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The Surprise Examination Paradox and the Second Incompleteness Theorem

Shira Kritchman and Ran Raz

Few theorems in the history of which the derivation is done; which is impossible have inspired mathematicians and philosophers by the second incompleteness theorem. as much as G¨odel’s incompleteness theorems. The first incompleteness theorem states that, for The First Incompleteness Theorem 1 2 any rich enough consistent mathematical , G¨odel’s original for the first incompleteness there exists a that cannot be proved or theorem [Gödel31] is based on the liar paradox. disproved within the theory. The second incom- The liar paradox: consider the pleteness theorem states that for any rich enough statement “this statement is .” consistent mathematical theory, the The statement can be neither true of the theory itself cannot be proved (or disproved) nor false. within the theory. G¨odel considered the related statement “this state- In this paper we give a new proof for G¨odel’s second incompleteness theorem, based on Kol- ment has no proof.” He showed that this statement mogorov complexity, Chaitin’s incompleteness can be expressed in any theory that is capable of theorem, and an that resembles the expressing elementary arithmetic. If the statement surprise examination paradox. has a proof, then it is false; but since in a consistent We then go the other way around and suggest theory any statement that has a proof must be that the second incompleteness theorem gives a true, we conclude that if the theory is consistent, possible resolution of the surprise examination the statement has no proof. Since the statement N paradox. Roughly speaking, we argue that the has no proof, it is true (over ). Thus, if the the- flaw in the derivation of the paradox is that it ory is consistent, we have an example for a true N contains a hidden assumption that one can prove statement (over ) that has no proof. the consistency of the mathematical theory in The main conceptual difficulty in G¨odel’s orig- inal proof is the self-reference of the statement Shira Kritchman did her M.Sc in applied mathematics at “this statement has no proof.” A conceptually the Weizmann Institute of Science and is currently work- simpler proof of the first incompleteness the- ing in design at ECI Telecom. Her email address is orem, based on Berry’s paradox, was given by [email protected]. Chaitin [Chaitin71]. Ran Raz is professor on the faculty of mathematics and Berry’s paradox: consider the ex- computer science at the Weizmann Institute of Science. pression “the smallest positive in- His email address is [email protected]. teger not definable in under eleven supported by the Israel Science Foundation (ISF), words.” This expression defines the Binational Science Foundation (BSF), and the Minerva that integer in under eleven words. Foundation. Article©2010 Shira Kritchman, Ran Raz. To formalize Berry’s paradox, Chaitin uses the 1We require that the theory can express and prove ba- sic arithmetical . In particular, ZFC and Peano notion of Kolmogorov complexity. The Kolmogorov arithmetic (PA) are rich enough. complexity K(x) of an integer x is defined to be the 2Here and below, we consider only first-order length (in bits) of the shortest computer program with recursively enumerable sets of . For simplic- that outputs x (and stops). Formally, to define K(x) ity, let us assume that the of axioms is computable. one has to fix a programming language, such as

1454 Notices of the AMS Volume 57, Number 11 LISP, Pascal or C++. Alternatively, one can define proofs only show the existence of a true statement K(x) by considering any universal Turing machine. (over N) that has no proof, without giving an Chaitin’s incompleteness theorem states that explicit example of such a statement. for any rich enough consistent mathematical A different proof for the second incomplete- theory, there exists a (large enough) integer L ness theorem, based on Berry’s paradox, was (depending on the theory and on the programming given by Kikuchi [Kikuchi97]. This proof is model- language that is used to define Kolmogorov com- theoretic and seems to us somewhat less intuitive plexity) such that, for any integer x, the statement for people who are less familiar with . “K(x) > L” cannot be proved within the theory. For previous model-theoretic proofs for the sec- The proof given by Chaitin is as follows. Let L be ond incompleteness theorem see [Kreisel50] (see a large enough integer. Assume for a also [Smory´nski77]). that, for some integer x, there is a proof for the statement “K(x) > L”. Let