General Computer Science 320201 GenCS I & II Lecture ... - Kwarc

means that if the assumption of the Subst rule was valid, then the conclusion is valid as well, i.e.
the validity property is preserved.
Soundness of Substitution
Lemma 318 Subst preserves validity.
Proof: We have to show that [B/X](A) is valid, if A is.
P.1 Let A be valid, B a formula, ϕ: Vo → {T, F} a variable assignment, and ψ :=
ϕ, [Iϕ(B)/X].
P.2 then Iϕ([B/X](A)) = I ϕ,[Iϕ(B)/X](A) = T, since A is valid.
P.3 As the argumentation did not depend on the choice of ϕ, [B/X](A) valid and we have
proven the assertion.
c○: Michael Kohlhase 186
The next theorem shows that the implication connective and the entailment relation are closely
related: we can move a hypothesis of the entailment relation into an implication assumption in the
conclusion of the entailment relation. Note that however close the relationship between implication
and entailment, the two should not be confused. The implication connective is a syntactic formula
constructor, whereas the entailment relation lives in the semantic realm. It is a relation between
formulae that is induced by the evaluation mapping.
The Entailment Theorem
Theorem 319 If H, A |= B, then H |= (A ⇒ B).
Proof: We show that Iϕ(A ⇒ B) = T for all assignments ϕ with Iϕ(H) = T whenever
H, A |= B
P.1 Let us assume there is an assignment ϕ, such that Iϕ(A ⇒ B) = F.
P.2 Then Iϕ(A) = T and Iϕ(B) = F by definition.
P.3 But we also know that Iϕ(H) = T and thus Iϕ(B) = T, since H, A |= B.
P.4 This contradicts our assumption Iϕ(B) = T from above.
P.5 So there cannot be an assignment ϕ that Iϕ(A ⇒ B) = F; in other words, A ⇒ B is
valid.
c○: Michael Kohlhase 187
Now, we complete the theorem by proving the converse direction, which is rather simple.
The Entailment Theorem (continued)
Corollary 320 H, A |= B, iff H |= (A ⇒ B)
Proof: In the light of the previous result, we only need to prove that H, A |= B, whenever
H |= (A ⇒ B)
P.1 To prove that H, A |= B we assume that Iϕ(H, A) = T.
P.2 In particular, Iϕ(A ⇒ B) = T since H |= (A ⇒ B).
P.3 Thus we have Iϕ(A) = F or Iϕ(B) = T.
P.4 The first cannot hold, so the second does, thus H, A |= B.
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