to a quantified formula x B(x) in terms of the associate (B(x))© of the subformula B(x), and analogously for .

xB(x) will of course be equivalent to x(B(x))©. And (B(x))© will be a truthfunctional compound of clear formulas A1, . . . , An , each of which either is atomic or begins with a quantifier. Consider a formula equivalent to (B(x))© that is in disjunctive normal form in the Ai . It will be a disjunction B1 · · · Br of formulas B j , each of which is a conjunction of some of the Ai and their negations. We may assume each B j has the form

C j,1 & . . . & C j,r & D j,1 & . . . & D j,s

where the Cs are the conjuncts in which the variable x does occur and the Ds those in which it does not; by clarity, the Ds will include all conjuncts that begin with or are the negations of formulas beginning with quantifiers, and the Cs will all be atomic. Then as x(B(x))© we may take the disjunction B1 · · · Br , where B j is

x(C j,1 & · · · & C j,r ) & D j,1 & . . . & D j,s .

(In the degenerate case where r = 0, B j is thus the same as B j .)

21.3 Dyadic Logic

Again we go straight to work.

Proof of Lemma 21.2: Lemma 21.1 tells us the satisfiability problem is unsolvable for predicate logic, and we want to show it is unsolvable for dyadic logic. It will be enough to show how one can effectively associate to any sentence of predicate logic a sentence of dyadic logic such that the former will be satisfiable if and only if the latter is. What we are going to do is to show how to eliminate one three-place predicate (at the cost of introducing new twoand one-place predicates). The same method will work for k-place predicates for any k ≥ 3, and applying it over and over we can eliminate all but twoand one-place predicates. The one-place ones can also be eliminated one at a time, since given a sentence S containing a one-place predicate P, introducing a new two-place predicate P* and replacing each atomic subformula Px by P*xx clearly produces a sentence S* that is satisfiable if and only if S is. Thus we can eliminate all but two-place predicates.

To indicate the method for eliminating a three-place predicate, let S be a sentence containing such a predicate P. Let P* be a new one-place predicate, and Qi for i = 1, 2, 3 a trio of new two-place predicates. Let w be a variable not appearing in S, and let S* be the result of replacing each atomic subformula of form Px1 x2 x3 in S by

w(Q1wx1 & Q2wx2 & Q3wx3 & P w).

We claim S is satisfiable if and only if S* is satisfiable. The ‘if’ direction is easy. For if S is unsatisfiable, then S is valid, and substitution (of a formula with the appropriate free variables for a predicate) preserves validity, so S* is valid, and S* is unsatisfiable.

For the ‘only if’ direction, suppose S has a model M. By the canonical domains theorem (Corollary 12.18) we may take the domain of Mbe the set of natural numbers. We want to show that S* has a model M*. We will take M* to have domain the set of natural numbers, and to assign to every predicate in S other than P the same denotation that M assigns. It will suffice to show that we can assign denotations to P* and the Qi in such a way that natural numbers a1, a2, a3 will satisfy w(Q1wx1 & Q2wx2 & Q3wx3 & P w) in M* if and only if they satisfy Px1 x2 x3 in M. To achieve this, fix a function f from the natural numbers onto the set of all triples of natural numbers. It then suffices to take as the denotation of P* in M* the relation that holds of a number b if and only if f (b) is a triple a1, a2, a3 for which the relation that is the denotation of P in M holds, and to take as the denotation of Qi in M* the relation that holds of b and a if and only if a is the ith component of the triple f (b).

Proof of Lemma 21.3: We want next to show that we can eliminate any number of two-place predicates P1, . . . , Pk in favour of a single three-place predicate Q. So given a sentence S containing the Pi , let u1, . . . , uk be variables not occurring in S, and let S* be the result of replacing each atomic subformula of form Pi x1 x2 in S by Qvi x1 x2, and let S† be the result of prefixing S* by v1 · · · vk . For instance, if S is

x y(P2 yx & z(P1 x z & P3zy))

then S† will be

v1 v2 v3 x y(Qv2 yx & z(Qv1 x z & Qv3zy)).

We claim S is satisfiable if and only if S† is satisfiable. As in the preceding proof, the ‘if’ direction is easy, using the fact that substitution preserves validity. (More explicitly, if there is a model M† of S†, some elements a1, . . . , ak of its domain satisfy the formula S*. We can now get a model M of S by taking the same domain, and assigning as denotation to Pi in M the relation that holds between b1 and b2 if and only if the relation that is the denotation of Q in M† holds among ai and b1 and b2.)

For the ‘only if’ direction, suppose M is a model of S. As in the preceding proof, we may take the domain of M to be the set of natural numbers, using the canonical domains theorem. We can now get a model M† of S†, also with domain the natural numbers, by taking as the denotation of Q in M† the relation that holds among natural numbers a and b1 and b2 if and only if 1 ≤ a ≤ k, and as the denotation of Pa in M the relation that holds between b1 and b2. From the fact that M is a model of S, it follows that 1, . . . , k satisfy S* in M†, and hence S† is true in M†.