w be the first proof Our Approach (say, according to the lexicographic order) for a We give a new proof for the second incomplete- statement of the form “K(x) > L”. Let z be the ness theorem, based on Chaitin’s incompleteness integer x such that w proves “K(x) > L”. It is easy theorem and an argument that resembles the to give a computer program that outputs z: the surprise examination paradox (also known as the program enumerates all possible proofs w, one by unexpected hanging paradox). one, and for the first w that proves a statement of The surprise examination para- the form “K(x) > L”, the program outputs x and dox: the teacher announces in stops. The length of this program is a constant + log L. Thus, if L is large enough, the Kolmogorov : “next week you are going complexity of z is less than L. Since w is a proof to have an exam, but you will not for “K(z) > L” (which is a false statement), we be able to know on which day of conclude that the theory is inconsistent. the week the exam is held until Note that the number of computer programs of that day.” The exam cannot be length L bits is at most 2L+1. Hence, for any integer held on Friday, because otherwise, L, there exists an integer 0 ≤ x ≤ 2L+1, such that the night before the students will K(x) > L. Thus, for some integer x, the statement know that the exam is going to be “K(x) > L” is a true statement (over N) that has no held the next day. Hence, in the proof. same way, the exam cannot be held A different proof for G¨odel’s first incomplete- on Thursday. In the same way, the ness theorem, also based on Berry’s paradox, was exam cannot be held on any of the given by Boolos [Boolos89] (see also [Vopenka66, days of the week. Kikuchi94]). Other proofs for the first incomplete- Let T be a (rich enough) mathematical theory, ness theorem are also known (for a recent survey, such as PA or ZFC. For simplicity, the reader can see [Kotlarski04]). assume that T is ZFC, the theory of all mathematics; thus any , and in particular any The Second Incompleteness Theorem proof in this paper, is obtained within T . The second incompleteness theorem follows di- Let L be the integer guaranteed by Chaitin’s rectly from G¨odel’s original proof for the first in- incompleteness theorem. Thus, for any integer x, theorem. As described above, G¨odel the statement “K(x) > L” cannot be proved (in the expressed the statement “this statement has no theory T ) unless the theory is inconsistent. Note, proof” and showed that, if the theory is consistent, however, that for any integer x such that K(x) ≤ L, this is a true statement (over N) that has no proof. there is a proof (in T ) for the statement “K(x) ≤ L”, Informally, because the proof that this is a true simply by giving the computer program of length at statement can be obtained within any rich enough most L that outputs x and stops and by describing theory, such as Peano arithmetic (PA) or ZFC, if the running of that computer program until it the consistency of the theory itself can also be stops. proved within the theory, then the statement can Let m be the number of integers 0 ≤ x ≤ 2L+1 be proved within the theory, which is a contra- such that K(x) > L. (The number m is analogous diction. Hence, if the theory is rich enough, the to the day of the week on which the exam is consistency of the theory cannot be proved within held in the surprise examination paradox.) Recall the theory. that because the number of computer programs Thus the second incompleteness theorem fol- of length L bits is at most 2L+1, there exists at lows directly from G¨odel’s original proof for the least one integer 0 ≤ x ≤ 2L+1 such that K(x) > L. first incompleteness theorem. However, the sec- Hence, m ≥ 1. ond incompleteness theorem doesn’t follow from Assume that m = 1. Thus there exists a sin- Chaitin’s and Boolos’s simpler proofs for the first gle integer x ∈ {0,..., 2L+1} such that K(x) > L, incompleteness theorem. The problem is that these and every other integer y ∈ {0,..., 2L+1} satisfies

