The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

G¨odel’sCompleteness Theorem as presented in his 1929

Juliette Kennedy

September 6, 2018 The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Statement of the theorem

Every valid formula expressible in the restricted functional calculus [i.e. without equality] can be derived from the axioms by means of a finite sequence of formal inferences. Equivalently: A first order sentence is either satisfiable or refutable.1

1A formula is refutable if its negation is provable. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Some aspects of the proof

• The proof system is explicitly mentioned: axioms are taken essentially from Whitehead and Russell 1910, notation from Hilbert and Ackerman 1928 (H.A. henceforth)

• Propositional variables X1, X2,... are included in the first order formulas, i.e symbols interpreted by >, ⊥. • Syntactic objects are meticulously distinguished from semantic objects. • There is no formal semantics, i.e. the concept of “true in a domain” is taken for granted.2

2Tarski gives an explicit recursive definition in 1934. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Some more aspects of the proof

• The formula in question is satisfied in the domain of the natural numbers. • K¨onig’stheorem is not cited by name but proved on the spot. • Proof is given first i.t.r.s. (in the restricted sense), i.e. for systems without equality. Then i.t.e.s. (in the extended sense), i.e. with equality. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Notation and terminology

Some of G¨odel’sterminology is nonstandard, as is to be expected. • Predicates are called functional variables. (In the 1930 paper, “variables for preperties and relations”.) The syntactic object Fi is interpreted by fi , which is a function defined on the natural numbers with range >, ⊥. The propositional variable Xi is interpreted by Ai , where Ai is one of >, ⊥. • A model in our sense is denoted by G¨odelas (f1,... fk ; a1,... ak ; A1,... Ak ), where ai are natural numbers. • The degree of a formula is the number of alternations of quantifiers in the quantifier prefix. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Useful theorems from HA

• Completeness of propositional logic.3 • Replacement rule. • Every logical expression is equivalent to one in normal form, i.e. a formula with a quantifier prefix followed by a quantifier free expression. • Arbitrary permutation of existential quantifiers in the existential part of the quantifier prefix of a formula of the form ∃x1∃x2,... ∃xnA(x1, x2,... xn) is allowed.

3This theorem is due to Post. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

More useful theorems from HA

• Theorem 5:

(p)F (x1,..., xn)&(q)G(y1,..., ym) ≡ (p)(q)[F (x1,..., xn)&G(y1,..., ym)], T provided {x1,..., xn} {y1,..., xm} = ∅, no xi appears in G and no yi appears in F , and the relative order of the quantifiers in p, q are preserved. • Theorem 6: [∀xF (x)&∃xG(x)] → ∃x[F (x)&G(x)] The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Soundness

“Since the logical axioms are vaild and the rules of inference are correct in that they preserve truth, it is clear that every provable formula is valid.” It remains to show that every valid logical expression is provable. Equivalently: Every logical expression is either satisfiable or refutable. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Part 1

Claim 1: If every expression of degree k is satisfiable or refutable, then so is every expression of degree k + 1. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Proof of Claim 1

Consider

0 (P)A = ∀x1...∀xk ∃y1...∃ys (P )A(x1, ..., xk ; y1, ..., ys ),

where the quantifier prefix (P) is of degree k + 1, and (P0), of degree k, results from P in the obvious way. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Skolemize!

Form the expression B:

∀u1 ... ∀ur ∃v1,... vs F (u1,... ur ; v1,... vs ) 0 &∀x1...∀xr ∀y1...∀ys [F (x1,... ys ) → (P )A],

where F is a new predicate symbol. Note that B → (P)A is valid. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Form the expression B00

∀u1 ... ∀ur ∀x1 ... ∀xr ∀y1...∀ys ∀z1...∀zp 00 ∃v1,... ∃vs ∃t1,... tq(P ) {F (u1,... ur ; v1,... vs )&[F (x1,... xr ; y1,... ys ) → A]}

where P00 is of degree k − 1. Note that B ≡ B00. Note also that B00 is of degree k, hence by the induction hypothesis B00 is either satisfiable or refutable.4

4All G did was move the x and y quantifiers in front of the existential v quantifiers, and then pull the t quantifers out of A as allowed by Theorems 5 and 6 from HA mentioned above. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Case 1: B00 is satisfiable. Then so is (P)A, since B ≡ B0 ≡ B00 and B → (P)A is provable, hence valid. Case 2: B00, and hence B, is refutable. Then (P)A is refutable, as can be seen by the replacement of F in ¬B by (P0)A. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Part 2

Claim 2: Every normal formula of degree 1 is either satisfiable or refutable. Note that together with Claim 1 this concludes the proof of the theorem. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Proof of Claim 2

Consider

(P)A = ∀u1...∀ur ∃v1...∃vs A(u1, ..., ur ; v1, ..., vs ),

where A(u1, ..., ur ; v1, ..., vs ) is quantifier free.

