Computability and Logic

178 PROOFS AND COMPLETENESS
whether or not ∼∼B is in Ɣ, Ɣ ∪ {∼∼B}⇒∅ is derivable. In case ∼∼B is in Ɣ,we
have Ɣ ∪ {∼∼B} =Ɣ, so what we actually show is something a little more general
than what we need:
.
Ɣ ∪{B} ⇒∅
Given
Ɣ ⇒{∼B}
(R2b)
Ɣ ∪ {∼∼B} ⇒∅.
(R2a)
Analogous remarks apply to (S3)–(S8) below.
(S3) If Ɣ ∪{B} ⇒∅ and Ɣ ∪{C} ⇒∅ are both derivable, then Ɣ ∪{B ∨ C} ⇒∅
is derivable.
Here we concatenate the two given derivations, writing one after the other:
.
Ɣ ∪{B} ⇒∅
.
Ɣ ∪{C} ⇒∅
Ɣ ∪{B ∨ C} ⇒∅.
Given
Given
(R4)
(S4) If either Ɣ ∪{∼B}⇒∅ or Ɣ ∪{∼C}⇒∅ is derivable, then
Ɣ ∪{∼(B ∨ C)}⇒∅ is derivable.
The two cases are exactly alike, and we do only the first:
.
Ɣ ∪{∼B} ⇒∅
Ɣ ⇒{B}
Ɣ ⇒{B, C}
Ɣ ⇒{B ∨ C}
Ɣ ∪{∼(B ∨ C)} ⇒∅.
Given
(R9a)
(R1)
(R3)
(R2a)
(S5) If Ɣ ∪{B(c)}⇒∅ is derivable, where c does not occur in Ɣ ∪{∃xB(x)}, then
Ɣ ∪{∃xB(x)}⇒∅ is derivable:
.
Ɣ ∪{B(c)} ⇒∅
Ɣ ∪{∃xB(x)} ⇒∅.
Given
(R6)
Note that the hypothesis that c does not occur in Ɣ or ∃xB(x) (nor of course in ∅)
means that the side conditions for the proper application of (R6) are met.