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INDEFINABILITY, UNDECIDABILITY, INCOMPLETENESS

Still, it is reasonable to understand ‘sufficiently strong’ as implying that this principle can be somehow expressed in the language of the theory, perhaps by a sentence

x(N (x) y(N (y) (L(x, y) x = y L(y, x))))

where N (x) appropriately expresses ‘x is a natural number’ and L(x, y) appropriately expresses ‘x is less than y’. Moreover, the sentence that thus ‘translates’ this or any axiom of Q should be a theorem of the theory. Such is the case, for instance, with the formal systems considered in works on set theory, such as the one known as ZFC, which are adequate for formalizing essentially all accepted mathematical proofs. When the notion of ‘translation’ is made precise, it can be shown that any ‘sufficiently strong’ formal system of mathematics in the sense we have been indicating is still subject to the limitative theorems of this chapter. In particular, if consistent, it will be incomplete.

Perhaps the most important implication of the incompleteness theorem is what is says about the notions of truth (in the standard interpretation) and provability (in a formal system): that they are in no sense the same.

17.2 Undecidable Sentences

A sentence in the language of a theory T is said to be disprovable in T if its negation is provable in T , and is said to be undecidable in or by or for T if it is neither provable nor disprovable in T . (Do not confuse the notion of an undecidable sentence with that of an undecidable theory. True arithmetic, for example, is an undecidable theory with no undecidable sentences: the sentences of its language that are true in the standard interpretation all being provable, and those that are false all being disprovable.) If T is a theory in the language of arithmetic that is consistent, axiomatizable, and an extension of Q, then T is an undecidable theory by Theorem 17.4, and there exist undecidable sentences for T by Theorem 17.7. Our proof of the latter theorem did not, however, exhibit any explicit example of a sentence that is undecidable for T . An immediate question is: can we find any such specific examples?

In order to do so, we use the fact that the set of sentences that are provable and the set of sentences that are disprovable from any recursive set of axioms is semirecursive, and that all recursive sets are definable by -rudimentary formulas. It follows that there are formulas PrvT (x) and DisprvT (x) of forms y PrfT (x, y) and y DisprfT (x, y) respectively, with Prf and Disprf rudimentary, such that T A

if and only if the sentence PrvT ( A ) is correct or true in the standard interpretation, and hence if and only if for some b the sentence PrfT ( A , b) is correct or—what is equivalent for rudimentary sentences—provable in Q and in T ; and similarly for disprovability. PrfT (x, y) could be read ‘y is a witness to the provability of x in T ’.

By the diagonalization lemma, there is a sentence GT such that

T GT y PrfT ( GT , y)

and a sentence RT such that

T RT y(PrfT ( RT , y) z < y Disprf( RT , z)).

17.2. UNDECIDABLE SENTENCES

225

Such a GT is called a Godel¨ sentence for T , and such an RT a Rosser sentence for T . Thus a Godel¨ sentence ‘says of itself that’ it is unprovable, and a Rosser sentence ‘says of itself that’ if there is a witness to its provability, then there is an earlier witness to its disprovability.

17.8 Theorem. Let T be a consistent, axiomatizable extension of Q. Then a Rosser sentence for T is undecidable for T .

Proof: Suppose the Rosser sentence RT is provable in T . Then there is some a that witnesses the provability of RT . Since T is consistent, RT is not also provable, and so no m witnesses the disprovability of RT , and in particular, no m < n does so. It follows that the rudimentary sentence

PrfT ( RT , n) & z < n DisprfT ( RT , z)

is correct and as such is provable from the axioms of Q, and hence from T . In other words, we have

T PrfT ( RT , n) & z < n DisprfT ( RT , z)

while, since RT is a Rosser sentence, we also have

T RT y(PrfT ( RT , y) z < y Disprf( RT , z)).

By pure logic it follows that

T RT .

But then T is inconsistent, both RT and RT being provable, contrary to assumption. This contradiction shows that RT cannot be provable.

Suppose the Rosser sentence RT is disprovable in T . Then there is some m that witnesses the disprovability of RT . Since T is consistent, RT is not also provable, and so no n witnesses the provability of RT , and in particular, no n m does so. It follows that the rudimentary formulas

DisprfT ( RT , m)

x((x < m x = m) PrfT ( RT , x))

are correct and hence provable in T . In other words we have

T DisprfT ( RT , m),

T y((y < m y = m) PrfT ( RT , y)).

By pure logic it follows from the former of these that

T y(m < y z < y DisprfT ( RT , z)).

As a theorem of Q we also have

T y(y < m y = m m < y).

