Computability and Logic

20.1. CRAIG’S THEOREM AND ITS PROOF 263
(S3) is just slightly different. Suppose Ɣ = Ɣ L ∪ Ɣ R is an unbarred division and
(D 1 ∨ D 2 )isinƔ. Again there are two cases, and we consider only the one where
(D 1 ∨ D 2 )isinƔ L . Each of the D i is of course a left sentence, since their disjunction
is. We claim that Ɣ ∪{D i }=(Ɣ L ∪{D i }) ∪ Ɣ R is an unbarred division for at least
one i. Towards showing this, note that if Ɣ L ∪{D 1 } is unsatisfiable, then Ɣ L implies
both (D 1 ∨ D 2 ) and ∼D 1 , and hence implies D 2 . In this case, the proof that (Ɣ L ∪
{D 2 }) ∪ Ɣ R is an unbarred division is just like the proof of the preceding paragraph.
Similarly, if Ɣ L ∪{D 2 } is unsatisfiable, then (Ɣ L ∪{D 2 }) ∪ Ɣ R gives an unbarred
division. So we are left to treat the case where Ɣ L ∪{D i } is satisfiable for both i.
In this case, (Ɣ L ∪{D i }) ∪ Ɣ R can fail to give an unbarred division only because
there is a sentence B i that bars the pair Ɣ L ∪{D i },Ɣ R . What we claim is that there
cannot exist such B i both both i = 1 and i = 2. For suppose there did. Then since
B i is implied by Ɣ L ∪{D i } for each i, and Ɣ L contains (D 1 ∨ D 2 ), it follows that
B = (B 1 ∨ B 2 ) is implied by Ɣ L . Moreover, since each ∼B i is implied by Ɣ R ,so
is ∼B. Finally, since each B i is both a left and a right sentence, the same is true
of B. Thus there is a sentence B that bars Ɣ L ,Ɣ R , contrary to hypothesis. This
contradiction completes the proof for the case where identity and function symbols
are absent.
Proof, Part II: We next consider the case where identity is present but function
symbols are still absent. Suppose A implies C. As in section 18.4, we introduce
the new two-place predicate ≡. We write E L for the conjunction of the
equality axioms and the congruence axioms for predicates in A, and E R for the
corresponding sentence for C. We write * to indicate replacing = by ≡. Since A
implies C, A & ∼C is unsatisfiable. What the proof of Proposition 19.13 tells us is
that therefore E L & E R & A*&∼C* is unsatisfiable. It follows that E L & A* implies
E R → C*. By the interpolation theorem for sentences without identity, there is a sentence
B* involving only ≡ and nonlogical predicates common to A and C, such that
E L & A* implies B* and B* implies E R → C*. It follows that E L & A*&∼B* and
E R & B*&C* are unsatisfiable. Then we claim B, the result of replacing ≡ by =
in B*, is the required interpolant between A and C. Certainly its nonlogical predicates
are common to A and C. Further, what the proof of Proposition 18.13 tell us
is is that A & ∼B and B & ∼C are unsatisfiable, and therefore A implies B and B
implies C. The treatment of function symbols is much the same, but using the machinery
of Proposition 19.12 rather than of Proposition 19.13. Details are left to the
reader.
In the remaining sections of this chapter we apply the interpolation theorem to
prove two results about theories: one about the conditions under which the union
of two theories is satisfiable, the other about the conditions under which definitions
are consequences of theories. (Throughout ‘theory’ is being used, as elsewhere
in this book, in a very broad way: A theory in a language is just a set of
sentences of the language that contains every sentence of that language that is a
logical consequence of the set. A theorem of a theory is just a sentence in that
theory.)