ Then α◦ = (0, 0), and α1◦ = 0 = 0, and α2◦ = 0 = 0, so MakePoint (α1◦, α2◦) = MakePoint (0, 0)=0= α, proving (i) for the case t = α. Similarly when t = β, with (1, 0) in place of (0,0). Now suppose t = γ. Since γ is not given speciﬁc coordinates the argument is diﬀerent. Then t◦ = (γ1,γ2), and t1◦ = X(γ); FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 213 similarly t2◦ = Y (γ). Hence MakePoint (t1◦, t2◦) = MakePoint (X(γ), Y (γ)) = γ, by Lemma 10.10, completing the case t = γ. That completes the basis case of the induction. The induction step has a case for each function symbol of ECG. Consider the case when t is pointOn1 (L). Then t◦ = (L1◦,L2◦), and

t◦ = (L1◦,L2◦)

= MakePoint (L1◦, L2◦) by the deﬁnition of a,b = MakePoint (L1,L2) by the induction hypothesis which completes this case. The cases when t is pointOn2 (L), or center (C), or pointOnCircle (C) are treated similarly. Conside the case when t = Line (a,b). Then t◦ = Append(a◦,b◦) = (a1◦,a2◦,b1◦,b2◦) and

t◦ = (a1◦,a2◦,b1◦,b2◦)

= Line (MakePoint (a1◦, a2◦), MakePoint (b1◦, b2◦)) = Line (MakePoint (a1,a2), MakePoint (b1,b2)) by the induction hypothesis = Line (a,b) byLemma10.10 = t which completes the proof for the case t = Line (a,b). Consider the case when t = Circle (a,b,c). (Remember Circle (a,b) is just an abbreviation for Circle (a,a,b).) By deﬁnition of t◦ we have

t◦ = (a◦,a◦, b◦ c◦ ) 1 2 | − | t◦ = Circle (MakePoint (a1◦, a2◦), MakePoint (x, y)) where x and y are calculated according to the last clause of Deﬁnition 12.49, namely, with r = b◦ c◦ , | − |

x = Add (a1◦, HilbertMultiply (γ3, r¯))

y = Add (a2◦, HilbertMultiply (SquareRoot (Add (1, Minus (γ3))), r¯))

By the induction hypothesis we have a1◦ = X(a) and a2◦ = Y (a). Then MakePoint (X(a), Y (a)) = a by Lemma 10.10.

It remains to show that Radius (t)= t3◦. We have

t◦ = b◦ c◦ 3 | − | t = b c 3◦ | ◦ − ◦| = SquareRoot (HilbertSquare (Add (b◦, Minus (c◦))))

= Distance (b◦, c◦) = Distance (b,c) by the induction hypothesis = Radius (t) by deﬁnition of Radius 214 MICHAEL BEESON

Now we have proved that t and t◦ have the same center and the same radius. Hence by Lemma 12.31, they coincide. Then by Lemma 12.45, they are equal. That completes the case when t is Circle (a,b,c). Consider the case when t is pointOnCircle (C). Let C◦ = (a◦,b◦,c◦). Then by deﬁnition, t◦ is (x, y) where

x = a◦ + γ3 b◦ c◦ · | − | 2 y = 1 γ3 b◦ c◦ q − · | − | 2 = γ4 b◦ c◦ where γ4 := 1 γ · | − | − 3 p Then t◦ = MakePoint (¯x, y¯), and

x¯ = Add (a◦, HilbertMultiply (γ3, Distance (b◦, c◦)))

= Add (a, HilbertMultiply (γ3, Distance (b,c))) by the induction hypothesis Similarly

y¯ = Add (b, HilbertMultiply (γ4, Distance (b,c)))

γ4 = SquareRoot (Add (1, Minus (HilbertSquare (γ3))))

Let p = pointOnCircle (Circle (0, 1)). Then by deﬁnition γ3 = X(p) and we also have γ4 = Y (p). Putting that in we have x¯ = Add (a, HilbertMultiply (X(p), Distance (b,c))) y¯ = Add (b, HilbertMultiply (Y (p), Distance (b,c)))

Using the deﬁnition of VectorAdd we then have

t◦ = VectorAdd (center (C), ScalarMultiply (Distance (b,c),p)) = VectorAdd (center (C), ScalarMultiply (Radius (C), pointOnCircle (Circle (0, 1)))) But by Lemma 12.46 the right hand side is equal to pointOnCircle (C). That completes the case when t is pointOnCircle (C). Consider the case when t = IntersectLines (L,K). Recall that t◦ is a list of two terms (x◦,y◦) giving the coordinates of the intersection point in terms of a = (L1◦,L2◦, b = (L3◦,L4◦), c = (K1◦,K2◦, and d = (K3◦,K4◦). Then

t◦ = ((x◦), (y◦) = (x, y) by the induction hypothesis

So in order to prove t = t◦, we have to check in ECG that the algebraic formula for the solution in terms of a,b,c, and d is correct. That is, first we have used ◦ to write a matrix equation for the coordinates (x◦,y◦) of the intersection point, in terms of the coordinates of two points a,b on L and two points c, d on K; that it works is a piece of algebra (linear algebra at that), plugging in x◦ and y◦ and simplifying until the sides are the same. Since we have verified the ring laws are preserved under φ φ¯, the same simplification can be done using HilbertMultiply and Add in ECG7→ . Consider the case when t = IntersectLineCircle1 (L, C) or IntersectLineCircle2 (L, C). FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 215

Then L◦ gives the coordinates of two points on the line, namely a = (L1◦,L2◦) and b = (L3◦,L4◦), and the center of the circle is e = (C1◦, C2◦), and its radius is r = C3◦. By deﬁnition, we have

t◦ = a + λ(b a) − where λ is given by equation (1), which we repeat here: (a e) (b a) ((a e) (b a))2 (b a)2((a e)2 r2) λ = − · − ± − · − − − − − − p (b a)2 − where the plus sign is taken for IntersectLineCircle2 and the minus sign for IntersectLineCircle1 . The term now of interest, t◦, is given by replacing the arithmetic of EF by geometric arithmetic. Thus ¯ ¯ t◦ = Add (¯a + HilbertMultiply (λ, Add (b, Minus (¯a)))) where in turn λ¯ is computed by the formula above, but using geometric arithmetic and with bars over a, b, e, and r. Those inner terms are evaluated like this:

a¯ = (L1◦,L2◦) = MakePoint (X(pointOn1 (L), Y (pointOn1 (L)))) by the induction hypothesis = pointOn1 (L) byLemma10.10 Similarly ¯b = pointOn2 (L) ande ¯ = center (C) andr ¯ = Radius (C). Now what has to be verified is free of any mention of the ◦ interpretation: it is just that the formulas above, interpreted using geometric arithmetic, do correctly give the intersection points of a line and circle. Looking at the derivation of equation (1), we see that what was used was the formula for a circle, and the fact that the points on Line (b,a) have the form a + λ(b a), plus the laws of field theory, which we have already verified. By Lemma− 12.31, the formula for a circle is provable in ECG. By Lemma 12.36, the points on Line (a,b) are all of the form x = VectorAdd (a + ScalarMultiply (λ, Add (b, Minus (a)))) for some λ depending on x, which is exactly what is required. Hence t◦ is a point lying on C and on L. To complete the argument, we still need to show that the positive sign for λ corresponds to IntersectCircles2 and the negative sign to IntersectCircles1 . That is, for z not on the line, Right (z,p,q) Right (z,a,b). Since linear transformations of positive determinant preserve↔ the sign of the cross product, and Right is given by the sign of a cross product, and linear transformations preserve the equation of a line, we can assume without loss of generality that a = 0 and b lies on Ray (0, 1), i.e. b > 0. The the formula says + x = HilbertMultiply (λ, b). Then let λ be the positive value of λ and λ− the negative value; and let q = HilbertMultiply (λ+,b)

p = HilbertMultiply (λ−,b) Then q > 0 and p < 0. Hence by Lemma 9.89, Right (z,a,b) is equivalent to Right (z,p,q). Hence p = IntersectLineCircle1 (L, C) and q = IntersectLineCircle2 (L, C) 216 MICHAEL BEESON as claimed. That completes the case when t = IntersectLineCircle1 (L, C) or t = IntersectLineCircle2 (L, C). The ﬁnal case to consider is when

t = IntersectCircles1 (C,K) or t = IntersectCircles2 (C,K).

Then C◦ and K◦ are triples of points giving the centers and radii of two circles, and the formulas given in the proof of Theorem 6.1 give the intersection points, if they are defined. The definition of t◦ gives an explicit formula for the intersection points. When we compute t◦ we replace the arithmetic operations +, , and √ · by Add , HilbertMultiply , and SquareRoot , and the induction hypothesis replaces e◦ by center (C) for the center e of a circle, and r◦ by Radius (C) for the radius r. Thus what has to be verified is that the formulas given in the proof of Theorem 6.1 (and used to define t◦) are provable in ECG, when geometric arithmetic is used for addition, multiplication, and square root. This is proved by just going over the proof of Theorem 6.1, and observing that nothing is used but calculations involving the laws of field theory. (That is why we gave the proof of that theorem in such detail.) That proves that t◦, when t is defined, lies on both circles. It remains to check that t◦ has been correctly defined, in the sense that the correct intersection points have been assigned to IntersectCircles1 (C,K) and IntersectCircles2 (C,K). That was done using the cross product in the proof of Theorem 12.21. But now, we have formalized the connection between Right and CrossProduct , so that argument can now be given in ECG as well. In case the reader has any doubt, we will do so. Namely, by Lemma ??, Right is invariant under linear transformations of positive determinant. Hence by a translation and rotation we can assume that C has its center at origin and K has its center on the positive x-axis. In that case IntersectCircles1 (C,K) lies in the (closed) lower half plane (if it is defined) and IntersectCircles2 (C,K) lies in the (closed) upper half plane (by Lemma 9.104 and Axiom H5). So we only need to check that in this special case, t2◦ has a positive sign when t = IntersectCircles2 (C,K) and a negative sign when t = IntersectCircles1 (C,K). That completes the proof of the theorem.

+ + Theorem 12.61. (i) EF proves ((φ¯)◦ φ) for every formula φ of EF . ↔ (ii) ECG proves (φ◦ φ) for every formula φ of ECGnot containing equality between terms of type Circle.↔ (iii) The interpretation φ φ◦ is faithful on φ with no equality between terms of type Circle. 7→ (iv) The interpretation φ φ¯ is faithful. 7→ + Proof. We first show that (i) and (ii) implies (iii). Suppose EF proves φ◦. Then by the soundness of the interpretation φ φ¯, ECGproves φ . By (ii) then ECG 7→ ◦ proves φ, so the interpretation φ φ◦ is faithful (on formulas without = between 7→ + terms of type Circle. Similarly, if ECGproves φ¯, then EF proves (φ¯)◦, so by (i), EF+proves φ. Hence the interpretation φ φ¯ is faithful. 7→ We turn to the proof of (i). Another fact we need in the proof is that (0 < x)◦ is equivalent in EF to P (x◦), where 0 < x is the defined relation in ECG, defined by Left (I, 0, x). FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 217

Let φ be a formula of EF. Then φ¯ expresses φ in terms of variables restricted to Line (0, 1) and using Add and HilbertMultiply for addition and multiplication, and 0 < x for P (x). Then we introduce coordinates X(b) and Y (b) in ECG, and use them to translate back into EF. We must show we get (something equivalent to) the original formula. We proceed by induction on the complexity of the formula φ. The base case is when φ is atomic. Then φ is of the form t = s or the form P (t), or of the form t , for terms t and s. Consider the case when φ is t = s. Then φ¯ is t¯ =s ¯, where↓t¯ is obtained from t by replacing + and by · Add and HilbertMultiply . Applying we find that φ¯◦ is t¯◦ =s ¯◦. Thus what has to be proved is ◦ t = s t¯◦ =s ¯◦. → Suppose t = s. Then t and s , so by Lemma 12.59 we have t¯◦ = t ands ¯◦ = s. ↓ ↓ Hence t¯◦ =s ¯◦ as desired. Suppose φ has the form t . Then we must prove ↓ t t¯◦ . ↓ → ↓ ¯ By Lemma 12.59 we have t ∼= t◦. By the definition of ∼=, that means that if either side is defined then both are defined and they are equal; in particular if t then t¯◦ . ↓Now we check↓ the case of an atomic formula P (t). Then