December 2010 Notices of the AMS 1455 K(y) ≤ L. In this case, one can prove that x satis- see, for example, [Kikuchi97], Theorem 1.2 and fies K(x) > L by proving that every other integer Section 2). + ∈ { L 1} ≤ + y 0,..., 2 satisfies K(y) L (and recall that (2) T ⊢ ∀y ∈ {0,..., 2L 1} ((K(y) ≤ L) there is a proof for every such statement). Because we proved that m ≥ 1, the only x for which we → PrT ( K(y) ≤ L )) . didn’t prove K(x) ≤ L must satisfy K(x) > L. Assume for a contradiction that T is consistent Thus if m = 1, then for some integer x the and T ⊢ Con(T ). Then, by Equation 1, statement “K(x) > L” can be proved (in T ). But (3) T ⊢ ∀x ∈ {0,..., 2L+1} ¬Pr (K(x) > L). we know that for any integer x the statement T “K(x) > L” cannot be proved (in T ) unless the We will derive a contradiction by proving by L+1 theory is inconsistent. Hence, if the theory is induction that, for every i ≤ 2 + 1, T ⊢ (m ≥ consistent, m ≥ 2. As we assume that T is a rich i + 1), where m is defined as in the previous L+1 enough theory, we can prove the last conclusion in section. Because, obviously, T ⊢ (m ≤ 2 + 1), T . That is, we can prove in T that if T is consistent, this is a contradiction to the assumption that T then m ≥ 2. is consistent and T ⊢ Con(T ). Because we already Assume for a contradiction that the consistency know that T ⊢ (m ≥ 1), we already have the of T can be proved within T . Thus we can prove base case of the induction. Assume (the induction L+1 in T the statement “m ≥ 2”. In the same way, we ) that for some 1 ≤ i ≤ 2 + 1, can work our way up and prove that m ≥ i + 1, T ⊢ (m ≥ i). for every i ≤ 2L+1 + 1. In particular, m > 2L+1 + 1, We will show that T ⊢ (m ≥ i + 1) as follows. Let which is a contradiction, as m ≤ 2L+1 + 1 (by the r = 2L+1 + 1 − i. definition of m). 1. By the definition of m, T ⊢ (m = i) → ∃ different y , . . . , y ∈ {0,..., 2L+1} r The 1 r j=1 (K(y ) ≤ L). To present the proof formally, one needs to be j 2. Hence, by Equation 2, T ⊢ (m = i) → able to express provability within T , in the lan- ∃ different y , . . . , y ∈ {0,..., 2L+1} r guage of T . The standard way of doing that is 1 r j=1 Pr (K(y ) ≤ L). by assuming that the language of T contains T j 3. For every different y , . . . , y ∈{0,..., 2L+1}, the language of arithmetics and by encoding ev- 1 r and every x ∈ {0,..., 2L+1}\{y , . . . , y }, ery formula and every proof in T by an integer, 1 r T ⊢ (m ≥ i) → r (K(y )≤L)→ (K(x) > usually referred to as the G¨odel number of that j=1 j formula or proof. For a formula A, let A be L) , (by the definition of m), and its G¨odel number. Let PrT (A) be the following hence by Hilbert-Bernays derivabil- formula: there exists w that is the G¨odel number of ity conditions, T ⊢ PrT (m ≥ i) → r a T -proof for the formula A. Intuitively, PrT (A) Pr (K(y ) ≤ L)→Pr (K(x)>L) . j=1 T j T expresses the provability of the formula A. For- 4. By the previous two items, T ⊢ ((m = mally, the formulas Pr (A) satisfy the so-called L+1 T i) ∧ PrT (m ≥ i)) → ∃x ∈ {0,..., 2 } Hilbert-Bernays derivability conditions (see, for PrT (K(x) > L). example, [Mendelson97]): 5. As T ⊢ (m ≥ i) (by the induction hypoth- 1. If T proves A, then T proves PrT (A). esis), T ⊢ PrT (m ≥ i). Hence, T ⊢ (m = L+1 2. T proves: PrT (A) → PrT (PrT (A)). i) → ∃x ∈ {0,..., 2 } PrT (K(x) > L). 3. T proves: PrT (A → B) → (PrT (A) → 6. Hence, by Equation 3, T ⊢ ¬(m = i). PrT (B)). 7. Hence, as T ⊢ (m ≥ i), T ⊢ (m ≥ i + 1). The consistency of T is usually expressed as the  formula Con(T ) ≡ ¬PrT (0 = 1). In all that comes below, T ⊢ A denotes “T proves A”. We will prove A Possible Resolution of the Surprise that T ⊢ Con(T ), unless T is inconsistent. Examination Paradox For our proof, we will need two about In the previous sections we gave a proof for G¨odel’s provability of claims concerning Kolmogorov com- second incompleteness theorem by an argument plexity. First, we need to know that Con(T ) → that resembles the surprise examination paradox. ¬PrT (K(x) > L). We will use the following In this section we go the other way around and form of Chaitin’s incompleteness theorem (see, suggest that the second incompleteness theorem for example, [Kikuchi97], Theorem 3.3). gives a possible resolution of the surprise exami- (1) T ⊢ Con(T ) → ∀x ∈ {0,..., 2L+1} nation paradox. Roughly speaking, we argue that the flaw in the derivation of the paradox is that it ¬Pr (K(x) > L). T contains a hidden assumption that one can prove Second, we need to know that (K(y) ≤ L) → the consistency of the mathematical theory in PrT (K(y) ≤ L). We will use the following form which the derivation is done, which is impossible (formally, this follows, as K(y) ≤ L is a 1 formula; by the second incompleteness theorem. Σ

