Hedman. A First Course in Logic, 2004 (Oxford)

contradiction. For definiteness, let ϕ be the sentence ¬(1 = 1). The G¨odel code for this sentence is

25 · 31 · 527 · 715 · 1127 · 133 = 97805776756277266140466823153062702475001841306686401367187500000.

For our sanity, let c denote this number. In this section, we consider the sentence ¬P rT (tc). G¨odel’s Second Incompleteness theorem states that this sentence cannot be derived from T .

For any theory T , T ¬(1 = 1) if and only if T is inconsistent. Since we are assuming that T is a theory, it is not the case that T ¬(1 = 1). By Proposition 8.24, N |= ¬P rT (tc). And so, ¬P rT (tc) is a true sentence that cannot be derived from T . In this sense, G¨odel’s Second Incompleteness is a special case of the First Incompleteness theorem. The Second Incompleteness theorem says more than the First Incompleteness theorem since it asserts that the sentence ¬P rT (tc) in particular cannot be derived from T .

To prove G¨odel’s Second Incompleteness theorem, we use the following Fixed Point lemma (often referred to as the Diagonalization lemma). The key to this lemma and its proof is a certain VN -formula D(x, y) that we now define. For any formula ϕ, let δϕ denote the formula x(x = tϕ ϕ). If ϕ happens to have x as its only free variable, then δϕ is equivalent to the sentence ϕ(tϕ) asserting that ϕ holds of its own G¨odel number. Given the G¨odel number for ϕ, we can compute the G¨odel number for δϕ in a primitive recursive manner. By Exercise 8.6, the relation between the G¨odel number of ϕ and that of δϕ is definable by a ∆0-formula. Let D(x, y) be a ∆0 formula that says x = [ϕ] and y = [δϕ] for some VN -formula ϕ.

Lemma 8.30 (Fixed Point lemma) Let T be a recursive set of VN -sentences such that ΓN T . Let ϕ(x) be a VN -formula having one free variable. There exists a VN -sentence ψ such that T ψ ↔ ϕ(tψ).

Proof Let ψ be the sentence δθ where θ(x) is the formula y(D(x, y) ϕ(y)). That is, ψ is the formula x(x = tθ θ(x)).

We see that ψ ≡ θ(tθ) ≡ y(D(tθ, y) ϕ(y)).

By Completeness, we have (*) T ψ ↔ y(D(tθ, y) ϕ(y)). Since D(x, y) is ∆0 and N |= D(tθ, tψ), we have T D(tθ, tψ).

The primitive recursive function that takes [ϕ] to [δϕ] is one-to-one. It follows that, for any model M of T ,

M |= y(D(tθ, y) ↔ y = tψ), and so T y(D(tθ, y) ↔ y = tψ). Substituting this into (*) yields:

T ψ ↔ y(y = tψ ϕ(y)), and so T ψ ↔ ϕ(tψ) as we wanted to show.

Corollary 8.31 Let T be a decidable VN -theory containing ΓN . There exists a VN -sentence γ such that T γ ↔ ¬P rT (tγ ).

Proof Apply the Fixed Point lemma to the formula ¬P rT (x).

Proposition 8.32 If T and γ are as in the previous corollary, then TN ¬P rT (tγ ) and T γ.

Proof The sentence γ asserts that it is not provable from T . If T γ, then T ¬P rT (tγ ). By Proposition 8.25, if T γ, then T P rT (γ). Since T is consistent, it must be the case that T γ. Since γ cannot be derived from T , ¬P rT (tγ ) TN by Proposition 8.24.

The previous proposition provides an alternative proof for G¨odel’s First Incompleteness theorem. For any recursive subset T of TN , the sentence γ that assert “I am not provable from T ” must be both true and not provable from T . This is the proof G¨odel originally gave for the First Incompleteness theorem. The Second Incompleteness theorem is deduced by showing that γ and ¬P rT (tc) are T -equivalent.

Theorem 8.33 (G¨odel’s Second Incompleteness theorem) If T is a decidable

VN -theory that contains ΓN , then T ¬P rT (tc).

Proof Let γ be as in Corollary 8.31. We show that γ and ¬P rT (tc) are T -equivalent.