(P (t))◦ (P¯(t¯)◦ ↔ (0 < t¯)◦ ↔ P (t¯◦) ↔ P (t) since t¯◦ = t ↔ That completes the proof of (i). We turn to the proof of (ii). Let φ be a formula of ECG. Then φ◦ expresses φ using coordinates and analytic geometry, and φ◦ translates back into ECG using Add and HilbertMultiply and 0 < x. We must prove the result is equivalent to φ. We do this by induction on the complexity of φ. Consider the atomic case when φ has the form t = s for terms t and s. Then φ t = s ◦ ↔ ◦ ◦ t = s ↔ ◦ ◦ t = s by Lemma 12.60 ↔ φ ↔ That completes the case φ is t = s. Now consider the case in which φ is t . Then ↓ φ◦ t◦ ↔ ↓ φ t ◦ ↔ ◦ ↓ t by deﬁnition of t ↔ ◦ ↓ ◦ ↓ t by Lemma 12.60 ↔ ↓ φ ↔ That completes the case in which φ is t . ↓ 218 MICHAEL BEESON

Now consider the case when φ is B(a,b,c). Then φ◦ is

a◦ = b◦ b◦ = c◦ b◦ a◦ + b◦ c◦ = a◦ b◦ . 6 ∧ 6 ∧ | − | | − | | − | and by Lemma 12.58, φ◦ is what we get by dropping the ◦ and using HilbertMultiply to multiply and SquareRoot to take square roots, and Add and Minus to add and subtract. That is, φ◦ is

Add (Distance (a◦, b◦), Distance (b◦, c◦)= Distance (a◦, b◦). By the induction hypothesis, that is equivalent to Add (Distance (a,b), Distance (b,c)) = Distance (a,c). By Lemma 12.44, that is equivalent to B(a,b,c), which is φ. That completes the case when φ is B(a,b,c). Now consider the case when φ is E(a,b,c,d), or ab = cd as we usually write 2 2 it. Then φ◦ is a◦ b◦ = c◦ d◦ , so | − | | − | φ a b 2 = c d 2 ◦ ↔ | ◦ − ◦| | ◦ − ◦| a b 2 = c d 2 ↔ | ◦ − ◦| | ◦ − ◦| Distance (a , b )= Distance (c , d ) ↔ ◦ ◦ ◦ ◦ Distance (a,b)= Distance (c, d) by the induction hypothesis ↔ E(a,b,c,d)by Lemma ?? ↔ That completes all the cases in which φ is atomic. Since both φ◦ and φ¯ commute with the propositional connectives, there is nothing to prove when φ is a disjunction, conjunction, implication, or negation. Consider the case when φ is x ψ, where x is a variable of type Point. The ∀ variables x◦ of ψ◦ are two variables belonging to the same “group” in Defi- nition 12.50, and so they have the same variable x∗ associated to them, and moreover that variable x∗ is the original variable x of ECG. Hence φ x ψ ◦ ↔ ∀ ◦ ◦ x∗ψ ↔ ∀ ◦ x∗ψ by the induction hypothesis ↔ ∀ xψ since x∗ is x ↔ ∀ φ ↔ That completes the case when φ is x ψ. The case when is changed to is the same throughout. ∀ ∀ ∃ The case when φ is x, ψ, where x is a variable of type Circle, is handled by ∀ the same calculation; the point is that the quantified variables x◦ belong to the same group and hence have the same x∗, which is the original variable. Also the case when φ is x, ψ and x of type Circle works the same way. The case when∃ φ is L ψ, where L is a variable of type Line, offers an additional wrinkle, because φ¯ restricts∀ the quantifier to the defined predicate . We have L φ ( L ψ) ◦ ↔ ∀ ◦ L ( (L ) ψ ) ↔ ∀ ◦ L ◦ → ◦ FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 219

Here L◦ is a list of four variables belonging to the same “group”, so they have the same corresponding variable L∗ in the Deﬁnition 12.50. Then we have

φ L∗( (L ) ψ ) ◦ ↔ ∀ L ◦ → ◦ L∗( (L ) ψ) by the induction hypothesis ↔ ∀ L ◦ → By Lemma 12.20, (L ) is provable in ECG, so we can drop it: L ◦ φ L∗ψ) ◦ ↔ ∀ φsince L∗ is L as before ↔ That completes the proof of the theorem.

§13. Classical geometry and constructive geometry compared. In this section we will ask and answer the question, what classes of formulas have the property that if they are classically provable, then they are constructively provable? We will prove that the negative formulas of ECG have this property; that covers formulas of the form, if some points, lines, and circles stand in certain relations to one another, and you perform a certain construction, then the result of the construction has certain properties (where the relations and properties in question do not involve or ). That covers all of Euclid’s theorems, I believe. At the end of the section∃ we∨ give an example to show that this result cannot be extended to formulas containing an existential quantifier. There are relatively simple cases (the weak Pasch axiom for example) in which a point that classically exists does not depend continuously on the parameters, and hence cannot be proved to exist constructively. 13.1. The double negation interpretation. In ??, Gödel defined his double- negation interpretation. One can also find it explained in [17]. Here is how it works: we start with a first-order theory T . It can be a many-sorted theory, and it can use the logic of partial terms. We associate to each formula A of T , another formula A− (the double-negation interpretation of A). The rules for defining A− are as follows:

( x A)− is x A− ∃ ¬ ∀ ¬ (A B)− is ( A− B−) ∨ ¬ ¬ ∧ ¬ P − is P for atomic P ¬¬ (A B)− is A− B− → → ( A)− is A− ¬ ¬ ( xA)− is xA− ∀ ∀ Informally, we put a double negation in front of , , and atomic formulas, and leave the rest of the formula unchanged. The actual∃ ∨ definition then converts to , and similarly for , because it is convenient that A− is always a ¬¬negative ∃ ¬∀¬ formula (i.e. does not contain∨ or . ∃ ∨ Theorem 13.1 (Gödel). If first-order classical logic proves A, then first-order intuitionistic logic proves A−. If theory T is such that T with intuitionistic logic 220 MICHAEL BEESON proves A− for every nonlogical axiom A of T , then intuitionistic T proves A− for every theorem of classical T .

The proof is a routine induction on the length of proofs. (Gödel’s genius was in thinking of the definition and theorem, not in proving it.) Although Gödel did not explicitly mention many-sorted theories, nothing new is required to check the validity of the theorem in that setting. We want to apply the theorem to theories in the logic of partial terms; so there is something to be checked. That can also easily be checked directly; but rather than go into the details, we check it indirectly.

Corollary 13.2. Suppose T is a theory in many-sorted predicate logic with the logic of partial terms. Supose that T proves φ− for each axiom φ of T , and also suppose T proves t t for each term t. Then the double-negation interpretation is also sound¬¬ ↓ for →T ; that↓ is, if T classically proves φ then T proves φ−.

Proof. We have already shown in Theorem 8.3 that the logic of partial terms LPT (with one sort or many sorts) can be reduced to the usual first-order logic (with one sort). The procedure described there assigns a formula φˆ to each formula φ. In the case of ECG, we would add three predicate symbols, Point, Line, and Circle. Variables of type Point become variables relativized to the unary predicate Point, and similarly for Line and Circle. For t a term of type Point, the formula t becomes Point (t). But there is nothing specific to ECG about this procedure.↓ It applies to any theory in LPT. The plan of the proof is this: first make this translation, then apply the double-negation interpretation. Then translate back into the logic of partial terms, replacing (t) by t . Now we carry out this plan in detail. ¶ ↓ Let Tˆ be the first-order theory axiomatized by the φˆ for φ an axiom of T . Since by hypothesis, T proves t , we will have Tˆ proves P (t) P (t). ¬¬ ↓ ¬¬ → Now we claim that the φˆ interpretation and the double-negation interpretation commute, i.e. if ψ = φ− then ψˆ = φˆ−. This is proved by induction on the complexity of φ. The only tricky case is when φ is atomic. If t is a term of type P , and φ is t , then ψ = φ− is t , and ψˆ is P (t). On the other hand φˆ ↓ ¬¬ ↓ ¬¬ is P (t), and φˆ− is P (t), so the conclusion is valid in this case. Then ¬¬

T classically proves φ Tˆ classically proves φˆ by Theorem 8.3 ↔ Tˆ proves (φˆ)− by G¨odel ↔ Tˆ proves (φˆ as proved above ↔ − T proves φ− by Theorem 8.3 ↔ Note that when the double-negation interpretation is applied in predicate calculus, we need P (t) P (t) to be provable in Tˆ; but by hypothesis, we do have that. That¬¬ completes→ the proof of the corollary. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 221

13.2. The double-negation interpretation applied to ECG. We finally arrive at the double-negation interpretation for ECG. To make it work we needed these things: (1) The axioms of ECG are negative, so they imply their double negation interpretations; (2) Betweenness and congruence are stable; (3) Definedness is stable, i.e. t t . We proved this in Theorem 10.16, and it required the strong parallel¬¬ postulate;↓ → ↓ and in fact it is equivalent to the strong parallel postulate. These are the the things that make the following theorem work. Since these are in hand, the proof is short. Theorem 13.3 (Double negation interpretation). Suppose ECG with classical logic proves A. Then ECG with intuitionistic logic proves A−. Proof. First we observe that A is equivalent to A for atomic A. This is an axiom for all atomic formulas¬¬ not of the form t , and was proved in Theorem 10.16 for formulas of the form t . Since the axioms↓ of ECG are quantifier-free ↓ and disjunction-free, it follows that A− is equivalent to A for axioms A of ECG. Now the theorem follows from Corollary 13.2. That completes the proof. Corollary 13.4. ECG with classical logic is conservative over ECG with intuitionistic logic for negative formulae.

Proof. For negative A, A− is identical to A. A typical theorem of Euclid has the form H A, where A will be quantifier- free when formulated in ECG, and H is a collection→ of hypotheses that certain points are distinct, or certain incidence relations hold or do not hold. As Proclus pointed out, sometimes this implies a theorem formulated by cases. For example, Euclid I.2 has one proof if A = C and another proof if A = C. Using the double-negation interpretation, we find a proof that A = C A6 = C implies the conclusion of Euclid 1.2, but without the law of the excluded∨ middle6 we cannot conclude the “uniform version” of Euclid I.2. Theorem 13.5. All negative theorems of Tarski’s geometry (formulated with classical logic and with line-circle continuity) are provable in ECG. Remark. One may use theory Tarski-B with strict betweenness, or theory Tarski- T with non-strict betweenness T , but then one must replace T with its negative definition in terms of B for the conclusion of the theorem to make sense.

Proof. Suppose φ is provable in Tarski-B with classical logic. Then φ◦ is provable + in EF with classical logic, so φ◦ is provable in ECG with classical logic. But by Theorem ??, φ◦ is equivalent to φ. Hence φ is provable in ECGwith classical logic. Hence φ− is provable in ECG. But since φ is negative, φ− is equivalent to φ. Hence φ is provable in ECG. That completes the proof of the theorem. Recall the example given in Section 2.3 of Euclid’s Prop. I.6, where we showed that a particular disjunction was eliminable from a proof, because the conclusion was negative, i.e. disjunction-free and -free. But all the theorems in Euclid are either already negative, or assert the∃ existence of some objects that can be 222 MICHAEL BEESON constructed using the terms of ECG; when formulated more explicitly, they are negative in the sense that they say that the result of a certain construction has certain (quantifier-free) properties. Hence the double-negation interpretation tells us that, once we have formulated the axioms of ECG in a quantifier-free, disjunction-free way, the work of constructivizing Euclid is done: we can then freely use classical logic to prove negative theorems, and all of Euclid’s theorems are negative. In particular, this gives us considerable freedom in choosing the betweenness and congruence axioms of ECG. As discussed above, considerable effort by Tarski, Gupta, Szmielew, and Schwabhauser went into finding a short and simple axiom system for geometry; and classical Euclidean geometry is now well-founded on this modern axiomatic system, based on equidistance and betweenness. Although betweenness does not occur explicitly in Euclid, we have chosen the same formal language, to permit an easy comparison of the classical and constructive theories. ECG has been designed in such a way that (i) The axioms are constructively valid. (ii) The axioms are quantifier-free and disjunction-free. (iii) Betweenness and congruence are stable. (iv) Definedness is stable, i.e. ,t t . ¬¬ ↓ → ↓ (v) The axioms are classically equivalent to the system given in [26] (with line-circle continuity rather than the full first-order continuity schema). Condition (v) follows immediately from our reduction of ECG to Euclidean field theory, since it is known that classically the models of Tarski’s theory are planes over Euclidean fields. (See [26], part II, or [29]). Our point here is that any axiomatization satisfying these five properties will also have the property in the theorem, that it proves all negative classical theorems of Tarski’s geometry.

