Computability and Logic

276 MONADIC AND DYADIC LOGIC
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 M be 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 Q i in such a way that natural numbers a 1 , a 2 , a 3 will satisfy ∃w(Q 1 wx 1 &
Q 2 wx 2 & Q 3 wx 3 & P ∗ w)inM* if and only if they satisfy Px 1 x 2 x 3 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*inM* the relation that holds
of a number b if and only if f (b) is a triple a 1 , a 2 , a 3 for which the relation that is the
denotation of P in M holds, and to take as the denotation of Q i 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 P 1 ,...,P k in favour of a single three-place predicate Q. So
given a sentence S containing the P i , let u 1 ,...,u k be variables not occurring in S,
and let S* be the result of replacing each atomic subformula of form P i x 1 x 2 in S by
Qv i x 1 x 2 , and let S † be the result of prefixing S*by∃v 1 ···∃v k . For instance, if S is
then S † will be
∀x∃y(P 2 yx & ∀z(P 1 xz& P 3 zy))
∃v 1 ∃v 2 ∃v 3 ∀x∃y(Qv 2 yx & ∀z(Qv 1 xz& Qv 3 zy)).
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 a 1 ,...,a k 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 P i in M the relation that holds between b 1 and b 2
if and only if the relation that is the denotation of Q in M † holds among a i and b 1
and b 2 .)
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 b 1 and b 2 if and only if 1 ≤ a ≤ k, and as the denotation of P a in M
the relation that holds between b 1 and b 2 . From the fact that M is a model of S, it
follows that 1, ..., k satisfy S*inM † , 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 u 1 , u 2 , u 3 , u 4 be variables not occurring in S, and let S* be
the result of replacing each atomic subformula of form P i x 1 x 2 x 3 in S by a certain
formula P ∗ (x 1 , x 2 , x 3 ), namely
∃u 1 ∃u 2 ∃u 3 ∃u 4 (∼Qu 1 u 1 & Qu 1 u 2 & Qu 2 u 3 & Qu 3 u 4
& Qu 4 u 1 & Qu 1 x 1 & Qu 2 x 2 & Qu 3 x 3 & ∼Qx 1 u 2 & ∼Qx 2 u 3 & ∼Qx 3 u 4 & Qu 4 x 1 ).