Since ¬(1 = 1) is contradictory, T ¬(1 = 1) → γ.

Let b = [¬(1 = 1) → γ]. By Proposition 8.25, T P rT (tb). Corollary 8.29 states that T (P rT (tc) P rT (tb)) → P rT (tγ ). Since T P rT (tb), we have T P rT (tc) → P rT (tγ ).

By contraposition, T ¬P rT (tγ ) → ¬P rT (tc).

This establishes T γ → ¬P rT (tc) by the definition of γ.

We now derive the converse from T .

Let θ be the sentence ¬γ ↔ P rT (tγ ). Let p = [P rT (tγ )]. Since ¬γ can be derived from θ and P rT (tγ ):

T (P rT (tθ) P rT (tp)) → P rT (t¬γ ) by Lemma 8.27. Now since T θ, we have T P rT (tθ) by Proposition 8.25.

We also have, by Proposition 8.26, T P rT (tγ ) → P rT (tp). By the previous three lines, T P rT (tγ ) → P rT (t¬γ )

Since ¬(1 = 1) can be derived from γ and ¬γ:

T (P rT (tγ ) P rT (t¬γ )) → P rT (tc) by Lemma 8.27.

Since we have shown T P rT (tγ ) → P rT (t¬γ ) it follows that

T P rT (tγ ) → P rT (tc) and so T ¬P rT (tc) → ¬P rT (tγ ). Finally, by the definition of γ, we have T ¬P rT (tc) → γ.

The sentence ¬P rT (tc) is commonly denoted Con(T ). This notation reflects the fact that N |= Con(T ) if and only if T is consistent. Now, suppose that we have a recursive set of V-sentences Γ and we want to determine whether or not T = Γ ΓN is consistent. Attempting to derive Con(T ) from T would be an extremely naive approach. The reason is that, if T happens to be inconsistent, then any VN -sentence can be derived from T . So if we are successful in deriving Con(T ) from T , then it is possible that T is inconsistent. By the Second Incompleteness theorem, it is not only possible, but necessary that T be inconsistent:

T Con(T ) if and only if T is inconsistent.

Whereas Con(T ) defines consistency semantically in N, it means precisely the opposite from the syntactic perspective of T .

8.6 Goodstein sequences

In this section, we describe a true statement regarding the natural numbers that can be formulated as a V-sentence but cannot be derived from ΓN . G¨odel’s First Incompleteness theorem guarantees the existence of such sentences, but does not provide an explicit example. Inexplicit examples are provided by the sentences γ from the previous section and β from the proof of Theorem 8.18. Since they are VN -sentences in TN , both γ and β are statements regarding the natural numbers. However, we do not know what these sentences express. The Fixed Point lemma implies the existence of γ such that ΓN γ ↔ ¬P rΓN (tγ ). This is all we know about of the sentence γ. Likewise, we know that the V-sentence β exists, but we do not know what it says about the natural numbers.

We consider sequences of non-negative integers known as Goodstein sequences. Given any natural number n, there is a unique Goodstein sequence that begins with n as its first term. Let us denote this sequence sn. The best way to describe these sequences is to provide an example. Suppose n = 14. Let a1, a2, a3, . . . denote the terms of the sequence s14. Then a1 = 14.

To find a2, first express the number 14 totally in base 2. That is, write 14 as a sum of powers of 2: 14 = 23 + 22 + 2. Moreover, write the exponents as sums of powers of two. Repeat this until every number occuring in the expression is a power of 2. In this case, write 14 as 2(2+1) + 22 + 2. To find a2, change each 2 in this expression to a 3 and then subtract 1:

a2 = (3(3+1) + 33 + 3) − 1 = (81 + 9 + 3) − 1 = 93 − 1 = 92.

To find a3, first express the number a2 totally in base 3: 92 = 3(3+1) +33 +2. The number 2 (which is not a power of 3) represents the sum 30 + 30. In general, when writing a number totally in base n, we allow numbers less than n as coe cients.

Now change each 3 to a 4 and subtract 1:

a3 = (4(4+1) + 44 + 2) − 1 = (1024 + 256 + 2) − 1 = 1282 − 1 = 1281,

and so forth. To find am, first express am−1 totally in base m, then change each m to m + 1, and then subtract 1 from the result. This rule generates each Goodstein sequence. If am equals 0 for some m, then the Goodstein sequence terminates.

