Computability and Logic

268 THE CRAIG INTERPOLATION THEOREM
by γ ′ . Let A(A ′ ) be the conjunction of the members of T 0 (T
0 ′ ). Then (1) is implied by
A & A ′ Let c 0 ,...,c k be constants not occurring in T ∪ T ′ , and hence not in A, A ′ ,
—α, x 0 ,...,x k —, or —α ′ , x 0 ,...,x k —. Then
—α, c 0 ,...,c k — ↔ —α ′ , c 0 ,...,c k —
is a consequence of (1) and therefore of A & A ′ . Here of course by —α, c 0 ,...,c k —
is meant the result of substituting c i for x i in —α, x 0 ,...,x k —for all i, and similarly
for —α ′ , c 0 ,...,c k —. It follows that
is valid, and hence that
(A & A ′ ) → (—α, c 0 ,...,c k — ↔ —α ′ , c 0 ,...,c k —)
(4)
A &—α, c 0 ,...,c k —
implies
(5)
A ′ → —α ′ , c 0 ,...,c k —.
We now apply the Craig interpolation lemma. It tells us that there is a sentence
B(c 0 ,... ,c k ) implied by (4) and implying (5), such that the nonlogical symbols of
B are common to (4) and (5). This means that they can include only the c i , which
we have displayed, and the β i . Since (4) implies B(c 0 ,...,c k ), A and therefore T
implies
—α, c 0 ,...,c k — → B(c 0 ,...,c k )
and since the c i do not occur in T , this means T implies
(6)
∀x 0 ···∀x k (—α, x 0 ,...,x k — → B(x 0 ,...,x k )).
Since B(c 0 , ... , c k ) implies (5), A ′ and therefore T ′ implies
B(c 0 ,...,c k ) → —α ′ , c 0 ,...,c k —
and since the c i do not occur in T ′ , this means T ′ implies
∀x 0 ···∀x k (B(x 0 ,...,x k ) → —α ′ , x 0 ,...,x k —).
Replacing each symbol γ ′ by γ ′ , it follows that T implies
(7)
∀x 0 ···∀x k (B(x 0 ,...,x k ) → —α, x 0 ,...,x k —).
But (6) and (7) together imply, and therefore T implies, the explicit definition
∀x 0 ···∀x k (—α, x 0 ,...,x k — ↔ B(x 0 ,...,x k )).
Thus, α is explicitly definable from the β i in T , and Beth’s theorem is proved.
Problems
20.1 (Lyndon’s interpolation theorem) Let A and C be sentences without constants
or function symbols and in negation-normal form. We say that an occurrence of
a predicate in such a sentence is positive if it is not preceded by ∼, and negative