226 INDEFINABILITY, UNDECIDABILITY, INCOMPLETENESS

It follows by pure logic that

T y(PrfT ( RT , y) z < y Disprf( RT , z))

and hence T RT , and T is inconsistent, a contradiction that shows that RT cannot be disprovable in T .

A theory T is called ω-inconsistent if and only if for some formula F(x), T x F(x) but T F(n) for every natural number n, and is called ω-consistent if and only if it is not ω-inconsistent. Thus an ω-inconsistent theory ‘affirms’ there is some number with the property expressed by F, but then ‘denies’ that zero is such a number, that one is such a number, than two is such a number, and so on. Since x F(x) andF(0), F(1), F(2), . . . cannot all be correct, any ω-inconsistent theory must have some incorrect theorems. But an ω-inconsistent theory need not be inconsistent. (An example of a consistent but ω-inconsistent theory will be given shortly.)

17.9 Theorem. Let T be a consistent, axiomatizable extension of Q. Then a Godel¨ sentence for T is unprovable in T , and if T is ω-consistent, it is also undisprovable in T .

Proof: Suppose the Godel¨ sentence GT is provable in T . Then the -rudimentary sentence y PrfT ( GT , y) is correct, and so provable in T . But since GT is a Godel¨ sentence, GT y PrfT ( GT , y) is also provable in T . By pure logic it follows that GT is provable in T , and T is inconsistent, a contradiction, which shows that GT is not provable in T .

Suppose the sentence GT is disprovable in T . Then y PrfT ( GT , y) and hencey PrfT ( GT , y) is provable in T . But by consistency, GT is not provable in T , and so for any n, n is not a witness to the provability of GT , and so the rudimentary sentencePrfT ( GT , n) is correct and hence provable in Q and hence in T . But this means T is ω-inconsistent, a contradiction, which shows that GT is not disprovable in T .

For an example of a consistent but ω-inconsistent theory, consider the theory T = Q + GQ consisting of all consequences of the axioms of Q together with GQ or y PrfQ(GQ, y). Since GQ is not a theorem of Q, this theory T is consistent. Of course y PrfQ(GQ, y) is a theorem of T . But for any particular n, the rudimentary sentence PrfQ(GQ, n) is correct, and therefore provable in any extension of Q, including T .

Historically, Theorem 17.9 came first, and Theorem 17.8 was a subsequent refinement. Accordingly, the Rosser sentence is sometimes called the Godel¨–Rosser sentence. Subsequently, many other examples of undecidable sentences have been brought forward. Several interesting examples will be discussed in the following, optional section, and the most important example in the next chapter.

17.3* Undecidable Sentences without the Diagonal Lemma

The diagonal lemma, which was used to construct the Godel¨ and Rosser sentences, is in some sense the cleverest idea in the proof of the first incompleteness theorem, and is heavily emphasized in popularized accounts. However, the possibility of implementing the idea of this lemma, of constructing a sentence that says of itself that it is unprovable, depends on the apparatus of the arithmetization of syntax and the

17.3*. UNDECIDABLE SENTENCES WITHOUT THE DIAGONAL LEMMA 227

representability of recursive functions. Once that apparatus is in place, a version of the incompleteness theorem, showing the existence of a true but unprovable sentence, can be established without the diagonal lemma. One way to do so is indicated in the first problem at the end of this chapter. (This way uses the fact that there exist semirecursive sets that are not recursive, and though it does not use the diagonal lemma, does involve a diagonal argument, buried in the proof of the fact just cited.) Some other ways will be indicated in the present section.

Towards describing one such way, recall the Epimenides or liar paradox, involving the sentence ‘This sentence is untrue’. A contradiction arises when we ask whether this sentence is true: it seems that it is if and only if it isn’t. The Godel¨ sentence in effect results from this paradoxical sentence on substituting ‘provable’ for ‘true’ (a substitution that is crucial for establishing that we can actually construct a Godel¨ sentence in the language of arithmetic). Now there are other semantic paradoxes, paradoxes in the same family as the liar paradox, involving other semantic notions related to truth. One famous one is the Grelling or heterological paradox. Call an adjective autological if it is true of itself, as ‘short’ is short, ‘polysyllabic’ is polysyllabic, and ‘English’ is English, and call it heterological if it is untrue of itself, as ‘long’ is not long, ‘monosyllabic’ is not monosyllabic, and ‘French’ is not French. A contradiction arises when we ask whether ‘heterological’ is heterological: it seems that it is if and only if it isn’t.