Continuing with the sequence s14, we have

a4 = (5(5+1) + 55 + 1) − 1 = (15625 + 3125 + 1) − 1 = 18751 − 1 = 18750, a5 = (6(6+1) + 66) − 1 = (279936 + 46656) − 1 = 326592 − 1 = 326591. Now 326591 = 5 · 6(6) + 5 · 65 + 5 · 64 + 5 · 63 + 5 · 62 + 5 · 6 + 5.

So a6 = (5 · 7(7) + 5 · 75 + 5 · 74 + 5 · 73 + 5 · 72 + 5 · 7 + 5) − 1 = 4215754.

So the sequence s14 begins 14, 92, 1281, 18750, 326591, 4215754, . . . . Clearly, the next few terms of this sequence get larger and larger. In this respect, there is nothing special about the number 14. The sequence sn gets quite large for most choices of n. This is not true for n = 1, 2, or 3 but it is true for n = 4 (try it). For some values of n, sn grows very rapidly. For example, let n = 18. Then, written in base 2, 18 = 25 + 2. Written totally in base 2, we have

a1 = 18 = 2(22+1) + 2, and so

a2 = (3(33+1) + 3) − 1 = 328 + 2 = 22876792454963.

Computing a Goodstein sequence without a computer would be unpleasant. Even with a computer, this computation may not be feasible. Each step of the computation consists of two parts: we must increase the base and subtract 1. Whereas the first part may increase our number greatly, the second part decreases the result slightly.

Although they may appear cumbersome, Goodstein sequences possess the following charming property.

Theorem 8.34 (Goodstein) Every Goodstein sequence converges to zero.

Proof Let sn = (a1, a2, a3, . . .) be an arbitrary Goodstein sequence.

We define a sequence b1, b2, b3, . . . of ordinals as follows. For each m N, let bm be the ordinal obtained by replacing each occurrence of (m + 1) with ω

in the total base (m + 1) representation of am. For example, if sn is s14, then:

Note that the sequence of bis is decreasing. Continuing, we see that

b5 = 5 · ω(ω) + 5 · ω5 + 5 · ω4 + 5 · ω3 + 5 · ω2 + 5 · ω + 5,

b6 = 5 · ω(ω) + 5 · ω5 + 5 · ω4 + 5 · ω3 + 5 · ω2 + 5 · ω + 4, and so forth.

Increasing the base in the sequence of ais has no e ect on the sequence of bis. Because we subtract 1 at each stage, bi+1 is necessarily smaller than bi. This observation proves the theorem.

For any Goodstein sequence a1, a2, a3 · · · , the corresponding sequence of ordinals b1 > b2 > b3 · · · is decreasing. By Exercise 4.22, this latter sequence must be finite. This is easily proved by induction on the ordinals. We conclude that the sequence a1, a2, a3, . . . must be finite. This only happens if am = 0 for some m.

Although each sequence eventually reaches zero, it may take a very long time for this to happen. For example, we know that the mth term of the sequence s18 is zero for some m. Since each step the sequence decreases by at most 1, the number m must be at least 22876792454963 (since this is the value of a2 for this sequence).

Note that the statement of Goodstein’s theorem can be formulated as a VN -sentence. Using the techniques of Section 8.2 we can define a VN -formula Good(l, m, k) that holds if and only if [l, m, k] codes the initial l nonzero terms of a Goodstein sequence. We can express that every Goodstein sequence is finite by saying that every such initial segment is contained in an initial segment having 1 as its final term. Clearly am = 1 if and only if am+1 = 0 and the sequence terminates. (So the fact that 0 is not in our vocabulary is not a problem.) Let ΦGood denote this sentence. In 1982, Kirby and Paris proved the following.

Theorem 8.35 ΓN ΦGood.

Clearly, our proof of Goodstein’s theorem, since it refers to the infinite ordinal ω, cannot be carried out in ΓN . Kirby and Paris’s theorem shows that no formal proof in ΓN can prove this theorem. Kirby and Paris show that Goodstein’s theorem is equivalent to an induction axiom that allows us to prove the consistency of ΓN . In this way, Theorem 8.35 can be deduced from G¨odel’s Second Incompleteness theorem. We refer the reader to [23] for the details of this proof.

