Computability and Logic

14.3. OTHER PROOF PROCEDURES AND HILBERT’S THESIS 181
to which may be added the step
Ɣ ∪{s = t}⇒{B(s)}∪
which follows by (R8a), and again we have no counterexample.
14.16 Corollary. Any sequent derivable using (R0)–(R9) is in fact derivable using only
(R0)–(R8).
Proof: Suppose there were a counterexample, that is, a derivation using (R0)–(R9)
the last step Ɣ ⇒ of which was not derivable using just (R0)–(R8). Among all such
derivations, consider a derivation that is as short as possible for a counterexample.
Ɣ ⇒ is not of the form {A} ⇒{A}, since any sequent of that form can be derived
in one step by (R0). So in the sequent Ɣ ⇒ is inferred by from one or more
premisses appearing as earlier steps. Since the derivation down to any earlier step is
too short to be a counterexample, for each premiss there is a derivation of it using
just (R0)–(R8). If there is only one premiss, let 0 be such a derivation of it. If there
are more than one premiss, let 0 be the result of concatenating such a derivation
for each premiss, writing one after the other. In either case, 0 is a derivation using
only (R0)–(R8) that includes any and all premisses among its steps. Let ′ be the
derivation that results on adding Ɣ ⇒ as one last step, inferred by the same rule as
in . If that rule was one of (R0)–(R8), we have a derivation of Ɣ ⇒ using only
(R0)–(R8). If the rule was (R9a), then is of the form {A}∪ ′ , where we have a
derivation of Ɣ ∪{∼A}⇒ ′ using only (R0)–(R8). In that case, the inversion lemma
tells us we have a derivation of Ɣ ⇒ , that is, of Ɣ ⇒{A}∪ ′ , using only (R0)–
(R8). Likewise if the rule was (R9b). So in any case, we have a derivation of Ɣ ⇒
using only (R0)–(R8), and our original supposition that we had a counterexample has
led to a contradiction, completing the proof.
14.17 Corollary. The proof procedure consisting of rules (R0)–(R8) is sound and
complete.
Proof: Soundness is immediate from the soundness theorem for (R0)–(R9), since
taking away rules cannot make a sound system unsound. Completeness follows from
completeness for (R0)–(R9) together with the preceding corollary.
(R10)
Instead of dropping (R9), one might consider adding the following.
Ɣ ⇒{(A → B)}∪
Ɣ ⇒{A}∪
.
Ɣ ⇒{B}∪
14.18 Lemma (Cut elimination theorem). Using (R0)–(R9), if there are derivations of
Ɣ ⇒{(A → B)}∪ and of Ɣ ⇒{A}∪, then there is a derivation of Ɣ ⇒{B}∪.
14.19 Corollary. Any sequent derivable using (R0)–(R10) is in fact derivable using
only (R0)–(R9).
14.20 Corollary. The proof procedure consisting of rules (R0)–(R10) is sound and
complete.