We order the r-tuples of elements of x1, x2,..., and restrict v1, ..., vs in the following way: The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Ordering of r-tuples

Consider this table of indices:

i1,1, ..., i1,r i1,r+1, ..., i1,r+s . . in,1, ..., in,r in,r+1, ..., in,r+s

The idea is that the first column lists all possible r-sequences of natural numbers and the second column lists just those that are needed as witnesses for the ∃v1...∃vs -part. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

To construct the table and to be sure that the first column really lists all r-sequences, we define an ordering < of the set K of all r-sequences (a1, ..., ar ) of numbers 1, 2, .... We first divide K into disjoint parts as follows:

Kt = {(a1, ..., ar ): a1 + ... + ar = t}.

As a first approximation of < we stipulate that sequences in Kt are less than sequences in Ks for t < s. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

The smallest set is Kr which contains only the sequence (1,...,1). The next is Kr+1, where the sequences consist of ones except for one elements which is 2. So there are the sequences

(1, ..., 1, 2), (1, ..., 2, 1), ..., (2, 1, ..., 1), ...

Like this: Kr < Kr+1 < ...

We define the order < inside each Kt to be the lexicographic order.

We also have to define s-sequences β1, ..., βk for the second column. β1 is the sequence (2, 3, ..., s). The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

We maintain the following conditions on the s-sequences:

(i) The numbers in βk are all different and all different from the numbers in α1, ..., αk , β1, ..., βk−1.

(ii) Of the sequence we could take as βk the one we choose has the smallest sum. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

For an experiment, let r = s = 1.

k αk βk 1 1 2 2 2 3 3 3 4 ... The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

For yet another experiment, let r = 2, s = 2.

k αk sum βk 1 1, 1 2 2, 3 2 1, 2 3 3, 4 3 2, 1 3 5, 6 4 1, 3 4 7, 8 5 2, 2 4 8, 9 5 3, 1 4 9, 10 ... The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

The general case

k αk βk 1 11...11 2, 3, ..., s + 1 2 11...12 s + 2, s + 3, ..., 2s + 2 3 11...21 2s + 3, 2s + 4, ..., 3s + 3 ... The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

i1,1, ..., i1,r = 1, 1, ..., 1, 1, 1 sum is r i2,1, ..., i2,r = 1, 1, ..., 1, 1, 2 sum is r + 1 i3,1, ..., i3,r = 1, 1, ..., 1, 2, 1 sum is r + 1 i4,1, ..., i4,r = 1, 1, ..., 2, 1, 1 sum is r + 1 . . in−1,1, ..., in−1,r = 2, 1, ..., 1, 1, 1 sum is r + 1 in,1, ..., in,r = 1, 1, ..., 1, 1, 3 sum is r + 2 in,1, ..., in,r = 1, 1, ..., 1, 2, 2 sum is r + 2 in,1, ..., in,r = 1, 1, ..., 2, 1, 2 sum is r + 2 . . The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

. . in,1, ..., in,r = 2, 1, ..., 1, 1, 2 sum is r + 2 in,1, ..., in,r = 1, 1, ..., 1, 3, 1 sum is r + 2 in,1, ..., in,r = 1, 1, ..., 3, 1, 1 sum is r + 2 . . in,1, ..., in,r = 3, 1, ..., 1, 1, 1 sum is r + 2 in,1, ..., in,r = 1, 1, ..., 1, 1, 4 sum is r + 3 . . . . The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Part 2.1

Let (Pn)An be formula

∃xi1,1 ...xin,r+s

(A(xi1,1 , ..., xi1,r ; xi1,r+1 , ..., xi1,r+s )∧

A(xi2,1 , ..., xi2,r ; xi2,r+1 , ..., xi2,r+s )∧ . .

