Maqk of (16), gives the conjunct Mbqk of (18). Finally, the fifth conjunct together with the last conjunct of (16) gives the last conjunct of (18). The reader will see now what we meant when we said at the outset, ‘Some facility at recognizing simple logical implications will be required.’]

Now the description of time 0 is one of the sentences in . By the foregoing lemma, if the machine does not stop at time 1, the description of time 1 will be a consequence of , and if the machine then does not stop at time 2, the description of time 2 will be a consequence of together with the description of time 1 (or, as we can more simply say, since the description of time 1 is a consequence of , the description of time 2 will be a consequence of ), and so on until the last time a before the machine halts, if it ever does. If it does halt at time a + 1, we have seen that the description of time a, which we now know to be a consequence of , implies D. Hence if the machine ever halts, implies D.

Hence we have established that if the decision problem for logical implication were solvable, the halting problem would be solvable, which (assuming Turing’s thesis) we know it is not. Hence we have established the following result, assuming Turing’s thesis.

11.2 Theorem (Church’s theorem). The decision problem for logical implication is unsolvable.

11.2 Logic and Primitive Recursive Functions

By the nullity problem for a two-place primitive recursive function f we mean the problem of devising an effective procedure that, given any m, would in a finite amount of time tell us whether or not there is an n such that f (m, n) = 0. We are going to show how, given f , to write down a certain finite set of sentences and a certain formula

D(x) in a language that contains the constants 0 and the successor symbol from the language of arithmetic, and therefore contains the numerals 0 , 0 , 0 , . . . or

1, 2, 3, . . . as we usually write them. And then we are going to show that for any m,implies D(m) if and only if there is an n such that f (m, n) = 0. It follows that if the decision problem for logical implication could be solved, and an effective method devised to tell whether or not a given finite set of sentences implies a sentence D, then the nullity problem for any f could be solved. Since it is known that, assuming Church’s thesis, there is an f for which the nullity problem is not solvable, it follows, again assuming Church’s thesis, that the decision problem for logical implication is unsolvable, or, as is said, that logic is undecidable. The proof of the fact just cited about the unsolvability of the nullity problem requires the apparatus of Chapter 8, but for the reader who is willing to take this fact on faith, this section otherwise presupposes only the material of Chapters 6–7 and 9–10.

To begin the construction, the function f , being primitive recursive, is built up from the basic functions (successor, zero, the identity functions) by the two processes of composition and primitive recursion. We can therefore make a finite list of primitive recursive functions f0, f1, f2, . . . , fr , such that for each i from 1 to r, fi is either the zero function or the successor function or one of the identity functions, or is obtained

from earlier functions in the list by composition or primitive recursion, with the last function fr being the function f . We introduce a language with the symbol 0, the successor symbol , and a function symbol fi of the appropriate number of places for each of the functions fi . In the standard interpretation of the language, the domain will be the natural numbers, 0 will denote zero, will denote the successor function, and each fi will denote fi , so that in particular fr will denote f .

The set of sentences will consist of one or two sentence for each fi for i > 0. In case fi is the zero function, the sentence will be

In case fi is the successor function, the sentence will be

(In this case fi will be another symbol besides for the successor function; but it does not matter if we happen to have two symbols for the same function.) In case fi is the identity function id nk , the sentence will be

(3) x1 · · · xn fi (x1, . . . , xn ) = xk .

In case fi is obtained from f j and fk , where j and k are < i, by primitive recursion, there will be two sentences, as follows.

[In (4) and (5) we have written x and x for x1 , . . . , xn and x1 · · · xn .] Clearly all these sentences are true in the intended interpretation. The formula D(x) will simply be y fr (x, y) = 0. For given m, the sentence D(m) will be true in the standard interpretation if and only if there is an n with f (m, n) = 0.

We want to show that for any m, D(m) will be implied by if and only if there is an n with f (m, n) = 0. The ‘only if’ part is easy. All sentences in are true in the standard interpretation, whereas D(m) is true only if there is an n with f (m, n) = 0. If there is no such n, we have an interpretation where also sentences in are true and D(m) isn’t, so does not imply D(m).

For the ‘if’ part we need one more notion. Call adequate for the function fi if whenever fi (a) = b, then fi (a) = b is implied by . (We have written a for a1 , . . . , an and a for a1 , . . . , an.) The presence of (1)–(3) in guarantees that it is adequate for any fi that is a basic function (zero, successor, or an identity function). What about more complicated functions?

11.3 Lemma

(a)If fi is obtained by composition from functions fk and f j1, . . . f j p for which is adequate, then is adequate also for fi .

(b)If fi is obtained by primitive recursion from functions f j and fk for which is adequate, then is adequate also for fi .

Proof: We leave (a) to the reader and do (b). Given a, b, and c with fi (a, b) = c, for each p ≤ b let cp = fi (a, p), so that cb = c. Note that since fi is obtained by primitive recursion from f j and fk , we have

plies fi (a, 2) = c2, which with (7b) for p = 2 implies fi (a, 3) = c3, and so on up to fi (a, b) = cb = c, which is what needed to be proved to show adequate for fi.

Since every fi is either a basic function or obtained from earlier functions on our list by the processes covered by Lemma 11.3, the lemma implies that is adequate for all the functions on our list, including fr = f . In particular, if f (m, n) = 0, then fr (m, n) = 0 is implied by , and hence so is y fr (m, y) = 0, which is D(m).

Thus we have reduced the problem of determining whether for some n we have f (m, n) = 0 to the problem of determining whether implies D(m). That is, we have established that if the decision problem for logical implication were solvable, the nullity problem for f would be solvable, which it is known, as we have said, that it is not, assuming Church’s thesis. Hence we have established the following result, assuming Church’s thesis.

11.4 Theorem (Church’s theorem). The decision problem for logical implication is unsolvable.

Problems

11.1The decision problem for validity is the problem of devising an effective procedure that, applied to any sentence, would in a finite amount of time enable one to determine whether or not it is valid. Show that the unsolvability of

11.9Add to the language of the theory of the proof of Theorem 11.4 the symbol <, and add to itself the sentence indicated in the preceding problem as well as the sentence

x 0 < x & x y(y < x ↔ (y < x y = x)).

Show that m < n is implied by whenever m <n and m < n is implied bywhenever m ≥ n, and that

y (y < n → y = 0 y = 1 . . . y = m) is implied by whenever n = m .

11.10Let f be a recursive total function, and f1 , . . . , fr a sequence of functions with last item fn = f , such that each is either a basic function or is obtainable by earlier functions in the sequence by composition, primitive recursion, or minimization, and all functions in the sequence are total. (According to Problem 8.13, such a sequence exists for any recursive total function f.) Construct and D as in the proof of Theorem 11.6, with the following modifications. Include the symbol < in the language, and the sentences indicated in the preceding two problems, and besides this, whenever fi is obtained from f j by minimization, include the sentence

x y((f j (x, y) = 0 & z(z < y → f j (x, z) = 0)) → fi (x) = y).

Show that the modified is adequate for fr = f .

11.11(Requires the material of Chapter 8.) Show that there is a two-place primitive recursive function f such that the nullity problem for f is recursively unsolvable, or in other words, such that the set of x such that y( f (x, y) = 0) is not recursive.

A distinction between unavoidable and lazy appeals to Church’s thesis was made at the end of section 7.2; though phrased there for recursive computability, it applies also to Turing computability.

11.12Distinguish the unavoidable from the lazy appeals to Turing’s thesis in section 11.1.

11.13Distinguish the unavoidable from the lazy appeals to Church’s thesis in section 11.2.