Computability and Logic

176 PROOFS AND COMPLETENESS
new interpretation and hence will have been true in old interpretation. This suffices
to show that Ɣ ∪{∃xA(x)}⇒ is secure.
For (R7), we suppose Ɣ ∪{s = s}⇒ is secure, and consider any interpretation
of a language containing all symbols in Ɣ and that makes all sentences in Ɣ true.
If there is some symbol in s not occurring in Ɣ or to which this interpretation fails
to assign a denotation, alter it so that it does. The new interpretation will still make
every sentence in Ɣ true by extensionality, and will make s = s true. By the security
of Ɣ ∪{s = s}⇒, the new interpretation will make some sentence in true, and
extensionality implies that the original interpretation already made this same sentence
in true. This suffices to show that Ɣ ⇒ is secure.
For (R8a), we suppose Ɣ ⇒{A(t)}∪ is secure and consider any interpretation
making all sentences in Ɣ ∪{s = t} true. Since it makes every sentence in Ɣ true, by
the security of Ɣ ⇒{A(t)}∪ it must make some sentence in {A(t)}∪ true. If
this sentence is one of those in , then clearly the interpretation makes a sentence
in {A(s)}∪ true. If the sentence is A(t), then note that since the interpretation
makes s = t true, it must assign the same denotation to s and to t, and therefore by
the must also make A(s) true by extensionality. Thus again it makes some sentence
in {A(s)}∪ true. This suffices to show that Ɣ ∪{s = t}⇒{A(s)}∪ is secure.
(R8b) is entirely similar.
(R9) is just like (R2), to finish the proof of soundness.
Now, for completeness, Theorem 14.2, according to which every secure sequent
is derivable. We begin with a quick reduction of the problem. Write ∼ for the set
of negations of sentences in .
14.14 Lemma. Ɣ ⇒ is derivable if and only if Ɣ ∪∼ is inconsistent.
Proof: If
is derivable, then
{C 1 ,...,C m }⇒{D 1 ,...,D n }
{C 1 ,...,C m , ∼D 1 }⇒{D 2 ,...,D n }
{C 1 ,...,C m , ∼D 1 , ∼D 2 }⇒{D 3 ,...,D n }
.
{C 1 ,...,C m , ∼D 1 ,...,∼D n }⇒∅
are derivable by repeated application of (R2b). If the last of these is derivable, then
{C 1 ,...,C m , ∼D 2 ,...,∼D n }⇒{D 1 }
{C 1 ,...,C m , ∼D 3 ,...,∼D n }⇒{D 1 , D 2 }
are derivable by repeated application of (R9a).
.
{C 1 ,...,C m }⇒{D 1 ,...,D n }