A(xin,1 , ..., xin,r ; xin,r+1 , ..., xin,r+s )) The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

where the following conditions hold:

(i) in,1, ..., in,r is the nth sequence in the assumed order.

(ii) The ik,r+1, ..., ik,r+s are all different.

(iii) The ik,r+1, ..., ik,r+s do not occur in the subscripts of variables in any

A(xit,1 , ..., xit,r ; xit,r+1 , ..., xit,r+s ),

where t < k.

(iv) {ik,1, ..., ik,r } ∩ {ik,r+1, ..., ik,r+s } = ∅. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

A simple case

Consider A = ∀x∃yφ(x, y), where φ(x, y) is quantifier-free. The An’s are defined:

A0 = φ(x0, x1)

A1 = φ(x0, x1) ∧ φ(x1, x2) . .

An = φ(x0, x1) ∧ ... ∧ φ(xn, xn+1) . .

G¨odelshows that from A we can derive for each n ∃x0 ... ∃xn+1An. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Lemma (P)A ` (Pn)An The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Proof of lemma Case n = 1 : Now (P1)A1 is just

∃xi1,1 ...xin,r+s A(xi1,1 , ..., xi1,r ; xi1,r+1 , ..., xin,r+s ) The inference we need is of the form

∀x∃yA(x, y) ` ∃xyA(x, y),

which is trivial. E.g.

∀x∃yA(x, y) ` ∃yA(x, y) ∀-elimination rule

∃yA(x, y) ` ∃xyA(x, y) ∃-introduction rule ` ∃yA(x, y) → ∃xyA(x, y) Deduction Theorem ∀x∃yA(x, y) ` ∃xyA(x, y) Modus Ponens The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

CASE n + 1: We assume the claim for n. Note that An+1 can be written as

An ∧ A(xin+1,1 , ..., xin+1,r ; xin+1,r+1 , ..., xin+1,r+s ). Let us write this as

An ∧ A(z1, ..., zr ; zr+1, ..., zr+s ).

By our assumption (iii)-(iv) the variables zr+1, ..., zr+s are different from z1, ..., zr and do not occur in An. The existential prefix (Pn) contains each of the variables z1, ..., zr , so it can be rearranged so that all z1, ..., zr are in the front (if they did not occur we would 0 have to add them). Let (Pn) be the remaining prefix. By a preceding HA theorem (on permuting existential quantifiers when the variables are distinct):

0 ` (Pn)An ≡ ∃z1...zr (Pn)An The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

By the replacement rule and theorems 5 and 6 from HA, reduce the provable formula

[(∀z1) ... (∀zr )(∃zr+1 ... ∃zr+s )A(z1,..., zr ; zr+1,... zr+s )

0 &∃z1 ... ∃zr (PnAn)] →

0 ∃z1 ... ∃zr [∃zr ∃zr+1 ... ∃zr+s A(z1 ... ∃zr+s )&(PnAn)] The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

to the equivalently provable formula:

(P)A&[(Pn)An → (Pn+1An+1)].

By the induction hypothesis, (P)A → (Pn)An is provable, so we are done. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Part 4

Back to the proof that any expression of degree 1 is either satisfiable or refutable.

We call An satisfiable if there is a model with universe i1,1, .., in,r+s

in which An is true, interpreting xij,k by the numeral ij,k . So the final model will have the natural numbers as its domain. There are two cases. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Case 1:

Some An is not satisfiable. For each conjunct substitute a new propositional variable, i.e.

φ(x0, x1) ∧ ... ∧ φ(xn, xn+1) is replaced by

X1 ∧ ... ∧ Xn+1 The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

G¨odelargues using the completeness theorem of propositional logic that ¬(X1 ∧ ... ∧ Xn+1) is provable and hence substituting back 0 0 the Ai s for the Xi s, that ¬An is provable, hence so is ∀x1 ... ∀x1+ns ¬An is (by theorem 3 of HA).

Thus ¬∃x1 ... ∃x1+ns An(xi1,1 ... xin,r+s ) is provable.

But this last is just ¬(Pn)An.(P)A → (Pn)An is provable, so we are done. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Satisfiability case.