§14. The relation of ECG to Tarski’s theories of geometry. In the previous section, we showed that negative theorems of Tarski’s geometry ( with line-circle continuity) are theorems of ECG. But one can still ask whether the intuitionistic versions of Tarski’s theories, which we have called Tarski-B and Tarski-T, are equivalent to ECG. The desired result here is that ECG is conservative over Tarski-B (for non-experts in logic, that means that if φ is a theorem in the common language of the two theories, then if ECG proves φ so does Tarski-B). A relatively trivial result is that Tarski-B and Tarski-T are equivalent, since one can define nonstrict betweenness in terms of strict betweenness and vice-versa. We have already interpreted ECG into EF+ and faithfully, so it only remains to interpret EF+ faithfully into (intuitionistic versions of) Tarski’s theories. The ideas used to interpret ECG into EF+ can also be used to interpret ECG directly into Tarski-B, i.e. circles can be represented as triples of points, and lines as pairs of points; but there seems to be no extra increment of understanding to be gained by writing out the details. Instead, we just interpret EF+ into Tarski-B. Of course, the geometric definitions of multiplication and addition and FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 223 square root are essentially the same as for ECG, but Tarski-B has no function symbols, and does not have the logic of partial terms, so there are some details to attend to. Theorem 14.1. ECG can be faithfully interpreted into intuitionistic Tarski- B. Proof sketch. Since we have already interpreted ECG faithfully in EF+, it suffices to interpret EF+ faithfully in Tarski-B. This is done as follows. We + assign to each term t of EF , a formula At of Tarski-B, with one more free variable than t. Let x = x1,... ,xn and let t have free variables x. Then At(y, x) is intended to express y = t, i.e. t is defined and has the value y. If t = x y, then At(z,x,y) should say that z is the value of HilbertMultiply (x, y). That is, there· is a circle passing through z, I, x, and y. That is, Strong Parallel Axiom2 We set aside three variableα ˆ, βˆ,γ ˆ to serve as the interpretations of the constants α, β, and γ. These variables will not be used otherwise. Lines are interpreted as pairs of points; specifically Line (a,b) is interpreted as (a,b). A variable L of type Line is interpreted as a pair of point variables L1 and L2. The terms pointOn1 (L) and pointOn2 (L) are interpreted as L1 and L2. The term pointOn1 (Line (t,s)) is interpreted as (t,ˆ sˆ), where tˆ is the interpretation of t. Circles are also interpreted as pairs of points, and center(Circle 3(a,b,c)) is interpreted as a, while pointOnCircle (Circle (a,b,c)) is interpreted as a point bc along the canonical line parallel to Line (α, β) that passes through a, and in the same direction from a as β is from α. The construction of the canonical parallel and the explanation of “same direction” will come later on; this is only a “proof sketch.” Incidence of points on lines is interpreted this way: on(p, Line (a,b) is interpreted as (B(ˆp, a,ˆ ˆb) pˆ =â B(â, p,ˆ ˆb) pˆ = ˆb B(â, ˆb, pˆ)) ¬¬ ∨ ∨ ∨ ∨ This is, of course, equivalent to a negative formula obtained by pushing the double negation in. For each term t with free variables x = x1,... ,xn, there is a formula At defining the interpretation of t, At(x, y), expressing that t is defined and has the value y. The interpretation of the formula t is y At(x). To interpret a formula not of the form t , one first writes the↓ formula∃ in the form φ(y := t), where y is a list of variables↓ and t a corresponding list of terms, and y := t indicates simultaneous substitution, and φ contains no function symbols or constants. Then the interpretation of φ(y := t) is given by

y(At(x, y) φˆ) ∀ → On formulas not containing function symbols and constants, the interpretation φ φˆ preserves logical connectives. Quantiﬁers are handled as follows. The interpretation7→ of xφ(y := t) is ∀ x y (At(x,z,y) φ) ∀ ∃ ∧ 224 MICHAEL BEESON

Note that x can occur free in At since some of the terms t may contain x; z is a list of other free variables of φ and t. The interpretation of zφ(y := t) is ∃ z,y (At(x,z,y) φ. ∃ ∧ That brings us to the interpretations of atomic formulas. The interpretation of t = s is y,z(At(x, y) As(x, z) y = z) ∃ ∧ ∧ where x on the left is the list of free variables of t and x on the right is the list of free variables of s, which may of course not be the same list. The interpretation of B(t,s,r) is

y,z,w(At(x, y) As(x, z) Ar(x, w) B(y,z,w) ∃ ∧ ∧ ∧ where as before, the list x of free variables may be diﬀerent at each occurrence. Similarly for the interpretation of E(t,s,p,q). We have to exhibit the formulas At and Bt corresponding to terms with one function symbol and variables for arguments. Without yet having done this, we can already prove a basic lemma: Lemma 14.2. If a target theory T in ordinary ﬁrst-order logic proves the interpretations of the axioms of a source theory S in the logic of partial terms, then T proves the interpretation of every theorem of S. Proof. We check the axioms and rules of inference of the logic of partial terms. Consider the axiom w φ t φ[w := t]. ∀ ∧ ↓ → Here t and φ may have other variables x not explicitly shown. The interpretation of this formula, in the special case when φ contains no compound terms, is

wφˆ yAt(x, y) y(At(x, y) φˆ w = y) ∀ ∧ ∃ → ∃ ∧ ∧ which is immediate. In the general case where the formula contains other terms, the axiom takes the form y φ[z := s] t φ[z := s,y := t] ∀ ∧ ↓ → where φ has no compound terms. The interpretation of this formula is Consider the axiom φ[y := t] t yφ) ∧ ↓ → ∃ The interpretation of this, in the special case when φ contains no compound terms, is ( y At(x, y) φˆ y At(x, y) y φˆ ∃ ∧ ∧ ∃ → ∃ which is again immediate. Consider the rule of inference, from B A infer B xA, with x not free in B. We write B = ψ(y := t) and A = φ(→z := s), where→ψ and ∀ φ do not contain function symbols or constants, and t and s are (lists of) terms. x does not occur free in ψ and is not among the variables z, but it could occur in As or Bs. To carry out induction on the lengths of proofs, we assume that the interpretation of B A is provable in T . That is → ( y At(w,y) ψ) z(As(q,z) φ) ∃ ∧ → ∀ → FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 225 where q are other variables besides x that may occur in A. From this we may infer y(At(w,y) ψ) x z(As(q,z) φ) ∀ → → ∀ ∀ → since x does not occur free in the antecedent. But this is the interpretation of B xA, the conclusion of the rule. The→ ∀ rule of inference, from A B infer xA B is treated similarly. That completes the proof of the lemma.→ ∃ → FINISH THIS 14.1. Interpretation of Tarksi’s theory in ECG. We will show in this section that the axioms of ECG suﬃce (at least with classical logic) to interpret Tarski’s axioms, as listed in [1], p. 745, where they are referenced to [30].30 Tarski’s B cannot be interpreted as the strict betweenness of ECG, because of his segment-extension axiom that says, given a segment qa and another segment bc (either of which might be a degenerate, one-point segment), there is an x such that B(q,a,x) ax = bc. Indeed, if both are degenerate segments, then we will need B(x, x, x)∧ to be true, so Tarski’s B cannot even be interpreted as non-strict betweenness on nondegenerate intervals. We must allow not only B(a,a,b) and B(a,b,b) but also B(a,a,a). Therefore we take as the interpretation of Tarski’s betweenness T(a,b,c) := (a = b b = c B(a,b,c)) ¬¬ ∨ ∨ Note that if T(a,b,a) then (a = b B(a,b,a) ¬¬ ∨ Theorem 14.3. If B(a,b,c) is replaced by T(a,b,c), then every axiom of Tarski’s geometry is provable in ECG plus classical logic. Remark. Since Tarski’s theory uses classical logic, this theorem implies, but is stronger than, the claim that ECG plus classical logic proves all the theorems of Tarski’s theory. Proof. We take Tarski’s axioms one at a time. Equidistance axiom E1. ab = ba. This is our Axiom 42. Equidistance axiom E2. ab = pa ab = rs pq = rs. This is our Axiom 43. ∧ → Equidistance axiom E3. ab = cc a = b. This is our Axiom 44. → Segment construction axiom (SC): x(T(qax) (ax = bc)). ∃ ∧ If we knew that q and a were distinct, we could take x = IntersectLineCircle2 (Line (q,a), Circle (a,b,c)). And of course if q = a then x can be any point on Circle (a,b,c). So we can verify the double negation of SC, but not SC itself. Five-segment axiom 5S. a = b T(a,b,c) T(p,q,r) ab = pq bc = qr 6 ∧ ∧ ∧ ∧ ad = ps bd = qs cd = rs ∧ ∧ →

30See page 190 of [30]; the system in question is EG2 with the continuity axiom 11 replaced by line-circle continuity. 226 MICHAEL BEESON

Pasch axiom. T(a,p,c) T(q,c,b) x(T(a, x, q) T(b,p,x). Let L = Line (a, q). If q∧= c then L→has ∃ the same points∧ as Line (a,c), and so we can take x = p. If q = b then we can take x = b. Hence we may assume B(b,q,c). We apply the version of the Pasch axiom in ECG to the three points b, c, and p, and the line L. Then L meets bp in some point x, as desired. Lower two-dimension axiom 2L. a,b,c( T(a,b,c) T(b,c,a) T(c,a,b)). Take a,b,c to be α,β,γ. Suppose∃ T(α,β,γ¬ ); we must∧ ¬ derive a∧ contradiction. ¬ By deﬁnition of B, we have (classically) α = β β = γ B(α,β,γ). By Axiom D4 we have α = β and β = γ. Hence B(α,β,γ).∨ Then by∨ Axiom B1-i, we have on(β, Line (α, γ6 )), contradicting6 axiom D1. That proves T(α,β,γ). Similarly one proves T(β,γ,α) and T(γ,α,β), using Axioms D2¬ and D3, respectively. ¬ ¬ Upper two-dimension axiom 2U. 3 (a = b) xia = xib T(x1, x2, x3) T(x2, x3, x1) T(x3, x1, x2) ¬ ∧ ∧i=1 → ∨ ∨ This says (using Axiom C5) that if three circles meet in two distinct points, then their centers are collinear. That is exactly the upper dimension axiom D5 of ECG. Parallel postulate PP. T(a,d,t) T(b,d,c) a = d x, y(T(a,b,x) T(a,c,y) T(y,t,x) ∧ ∧ 6 → ∃ ∧ ∧ First we deal with the degenerate cases: if b = a or c = a then Line (a,b) and Line (c, d) have the same points, so since B(b,d,c), the point d is also on that line, by Axiom B1-i. Then we can take x = y = t. If d = b then we take x = t, and y can be taken to be any point on Line (a,c) with T(a,c,y), for example, y = c will do. Similarly if d = c. If d = t then we take x = b and y = c. That is all pairs of possibly equal points, so we may assume the points mentioned in the hypothesis are distinct and the betweenness relations given are strict. Moreover, we may assume that a, b, and c are not collinear, since if they are, then Line (a, x) and Line (a,y) coincide, and since d is between b and c, then by Axiom B1-i, d is also on that line, so Line ,a,d) coincides with the other two, and we can take x = y = t. We claim point t is not on Line (b,c). To prove this, suppose t is on Line (b,c). Since t = d, then Line (b,c) and Line (a,t) have the same points. Then a, b, and c are collinear,6 contradiction. Hence t is not on Line (b,c). Now (without using a parallel axiom) we can construct line L through t parallel to Line (b,c). We use the parallel axiom this way: from the parallel axiom we can prove the transitivity of the relation of being parallel. Now Line (b,c) is not parallel to Line (a,b), since they meet at b. Then L is not parallel to Line (a,b). Hence L meets Line (a,b) in some point x. Consider triangle atx; note that Line (b,c) meets side at at d, but does not meet side xt, and is not equal to x, since Line (b,c) is parallel to Line (x, t). Then by Pasch, either x = b or Line (b,c) meets ax. But the meeting point is b; hence B(a,b,x). Similarly, L meets Line (a,c) in some point y, and B(a,c,y). That completes the veriﬁcation of Tarski’s PP in ECG. Line-circle continuity LC

ax = ax′ az = az′ T(a,x,z) T(x,y,z) y′(ay = ay′ T(x′,y′,z′) ∧ ∧ ∧ → ∃ ∧ FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 227

Let L = Line (x′,y′) and C = Circle (a,y). Take y′ = IntersectLineCircle2 (L, C). Then y′ is deﬁned, by Axiom Cont 1, and by Axiom I7, we have on(y′,L) and On(y, C). It remains to show T(x′,y′,z′). We have ax = ax′ and ay = ay′ and az = az′. Therefore by Lemma ??, we have x′y′ = xy and y′z′ = yz and x′z′ = xz. It follows that T(x′,y′,z′). FINISH THIS

