Computability and Logic

180 PROOFS AND COMPLETENESS
possible, so that no strictly shorter derivation would be a counterexample. So suppose
that Ɣ ∪{∼A}⇒ is the sequent derived in such a shortest possible counterexample.
We ask by what rule the last step Ɣ ∪{∼A}⇒ could have been justified.
Could it have been (R0)? If that were so, the counterexample would simply be
the one-step derivation of {∼A}⇒{∼A}, and we would have Ɣ = ∅, ={∼A}.
The sequent Ɣ ⇒{A}∪ for which supposedly no derivation exists would then just
be ⇒{A, ∼A}. But there is a derivation of this sequent, in two steps, starting with
{A}⇒{A} by (R0) and proceeding to ⇒{A, ∼A} by (R2a). So (R0) is excluded,
and Ɣ ∪{∼A}⇒ must have been inferred from some earlier step or steps by one
of the other rules.
Could it have been (R3)? If that were so, the counterexample would be a derivation
of
where the last step was obtained from
Ɣ ∪{∼A}⇒{(B ∨ C)}∪ ′
Ɣ ∪{∼A}⇒{B, C}∪ ′ .
But then, since the derivation down to this last-displayed sequent is too short to be a
counterexample, there will be a derivation of
and by applying (R3) we can then get
Ɣ ⇒{A}∪{B, C}∪ ′ ,
Ɣ ⇒{A}∪{(B ∨ C)}∪ ′ ,
which is precisely what we are supposed not to be able to get in the case of a
counterexample to the lemma. Thus (R3) is excluded. Moreover, every case where
∼A is not an entering sentence is excluded for entirely similar reasons.
There remain to be considered three cases where ∼A is an entering sentence. One
case where ∼A enters arises when Ɣ ∪{∼A}⇒ is obtained by (R1) from Ɣ ′ ⇒ ′ ,
where Ɣ ′ is a subset of Ɣ not containing ∼A and ′ is a subset of . But in this case
Ɣ ⇒{A}∪ equally follows by (R1) from Ɣ ′ ⇒ ′ , and we have no counterexample.
If ∼A enters when Ɣ ∪{∼A}⇒ is obtained by (R2b), the premiss must be
Ɣ ⇒{A}∪ itself or Ɣ ∪{∼A}⇒{A}∪, and in the latter case, since the derivation
of the premiss is too short to be a counterexample, there must exist a derivation
of Ɣ ⇒{A}∪{A}∪ or Ɣ ⇒{A}∪; so we have no counterexample.
The other case where ∼A enters arises when ∼A is of the form ∼B(s) and the last
lines of the derivation are
using (R8b).
Ɣ ∪{∼B(t)}⇒
Ɣ ∪{s = t, ∼B(s)}⇒