Each An is satisfiable. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Case 2 Each An is satisfiable. Definition A satisfying system of level n for (P)A will be a system of i1 ip functions, f1 ,... fp (defined in the domain of the natural numbers x, with 1 ≤ x ≤ 1 + ns), and of propositions U1,... Uq, such that a true proposition results when we put the functions f1,... fp for the functional variables F1,... Fp, the propositions U1,... Uq for the propositional variables X1,... Xq, and, for each xi , the corresponding numeral i. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Note that for each n, there are only finitely many possible satisfying systems (models henceforth) of An with universe {i1,1...in,r+s }. We denote this finite set by Sn. Note that by allowing n to take on all S finite values, there are infinitely many models, i.e. n Sn is infinite. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

For M, M0 ∈ T we define

M < M0

if and only if

(1) M ∈ Sn for some n. 0 (2) M ∈ Sm for some m > n. (3) The model M is a submodel of M0, M ⊆ M0.5

5I.e. the domain of M (call it A) is a subset of the domain of M0; 0 RM ⊆ RM ∩ An, where R has arity n. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

In this way we get a tree T which is finitely branching but infinite

Figure: The tree T . The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

T is finitely branching because immediately above a model M in Sn we can only put models from the finite set Sn+1. By K¨onig’s Lemma (which G¨odeldoes not explicitly mention but rather proves on the spot) there is an infinite branch B in this tree.6

Figure: The tree T has an infinite branch.

6Note that K¨onig’slemma is provable in ZF , where the lemma states: if T is an infinite tree and finitely branching, then T has an infinite branch. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

On this branch B any two elements M, M0 satisfy M ⊆ M0 or M0 ⊆ M. Therefore the union of the models on B form a model M with universe {1, 2,...} such that every model in B is a submodel of M. Since each An is quantifier free, and the models in Sn all satisfy An, the union M satisfies each An, when each xm is interpreted as the numeral m. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Claim: The original sentence (P)A is true in M. Recall that

(P)A = ∀u1...∀ur ∃v1...∃vs A(u1, ..., ur ; v1, ..., vs ).

Let us take an arbitrary interpretation of u1, ..., ur in M, say i1, ..., ir . Let j1, ..., js be some sequence of distinct natural numbers so that hi1, ..., ir , j1, ..., js i is in G¨odel’stable as row n for some n.

Then A(xi1 , ..., xir ; xj1 , ..., xjs ) is one of the conjuncts of An and is therefore true in M when the variables are interpreted by their corresponding numerals. Thus A is true in M. So A is satisfiable QED The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

The Equality Case

Replace x = y by G(x, y). • Again we have a logical expression A and we show it is either provable or its negation is satisfiable. • Replace the identity symbol everywhere it occurs in A by a new symbol G, i.e. replacing everywhere x = y by G(x, y). • In this way we obtain from A a new expression A0. • Write down a sentence K 0 stating the Identity Axioms, but in terms of G. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Conditions on G The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

• The sentence K 0 → A0 does not contain identity. • Can apply the previous proof: K 0 → A0 is either provable or its negation is satisfiable. • We consider the two cases separately, as above. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Case 1: K 0 → A0 is provable.

• Replace G by = in K 0, to obtain K. • In logic with identity we assume some Identity Axioms (Axioms 7 and 8 in G). • One can prove K readily from these Identity Axioms. • Since K 0 → A0 is provable, so is K → A. (This is a subtle logical point!) • Hence A is provable. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Case 2: The negation of K 0 → A0 is satisfiable.

• We show that in this case the negation of A is satisfiable. m 0 0 • Let (g, fl ) be a system (with domain N) satisfying K ∧ ¬A . m Here g is the interpretation of G and fl are as usual. • Clearly, g is an equivalence relation on N. • Let N0 be the set of equivalence classes of g. 0 0m • Let g and fl be defined canonically on the equivalence classes:

0 g ([ai1 ], [ai2 ]) holds iff g(ai1 , ai2 ) holds.