§15. Interpretation of ECG in Hilbert’s and Tarski’s theories. 15.1. Right and left handedness. Since ECG, following Euclid, has no disjunctive axioms (and hence no disjunctive theorems), we cannot prove that two points p not on L are either on the same side or the opposite side of L; that is constructively all right since to be given a point p “on a given side of L(a,b)” means that we are told either that abp is a right turn, or that it is a left turn. Nevertheless we can, in constructive geometry, reduce the question of whether p and q are on the same sides of L to the case when L = Line (α, β) and q is the point γ, where α, β, and γ are the three fixed constants mentioned in the handedness axioms H1 and H2. The idea is that handedness is invariant under translations, rotations, and dilations, and using these operations we can bring L to coincide with Line (α, β). Lemma 15.1. (In ECG)Let ABC be any triangle. Then we can determine the handedness of the turn ABC, in the following sense. Let α, β, and γ be the three fixed non-collinear points mentioned in the axioms of ECG, so that αβγ is a left turn by definition. Let L = Line (α, β). Then we can construct a point R such that ABC is a left turn if and only if R is on the same side of L as γ, and a right turn if and only if R is on the opposite side of L from γ. Proof. The idea of the proof is this: by a series of “moves” (applications of the axioms for Left and Right , which correspond to translations, dilations, and rotations), the triangle ABC is “reduced” to either triangle αβγ or triangle αγβ, preserving the handedness of the turns; but Axioms 61 and 62 directly specify the handedness of the turns αβγ and αγβ. We now give the details. By a “move”, applied to a triple of non-collinear points P QR, we mean a construction of a new triple UV W such that, according to the axioms for Left and Right , if P QR is a left turn then so is UV W , and if P QR is a right turn then so is UV W . This latter condition we describe for short by saying that the move “preserves handedness”. The axioms describe several types of moves that preserve handedness, specifically, moving P along the ray QP , moving R along the ray QR, rotating P R or QR in such a way that the points P ,Q, and R never become collinear, and translating the whole triple. We first show that the number of moves (applications of these axioms) required to perform a given rotation is bounded by a fixed constant. Consider the following procedure: First move P to decrease the angle to less than a right angle. Then we can rotate P QR by any angle up to a right angle, using two moves (one moves PQ and one moves QR). An arbitrary rotation can be performed by performing at most four rotations of less than ninety degrees. Thus in four or fewer rotations (requiring eight or fewer moves) we can bring one of its sides onto the desired (“target”) ray. Then we can move P by the same amount as in the first step resulting in 228 MICHAEL BEESON the desired rotation. Hence ten or fewer applications of the above axioms suffice to perform any rotation. Now consider three non-collinear points P , Q, and R. We give a procedure for determining whether P QR is a left turn or a right turn. First translate P QR so that Q coincides with the point β (given by a constant of ECG). Then rotate it so that P lies on Ray (β, α). Then move P to α. By the axioms above, all these steps preserve the handedness of P QR. Now, if R is on the same side of Line (α, β) as γ, then by Axioms 65 and 69, P QR is a left turn, since αβγ is a left turn by definition. And if R (after the moves described) is on the opposite side of Line (α, β), then we claim that P QR is a right turn. To see this, let γ′ be a point on the same side of Line (α, β) as γ, and on the same side of Line (Q, R) as α. (Such a point can be constructed by bisecting the angle formed by Ray (β, α) and the opposite ray to Ray (Q, R).) Then we can move P to γ′ without changing the handedness of P QR, and then we can move R to α without changing the handedness of P QR. But now P QR coincides with γβα, which by definition is a right turn. That completes the proof of the lemma. Remark. With classical logic, we could prove that ABC is either a right turn or a left turn. To reach that conclusion, we would need to know that if point R is not on L = Line (α, β), then either R is on the same side of L as γ or on the opposite side. The constructive status of this statement is discussed below. Lemma 15.2. The predicates Right (A, B, C) and Left (A, B, C) are definable in Greenberg’s theory G, relative to an arbitrary choice of Left (α,β,γ) and Right (α,γ,β) for some triple of non-collinear points α, β, and γ. This can even be done with intuitionistic logic. Proof. It will suffice to define the relation T (A,B,C,P,Q,R) with the meaning “ABC and P QR have the same handedness.” First we note that it is possible to define the notion of one triangle being a translation of another; namely, ABC is a translation of P QR if the two triangles are congruent and AP = BQ = CR. It is also possible to define the notion of ABC being a rotation of PBQ (when the two angles share vertex B). This requires twenty variables to express, so it is too complex to write down intelligibly, but the definition in question says there exist twenty points representing ten “moves” according to the ECG axioms for rotations given above. Then ABC and P QR have the same handedness if there exist P ′ and Q′ such that P ′BR′ is a translation of P QR and there is a rotation P ′′BR′′ of P QR with R′′ on Ray (B, C) and P ′′ is on the same side of Line (B, C) as A. 15.2. Apartness. The absence of disjunction in Euclid is crucial to our axiomatization. In order to apply the methods of proof theory, it will be very helpful that ECG is axiomatized in a quantifier-free, disjunction-free way. That is possible, while still providing the ability to prove all Euclid’s theorems, because none of Euclid’s theorems involves disjunction in an essential way. Here is an example, due to Mandelkern [19], of a theorem that requires disjunction to state and to prove, which some might consider constructively valid. Given two lines L and K that are not parallel, and a third line M, then either M meets L or M meets K. This theorem cannot be proved in ECG, since the intersection point of M with one of the two lines K and L will not depend FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 229 continuously on line M. For the same reason, the intersection point cannot be constructed by ruler and compass. It is not hard to see that Mandelkern’s theorem (which he takes as an axiom) is essentially a version of “apartness”. Stated arithmetically, the principle of apartness is that for all numbers x we have x< 1 or x > 0. That principle is constructive, if we think that we can compute any number to any accuracy; but that “computation” cannot be done with ruler and compass. Unlike previous intuitionistic geometries, ECGdoes not have apartness. Un- like previous algebraic theories of geometric constructions, it does not have a test-for-equality construction. In both these respects, ECG is like Euclid. In order to arrive at such a metatheorem, we first formulate the principle in question in the language of ECG, which does not contain < as a primitive. Our formulation is as follows: If two unequal points B and C are both between A and D, then either B is between A and C or C is between A and B. Formally that is B = C B(A,B,D) B(A,C,D) B(A, B, C) B(A,C,B) (Axiom 75) 6 ∧ ∧ → ∨ The point D is a matter of convenience; the axiom is really about A, B, and C and their positions on a ray emanating from A, but ECG does not have rays as primitive, so we need point D to express this in ECG. We defined AB < CD in Definition 48. We write AB > CD to mean CD < AB. Lemma 15.3. (in ECG) Axiom 75 implies that if AB = CD, then either AB >CD or AB CD. Then according to Axiom 75, if Q = B then either Q is between A and B or B is between Q and A. In the first case6 we have AB >CD. In the second case, B is inside the circle of radius CD about A. It follows that D is outside the circle of radius AB and center C. That completes the proof of the lemma. The theory ECG has quantifier-free, disjunction-free axioms. It follows (as we will prove in Theorem 16.3) that no non-trivial disjunction can be proved in ECG. That is, if P is negative and ECG proves x P (x) A(x) B(x), then ECG proves x P (x) A(x) or ECG proves ∀x P (x) → B(x).∨ Hence, Axiom 75 is not provable∀ in→ECG. ∀ → When we add Axiom 75, we will also introduce a new construction term, which we write if (AB > CD,P,Q). This abbreviates the official version, which is if (A,B,C,D,P,Q). Provided AB = CD, this term constructs a point which is equal either to P or to Q, depending6 on whether AB > CD or AB < CD . The axiom expressing this is (AB >CD if (AB > CD,P,Q)= P ) (Axiom 76) (AB CD,P,Q)= Q) ∧ → Note that Axiom 76 does not contain disjunction, but that by Axiom 75, we have AB = CD if(AB > CD,P,Q) . 6 → ↓ 230 MICHAEL BEESON

Definition 15.4. The theory ECGD is ECG plus the new function symbol “if ”, with Axioms 75 and 76. Remark. The “D” in ECGD is for “disjunction”. The following lemmas give two appealing theorems of ECGD that cannot be proved in ECG (because they are non-trivial disjunctions). Lemma 15.5. (in ECGD) Let P be point not on line L. Then any point Q is either on the same side of L as P , or on the opposite side. Proof. Drop a perpendicular K from P to L, meeting L at point R. Projecting Q to point Q′ on K. Extend segment P R past R by the amount RQ′ twice, arriving at point D on K. Then both R and Q′ are between P and D, so by Axiom 75, either R is between P and Q′ or Q′ is between P and R. In the ﬁrst case, Q′ and P are on the same side of L, and in the second case, they are on opposite sides. But Q and Q′ are on the same side of L. Hence Q and P are on the same side, or on opposite sides, of L. That completes the proof of the lemma. Lemma 15.6. (in ECGD) In any triangle ABC, either ABC is a left turn or ABC is a right turn. Proof. In Lemma 15.1, we have already shown how to construct a point Q not on line L = Line (α, β), such that ABC is a left turn or a right turn according as Q lies on the same side of L as γ, or on the opposite side. In ECGD by Lemma 15.5, Q must lie on one side or the other of L. Hence ABC is either a left turn or a right turn. That completes the proof. Terms of ECGD that involve the new symbol if represent geometrical constructions that can proceed by cases, with comparisons between constructed (unequal) lengths determining the next construction steps. Given Euclid’s cav- alier approach to case splits, the fact that such constructions are not explicitly mentioned in Euclid does not necessarily mean that they are not required to give a correct and complete version of Euclid. The question thus arises, whether Axiom 75 (or more generally, disjunctive axioms of any kind) are required to formalize Euclid. But because the theorems of Euclid do not mention disjunction in any essential way, we can simply take the double-negation interpretation, and eliminate Axioms 75 and 76, as will be shown below. Thus what happened in the example of Proposition I.6 happens necessarily in all examples of similar logical form

§16. Metatheorems. 16.1. Things proved to exist in ECG can be constructed. In this section we take up our plan of doing for ECG what cut-elimination and recursive realizability did for intuitionistic arithmetic and analysis, namely, to show that existence proofs lead to programs (or terms) producing the object whose existence is proved. In the case of ECG we want to produce geometrical constructions, not just recursive constructions (which could already be produced by known techniques, since ECG is interpretable in Heyting’s arithmetic of ﬁnite types). Terms of ECG correspond in a natural way to straightedge and compass constructions. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 231

Theorem 16.1 (Geometric constructions extracted from intuitionistic proofs). (i)Suppose ECG proves P (x) y φ(x, y) where P is negative (does not contain or ). Then there is a→ term ∃ t(x) of ECG (representing a geometric construction)∃ ∨ such that P (x) (φ(x, t(x)) is also provable in ECG. → (ii) Same as (i) but with ECGD in place of ECG. (iii) Let ECG + DE be ECG, augmented with a constant D and the axiom saying D is a test-for-equality function. Then the analogue of (i) holds for ECG + DE. Proof. We use cut-elimination.31 Since our axiomatization is quantifier-free, if ψ y φ is provable in constructive ECG, then there is a list Γ of quantifier-free axioms→ ∃ such that Γ, ψ y φ is provable by a cut-free (hence quantifier-free) proof. Since our axiomatization⇒ ∃ is disjunction-free, by [18] (or rather, by its adaptation to multi-sorted logic with the logic of partial terms) we can permute the inferences so that the existential quantifier is introduced at the last step. Then we obtain the desired proof just by omitting the last step of the proof. That completes the proof of part (i). All the work was in arranging the axiom system to be quantifier-free and disjunction-free. Part (iii) is proved in the same way, noting that the axioms for D are also disjunction-free. Part (ii) requires a bit more work. Since Axiom 75 contains a disjunction, there is an issue about permuting the inferences in a cut-free proof. Suppose that we have a cut-proof of Γ ψ y φ, where Γ is a list of axioms (or subformulas of axioms) of ECGD.⇒ Among→ ∃ Γ there may be occurrences of Axiom 75, which contains a disjunction. We prove by induction on the number of disjunctions in Γ that there exists a term t of ECGD and a list ∆ of axioms of ECGD such that Γ, ∆ φ(t) is provable. The basis case, when there are no disjunctions, is part (i)⇒ of the theorem. Now for the induction step. If the on the right is introduced at a lower level (nearer the end-sequent) than the lowest∃ introduction of disjunction on the left, then we can complete the proof as above, since the line just before the is introduced will contain the desired term, and the - introduction can just∃ be postponed until the end. Otherwise there is a part∃ of the proof that looks like this:

AB > AC, Γ1 y φ AB < AC, Γ2 y φ ⇒ ∃ ⇒ ∃ AB > AC AB < AC, Γ1, Γ2 y φ ∨ ⇒ ∃ By induction hypothesis, there are terms t1 and t2 such that AB > AC, Γ1 ⇒ φ(t1) is provable and AB < AC, Γ2 φ(t2) is provable. Let t be the term ⇒ if (AB > AC,t1,t2). Then ECGD proves AB > AC t = t1 (by Axiom 76), → and ECGD proves AB < AC t = t2. Hence, for some list ∆ of axioms → of ECGD, there is a cut-free proof of AB > AC, ∆, Γ1 φ(t), and a cut-free ⇒ proof of AB < AC, ∆, Γ2 φ(t). These two proofs can then be combined as follows: ⇒ AB > AC, ∆, Γ1 φ(t) AB < AC, ∆, Γ2 φ(t) ⇒ ⇒ AB > AC AB < AC, ∆, Γ1, Γ2 φ(t) ∨ ⇒ 31In fact, we use cut-elimination for many-sorted logic with the logic of partial terms. The details of the cut-elimination theorem for such logics have not been published, but they are not significantly different from Gentzen’s formulation for first-order logic. 232 MICHAEL BEESON

That completes the induction step, and with it, the proof of the lemma.