Exercises

8.1.Explain why G¨odel’s Incompleteness theorems do not contradict the Completeness theorem (also proved by Kurt G¨odel).

8.2.Encode the following finite sequences as a triple [l, m, k] using the method described in Section 8.2: (a) (1, 1, 1); (b) (3, 3, 3, 3, 3); (c) (1, 2, 3).

8.3.What finite sequence is coded by the triple [4, 5, 373777]?

(b)Show that ΓN xϕ(x) where ΓN is the set of axioms from Section 8.1.

8.6.Prove that a definable subset D of N is definable by a ∆0 VN -formula if and only if D is primitive recursive.

8.7.Show that the following sets of natural numbers are primitive recursive by describing a ∆0 formula that defines the set:

(a)T = {n|n is the G¨odel code for a VN -term }

(b)F = {n|n is the G¨odel code for a VN -formula }

(c)S = {n|n is the G¨odel code for a VN -sentence }.

8.8.Let T be recursive VN -theory containing ΓN . Show that the set {n|N |= P rT (n)} is not primitive recursive.

8.9.Show that the decision problems corresponding to each of the four sets defined in the previous two problems are in NP. If P =NP, then which these problems are in P?

•for some M |= T , each recursive subset of D2 is a definable subset of M .

•for each m N there exists a term tn D such that n is more than m times the length of tn. (i.e., there exist terms tn D that are arbitrarily short relative to n.) Prove that T is undecidable.

8.12. (L¨ob) Let T be a recursive subset of TN . Show that there exists a VN -sentence ϕ so that T ϕ ↔ P rT (tϕ). Show that, unlike the sentence

γ that asserts its own unprovability, ϕ can be derived from T .

8.13.(Tarski) Let V = {[ϕ]|N |= ϕ}. Use the Fixed Point lemma to show that V is not a definable subset of N.

8.14.Show that for any two VN -formulas ϕ1(x) and ϕ2(x) in one free variable there exist VN -sentences ψ1 and ψ2 such that

Tψ1 ↔ ϕ1(tψ2 ) and T ψ2 ↔ ϕ2(tψ1 ).

8.15.Let T be a recursive set of V-sentences.

Let T0 = {(ϕ) (ϕ) . . . (ϕ)|T ϕ} and let C0 = {[ϕ]|ϕ T0}.

nϕ times

Recall the primitive recursive function bin(e, x, n, 1) from the proof of Theorem 7.33. Show that there exists a T ++ program Pe such that bin(e, x, x, 1) is the characteristic function of C0.

8.16.Let T be a deductively closed VN -theory and let CT N be the set of G¨odel codes of sentences in T . Show that the following are equivalent.

(i)T is decidable.

(ii)CT is recursively enumerable.

(ii)T is axiomatized by a primitive recursive set of sentences.

8.17.Let T be a recursive set of VN -sentences that contains ΓN . Show that if T P rT (tϕ) then T ϕ.

8.18.Let T be a recursive set of VN -sentences such that ΓN T . Let f (¯x) be a primitive recursive function on N and let ϕf (¯x, y) be a VN -formula

expressing that f (¯x) = y. Show that T ϕf (¯x, y) → P rT (td), where d is the G¨odel code for ϕf (¯x, y). (Proceed by induction on the primitive recursive function f .)

8.19.Sketch a proof for Lemma 8.26.

8.20.Let Tn be the set of Σn sentences in TN . Let Sn = {[ϕ]|ϕ Tn}. Show that Sn is a Σn set that is not Πn (see Exercise 7.29).

8.21.Let Tn be the set of Σn sentences in TN . Show that, for any n N, Tn is incomplete.

8.22.(Rosser) Let T be a recursive VN -theory that contains ΓN .

Let Y = {[ϕ]|T ϕ} and N = {[ϕ]|T ¬ϕ}. Show that Y and N are recursively inseparable (as defined in Exercise 7.24).