0m m fl ([ai1 ], ..., [aim ]) holds iff fl (ai1 , ..., aim ) holds. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

• Clearly, g 0 is the identity on N0. m • If a (quantifier-free) formula containing (only) g, fl and

ai1 , ..., ain ∈ N holds, then also the same formula holds if g is 0 m 0m replaced by g , fl is replaced by fl and am is replaced by [am]. This is true by definition for atomic formulas and the rest is induction on ∧, ∨, ¬. • Hence ¬A is true under this interpretation, as ¬A0 is true under the equivalence class interpretation. • We have shown that ¬A is satisfiable. QED The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Extension to countably many formulae

Task. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

G¨odelmakes an observation on the Proof

Doesn’t the use of the law of excluded middle in its proof “invalidate the entire completeness proof”? The Completeness Theorem asserts: “...a kind of decidability, namely every quantificational formula is either provable or a counterexample to it can be given, whereas the principle of the excluded middle seems to express nothing other than the decidability of every problem.” The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Thus the proof may be circular: one uses the decidability of every question in order to prove something weaker, namely “a kind of decidbaility.” But, G¨odelremarks, what he has shown is the provability of a valid formula from “completely specified, concretely enumerated inference rules,” not merely from all rules imaginable; whereas the law of excluded middle is used informally: the notion of decidability or solvability asserted by the law is left unspecified. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

As G¨odelputs it: “. . . what is affirmed (by the law of excluded middle) is the solvability not at all through specified means but only through all means that are in any way imaginable . . .” The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

G¨odeltells us that the step forward provided by the Completeness Theorem is that if we assume solvability by all means imaginable, then we have, in the case of a sentence of first-order predicate calculus, a reduction to solvability by very specific means laid out beforehand. Extraordinary sensitivity to language/metalanguage distinction. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

This and the below observation about incompleteness from G¨odel’s thesis is not included the 1930 paper for the Monatshefte. See footnote 1 of G¨odel’s1930 paper: “I am indebted to Professor H. Hahn for several valuable suggestions that were of help to me in writing this paper.” The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

G¨odeland Skolem

As for G¨odel’scontribution to the Completeness Theorem, it was simply to solve the infinitary Hilbert-Ackermann question. This required seeing that both the refutation procedure and the satisfaction procedure can be incorporated into a single framework, as distinct from L¨owenheim, Skolem and Herbrand, who saw these as cases to be handled separately, if at all. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

It should also be noted that G¨odelsolved the Hilbert-Ackermann question in such a definitive way as to dampen objections concerning the infinitary aspects either of the statement of the theorem, or of its proof—objections which could easily have arisen in the general logical community, given the times. This was to become G¨odel’strademark in his logical work: the vindication of infinitary methods in metamathematics by showing they can be handled in an exact way, that is to say: mathematically. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

G¨odelput it this way: “The Completeness Theorem, mathematically, is indeed an almost trivial consequence of Skolem 1923a. However, the fact is that, at that time, nobody (including Skolem himself) drew this conclusion (neither from Skolem 1922 nor, as I did, from similar considerations of his own ...This blindness (or prejudice, or whatever you may call it) of logicians is indeed surprising. But I think the explanation is not hard to find. It lies in the widespread lack, at that time, of the required epistemological attitude toward metamathematics7 and toward non-finitary reasoning.”8

7i.e. reasoning about “all models of a theory” 8Letter to Wang, Dec. 7, 1967. See [Collected Works, 2003b] The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

As an aside, completeness as such seems not to have been very much in the air in the early and mid-1920s. For example Fraenkel in his 1923 review of Skolem 1922 does not mention completeness at all, never mind that it might follow from Skolem’s paper. Also, the very fact that completeness is stated as an open question in Hilbert and Ackermann 1928 implies that, if Hilbert and Ackermann read the relevant papers of L¨owenheim and Skolem before they published their text, they obviously did not obtain from them a solution of their problem. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Indeed Skolem himself, in his 1938 review of the second edition of Hilbert-Ackermann, notes their inclusion of theorems proved since the previous 1928 edition appeared. Among these new theorems he includes the Completeness Theorem!! The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

A few immediate observations

If a first order sentence holds in all models, it must be because of a uniform reason (proof) and not for an accidental reason particular to each model, which happens in each model or in each case for a different reason.9

9A. Blass, “Some semantical aspects of linear logic,” J. Interest Group in Pure and Applied Logic, 5:115-126, 1998. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

On informal rigour

The Completeness Theorem gives the practitioner of informal rigour a powerful new tool: to check whether a sentence is provable in a theory, one needs only to check whether the sentence holds in an arbitrary model of the theory. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Mysterious result: Semantics enjoys a “speed-up” over syntax

