Computability and Logic

14.2. SOUNDNESS AND COMPLETENESS 177
Since it is easily seen that Ɣ secures if and only if Ɣ ∪∼ is unsatisfiable,
proving that if Ɣ secures , then Ɣ ⇒ is derivable, which is what we want to do,
reduces to showing that any consistent set is satisfiable. (For if Ɣ secures , then
Ɣ ∪∼ is unsatisfiable, and supposing we have succeeded in showing that it would
be satisfiable if it were consistent, it follows Ɣ ∪∼ is inconsistent, and so by the
preceding lemma Ɣ ⇒ is derivable.) By the main lemma of the preceding chapter,
in order to show every consistent set is satisfiable, it will suffice to show that the set S
of all consistent sets has the satisfiability properties (S0)–(S8). (For any consistent
set Ɣ will by definition belong to S, and what Lemma 13.3 tells us is that if S has the
satisfaction properties, then any element of S is satisfiable.) This we now proceed to
verify, recalling the statements of properties (S0)–(S8) one by one as we prove S has
them.
Consider (S0). This says that if Ɣ is in S and Ɣ 0 is a subset of Ɣ, then Ɣ 0 is in S.
So what we need to prove is that if Ɣ ⇒ ∅ is not derivable, and Ɣ 0 is a subset of Ɣ,
then Ɣ 0 ⇒ ∅ is not derivable. Contraposing, this is equivalent to proving:
(S0) If Ɣ 0 ⇒ ∅ is derivable, and Ɣ 0 is a subset of Ɣ, then Ɣ ⇒ ∅ is derivable.
We show this by indicating how to extend any given derivation of Ɣ 0 ⇒ ∅ to a
derivation of Ɣ ⇒ ∅. In fact, only one more step need be added, as follows:
.
Ɣ 0 ⇒ ∅
Ɣ ⇒ ∅.
Given
(R1)
(Here the three dots represent the earlier steps of the hypothetical derivation of
Ɣ 0 ⇒ ∅.)
For each of (S1)–(S8) we are going to give a restatement, in contraposed form, of
what is to be proved, and then show how to prove it by extending a given derivation
to a derivation of the sequent required. First (S1)
(S1) If A and ∼A are both in Ɣ, then Ɣ ⇒ ∅ is derivable.
The hypothesis may be restated as saying that {A, ∼A} is a subset of Ɣ. We then have
{A} ⇒{A}
{A, ∼A} ⇒∅
Ɣ ⇒ ∅.
(R0)
(R2a)
(R1)
As for (S2), literally, this says that:
(S2) If Ɣ ⇒ ∅ is not derivable and ∼∼B is in Ɣ, then Ɣ ∪{B}⇒∅ is not derivable.
Contraposing, this says that if Ɣ ∪{B}⇒∅ is derivable and ∼∼B is in Ɣ, then
Ɣ ⇒ ∅ is derivable. What we actually show is that if Ɣ ∪{B}⇒∅ is derivable, then