Theorem 16.2 (Geometric constructions extracted from classical proofs). Suppose ECG with classical logic proves P (x) y φ(x, y) where P is quantiﬁer-free and disjunction-free. → ∃ Then there are terms t1(x),... ,tn(x) of ECG such that

P (x) φ(x, t1(x)) . . . φ(x, tn(x)) → ∨ ∨ is also provable in ECG with classical logic. Proof. This is a special case of Herbrand’s theorem. Example 2. Euclid’s proof of Book I, Proposition 2 provides us with two such constructions, t1(A, B, C) = C and t2(A, B, C) the result of Euclid’s construction of a point D with AD = BC, valid if A = B. Classically we have 6 A, B, C D(AD = BC), but we need two terms t1 and t2 to cover all cases. ∀ Example∃ 3. Let P and Q be distinct points and L a given line, and A, B, and C points on L, with A and B on the same side of C. Then there exists a point D which is equal to P if B is between A and C and equal to Q if A is between B and C. The two terms t1 and t2 for this example can be taken to be the variables P and Q. One term will not suﬃce, since D cannot depend continuously on A and B, but all constructed points do depend continuously on their parameters. This classical theorem is therefore not constructively provable. 16.2. Disjunction properties. We mentioned above that ECG cannot prove any non-trivial disjunctive theorem. That is a simple consequence of the fact that its axioms contain no disjunction. We now spell this out: Theorem 16.3 ( ECG cannot prove a nontrivial disjunctive theorem). Sup- pose ECG proves H(x) P (x) Q(x), where H is negative. Then either ECG proves H(x) P (x) or→ ECG∨proves H(x) Q(x). (This result depends only on the lack→ of disjunction in the axioms of →ECG.) Proof. Consider a cut-free proof of Γ, H(x) P (x) Q(x), where Γ is a list of some axioms of ECG. Tracing the disjunction→ upwards∨ in the proof, if we reach a place where the disjunction was introduced on the right before reaching a leaf of the proof tree, then we can erase the other disjunct below that introduction, obtaining a proof of one disjunct as required. If we reach a leaf of the proof tree with P (x) Q(x) still present on the right, then it occurs on the left, where it appears positively.∨ Its descendants will also be positive, so it cannot participate in in application of the rule for proof by cases (which introduces in the left side of a sequent); and it cannot reach left side of the bottom sequent,∨ namely Γ, H(x), as these formulas contain no disjunction. But a glance at the rules of cut-free proof, e.g. on p. 442 of [18], will show that these are the only possibilities. That completes the proof. Theorem 16.4 (Disjunction Properties for ECG and ECGD). Suppose ECGD proves H(x) P (x) Q(x) → ∨ FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 233 where H is negative. Then there is a term t(x) of ECGD such that ECG proves H(x) t(x)= α t(x)= β → ∨ and ECGD proves t(x)= α P (x) t(x)= β Q(x). → ∧ → (Here α and β are two constants of ECGD, with α = β an axiom.) 6 Proof. Suppose ECGD proves H(x) P (x) Q(x). Then also ECGD proves → ∨ H(x) y ((y = α P (x)) (y = β Q(x))). → ∃ → ∧ → The formula on the right is disjunction-free, so by Theorem 16.1, there is a term t as required.

§17. Independence results for the parallel axioms. By the interpretations given in the previous section, questions about the logical relations between different parallel axioms are reduced to the corresponding questions of logical relations between different axiom systems for Euclidean fields. For example, those results explain why Axiom 58 implies Euclid’s Postulate 5: it is just that if reciprocals of non-zero elements exist, then of course reciprocals of positive elements exist. In fact, technically the interpretations given above already prove that Axiom 58 implies Euclid’s Postulate 5, but we still felt it worthwhile to include the direct proof in the previous section. One might also ask what the field-theoretic version of Playfair’s axiom is. Recall that Playfair says, if P is not on L and K is parallel to L through P , that if line M through P does not meet L then M = K. Since M = K M = K, Playfair is just the contrapositive of Axiom 58, which says¬¬ that if M→= K then M meets L. Hence it corresponds to the contrapositive of x = 0 6 1/x ; that contrapositive says that if x has no multiplicative inverse,6 then→ x =↓ 0. Thus Playfair geometries correspond to ordered fields in which elements without multiplicative inverses are zero. Thus, constructively, Axiom 58 implies Euclid 5 implies Playfair; we wish to show that neither implication is reversible. Since classically, the implications are reversible, we cannot hope to give counterexamples. In terms of field theory, we won’t be able to construct a Euclidean ring in which positive elements have reciprocals but nonzero elements do not. We must use some tools of logic. Among the possible techniques for proving that a constructive implication is not reversible is the method of Kripke models. We shall not try to explain this technique in its full generality, but only in the case of theories based on (ordered) ring theory. A full explanation can be found in [32], but our presentation here is self-contained. 17.1. Kripke models of ring theory. It will be essential to understand the notions “term” and “formula” (of ordered ring theory) as logicians use them. A term is, intuitively, an expression meant to denote an element of a ring. A variable (such as x, y, etc.) is a term, and so are the constants 0 and 1. If t and s are terms then so are (t s) and (t + s), as well as ( t) and (1/t). In informal usage, many parentheses are· left unwritten, according− to the usual conventions. 234 MICHAEL BEESON

For example, (1/((x+1) (y +1))) is a term. Note that 1/0 is technically a term; not every term necessarily· denotes something (“is defined”). Next we explain the notion “formula”. This notion is defined recursively: formulas are built up by combining smaller formulas using the logical connectives (and), (or), (not), and (implies), as well as the quantifiers x and ∧x, according∨ to¬ some rules which→ we will not spell out in detail. The base∀ case of∃ this definition is the “atomic formula”, which is either of the form P (t) for some term t, or of the form t = s for some terms t and s, or of the form t for some term t. The formula t is read “t is defined.” ↓ An (ordered) ring is a model↓ of (ordered) ring theory; a Kripke model of (ordered) ring theory is a slightly more complicated thing. It is a collection of rings, Rα, where the subscripts α come from some partial ordering (K,<). Each Rα must have a notion of “positive”, i.e. a subset of so-called positive elements to interpret the predicate symbol P (x), but it is not required that Rα be an ordered ring. The ordering on the index set K has nothing to do with the ordering on the ring, which is given by a predicate P (x) defining the positive elements. It is usually required that if α β, then Rα is a subring of Rβ. It will be convenient ≤ to generalize this requirement by allowing Rα to be embedded in Rβ by means of a one-to-one function jαβ (which would be the identity if Rα Rβ). These ⊆ functions must compose according to the law jαβ jβγ = jαγ . Using some abstract nonsense, we could replace each Rα by a suitable copy to ensure that the jαβ are all the identity and Rα Rβ, but it will convenient not to require that. There ⊆ is also a requirement of “persistence”: if x is positive in Rα, then jαβx has to be positive in Rβ. But note, there can also be positive elements in Rβ that do not arise in that way. The Kripke model is technically the function α Rα with domain K, R 7→ though one often thinks of it as the collection of the Rα. Usually the index set K has a least element, the “root”. The elements of K are called “nodes”. Usually (and in all the models we will use) the set K is a tree, i.e., the set of nodes less than any given α is linearly ordered. What we have defined so far is a “Kripke structure”; to be a Kripke model of (ordered) ring theory, or of ordered field theory, the structure must “satisfy the axioms” of the theory. We next define that concept. We consider valuations σ; these are functions that assign an element xσ of Rα to each variable x. (Logicians usually write valuations on the right as xσ, rather than σ(x).) If σ is a valuation, it starts out as a function defined on variables, but is easily extended to a function defined on terms. For example, (t + s)σ = tσ + sσ, where the + on the right is addition in Rα, and the + on the left is just a symbol. This extended function is a partial function, because (1/t)σ is undefined if tσ is 0. If t belongs to the domain of the extended valuation σ then we say t is defined in Rα. If α<β and σ is a valuation into Rα and τ is a valuation into Rβ then we say τ agrees with σ at x if xσ = (jαβx)τ. FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 235

We next define the notion “Rα satisfies formula A under valuation σ”, which is written Rα =σ A. The rules for this definition are as follows: | Rα =σ P (x) if and only if xσ is positive in the ordered ring Rα | Rα =σ t = s if and only if terms tσ and sσ are equal (equivalent) elements of Rα | Rα =σ t if and only if tσ is defined in Rα | ↓ Rα =σ (A B) if and only if Rα =σ A and Rα =σ B | ∧ | | Rα =σ (A B) if and only if Rα =σ A or Rα =σ B | ∨ | | Rα =σ ( A) if and only if for all β α and valuations τ on Rβ, | ¬ ≥ if τ agrees with σ then not Rβ =σ A | Rα =σ (A B) if and only if for all β α, Rβ =σ A implies γ βRγ =σ B | → ≥ | ∃ ≥ | Rα =σ ( x A) if and only if for some τ that agrees with σ except perhaps on x, Rα =τ A | ∃ | Rα =σ ( x A) if and only if for all β > α, if x Rβ and | ∀ ∈ τ is a valuation on Rβ that agrees with σ except perhaps on x, then Rβ =τ A | A Kripke model of a theory T is one such that all the axioms of T are satisfied at every node of the model, i.e. Rα =σ A for every axiom A and every valuation. The Kripke completeness theorem| says that the formulas A that are provable from T with intuitionistic logic are exactly the formulas satisfied in all Kripke models of T . In particular, if one can construct a Kripke model of T in which some formula B is not satisfied, then B is not a consequence of T with intuitionistic logic.32 Such a model is called a Kripke countermodel to B. We will apply this technique to settle the question of the reversibility of the implications between the three parallel postulates. 17.2. Euclid 5 does not imply Axiom 58. The following concepts will be used in the next proof, because we will make Kripke models in which the “points” are functions. Definition 17.1. A function f from R to R is called positive definite if f(x) > 0 for every x. f is called positive semidefinite if f(x) 0 for all x. f is called strongly positive semidefinite≥ if it is positive semidefinite and is not zero on any open interval. Definition 17.2. A Pusieux series in t is a power series in a rational power of t, convergent in a neighborhood of 0. A Pusieux series at a in t is a power series in a rational power of (t a), convergent in a neighborhood of a. − A generalized Pusieux series in t is a Pusieux series in t or a power series in a rational power of t, times t , convergent in a neighborhood of 0. A generalized Pusieux series| | in t at a is a Pusieux series in (t a) or a power series in a rational power of (t a), times t a , convergent− in a neighborhood of a. − | − |

32See [32] for a proof of the Kripke completeness theorem, and Exercise 4, p. 99 of [2] for the extension to the logic of partial terms. 236 MICHAEL BEESON

