Logical Labyrinths

288 VI. More on First-Order Logic
Theorem 24.2. If theories T 1 and T 2 are conservative extensions of a theory T,
and if every sentence that is in the language of both T 1 and T 2 is in the language
of T, then the closure of T 1 ∪T 2 is also a conservative extension of T.
Proof: Assume the hypotheses. Now suppose that X is in the closure of
T 1 ∪T 2 and is in the language of T. We are to show that X is in T. Now,
X, being in (T 1 ∪T 2 ) ∗ , is a logical consequence of T 1 ∪T 2 and therefore
the set (T 1 ∪T 2 )∪{∼X} is not satisfiable (because ∼X is obviously a consequence
of this set, and no sentence and its negation can both be consequences
of a satisfiable set), and therefore the set T 1 ∪(T 2 ∪{∼X}) ∗ is unsatisfiable.
(Obviously , T 2 ∪{∼X} is a subset of its closure (T 2 ∪{∼X}) ∗ ;
hence T 1 ∪(T 2 ∪{∼X}) is a subset of T 1 ∪(T 2 ∪{∼X}) ∗ ; hence the set
T 1 ∪(T 2 ∪{∼X}) ∗ is unsatisfiable.) Therefore, by Theorem 22.1, there is a
sentence Y in T 1 whose negation is in (T 2 ∪{∼X}) ∗ . The sentence Y is in
the language of T 1 , hence so is ∼Y. The sentence ∼Y must also be in the
language of T 2 (because X, being in the language of T, is obviously in the
language of its extension T 2 , hence the languages of T 2 and T 2 ∪{∼X}
are the same, and ∼Y is certainly in the language of T 2 ∪{∼X}, being in
(T 2 ∪{∼X}) ∗ ). Thus ∼Y is in the language of both T 1 and T 2 , hence is in
the language of T (by hypothesis). Since X is also in the language of T,
so is ∼X, and therefore (∼X⇒∼Y) is in the language of T.
Now, ∼Y is a logical consequence of T 2 ∪{∼X} (being a member of
(T 2 ∪{∼X}) ∗ ); hence (∼X⇒∼Y) is a logical consequence of T 2 (by Problem
24.2!), and hence is a member of T 2 (since T 2 is logically closed). Thus
(∼X⇒∼Y) is in T 2 and also is in the language of T, hence is a member
of T (since T 2 is a conservative extension of T). Thus Y and (∼X⇒∼Y)
are both members of T, hence so is X (since X is a consequence of the set
{Y, (∼X⇒∼Y)}, and hence of the set T, and T is logically closed).
Problem 24.8. Prove the following two facts:
(1) Any conservative extension of a satisfiable theory is satisfiable.
(2) Any satisfiable extension of a complete theory is conservative.
Problem 24.9. Now complete the proof of Robinson’s Consistency Theorem.
Solutions
24.1. It is the second statement that is true: Suppose X is implied by S 1 .
Thus X is true under every interpretation that satisfies S 1 . Now
let I be any interpretation that satisfies S 2 . Then I obviously also