Imagine that we want to be convinced of a basic logical consequence φ ` ψ. One possibility is to try to derive ψ from φ in some standard proof system. However, by the Completeness Theorem, φ ` ψ is equivalent to φ |= ψ i.e. to the statement that every model of φ is a model of ψ. Thus to prove φ ` ψ it suffices to take an arbitrary model of φ and show that it is a model of ψ. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

G¨odel’s1936 ‘Speed-up’ theorem, published in an abstract “On the length of proofs”,10 says that while some sentences of arithmetic are true but unprovable, there are other sentences which are provable, but even the shortest proof is longer than any bound given in advance as a recursive function of the sentence. More exactly:

10G¨odel“Uber¨ die L¨angevon Beweisen”, Ergebnisse eines mathematischen Kolloquiums, 7: 23-24. Reprinted in CW1, pp. 395-399. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Here’s the compression

Theorem Let n be a natural number > 0. If f is a computable function, then 11 there are infinitely many formulas A, provable in Sn , such that if k is the length of the shortest proof of A in Sn and l is the length of the shortest proof of A in Sn+1, then k > f (l). Thus, passing to the logic of the next higher order has the effect, not only of making provable certain propositions that were not provable before, but also of making it possible to shorten, by an extraordinary amount, infinitely many of the proofs already available.12

11 Sn is nth order logic 12G¨odel,ibid The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Corollary

This means is that proof via the semantic method enjoys a speed-up over proof obtained by exhibiting a proof in a proof system. Intuitively, this is because you can define semantic consequence |= for a logic of order n in the higher order logic at level n + 1 (or above.)13

13See also V¨a¨an¨anen-Vardi, unpb. manuscript The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

A little story about G¨odel,Skolem and the history of the Completeness Theorem

The L¨owenheim-Skolem Theorem as given especially in Skolem 1922, has raised a number of questions as to when G¨odelmight have come across that paper. If one might speculate that it was not known to Hilbert and Ackermann, as in the above, then one might speculate in turn that it was also not known to G¨odel. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Library slips from the G¨odelNachlass (which have been reexamined just recently) indicate that G¨odelrequested the proceedings volume containing Skolem 1922 once in November 1928, before he turned in his dissertation in July 1929, and twice afterwards. He was not able to obtain Skolem 1922 from any of the libraries he requested the volume from before November 1929 at the earliest, though. Indeed the proceedings volume which contains the paper was not owned by the National or University libraries in Vienna at the time (nor do they own it now).14

14See van Atten, “On G¨odel’sawareness of Skolem’s Helsinki lecture.” The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Back to consequences of the theorem

We have now separated first and second order logic, on grounds of categoricity. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Theorem The following are equivalent: 1. Completeness Theorem 2. If a FO theory T is consistent15, it has a model.

Proof of the equivalence: Suppose T cannot prove ⊥. To show: T has a model. If T has no models, then ⊥ is true in every model of T , as there are none. By Completeness Theorem there is a proof of ⊥ from T , contrary to the assumption. (Other direction even easier.)

15i.e. T cannot prove φ ∧ ¬φ The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Definition The Compactness Theorem is the statement that if every finite subset of a theory has a model then the whole theory has a model. Theorem The following are equivalent: (1) Completeness Theorem. (2) Compactness Theorem. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Application: There are nonstandard models of Peano Arithmetic!

Malcev 1936: Let T be the theory PA ∪ {0¯ < c¯, 1¯ < c¯, 2¯ < c¯,...}. Note that LT = LPA ∪ {c¯}. T is finitely consistent, hence consistent. Hence it has a model M in whichc ¯M is a “non-standard” integer, i.e. an integer which is above than any natural number, i.e. element of N. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Notion of “general model” is not in view

The existence of non-standard models of the Peano axioms was not pointed out by G¨odelin any of the publications connected with the Completeness Theorem from that time, and it seems not to have been noticed by the general logic community until much later. Skolem’s definable ultrapower construction from 1933 gives a direct construction of a non-standard model of True Arithmetic (which of course extends Peano Arithmetic). But Skolem never mentions that the existence of such models follows from the completeness and compactness theorems. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

G¨odelin his review of Skolem 1933 also does not mention this fact, rather observing that the existence of non-standard models of True Arithmetic follows from the incompleteness theorem. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

In Warsaw in the 1920s there were logic seminars on particular relational structures, e.g linear orders. But the general notion of model emerged in logic only slowly. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

The concept: “model of a theory” was given by Tarski in 1933,16 made precise in the early 1950s, and made completely precise in the paper Tarski-Vaught in 1953.17 The idea that emerged is set-theoretic: Definition A model M =< S, R1,... Rk , f1,... fn, c1,... cm > is thought of as a set S (the domain of M) together with relations R ⊆ Si where the arity of R is i, and functions f ⊆ Sj+1, where the arity of f is j, together with distinguished elements cn of M. The arity of R is i, and the arity of f is j. In other words, a model is an object belonging to V , the cumulative hierarchy of sets.