Proof of Theorem 21.4: We want next to show that we can eliminate a single threeplace predicate P in favour of a single two-place predicate Q. So given a sentence S containing the P, let u1, u2, u3, u4 be variables not occurring in S, and let S* be the result of replacing each atomic subformula of form Pi x1 x2 x3 in S by a certain formula P (x1, x2, x3), namely

u1 u2 u3 u4 ( Qu1u1 & Qu1u2 & Qu2u3 & Qu3u4

& Qu4u1 & Qu1 x1 & Qu2 x2 & Qu3 x3 & Qx1u2 & Qx2u3 & Qx3u4 & Qu4 x1).

We then claim S is satisfiable if and only if S* is satisfiable. As in the preceding proofs, the ‘if’ direction is easy, and for the ‘only if’ direction what we need to do is to show, given a model M of S, which may be taken to have domain the natural numbers, that we can define an interpretation M*, also with domain the natural numbers, which will assign as denotation to Q in M* a relation such that for any natural numbers b1, b2, b3, those numbers will satisfy P (x1, x2, x3) in M* if and only if those numbers satisfy P(x1, x2, x3) in M. To accomplish this last and so complete the proof, it will be enough to establish the following lemma.

21.13 Lemma. Let R be a three-place relation on the natural numbers. Then there is a two-place relation S on the natural numbers such that if a, b, c are any natural numbers, then we have Rabc if and only if for some natural numbers w, x, y, z we have

(1)

Sww & Swx & Sx y & Syz & Szw &

Swa & Sxb & Syc & Sax & Sby & Scz & Sza.

Proof: One of the several ways of enumerating all triples of natural numbers is to order them by their sums, and where these are the same by their first components, and where these also are the same by their second components, and where these also are the same by their third components. Thus the first few triples are

(0, 0, 0) (0, 0, 1) (0, 1, 0) (1, 0, 0) (0, 0, 2) (0, 1, 1) (0, 2, 0) (1, 0, 1) (1, 1, 0) (2, 0, 0) (0, 0, 3)

.

.

.

·

Counting the initial triple as the first rather than the zeroth, it is clear that if the nth triple is (a, b, c), then a, b, c are all <n. It follows that if w, x, y, and z are respectively 4n + 1, 4n + 2, 4n + 3, and 4n + 4, then a, b, c are all less than w − 4, x − 4, y − 4, and z − 4. (For instance, a < n implies a + 1 ≤ n, which implies 4a + 4 ≤ 4n < 4n + 1.)

Now to define S. If the nth triple is (a, b, c), we let Svu hold in each of the following four cases:

Svu is not to hold in any other cases. Note that

(2)if Svu, then v + 1 ≥ u

(3)there is at most one u < v − 4 such that Svu.

We must now show that Rabc holds if and only if there are w, x, y, z such that

(1) above holds. The ‘only if’ direction is immediate: if Rabc, take w, x, y, z to be 4n + 1, 4n + 2, 4n + 3, 4n + 4, where (a, b, c) is the nth triple, and (1) will hold. [ Sax holds because a < x − 4 holds, so a + 1 ≥ x fails, so Sax fails by (2); similarly for the other negations in (1).]

Now suppose (1) holds for some w, x, y, z. We must show that Rabc. To begin with, (1) gives us Sww, so w must be of the form 4n + 1 for some n ≥ 1.

Also, (1) gives us Swx, Sx y, Syz, Szw. Therefore, by (2), x + 3 ≥ y + 2 ≥ z + 1 ≥ w, whence x ≥ w − 3. Similarly, y ≥ x − 3 and z ≥ y − 3. So neither x < w − 4 nor y < x − 4 nor z < y − 4. Since Swx holds while x < w − 4 fails, we must have x = w + 1 = 4n + 2.

Also, since Sxy holds while y < x − 4 fails, either y = x or y = x + 1. Similarly, either z = y or z = y + 1. But if either y = x or z = y, then z = w + 1 = 4n + 2 or z = w + 2 = 4n + 3. But this is impossible, since Svu never holds for u = 4m + 1 and v = 4n + 2 or 4n + 3, whereas we have Szw. It follows that y = x + 1 = 4n + 3 and z = y + 1 = 4n + 4.

If we can show that the nth triple is (a, b, c), then we can conclude that Rabc: for if (a, b, c) is the nth triple, then Sza if and only if Rabc, and (1) gives Sza.

We have Swa, Swb, Syc and Sax, Sby, Scz from (1). And since we know w = 4n + 1, x = 4n + 2, y = 4n + 3, z = 4n + 4, we also have Sxx, Sxy, Syy, Syz, and Szz from the definition of S. So a = x, b = x, b = y, c = y, and c = z. So we have Swa and a < w − 4, Sxb and b < x − 4, and Syc and c < y − 4. If the nth triple is (r, s, t), then we also have Swr, Sxs, Syt and r < w − 4, s < x − 4, t < y − 4. So by (3) above we must have r = a, s = b, and t = c. So (a, b, c) is the nth triple, and the proof is complete.

Problems

21.1Prove Lemma 21.8 directly, without deriving it from Lemma 21.9.

21.2Show that the estimates 2k and 2k · r in Lemmas 21.8 and 21.9 cannot be improved.

21.3What happens if constants are added to monadic logic with identity?

21.4The language of set theory has a single nonlogical symbol and two-place predicate . ZFC is a certain theory in this language, of which it was asserted towards the end of section 17.1 that it is ‘adequate for formalizing essentially all accepted mathematical proofs’. What is the bearing of this fact on Theorem 21.4?