9Beyond first-order logic

We consider various extensions of first-order logic. Informally, a logic L is an extension of first-order logic if every sentence of first-order logic is also a sentence of L. We also require that L is closed under conjunction and negation and has other basic properties of a logic. In Section 9.4, we list the properties that formally define the notion of an extension of first-order logic. Prior to Section 9.4, we provide various natural examples of such extensions. In Sections 9.1–9.3, we consider, respectively, second-order logic, infinitary logics, and logics with fixed-point operators.

We do not provide a thorough treatment of any one of these logics. Indeed, we could easily devote an entire chapter to each. Rather, we define each logic and provide examples that demonstrate the expressive power of the logics. In particular, we show that none of these logics has compactness.

In the final Section 9.4, we prove that if a proper extension of first-order logic has compactness, then the Downward L¨owenheim–Skolem theorem must fail for that logic. This is Lindstr¨om’s theorem. The Compactness theorem and Downward L¨owenheim–Skolem theorem are two crucial results for model theory. Every property of first-order logic from Chapter 4 is a consequence of these two theorems. Lindstr¨om’s theorem implies that the only extension of first-order logic possessing these properties is first-order logic itself.

9.1 Second-order logic

Second-order logic is the extension of first-order logic that allows quantification of relations. The symbols of second-order logic are the same symbols used in firstorder logic. The syntax of second-order logic is defined by adding one rule to the syntax of first-order logic. The additional rule makes second-order logic far more expressive than first-order logic. Specifically, the syntax of second-order logic is defined as follows. Any atomic first-order formula is a formula of second-order logic. Moreover, we have the following four rules:

(R1) If ϕ is a formula then so is ¬ϕ.

(R2) If ϕ and ψ are formulas then so is ϕ ψ.

(R3) If ϕ is a formula, then so is xϕ for any variable x.

(R4) If ϕ is a formula, then so is Rnϕ for any n and n-ary relation R.

Definition 9.1 A string of symbols is a second-order formula if and only if it can be built up from atomic formulas using these four rules.

Recall that rules (R1), (R2), and (R3) define the syntax for first order logic. These rules regard the primitive symbols ¬, , and . We allow the same abbreviations , , →, and ↔ from first-order logic. For any formula ϕ, we naturally define Rnϕ to be the formula ¬ Rnϕ. We define a second-order sentence in the same manner that we defined the notion of a first-order sentence.

Definition 9.2 A second-order sentence is a formula having no free variables.

This definition does not refer to relations. In second-order logic, relations like variables may have free or bound occurences within a formula. A second-order sentence may have free relations but not free variables.

Let V be a vocabulary. A second-order sentence ϕ is a V-sentence if each constant and function occurring in ϕ is in V and each relation having free occurrence in ϕ is in V. A second-order V-sentence may contain relations that are not in V provided that they are bound by a quantifier.

Example 9.3 Let V = {P , Q} be the vocabulary consisting of unary relations P and Q. Consider the following sentence:

x y(R(x, y) → (P (x) Q(y)))

x(P (x) → yR(x, y))

x y z(R(x, y) R(x, z) → y = z).

Call this sentence ϕ0. This is a first-order sentence, but it is not a V-sentence since the relation R is not in V. Now let ϕ be the second-order sentence R2ϕ0. This is a V-sentence since the free relations are in ϕ.

We now define the semantics of second-order logic. Let M be a V-structure and let ϕ be a second-order V-sentence. We must say what it means for M to be a model of ϕ. We use the notation M |= ϕ to express that M is a model of ϕ. We define this concept by induction on the complexity of ϕ. If ϕ is firstorder, then M |= ϕ is as defined in Section 2.3. Now, suppose M |= ϕ0 has been defined and let ϕ have the form Rnϕ0. Then M |= ϕ if and only if there exists an interpretation of R on the universe of M which makes ϕ0 true. Put another way, M |= ϕ if and only if there exists an expansion M of M to the vocabulary V {R} such that M |= ϕ0.

Example 9.4 Let V = {P , Q} and let M be a V-structure. Let ϕ0 and ϕ be as in Example 9.3. It makes no sense to ask whether M models ϕ0 since M is an V-structure and ϕ0 contains the relation R which is not in V. Whether a structure models ϕ0 depends on how R is interpreted. The sentence ϕ asserts