16Tarski, “On the Concept of Logical Consequence” 17Tarski, A. and Vaught, R. 1956, “Arithmetical extensions of relational systems”, Compositio Mathematica, 13: 81-102. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Back to PA: the “slippage” is as bad as it can possibly be Theorem There are 2ℵ0 non-isomorphic countable models of PA. Add a constant symbol c to the language of PA. Let A be a subset of the prime numbers (possibly infinite). Consider the axioms of PA together with the following axioms {p|c | p ∈ A} S{p 6 |c | p ∈/ A}. The set of axioms is finitely consistent, hence consistent. Hence these axioms have a model. Let MA denote this model. Subclaim: Let A and B be sets of prime numbers. If A 6= B then MA is not isomorphic to MB . Proof of theorem: There are 2ℵ0 distinct subsets of prime numbers, hence there are 2ℵ0 non-isomorphic models of PA. (At least!) Moreover, these models can always be taken to be countable. FACT: there cannot be more than 2ℵ0 countable non-isomorphic models. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Fun fact: You can arrange things so that you cannot distinguish these models by a FO statement provable in one but not the other. They could all be so-called “elementarily equivalent”, i.e. satisfy the same sentences in the language of PA. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Another fun fact about nonstandard models of PA: The natural numbers are not definable in them Theorem Let M be a nonstandard model of PA. Then for no φ do we have {x ∈ M|M |= φ(x)} = N. Proof. By overspill. Suppose to the contrary that φ defined the standard model N in the nonstandard model M. The induction axiom schema implies that

φ(0) ∧ [∀x(φ(x) → φ(x + 1)) → ∀(x)φ(x)].

Thus M = N, contradicting the fact that M is nonstandard. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Another fun fact: Countable nonstandard models of PA are not recursive18

Theorem Let M be a countable nonstandard model of PA. Regard +M and

·M as ternary relations R+M , R·M on the natural numbers. I.e., 3 R+M , R·M ⊆ N . Then R+M , R·M cannot be recursive. For a nice, easy proof of this theorem see http://web.mat.bham.ac.uk/R.W.Kaye/papers/tennenbaum/tennrosser

18recursive = computable by a Turing machine. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

In sum...

• Referential indiscernibility: The FO Peano axioms do not 19 uniquely describe the natural numbers (denoted N). Moreover, as we showed, the axioms have the maximum number of (even countable) models possible. • These (nonstandard) models are (in a sense) wild, i.e. nonrecursive. • N is always a so-called initial segment of any model M of PA, however M can never “see” the structure N. • And of course, PA is incomplete. Not only are there nonstandard interpretations of the Peano axioms, but we cannot separate the standard from the nonstandard interpretations in a FO way. 19in the terminology: the first order theory PA is not categorical. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

In other words...

We need a general theory of models! The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

G¨odelhints at future developments

G¨odeltalks about incompleteness in the beginning of his thesis. Systematically, so to speak, whether G¨odelrecognized this explicitly or only sensed it, G¨odel’s1929 thesis was precisely the place to bring up incompleteness. This is because once the Completeness Theorem has been proved, it is now, and only now, that it is meaningful to talk about unsolvability. Why? The Completeness Theorem shows that first-order logic is maximal, in the sense of being a proof system strong enough to prove any statement that holds in all models of a particular first-order theory. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Therefore it becomes meaningful to ask, can one now decide every proposition in the language of the theory? If we lack a completeness theorem (as in the case of the full second order logic), then there are consistent theories without a model. Proving an incompleteness theorem would be irrelevant. It is only in the context of a completeness theorem, when we know that every consistent theory has a model, that it becomes relevant whether the model is unique or do we actually always have several essentially different models. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Another observation of G¨odel:Incompleteness Another remark in the introduction to his 1929 thesis that not included in the article G¨odelpublished, based on the thesis. “L.E. Brouwer, in particular, has emphatically stressed that from the consistency of an axiom system we cannot conclude without further ado that a model can be constructed. But one might perhaps think that the existence of the notions introduced through an axiom system is to be defined outright by the consistency of the axioms and that, therefore, a proof [that a model exists] has to be rejected out of hand. This definition (if only we impose the self-evident requirement that the notion of existence thus introduced obeys the same operation rules as does the elementary one), however, manifestly presupposes the axiom that every mathematical problem is solvable.” The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