1456 Notices of the AMS Volume 57, Number 11 The important step in analyzing the paradox Wednesday? No, because they can also use S to is the translation of the teacher’s announcement prove that the exam will be held on Thursday. Thus into a mathematical language. The key point lies we conclude that, since S is a contradiction, prov- in the formalization of the notions of surprise and ability from S doesn’t imply knowledge. Recall, knowledge. however, that the very intuition behind the formal- As before, let T be a rich enough mathematical ization of the teacher’s announcement as S was theory (say, ZFC). Let {1,..., 5} be the days of that the notion of knowledge can be replaced by the week and let m denote the day of the week the notion of provability. But if provability from S on which the exam is held. Recall the teacher’s doesn’t imply knowledge, the statement S doesn’t announcement: “next week you are going to have seem to be an accurate translation of the teacher’s an exam, but you will not be able to know on announcement into a mathematical language. which day of the week the exam is held until Is there a better way to formalize the teacher’s that day.” The first part of the announcement is announcement? To answer this question, let us an- formalized as m ∈ {1,..., 5}. A standard way that alyze the situation from the students’ point of view appears in the literature to formalize the second on Tuesday night. There are three possibilities: part is by replacing the notion of knowledge by 1. On Tuesday night, the students are not the notion of provability [Shaw58, Fitch64] (for a able to prove that the exam will be held on recent survey see [Chow98]). The second part is Wednesday. rephrased as “on the night before the exam you 2. On Tuesday night, the students are able will not be able to prove, using this statement, to prove that the exam will be held on that the exam is tomorrow,” or, equivalently, “for Wednesday, but they are also able to prove every 1 ≤ i ≤ 5, if you are able to prove, using this for some other day that the exam will be statement, that (m ≥ i) → (m = i), then m = i.” held on that day. This can be formalized as the following statement (Note that this possibility can only occur that we denote by S (the statement S contains both if the system is inconsistent and is in parts of the teacher’s announcement): equivalent to the inconsistency of the system). S ≡ [m ∈ {1,..., 5}] 3. On Tuesday night, the students are able 1≤i≤5 to prove that the exam will be held on Pr (m ≥ i → m = i) → (m ≠ i) , T ,S Wednesday, and they are not able to prove where PrT ,S (A) expresses the provability of a for any other day that the exam will be formula A from the formula S in the theory T (for- held on that day. mally, PrT ,S (A) is the formula: there exists w that We feel that only in the third case is it fair to say is the G¨odel number of a T -proof for the formula A that the students know that the exam will be held from the formula S). Note that the formula S is on Wednesday. They know that the exam will be self-referential. Nevertheless, it is well known that held on Wednesday only if they are able to prove this is not a real problem and that such a formula that the exam will be held on Wednesday, and they S can be formulated (see [Shaw58, Chow98]; for are not able to prove for any other day that the more about this issue, see below). exam will be held on that day. Let us try to analyze the paradox when the We hence rephrase the second part of the teacher’s announcement is formalized as the above teacher’s announcement as “for every 1 ≤ i ≤ 5, if statement S. We will start from the last day. The one can prove (using this statement) that (m ≥ i) → statement m ≥ 5 together with m ∈ {1,..., 5} (m = i), and there is no j = i for which one can prove (using this statement) (m ≥ i) → (m = j), imply m = 5. Hence, PrT ,S (m ≥ 5 → m = 5), and = then m = i”. Thus the teacher’s announcement is by S we can conclude m 5. Thus S implies 3 m ∈ {1,..., 4}. In the same way, working our way the following statement: down, we can prove PrT ,S (m ≥ 4 → m = 4), and  by S we can conclude m = 4. In the same way, S ≡ [m ∈ {1,..., 5}] PrT ,S (m ≥ i → m = i)  m = 3, m = 2, and m = 1. In other words, S implies 1≤i≤5  m ∉ {1,..., 5}. Thus S contradicts itself.   The fact that S contradicts itself gives a certain ¬Pr m ≥ i → m = j → (m ≠ i) . T ,S explanation for the paradox; the teacher’s an- 1≤j≤5,j≠i   nouncement is just a contradiction. On the other Let us try to analyze the paradox when hand, we feel that this formulation doesn’t fully the teacher’s announcement is formalized as explain the paradox: Note that, because S is a con- tradiction, it can be used to prove any statement. 3This statement is equivalent to one of the suggestions So, for example, on Tuesday night the students (the statement I5) made by Halpern and Moses [HM86]. can use S to prove that the exam will be held However, the analysis of the paradox there is different on Wednesday. Is it fair to say that this means from the one shown here and makes no use of G¨odel’s that they know that the exam will be held on second incompleteness theorem.

