Computability and Logic

264 THE CRAIG INTERPOLATION THEOREM
We begin with a preliminary result.
20.2 Robinson’s Joint Consistency Theorem
20.5 Lemma. The union T 1 ∪ T 2 of two theories T 1 and T 2 is satisfiable if and only if
there is no sentence in T 1 whose negation is in T 2 .
Proof: The ‘only if’ part is obvious: if there were a sentence in T 1 whose negation
was T 2 , the union could not possibly be satisfiable; for there could be no interpretation
in which both the sentence and its negation were true.
The ‘if’ part follows quickly from the compactness theorem and Craig’s theorem:
Suppose the union of T 1 and T 2 is unsatisfiable. Then by the compactness theorem,
there is a finite subset S 0 of the union which is unsatisfiable. If there are no members
of S 0 that belong to T 1 , then T 2 is unsatisfiable, and so ∀x x= x is a sentence in T 1
whose negation is in T 2 ; if no members of S 0 belong to T 2 , then T 1 is unsatisfiable,
and so ∼∀x x= x is a sentence in T 1 whose negation is in T 2 . So we may suppose
that S 0 contains some members both of T 1 and of T 2 . Let F 1 ,...,F m be the members
of S 0 that are in T 1 ; let G 1 ,...,G n be the members of S 0 that are in T 2 .
Let A be (F 1 & ... & F m ), and let C be ∼(G 1 & ... & G n ). A implies C. By
Craig’s theorem, there is a sentence B implied by A, implying C, and containing
only nonlogical symbols contained in both A and C. B is therefore a sentence in
the languages of both T 1 and T 2 . Since A is in T 1 and implies B, B is in T 1 . Since
(G 1 & ... & G n )isinT 2 ,sois∼B, as(G 1 & ... & G n ) implies ∼B. SoB is a
sentence in T 1 whose negation is in T 2 .
An extension T ′ of a theory T is just another theory containing it. The extension
is called conservative if every sentence of the language of T that is a theorem of T ′
is a theorem of T . We next prove a theorem about conservative extensions.
20.6 Theorem. Let L 0 , L 1 , L 2 be languages, with L 0 = L 1 ∩ L 2 . Let T i be a theory
in L i for i = 0, 1, 2. Let T 3 be the set of sentences of L 1 ∪ L 2 that are consequences
of T 1 ∪ T 2 . Then if T 1 and T 2 are both conservative extensions of T 0 , then T 3 is also a
conservative extension of T 0 .
Proof: Suppose B is a sentence of L 0 that is a theorem of T 3 . We must show that
B is a theorem of T 0 . Let U 2 be the set of sentences of L 2 that are consequences of
T 2 ∪{∼B}. Since B is a theorem of T 3 , T 1 ∪ T 2 ∪{∼B} is unsatisfiable, and therefore
T 1 ∪ U 2 is unsatisfiable. Therefore there is a sentence D in T 1 whose negation ∼D
is in U 2 . D is a sentence of L 1 , and ∼D of L 2 . Thus D and ∼D are both in L 0 , and
hence so is (∼B →∼D). Since D is in T 1 , which is a conservative extension of T 0 ,
D is in T 0 . And since ∼D is in U 2 , (∼B →∼D) isinT 2 , which is a conservative
extension of T 0 . Thus (∼B →∼D) is also in T 0 , and therefore so is B, which follows
from D and (∼B →∼D).
An immediate consequence is
20.7 Corollary (Robinson’s joint consistency theorem). Let L 0 , L 1 , L 2 be languages,
with L 0 = L 1 ∩ L 2 . Let T i be a theory in L i for i = 0, 1, 2. If T 0 is complete, and T 1 and
T 2 are satisfiable extensions of T 0 , then T 1 ∪ T 2 is satisfiable.