“Or, more precisely, it presupposes that we cannot prove the unsolvability of any problem. For, if the unsolvability of some problem (in the domain of real numbers, say) were proved, then, from the definition above, there would follow the existence of two non-isomorphic realizations of the axiom system for the real numbers, while on the other hand we can prove the isomorphism of any two realizations.” The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

So there might be constraints on the principle that consistency implies existence. (Hilbert’s Principle in his 1899 correspondence with Frege.) The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

The First Incompleteness theorem is announced in September 1930 in K¨onigsberg. But G¨odelmust have started thinking about the theorem at least from March 1928, after hearing Brouwer’s two lectures in Vienna (on March 10th, on intuitionistic foundations, and on March 14th, on the intuitionistic continuum). The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

We know this from an entry in Carnap’s diary dated December 12, 1929, in which he records a conversation he had with G¨odelthat day about Brouwer’s lecture(s) of the previous year. In his entry Carnap states that G¨odeltalked “about the inexhaustibility of mathematics (see separate sheet). He was stimulated to this idea by Brouwer’s Vienna lecture. Mathematics is not completely formalizable. He appears to be right.” The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

We admit as legitimate mathematics certain reflections on the grammar of a language that concerns the empirical. If one seeks to formalize such a mathematics, then with each formalization there are problems, which one can understand and express in ordinary language, but cannot express in the given formalized language. It follows (Brouwer) that mathematics is inexhaustible: one must always again draw afresh from the fountain of intuition. There is, therefore, no characteristica universalis for the whole mathematics, and no decision procedure for the whole mathematics. In each and every closed language there are only countably many expressions. The continuum appears only in the whole of mathematics... If we have only one language, and can only make elucidations about it, then these elucidations are inexhaustible, they always require some new intuition again. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Coming attractions for next lecture... The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Mysterious result: Semantics enjoys a “speed-up” over syntax

Imagine that we want to be convinced of a basic logical consequence φ ` ψ. One possibility is to try to derive ψ from φ in some standard proof system. However, by the Completeness Theorem, φ ` ψ is equivalent to φ |= ψ i.e. to the statement that every model of φ is a model of ψ. Thus to prove φ ` ψ it suffices to take an arbitrary model of φ and show that it is a model of ψ. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

G¨odel’s1936 ‘Speed-up’ theorem, published in an abstract “On the length of proofs”,20 says that while some sentences of arithmetic are true but unprovable, there are other sentences which are provable, but even the shortest proof is longer than any bound given in advance as a recursive function of the sentence. More exactly:

20G¨odel“Uber¨ die L¨angevon Beweisen”, Ergebnisse eines mathematischen Kolloquiums, 7: 23-24. Reprinted in CW1, pp. 395-399. The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Here’s the compression

Theorem Let n be a natural number > 0. If f is a computable function, then 21 there are infinitely many formulas A, provable in Sn , such that if k is the length of the shortest proof of A in Sn and l is the length of the shortest proof of A in Sn+1, then k > f (l). Thus, passing to the logic of the next higher order has the effect, not only of making provable certain propositions that were not provable before, but also of making it possible to shorten, by an extraordinary amount, infinitely many of the proofs already available.22

21 Sn is nth order logic 22G¨odel,ibid The Proof G¨odelmakes a remark on the Proof G¨odel,Skolem, immediate observations Consequences G¨odeldrops a hint

Corollary

This means is that proof via the semantic method enjoys a speed-up over proof obtained by exhibiting a proof in a proof system. Intuitively, this is because you can define semantic consequence |= for a logic of order n in the higher order logic at level n + 1 (or above.)23

23See also V¨a¨an¨anen-Vardi, unpb. manuscript