December 2010 Notices of the AMS 1457 the new statement S. As before, m ≥ 5 to- Let q be the G¨odel number of the formula Q(x). gether with m ∈ {1,..., 5} imply m = 5. Hence, The statement S is formalized as S ≡ Q(q). To PrT ,S (m ≥ 5 → m = 5). However, this time see that this statement is the one that we are one cannot use S to conclude m = 5, as it interested in, denote by s the G¨odel number of S is possible that for some j = 5 we also have and note that s = Sub(q, q). Thus Pr m ≥ 5 → m = j . This happens iff the T ,S +  system T S is inconsistent. Formally, this time S ≡ [m ∈ {1,..., 5}] Pr (s ⇒ v ) = T ii one cannot use S to deduce m 5, but rather the 1≤i≤5  formula   Con(T , S) → (m = 5), ¬Pr s ⇒ v → (m ≠ i) . T ij where Con(T , S) ≡ ¬PrT ,S (0 = 1) expresses the 1≤j≤5,j≠i   consistency of T + S. Because by the second in- completeness theorem one cannot prove Con(T , S) References within T + S, we cannot conclude that S implies [Boolos89] G. Boolos, A new proof of the Gödel m = 5 and hence cannot continue the argument. incompleteness theorem, Notices Amer. More precisely, because S doesn’t imply m ∈ Math. Soc. 36 (1989), 388–390. {1,..., 4}, but rather Con(T , S) → m ∈ {1,..., 4}, [Chaitin71] G. J. Chaitin, Computational complex- when we try to work our way down we do not get ity and Gödel’s incompleteness theorem, ACM SIGACT News 9 (1971), 11–12. the desired formula Pr (m ≥ 4 → m = 4) but T ,S [Chow98] T. Y. Chow, The surprise examination rather the formula or unexpected hanging paradox, Amer. Math. Monthly 105 (1998), 41–51. PrT ,S (Con(T , S) ∧ (m ≥ 4) → m = 4), [Fitch64] F. Fitch, A Gödelized formulation of the which is not enough to continue the argument. prediction paradox, Amer. Phil. Quart. 1 Thus our conclusion is that if the students (1964), 161–164. believe in the consistency of T +S, the exam cannot [Gödel31] K. Gödel, Über formal unentscheid- be held on Friday, because on Thursday night the bare Sätze der Principia Mathematica students will know that if T + S is consistent the und verwandter Systeme I, Monatshehte exam will be held on Friday. However, the exam für Mathematik und Physik 38 (1931), can be held on any other day of the week because 173–198. [HM86] J. Halpern and Y. Moses, Taken by sur- the students cannot prove the consistency of T +S. prise: The paradox of the surprise test Finally, for completeness, let us address the revisited, Journal of Philosophical Logic issue of the self-reference of the statement S. The 15 (1986), 281–304. issue of self-referentiality of a statement goes back [Kikuchi94] M. Kikuchi, A note on Boolos’s proof of to G¨odel’s original proof for the first incomplete- the incompleteness theorem, Math. Logic ness theorem. The self-reference is what makes Quart. 40 (1994), 528–532. G¨odel’s original proof conceptually difficult and [Kikuchi97] , Kolmogorov complexity and the what makes the teacher’s announcement in the second incompleteness theorem, Arch. surprise examination paradox paradoxical. Math. Logic 36 (1997), 437–443. [Kotlarski04] H. Kotlarski, The incompleteness the- To solve this issue, G¨odel introduced the tech- orems after 70 years, Ann. Pure Appl. nique of diagonalization. The same technique can Logic 126 (2004), 125–138. be used here. To formalize S, we will use the nota- [Kreisel50] G. Kreisel, Notes on arithmetical models tion a ⇒ b to indicate implication between G¨odel for consistent formulae of the predi- numbers a and b. That is, a ⇒ b is a statement cate calculus, Fund. Math. 37 (1950), indicating that a is a G¨odel number of a statement 265–285. A and b is a G¨odel number of a statement B, [Mendelson97] E. Mendelson, Introduction to Mathe- such that A → B. We will also need the matical Logic, CRC Press, 1997. Sub(a, b) that represents substitution of b in the [Shaw58] R. Shaw, The unexpected examination, Mind 67 (1958), 382–384. formula with G¨odel number a. That is, if a is a [Smory´nski77] C. A. Smory´nski, The incompleteness G¨odel number of a formula A(x) with free variable theorem. In: Introduction to Mathemati- x and b is a number, then Sub(a, b) is the G¨odel cal Logic (J. Barwise, ed.), North-Holland, number of the statement A(b). 1977, 821–865. Let vij ≡ m ≥ i → m = j. Denote by Q(x) the [Vopenka66] P. Vopenka, A new proof on the Gödel’s formula result of nonprovability of consistency, Bull. Acad. Polon. Sci. 14 (1966), 111–116.  Q(x)≡[m∈{1,..., 5}] PrT (Sub(x, x)⇒vii ) 1≤i≤5

  ¬Pr Sub(x, x) ⇒ v → (m ≠ i) . T ij 1≤j≤5,j≠i  

1458 Notices of the AMS Volume 57, Number 11