A Pusieux series at is a Pusieux series convergent in a neighborhood of ; similarly for a generalized∞ Pusieux series at . ∞ ∞ For example, √t3 = t t has a generalized Pusieux series at 0, but not a Pusieux series. | | Theorem 17.3. (i) If the strong parallel axiom SPP in ECG is replaced by Euclid’s Postulate 5, then SPP is not provable. (ii) In Euclidean field theory, the existence of reciprocals of positive elements does not imply the existence of reciprocals of nonzero elements, i.e. Axiom EF7 does not imply Axiom EF1 (using axioms EF0 and EF2-EF6). Proof. We first show that (ii) implies (i). Suppose Euclid 5 plus neutral ECG does imply SPP. Then, by Theorem 12.21, the theory of weakly Euclidean fields (i.e., EF with Axiom EF1 replaced by EF7, that positive elements have reciprocals) proves the interpretation SPP◦ of SPP. Now we argue that nonzero elements have reciprocals. Assume m = 0 and consider the line through (0, 1) with slope 6 m. Then by SPP◦, the line meets the x-axis. The meeting point (a, 0) lies on the line, so 1 = am. But then m has multiplicative inverse a. Hence nonzero elements have reciprocals. But that contradicts (ii). Hence (ii) implies (i) as claimed. We now proceed to part (ii). Let K be the field of “constructible numbers”, which is the least subfield of R closed under taking square roots. Let 0 be the A ring of polynomial functions from R to R with coefficients in K. 0 is not a field, A since for example 1/t does not belong to 0. For each nonnegative integer n, we A define the ring n+1 to be the least set of real-valued functions containing n A A together with all sums and differences of members of n, together with all square A roots of positive semidefinite members of n, and reciprocals of all strongly A positive semidefinite members of n. These square roots and reciprocals are defined on dense subsets of R, as weA will show below. For example, the functions 2 2 √1+ t and 1/(1 + t ) are in 1, and A 2 4 1 1+ t + 1+ t + 2 qp p 1+ t 2 is in 2. Also √x = x is in 2, and x x is in 3; that function is zero on A | | A | |− A the positive real axis. Now define to be the union of the n. Then is a ring of functions from R to R. AnotherA way of describing isA to say thatA it is the least ring of functions containing K[x] and closed underA taking square roots and reciprocals of positive semidefinite functions. We claim that each f in n (i) is defined at all but a finite number of points in R, and has a Pusieux seriesA at all but a finite number of points a, and also has a Pusieux series at , and a generalized Pusieux series at the remaining points of its domain, and (ii)∞ f is zero on a finite number of closed intervals and a finite number of isolated points. For example x = √x2 does not have a Pusieux series at 0; but such points | | occur only at the zeroes of functions belong to n 1, as we will show in detail below. A − We prove (i) and (ii) simultaneously by induction on n. First we show that (i) at n follows from (ii) at n 1. Case 1, f is a square root, f = √g and − FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 237 g(x) 0 for all x. Then at points where g(x) > 0, f has a Pusieux series, and at the finitely≥ many points (by (ii) for n 1) where g(x) = 0, f has a generalized Pusieux series; and f is everywhere defined− and has a Pusieux series at . Case ∞ 2, f is a reciprocal of a strongly positive semidefinite function g in n 1. Then by (ii) for n 1, g has only finitely many zeroes (since there areA no− intervals of zeroes because− g is strongly positive definite), so the domain of f omits only finitely many points, and at each of those points where g is positive, f is defined and has a Pusieux series, and at the finitely many points where g is zero, it is undefined; so (i) for n follows from (ii) for n 1 also when f is a reciprocal. When f is a sum or difference, i for n follows from− (i) for n 1. Now we show − that (ii) at n follows from (i) at n. By (i) at n, f in n has a Pusieux series at a for all but finitely many points a; and at the pointsA where f has a Pusieux series, if f(a) = 0 then the zero is isolated, or the function is identically zero in an interval about a. The endpoints of these intervals of zeroes will be among the finitely many points where f does not have a Pusieux series, so there are finitely many of these intervals. Hence the zeroes of f consist of (at most) finitely many points where f does not have a Pusieux series, plus finitely many intervals, plus finitely many points where f does have a Pusieux series–altogether finitely many. So (ii) for n does follows from (i) for n, as claimed. That completes the inductive proof of (i) and (ii). We call a zero of f “half-isolated” if it is the endpoint of one of the intervals on which f is zero. By (i) and (ii), there is a countable set of reals that includes all the isolated or half-isolated zeroes and all the singularities of all the functions in . Define Ω to be the complement of that set; thus for each f in , if f(x) is zeroA for any x in Ω, then f is identically zero on some interval aboutA x. Note that, since the complement of Ω is countable, Ω is dense in R. We will exhibit a Kripke model in which positive elements have reciprocals, but the model does not satisfy that all nonzero elements have reciprocals. The partially ordered set K is 0 Ω, ordered so that 0 is the root node, and 0 < α for each α Ω, and the different{ } ∪ elements of Ω are incomparable. The root node of this model∈ is the ring , with P (x) interpreted to mean “x is strongly positive A semidefinite.” Like 0, is not a field because 1/t is not in ; but also is not even an orderedA ring,A because there are polynomials x suchA that neitherA x nor x is strongly positive semidefinite. But we can still use as the root of our Kripke− model. A For α Ω, the structure α at the node α is the quotient field of , with P (x) interpreted∈ to mean that xA(α) > 0, and x = y interpreted to meanA that x and y are equal on some neighborhood of a. In other words, the elements of (α) are equivalence classes of quotients of members of , where f/g is equivalentA to u/v if vf and ug agree in some neighborhood of α.A Since Ω does not include any singularities or zeros of members of , x(α) is defined for each x in α. A A We note that α is isomorphic to the least Euclidean subfield of R containing α. The isomorphismA is given by x x(α). It is an isomorphism because its kernel is trivial, since x(α) = 0 only if7→x is identically zero on some neighborhood of α. Moreover, if P (x) holds in , then x is strongly positive semidefinite and not identically zero; hence for α inA Ω, we have x(α) > 0, since x(α) = 0 for α in 6 Ω. Hence P (x) holds in α. A 238 MICHAEL BEESON

The ring axioms are satisfied in this Kripke structure, since all these structures ( and the σ) are rings. The ordered field axioms are satisfied at the leaf nodes A C α, since these structures are classical ordered fields. We therefore only need to Averify the reciprocal and order axioms at the root node . Consider Axiom EF2, which says that sums and productsA of positive elements are positive. This holds at since the sum and product of strongly positive semidefinite functions are alsoA strongly positive semidefinite. Consider Axiom EF3,which says that not both x and x are positive. Suppose both x and x are strongly positive semidefinite members− of . Then for each α Ω we have− x(α) > 0 and x(α) > 0, since x(α) is not zeroA for α in Ω. But that∈ is a contradiction; hence− Axiom EF3 holds at . Consider Axiom EF4, which says that if both x andA x are not positive, then x is zero. Suppose both x and x are satisfied at −to be not positive. That − A means that for every node α, x and x are not positive at α. That means that for every α Ω x(α) A 0 and −x(α) 0. Hence, x(α)A = 0 for every α in Ω. But if x is∈ not identically≤ zero,− then ≤x(t) = 0 for some t, and since all the functions in are continuous on their domains,6 xα) = 0 for some α in Ω, contradiction. HenceA satisfies Axiom EF4. 6 Consider Axiom EF5,A which says that if x is not positive, then x has a square root. If x is identically zero there is nothing− to prove, so we may assume that x is not identically zero. If satisfies that x is not positive, that means A − that x is not positive in any α; that is, x(α) 0 for all α Ω. Then x(α) − 0. Since this is true forA every α Ω,− and since≤ Ω is dense∈ in R, and x ≥ ∈ is continuous, it follows that x is positive semidefinite. Hence x(σ) belongs to , by construction of . Hence satisfies Axiom EF5. p A We now consider Markov’sA principleA EF6. Suppose that P (x) is satisfied at the root node . Then for every α in Ω, P (x) is satisfied¬¬ at the leaf node A α; that means that x(α) > 0 for each α in Ω. As shown in the verification of E5,A this implies that x is positive semidefinite; and it is not identically zero on a neighborhood of α, since x(α) > 0. Hence P (x) is satisfied at the root node . Now we consider Axiom EF7, which says positive elements have reciprocals.A Suppose P (x) is satisfied at the root node; then x is strongly positive semidefinite and not identically zero, and x belongs to some n; so 1/x belongs to n+1. Therefore the Kripke structure is a model ofA all the axioms EF2-EF7.A We will now show that it does not satisfyK EF1, which says that all nonzero elements have reciprocals. Consider the element of given by the identity function, i(t)= t. Suppose satisfies i y = 1, where 1A is the constant function with value 1. Then for eachA real number· t we have ty(t) = 1. But this is a contradiction when t = 0. Hence does not satisfy EF1. That completes theA proof.

Remark. In the Kripke model , the line parallel to the x-axis, passing through the point (0, 1) with slope t, whereK t is the element of representing the identity function, will intersect the x-axis if t is positive or negative,A but if we only assume that the slope t is nonzero, we cannot construct the intersection point. Indeed, in different extensions σ of , t might be positive, or it might be negative, so the intersection point (forC smallA t) will be a point on the x-axis, far from the FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 239 origin, but it will be a large positive value in some of the σ and a large negative value in others. C 17.3. Playfair does not imply Euclid 5. Recall that “Playfair field theory” is the theory of ordered fields satisfying Axioms EF0, EF2-EF6, and EF9- EF10. That is, without the axiom of reciprocals (EF7), but with the axiom (EF9) that elements without reciprocals are zero, and the axiom (EF10) that says if x is greater than a positive invertible element, then x is invertible. Theorem 17.4. (i) If the parallel axiom (Axiom 58) in ECG is replaced by Playfair’s Axiom, then Euclid’s Proposition 5 is not provable, and (hence) Axiom 58 is not provable. (ii) Playfair field theory does not imply that reciprocals of positive elements exist. Proof. By the results of the previous section, part (ii) implies part (i). We therefore proceed to part (ii). To say that y (y x = 1) holds at a node of a Kripke model is to say that no node above¬∃ ·contains an inverse of x.A If one of the leaf nodes above is a (classical) field,A then x must be zero in that field, and hence in also. HenceA the axiom that elements without reciprocals are zero will hold inA any Kripke model, all of whose leaf nodes are fields. What we need, then, is a Kripke model in which all the leaf nodes are ordered fields, and the root node has a positive element without a reciprocal. We say that x is “almost positive definite” if it is positive semidefinite, and it is not zero on any open interval. We construct a model similar to the one in the preceding proof, except that when constructing , we throw in square roots of positive semidefinite functions, but reciprocals onlyA of functions bounded below by functions that already have reciprocals, instead of reciprocals of all positive semidefinite functions. More precisely, Let 0 be the ring of polynomial functions from R to R with coefficients in K. For eachA nonnegative integer n, we define the ring n+1 to be the least set of real-valued functions containing n together A A with all sums and differences of members of n, together with all square roots A of positive semidefinite members of n, together with 1/f for each function A f in n such that for some function g with both g and 1/g in n, we have f(x) A g(x) 0 for all x in the domain of both f and g. TheseA square roots and reciprocals≥ ≥ are defined on dense subsets of R, as we will show below. Now define to be the union of the n. Then is a ring of functions from R to R. AnotherA way of describing Ais to say thatA it is the least ring of functions containing K[x] and closed underA taking square roots of positive semidefinite functions, and closed under the rule that if g f and 1/g is in then 1/f is in . ≤ A A The only invertible elements in 0 are the nonzero constants, so A1 will contain the reciprocals of functions thatA are bounded below by some positive constant, and A2 will contain the reciprocals of functions bounded away from zero by some positive constant. As in the previous proof, all the members of have Pusieux series except at finitely many points, and their zero sets haveA the same structure as before, so we can again find a countable set containing all the singularities and all the isolated and half-isolated zeroes. let Ω be the complement of this countable set, 240 MICHAEL BEESON and define a Kripke model as before, with index set 0, Ω and R0 = and for { } A σ Ω, Rσ = Cσ. In this model, we claim that all members of are continuous on∈ the real line. To prove this, we proceed by induction on n toA prove that the members of n are continous on the whole real line. That is true for n = 0 since A the elements of 0 are polynomials. The square root of a positive semidefinite function continuousA on R is also continuous on R, as are sums and differences of such functions. Suppose f(x) g(x) 0 and f and g and 1/g are in n. ≥ ≥ A Then 1/f will be in n+1. By induction hypothesis, both g and 1/g are defined everywhere; hence gA(x) is everywhere positive. Hence f(x) is also everywhere positive; hence 1/f is everywhere defined and continuous. That completes the proof that all members of are continuous on R. We define P (x) to holdA at the root node if and only if x is strongly positive definite. The verification of axioms E2 through E6 is the same as in the previous proof. Now we verify Axiom EF10. It is automatic at the leaf nodes since there the structures are strong Euclidean fields. We check the root node. Suppose P (g) is satisfied at the root; then g is strongly positive definite. Suppose also P (g f) is satisfied; then f(x) g(x) for all x. Suppose g has a reciprocal in the¬ model.− ≥ Then for some n, f, g, and 1/g belong to n. Then by construction of the A model, 1/f belongs to n+1. Hence Axiom EF10 holds. It only remains to showA that EF7 is not satisfied at the root node. If EF7 were satisfied at , then every strongly positive definite element of would have its reciprocal in A. To refute this, we must exhibit a strongly positiveA definite A element of whose reciprocal is not in . Take f(x) = x . Since x = √x2, A A | | | | this function is in 1. But 1/f is not defined at x = 0, and we have proved above that all membersA of are defined and continuous on the whole real line. Hence 1/ x is not in . ThatA completes the proof. | | A 17.4. Independence of Markov’s principle. Theorem 17.5. Markov’s principle is independent of the other axioms of ECG. Proof. We construct a model similar to the ones used above. This time, let be the least class of functions containing the polynomials, and closed under Asquare roots of positive semidefinite functions and reciprocals of not-identically- zero functions. Interpret P (x) at the root as “x is positive definite”, instead of as “x is positive semidefinite and not identicallyA zero.” Then the function x(t)= t2 is not satisfied to be positive at the root node, but since 0 is not in Ω, we have x(α) > 0 for all α in Ω, so P (x) is satisfied at every leaf node. Hence P (x) is satisfied at the root node. Hence P (x) P (x) is not satisfied at the¬¬ root. That completes the proof. ¬¬ →

§18. Apartness. What happens if we add apartness? 18.1. Constructions and Apartness. Mandelkern’s line chooser; other apartness constructors; 18.2. Euclidean ﬁelds and apartness. 18.3. Apartness and the parallel axioms. In particular are the parallel axioms still independent? FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 241

