Computability and Logic

262 THE CRAIG INTERPOLATION THEOREM
(Lemma 13.3). This tells us that if L is a language containing all the nonlogical
symbols of A and of C, and if L* is a language obtained by adding infinitely many
new constants to L, then {A, ∼C} will be satisfiable provided it belongs to some set S
of sets of sentences of L* having certain properties. For the present situation, where
identity and function symbols are absent, these properties are as follows:
(S0) If Ɣ is in S and Ɣ 0 is a subset of Ɣ, then Ɣ 0 is in S.
(S1) If Ɣ is in S, then for no sentence D are both D and ∼D in Ɣ.
(S2) If Ɣ is in S and ∼∼D is in Ɣ, then Ɣ ∪{D} is in S.
(S3) If Ɣ is in S and (D 1 ∨ D 2 )isinƔ, then Ɣ ∪{D i } is in S for either i = 1ori = 2.
(S4) If Ɣ is in S and ∼(D 1 ∨ D 2 )isinƔ, then Ɣ ∪{∼D i } is in S for both i = 1 and
i = 2.
(S5) If Ɣ is in S and {∃xF(x)} is in Ɣ, and then Ɣ ∪{F(b)} is in S for any constant b
not in Ɣ or ∃xF(x).
(S6) If Ɣ is in S and ∼∃xF(x)isinƔ, then Ɣ ∪{∼F(b)} is in S for any constant b at
all.
What we need to do is to define a set S, use the hypothesis that there is no interpolant
B to show {A, ∼C} is in S, and establish properties (S0)–(S1) for S. Towards defining
S, call a sentence D of L* aleft formula (respectively, a right formula) if every
predicate in D is in A (respectively, is in C). If Ɣ L is a satisfiable set of left sentences
and Ɣ R a satisfiable set of right sentences, let us say that B bars the pair Ɣ L ,Ɣ R if
B is both a left and a right sentence and Ɣ L implies B while Ɣ R implies ∼B. Our
assumption that there is no interpolant, restated in this terminology, is the assumption
that no sentence B of L bars {A}, {∼C}. It follows—by a proof quite like that of
Proposition 20.1—that no sentence B of L* bars {A}, {∼C}. Let S be the set of all Ɣ
that admit and unbarred division in the sense that we can write Ɣ as a union Ɣ L ∪ Ɣ R
of two sets of sentences where Ɣ L consists of left and Ɣ R of right sentences, each of
Ɣ L and Ɣ R is satisfiable, and no sentence bars the pair Ɣ L ,Ɣ R . Then what we have
said so far is that {A, ∼C} is in S. What remains to be done is to establish properties
(S0)–(S6) for S.
(S0) is easy and left to the reader. For (S1), if Ɣ = Ɣ L ∪ Ɣ R is an unbarred division,
then the assumptions that Ɣ L is satisfiable implies that there is no sentence D with
both D and ∼D in Ɣ L . Similarly for Ɣ R . Nor can there be a D with D in Ɣ L and
∼D in Ɣ R , for in that case D would be both a left sentence (since it belongs to Ɣ L )
and a right sentence (since it belongs to Ɣ R ) and therefore would be a sentence that
is implied by Ɣ L and whose negation is implied by Ɣ R , and so would bar Ɣ L ,Ɣ R .
Similarly, the reverse case with ∼D in Ɣ L and D in Ɣ R is impossible, since ∼D would
bar the pair Ɣ L ,Ɣ R .
For (S2), suppose Ɣ = Ɣ L ∪ Ɣ R is an unbarred division and ∼∼D is in Ɣ. There
are two cases according as ∼∼D is in Ɣ L or in Ɣ R , but the two are just alike, and
we consider only the former. Then since ∼∼D is a left formula, so is D, and since
D is implied by Ɣ L , adding D to Ɣ L cannot make it unsatisfiable if it was satisfiable
before, nor can it make any sentence B a consequence that was not a consequence
before. So Ɣ ∪{D} =(Ɣ L ∪{D}) ∪ Ɣ R is an unbarred division, and Ɣ ∪{D} is in S.
(S4)–(S6) are very similar.