Let us modify the definition of heterologicality by substituting ‘provable’ for ‘true’. We then get the notion of self-applicability: a number m is self-applicable in Q if it is the Godel¨ number of a formula μ(x) such that μ(m) is provable in Q. Now the same apparatus that allowed us to construct the Godel¨ sentence allows us to construct what may be called the Godel¨–Grelling formula GG(x) expressing ‘x is not self-applicable’. Let m be its Godel¨ number. If m were self-applicable, then GG(m) would be provable, hence true, and since what it expresses is that m is not self-applicable, this is impossible. So m is not self-applicable, and hence GG(m) is true but unprovable.

Another semantic paradox, Berry’s paradox, concerns the least integer not namable in fewer than nineteen syllables. The paradox, of course, is that the integer in question appears to have been named just now in eighteen syllables. This paradox, too, can be adapted to give an example of a sentence undecidable in Q. Let us say that a number n is denominable in Q by a formula φ(x) if x(φ(x) x = n) is (not just true but) provable in Q.

Every number n is denominable in Q, since if worse comes to worst, it can always be denominated by the formula x = n, a formula with n + 3 symbols. Some numbers n are denominable in Q by formulas with far fewer than n symbols. For example, the number 10 10 is denominable by the formula φ(10, 10, x), where φ is a formula representing the super-exponential function . We have not actually written out this formula, but instructions for doing so are implicit in the proof that all recursive functions are representable, and review of that proof reveals that writing out the formula would not take more time or more paper than an ordinary homework assignment. By contrast, 10 10 is larger than the number of particles in the visible universe. But while big numbers can thus be denominated by comparatively short formulas, for any fixed k, only finitely many numbers can be denominated by formulas with fewer than

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k symbols. For logically equivalent formulas denominate the same number (if they denominate any number at all), and every formula with fewer than k symbols is logically equivalent, by relettering bound variables, to one with only the first k variables on our official list of variables, and there are only finitely many of those.

Thus, there will be numbers not denominable using fewer than 10 10 symbols. The usual apparatus allows us to construct a Godel¨–Berry formula GB(x, y), expressing ‘x is the least number not denominable by a formula with fewer than y y symbols’. Writing out this formula would involve writing out not just the formula representing the super-exponential function , but also the formulas relating to provability in Q. Again we have not actually written out these formulas, but only given an outline of how to do so in our proofs of the arithmetizability of syntax and the representability of recursive functions in Q. Review of those proofs reveals that writing out the formula GB(x, y) or GB(x, 10), though it would require more time and paper than any reasonable homework assignment, would not require more symbols than appear in an ordinary encyclopedia, which is far fewer than the astronomical figure 10 10. Now there is some number not denominable by a formula with fewer symbols than that astronomical figure, and among such numbers there is one and only one least, call it n. Then GB(n, 10) and x(GB(x, 10) x = n) are true. But if the latter were provable, the formula GB(x, 10) would denominate n, whereas n is not denominable except by formulas much longer than that. Hence we have another example of an unprovable truth.

This example is worth pressing a little further. The length of the shortest formula denominating a number may be taken as a measure of the complexity of that number. Just as we could construct the Godel¨ –Berry formula, we can construct a formula C(x, y, z) expressing ‘the complexity of x is y and y is greater than z z’, and using it the Godel¨–Chaitin formula GC(x) or yC(x, y, 10), expressing that x has complexity greater than 10 10. Now for all but finitely many n, GC(n) is true. Chaitin’s theorem tells us that no sentence of form GC(n) is provable.

The reason may be sketched as follows. Just as ‘y is a witness to the provability of x in Q’ can be expressed in the language of arithmetic by a formula PrfQ(x,y), so can ‘y is a witness to the provability of the result of subsituting the numeral for x for the variable in GC’ be expressed by a formula PrfGCQ(x,y). Now if any sentence of form GC(n) can be proved, there is a least m such that m witnesses the provability of GC(n) for some n. Let us call m the ‘lead witness’ for short. And of course, since any one number witnesses the provability of at most one sentence, there will be a least n—in fact, there will be one and only one n—such that the lead witness is a witness to the provability of GC(n). Call n the number ‘identified by the lead witness’ for short.

If one is careful, one can arrange matters so that the sentences K (m) and L(n) expressing ‘m is the lead witness’ and ‘n is the number identified by the lead witness’ will be -rudimentary, so that, being true, K (m) and L(n) will be provable. Moreover, since it can be proved in Q that there is at most one least number fulfilling the condition expressed by any formula, x(x = m K (x)) and x(x = n L(x)) will also be provable. But this means that n is denominated by the formula L(x), and hence has complexity less than the number of symbols in that formula. And though it might take an encyclopedia’s worth of paper and ink to write the formula down, the number