Lemma 18.1. ECGD proves x =0 x> 0 x< 0. 6 → ∨ Proof. Since x, 0, and I are distinct and not collinear, xyI is either a left turn or a right turn, as shown in [??], Lemma 9.3. Note that Axiom 75 is required for that lemma, so we need ECGD, not just ECG, in the theorem. That completes the proof. 242 MICHAEL BEESON

§19. Appendix A: Roads not taken. In the course of developing ECG, we made several decisions, in some cases exploring alternative formulations to some depth. We chose a rigid compass (i.e. construction of circles from center and radius); we chose to allow degenerate circles; we rejected a primitive operation of projection on a line; we chose to have a term center(C) for the center of a circle. In this appendix we justify, or at least explain, those choices, and mention a few related problems, some but not all of which we solve. 19.1. Rigid compass undefinable from the collapsible compass. Book I, Proposition 2 has been discussed in the introduction. The question it addresses concerns the constructor Circle3 (A, B, C), which constructs a circle with center A and radius BC. As discussed in the introduction, Euclid gives a term that accomplishes this aim under the assumptions, not only that B = C, but also that A = C B = C. We consider the stronger theorem “uniform6 Euclid I.2”, which asserts6 ∨ that6 for every A and BC with B = C, there exists a D with AD = BC. The “uniformity” refers to the missing assumption6 A = B. To “realize” uniform Euclid I.2, we would need a term e(A, B, C) that produces6 D uniformly, whether A = B or not. If we had such an e, then of course we could define Circle3 (A, B, C)= Circle (A, e(A, B, C)). Conversely, such an e can be defined from Circle3 like this: e(A, B, C)= pointOnCircle (Circle3 (A, B, C)) But the mere fact that Euclid’s own construction does not suffice to define Circle3 does not show that some other construction won’t do the job. In this section we will prove that it is really true that no other construction can do the job: Theorem 19.1. Circle3 is not definable in the standard plane R2 from the (other) elementary constructions. We will work up to this theorem gradually, beginning by showing that the collapsible compass cannot simulate the rigid compass if degenerate circles are not allowed, and then elaborating the proof to allow degenerate circles. 2 2 For that we consider the model R− , which is the same as R except that Circle (A, A) is undefined. Theorem 19.2 (with Freek Wiedijk). No total unary point-valued function 2 (other than the identity) is definable in the standard model R − (with degenerate circles undefined) from the elementary geometrical constructions excluding Circle3 . More precisely, let t be a compound term containing exactly one variable A of type Point, and no other variables, but possibly containing some constants α,β,... of type Point. Suppose that t does not contain Circle3. Let α be a constant contained in t, or if t has no constants, let α be any constant. Let the constants be interpreted as certain (fixed) distinct points α,¯ β,...¯ in the standard plane, and let t¯ be the interpretation of t when the variable A is interpreted as α¯. Then t¯ is undefined. Proof. We start by eliminating the “overloaded” versions of the elementary constructions from t. For instance, if t contains a subterm IntersectLines (a,b,c,d), FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 243 we replace it by a term using the “fundamental” form of the construction, IntersectLines (Line (a,b), Line (c, d)). The result of such replacements is a term with the same value as t under any assignment of a value to the variable A, and containing no variables of types other than Point . We proceed by induction on the complexity of terms t as in the theorem, but also containing only the fundamental versions of the elementary constructions (no overloaded versions). Since the theorem only applies to compound terms, the basis case occurs when t has only variables or constants for arguments. We note that Circle (A, A), Line (A, A), Ray (A, A), Segment (A, A), Arc (A, A, A) are undefined. But we also have to consider the possibility that the constants β or γ occupy one of the argument places. For example, Circle (A, α) is undefined when A takes the value α¯. In the rest of the proof Similarly, when A takes the valueα ¯, Circle (α, A), Line (A, α) and Line (α, A) are all undefined. Note that this argument does not work for Circle3 , since Circle3 (A,β,γ) is always defined, but by hypothesis, t does not contain Circle3 . Since t does not contain any overloaded constructions, the basis case is finished, as there is no fundamental construction that takes only arguments of type Point. Specifically, there are just five fundamental constructions for producing the intersections of lines and lines, lines and circles, or circles and circles, and they each need arguments of type Line or Circle. Now consider the induction step. If the main symbol of t is a constructor, such as Line, then t has the form Line (a,b). One of a or b must contain a variable, and hence be somewhere undefined. Hence t is also somewhere undefined, and indeed the same assignment of a value to A that makes a or b undefined will work. Similarly for the other constructors (since Circle3 , to which this argument does not apply, is not allowed). Next consider the case when t is IntersectLines (p, q). Then p and q have type Line. The only terms of type Line are those of the form Line (a,b), or variables of type Line. But t is not allowed to contain variables of type Line. Therefore t must have the form, IntersectLines (Line (a,b), Line (c, d)) Since t contains a variable, one of s = Line (a,b) or s = Line (c, d) must contain a variable, and by induction hypothesis the term sσ is undefined when substitution σ assigns the variable of s to one of the constants in s (or any constant if s has no constants). Hence tσ is also undefined. Next consider the case when t is IntersectLineCircle1 (p, q). Since t contains the variable A, then one of the two arguments of t contains A. By induction hypothesis, whichever it is, its interpretation (be it p¯ orq ¯) is undefined, and that makes t¯ undefined as well. Similarly when t is IntersectLineCircle2 , IntersectCircles1 , or IntersectCircles2 . There is also the possibility that t has the form f(p), where f is one of the access functions pointOnCircle , pointOn1 , or pointOn2 ). Then p must contain the variable A, and by induction hypothesis, p¯ is undefined. That completes the proof. The following corollaries show that the rigid compass is not definable from the collapsible compass, at least if the collapsible compass cannot be used to 244 MICHAEL BEESON draw circles of zero radius. The first corollary shows the serious limitations of geometry with only a collapsible compass and no equality-test constructor. Corollary 19.3. Let e(A) be any term built up from the elementary constructions, not containing Circle3, having type Point, and containing exactly 2 one variables A of type Point. Then it is not the case that in R −, we always have e(A) = A. 6 Proof. According to the theorem, e(A) has to be undefined for some value of the variable A. Corollary 19.4. Let e(A, B, C) be any term built up from the elementary constructions, not containing Circle3, having type Point, and containing exactly 2 three variables A, B, and C of type Point. Then it is not the case that in R −, whenever B = C then e(A, B, C) is defined and is a point D such that AD = BC. 6 Proof. Let us invent two constants β and γ, and interpret them as two distinct points β¯ andγ ¯ (fixed for the rest of the proof). Then let f(A) = e(A,β,γ). Suppose, for proof by contradiction, that e(A, B, C) is defined whenever B = C and is a point D such that AD = BC. Then f(A) is defined for all A and6 is always different from A, contradicting the previous lemma. 2 Corollary 19.5. Circle3 is not definable in the model R − from the (other) elementary constructions. Proof. Circle3 is a term e fulfilling the hypotheses of the previous corollary. 19.2. Rigid compass not definable, even with degenerate circles. Next we will extend these results to the standard model R2, that is, we will remove the restriction that circles must have positive radius. Theorem 19.6. let t be a compound term containing exactly one variable A of type Point, and no other variables, but possibly containing some constants α,β,... of type Point. Let α,¯ β,...¯ be the interpretstions of these constants. Suppose that t does not contain Circle3. Let t¯ be the interpretation of t in R2. Let α be any constant contained in t, or if t does not contain any constant, let α be any constant. Let A be interpreted as α¯. Then either (i) t¯ is undefined, or (ii) t is of type Point, and t¯ is equal to α¯; or (iii) t is of type Circle, and t¯ is a degenerate circle with center equal to α¯. Remark. The restriction that the term not contain Circle3 is certainly necessary; not only does the proof depend on it, but we will show in the next section that circles of zero radius are actually useful in connection with Circle3 , while this theorem shows that they are essentially useless if we only have the collapsible compass constructor Circle . Proof. As in the proof of the previous theorem, we proceed by induction on the complexity of terms t with variables only of type Point, and built up using points, lines, and circles only (not rays and arcs). Since the theorem only applies to compound terms, the basis case occurs when t has only variables or constants for arguments. Now Circle (A, A) is no longer undefined; but it is a degenerate FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 245 circle, and when A is interpreted asα ¯, then the conclusion of the theorem is true, that Circle (A, A) is interpreted as a degenerate circle with centerα ¯. We also have to consider Circle (α, A) and Circle (A, α), but both of these degenerate as required when A = α. We do not have to consider Circle (α, β) since t is required to contain at least one variable. Line (A, A) is undefined, as before, no matter how A is interpreted. Line (α, A) and Line (A, α) become undefined when A is interpreted as α. Line (α, β) does not contain a variable and so does not need to be considered. That completes the basis case. Now consider the induction step. If the main symbol of t is Line, then t has the form Line (a,b). One of a or b must contain the variable A of t, say a. Since a has type Point, the interpretationa ¯ of a is either undefined, or equal toα ¯. If it is undefined then so is Line (a,b) and we are done. So we may supposea ¯ =α ¯. Similarly, we may suppose ¯b =α ¯. Then Line (a,b) is interpreted as Line (¯α, α¯, which is undefined. Next consider the case when t is IntersectLines (p, q). Then t must have the form,

IntersectLines (Line (a,b), Line (c, d))

Since t contains a variable, one of s = Line (a,b) or s = Line (c, d) must contain a variable, and by induction hypothesis the terms ¯ is undefined, since alternatives (ii) and (iii) cannot apply to a term of type Line. But then t¯ is also undefined, since it has an undefined argument. Next consider the case when t is IntersectLineCircle1 (Line (a,b), Circle (c, d)). If Line (a,b) contains the variable A, then its interpretation is undefined, and hence t is undefined. So we may assume that Line (a,b) does not contain A; then Circle (c, d) contains A. By induction hypothesis its interpretation is undefined (in which case we are done with this case) or it is a degenerate circle with centerα ¯. In that case, the intersection points of Line (a,b) are equal and both of them are equal toα ¯, so alternative (ii) of the theorem holds. Similarly for IntersectLineCircle2 . Next consider the case when t is IntersectCircles1 (Circle (a,b), Circle (c, d)). If the interpretation of either argument is undefined, then t¯ is undefined too and we are done with this case. So we may assume that the arguments are two degenerate circles with the same centerα ¯. But the intersection points of two circles with the same center are undefined (whether or not the circles are degenerate), so t¯ is undefined. Similarly for IntersectCircles2 . That takes care of the five constructors; but there are also the access functions to consider. Consider the case when t is pointOnCircle (Circle (a,b)). By induction hypothesis, the interpretation of Circle (a,b) is either undefined, in which case we are done with this case, or a degenerate circle with centerα ¯. In that case the interpretation of pointOnCircle (Circle (a,b)) is alsoα ¯, as that is the only point on the degenerate circle. Hence alternative (ii) of the theorem holds. Consider the case when t is pointOn1 (Line (a,b). The interpretation of this term isa ¯, which by induction hypothesis is either undefined orα ¯, so we are done with this case. Similarly for pointOn2 (Line (a,b)). That completes the proof. 246 MICHAEL BEESON

Corollary 19.7. Without Circle3 , we cannot even define a construction f such that for each point A, f(A) is a point different from A, even if we allow degenerate circles. Proof Any definable function f will either be undefined somewhere, or will fix a constant α that is used in the definition. Finally we reach the theorem announced at the beginning of this section: Theorem 19.8. Circle3 is not definable in the standard plane R2 from the (other) elementary constructions. Proof. If it were definable, then let α and β be distinct constants, and define f(A)= Circle3 (A, α, β). But according to the above theorem, f(A) is undefined or a degenerate circle when A is interpreted as α. That is a contradiction, since Circle 3(A, α, β) is defined and not degenerate, sinceα ¯ = β¯. That completes the proof. 6 2 2 The fact that Circle3 is not definable in R− or in R means that, if we do not include Circle3 as a primitive construction, we shall not be able to define 2 2 it in any axiomatic theory that has R − or R for a model. It seems clear 2 that Euclid’s book does have R− as a model. Although Euclid does not use degenerate circles, we have now proved that they cannot help; so the uniform version of Book I, Proposition 2 is essentially non-constructive, using Euclid’s non-rigid compass. We therefore add Circle3 as a fundamental construction (rendering I.2 a triviality) and give a constructive theory that works for the rest of Euclid. In the next section we give further justifications for adding the rigid compass Circle3 . 19.3. Circles of zero radius, the rigid compass, and projection. In this section, we address the two issues of whether our basic theory ought to include the rigid compass, represented by Circle3 , and whether we ought to allow circles of zero radius. We conclude that both should be allowed. Our approach is to consider what is required to achieve a formalization of Euclid using intuitionistic logic, with no test-for-equality construction needed. We exhibit three important constructions that are possible with Circle3 , and at least one of them (projection on a line) is definitely not possible without Circle3 . Projection is absolutely necessary in order to reduce geometry to algebra. We want to pick a line, call it X, and erect a perpendicular Y to X, and project each point P onto its coordinates x = Project(P,X) and y = Project(P, Y ). We show below that we can define projection using Circle3 , if circles of zero radius are allowed. This is the main reason why we want to allow circles of zero radius. We considered replacing Circle3 by projection, but Circle3 seems more fundamental. Here are some possible constructions we wish to consider. Euclidean Extension. One of the Euclidean axioms says that we can extend a given segment AB by a segment CD. More precisely, we can construct a point P = Extend (A,B,C,D) such that BP = CD and B is between A and P . The assumptions here are that A = B and C = D, but it is not assumed that B = C or B = D. 6 6 6 Strong6 Extension. Extend (A,B,C,D) is the unique continuous extension of Extend (A,B,C,D) that is defined when A = B, i.e. without assuming C = D. 6 6 FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 247

When C = D, we have Extend (A,B,C,D)= B. We will show below how to deﬁne Extend in terms of Circle3 . The following lemma helps make the case for allowing Circle3 , since we need projection. The lemma shows that we can’t deﬁne projection without Circle3 , so the alternative to Circle3 would be to take projection itself as primitive.

2 Lemma 19.9. In R − (where circles of zero radius do not exist), as well as in R2 (where they do exist) projection is not definable in terms of the elementary constructions without Circle3. Proof. Let α and β be constants, whose values will be two distinct (fixed) points. Let t(A) be the term Project(A, Line (α, β)). Then t(A) is defined for all values of A. By Theorem 19.2, t(A) is not definable in terms of the elementary 2 constructions without Circle3 , in the model R −. That completes the proof. 19.4. Constructing the center of a circle. We chose to include an accessor function center producing the center of a circle. There is a question whether including center is faithful to Euclid, since Euclid III.1 provides a way to construct center(C) from two points on circle C. Since circle C could only have been drawn using its center, in III.1 Euclid appears to tell us, if somehow the center of C got accidentally rubbed out of the diagram, here is how you can recover it. In any event, the existence of Prop. III.1 may be taken to indicate that we could omit the function symbol center(C) and replace it with the term for the point constructed in III.1? There are two answers to this suggestion. The first one is that the proof of III.1 is erroneous: it implicitly requires the use of III.2, that a chord of a circle lies inside the circle, in order to know that point E in III.1 exists. But the proof of III.2 requires III.1, that the center of the circle be known. However, as Heath’s commentary on this proposition [8] shows, there is another way to construct the center, due to de Morgan, which is correct. The second answer is that Euclid’s construction in III.1 requires two points A and B on circle C. One point, say A, can be obtained as pointOnCircle (C). But there is no good (uniform) way to obtain a second point, without knowing the center of C. The second point can’t be obtained as the intersection of some line through A with C, because we have no good candidate for a second point on that line. Could we obtain it as the intersection of a circle with center A with C? To know that circle K with center A meets C we would have to know that K contains a point inside C. That is hard to do without having the center of C. Hence, even if Euclid III.1 were correctly proved, which it is not, we still would need the term center(C), or a way to get a second point. Finally, Heath’s commentary shows that de Morgan’s method requires three distinct points on C. The conclusion is this: we could eliminate the function symbol center, but we would then need three point-access functions guaranteed to produce three distinct points on a circle. The question would naturally arise, how do we construct these three points in a uniform way? We feel that it is more natural to have the symbol center, and regard (de Morgan’s repaired) version of Euclid III.1 as a reconstruction of the center from three points, rather than a construction from a circle given deus ex machina. 248 MICHAEL BEESON

We can pose technical questions in this connection: Is center deﬁnable in terms of the rest of the elementary constructions? In other words, if we delete the function center, is there a term t(C) with C of type Circle such that t(C) is always the center of circle C? We conjecture there is no such term. We know by de Morgan that there is a term t(p,q,r,C) such that if p, q, and r are distinct points on C then t(p,q,r,C) is the center of C. What about with two points instead of three? We conjecture it is not possible.

§20. Appendix B: Possible reductions in the primitive construction mechanisms. While it is natural to take as primitives the operations of intersecting a line and a circles, and intersecting two circles, it is known that this is not a primitive set of constructions. In this section we summarize the known results, and show that in one case, the known result cannot be improved. 20.1. The Mohr-Mascheroni theorem. In the “algebraic framework”, where instead of logical theories one just considers “algebras” with operations for the geometric constructions, one can ask for a minimal set of primitives. The Mohr- Mascheroni theorem shows that, in classical geometry, “every geometric construction carried out by compasses and ruler can be done without ruler.” More precisely, Theorem 20.1 (Mohr-Mascheroni). All points constructible using the ﬁve elementary constructions can be constructed using Circle3 , IntersectCircles1 and IntersectCircles2 only. Proof. See [16], which also explains the history of this theorem, which goes back to the time of Galileo. But the proof of the Mohr-Mascheroni theorem uses a case distinction as to whether a line passes through the center of a circle or not. Hence, it will not produce a single term in the algebra of constructions that will deﬁne IntersectLineCircle1 , but rather, two terms, one of which does the job when the line passes through the center of the circle, and the other when it does not. That implies that the proof is non-constructive. Is the theorem non-constructive, or only the known proof? One approach to making this question precise is to ask whether there is a single term of ECG that produces the same result as IntersectLineCircle1 ; or a single term i(L, C) that always produces an intersection point of a line L and circle C if they have one. Another approach is to remove the axioms for IntersectLineCircle1 and IntersectLineCircle2 from ECG and ask if the resulting theory can prove the line-circle continuity axiom of Tarski’s theory. The answers to these questions are presently unknown. 20.2. Strommer’s theorem. The Mohr-Mascheroni theorem shows us how to eliminate IntersectLineCircle1 and IntersectLineCircle2 in favor of IntersectCircles1 and IntersectCircles2 , at least classically. On the other hand, it is also possible to eliminate IntersectCircles1 and IntersectCircles2 in favor of IntersectLineCircle1 and IntersectLineCircle2 , as Strommer showed in [28]. Apparently this fundamental result was not known until this 1973 publication! while the Mohr-Mascheroni theorem has 450 years of history. Strommer’s construction is uniform, i.e., the same construction term applies in all cases, and FOUNDATIONS OF EUCLIDEAN CONSTRUCTIVE GEOMETRY 249 works even when the two intersection points of the line and circle coincide, en- abling us to construct the common tangent of the circles.

REFERENCES

[1] Jeremy Avigad, Edward Dean, and John Mumma, A formal system for euclid’s elements, Review of Symbolic Logic, vol. 2 (2009), pp. 700–768. [2] Michael Beeson, Foundations of constructive mathematics, Springer-Verlag, Berlin Heidelberg New York, 1985. [3] , Constructive geometry, Proceedings of the tenth asian logic colloquium, kobe, japan, 2008 (Singapore) (Tosiyasu Arai, editor), World Scientific, 2009, pp. 19–84. [4] Errett Bishop, Foundations of constructive analysis, McGraw-Hill, New York, 1967. [5] Karol Borsuk and Wanda Szmielew, Foundations of geometry: Euclidean and bolyai-lobachevskian geometry, projective geometry, North-Holland, Amsterdam, 1960, translated from Polish by Erwin Marquit. [6] Max Dehn, Die grundlegung der geometrie in historischer entwicklung, B. G. Teub- ner, 1926, Appendix in [24]. [7] Rene´ Descartes, The geometry of rene descartes, with a facsimile of the first edition, published in 1637 as an appendix to discours de la methode., Dover, New York, 1952, Original title, La Geometrie. Facsimile with English translation by David Eugene Smith and Marcia L. Latham. [8] Euclid, The thirteen books of the elements, Dover, New York, 1956, Three volumes. Includes commentary by the translator. [9] , The bones: A handy, where-to-find-it pocket reference companion to euclid’s elements, Green Lion Press, Santa Fe, New Mexico, 2007, Statements and diagrams only, no proofs. [10] , Euclid’s elements, Green Lion Press, Santa Fe, New Mexico, 2007, All thirteen books in one volume, without commentary. [11] Kurt Godel¨ , Zur intuitionistischen arithmetik und zahlentheorie, Kurt gödel, col- lected works volume i, Oxford University Press, 1933, with English translation, pp. 286–295. [12] Marvin Jay Greenberg, Euclidean and non-euclidean geometries: Development and history, fourth ed., W. H. Freeman, New York, 2008. [13] Haragauri Narayan Gupta, Contributions to the axiomatic foundations of geometry, Ph.D. thesis, University of California, Berkeley, 1965. [14] Robin Hartshorne, Geometry: Euclid and beyond, Springer, 2000. [15] David Hilbert, Foundations of geometry (grundlagen der geometrie), Open Court, La Salle, Illinois, 1960, Second English edition, translated from the tenth German edition by Leo Unger. Original publication date, 1899. [16] Norbert Hunberbuhler¨ , A short elementary proof of the mohr-mascheroni theorem, The American Mathematical Monthly, vol. 101 (1994), no. 8, pp. 784–787. [17] Stephen C. Kleene, Introduction to metamathematics, van Nostrand, Princeton, 1952. [18] , Permutability of inferences in Gentzen’s calculi LK and LJ, A.M.S. Memoirs, vol. 10, pp. 1–26, A.M.S. Memoirs, American Mathematical Society, Providence, R. I., 1952, pp. 1–26. [19] Mark Mandelkern, Constructive coordinatization of desarguesian planes, Beiträge zur Algebra und Geometrie ( Contributions to Algebra and Geometry), vol. 48 (2007), no. 2, pp. 547–589. [20] Elena Anne Marchisotto and James T. Smith, The legacy of mario pieri in geometry and arithmetic, Birkhauser, Boston, Basel, Berlin, 2007. [21] Julien Narboux, Automated deduction in geometry: 6th international workshop, adg 2006, pontevedra, spain, august 31-september 2, 2006, revised papers, (Francisco Botana and Tomas Recio, editors), Lecture Notes in Artificial Intelligence, Springer, 2008, pp. 239–156. 250 MICHAEL BEESON

[22] Victor Pambuccian, Axiomatizations of hyperbolic and absolute geometries, Non- euclidean geometries, jános bolyai memorial volume. papers from the international conference on hyperbolic geometry held in budapest, july 6–12, 2002 (New York) (András Prékopa and Emil Molnár, editors), Mathematics and Its Applications, vol. 581, 2006, pp. 119– 153. [23] Moritz Pasch, Vorlesung über neuere geometrie, Teubner, Leipzig, 1882. [24] Moritz Pasch and Max Dehn, Vorlesung über neuere geometrie, B. G. Teubner, Leipzig, 1926, The first edition (1882), which is the one digitized by Google Scholar, does not contain the appendix by Dehn. [25] Mario Pieri, Della geometria elemetare come sistema ipotetico deduttivo: Monografia del punto e del moto, Memorie della Reale Accademia delle Scienze di Torino (series 2), vol. 49 (1900), pp. 173–222, Reprinted in [??], pp. 183–234. [26] W. Schwabhauser,¨ Wanda Szmielew, and Alfred Tarski, Metamathematische methoden in der geometrie: Teil i: Ein axiomatischer aufbau der euklidischen geometrie. teil ii: Metamathematische betrachtungen (hochschultext), Springer, 1983. [27] Dana Scott, Dimension in elementary euclidean geometry, The axiomatic method with special reference to geometry and physics, proceedings of an internation symposium, held at the university of california, berkeley, december 26, 1957-january 4, 1958. (Amsterdam) (Leon Henkin, Patrick Suppes, and Alfred Tarski, editors), Studies in logic and the foundations of mathematics, North-Holland, 1959, pp. 53–67. [28] J. Strommer, Konstruktionen mit dem rechten und schiefen zeichenwinkel in der absoluten geometrie, Periodica Mathematica Hungarica, vol. 6 (1975), no. 1, pp. 87–95. [29] Alfred Tarski, What is elementary geometry?, The axiomatic method, with special reference to geometry and physics. proceedings of an international symposium held at the univ. of calif., berkeley, dec. 26, 1957–jan. 4, 1958 (Amsterdam) (Leon Henkin, Patrick Suppes, and Alfred Tarksi, editors), Studies in Logic and the Foundations of Mathematics, North-Holland, 1959, Available as a 2007 reprint, Brouwer Press, ISBN 1-443-72812-8, pp. 16– 29. [30] Alfred Tarski and Steven Givant, Tarski’s system of geometry, The Bulletin of Symbolic Logic, vol. 5 (1999), no. 2, pp. 175–214. [31] Anne Troelstra, Metamathematical investigation of intuitionsitic arithmetic and analysis, Lecture Notes in Mathematics, no. 344, Springer, Berlin, Heidelberg, New York, 1973. [32] Dirk van Dalen, Intuitionistic logic, The handbook of philosophical logic, vol. 3, Reidel, 1986, pp. 225–339.