Lecture9

# Lecture9 - Lecture 9 Completeness of Propositional...

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We are half way through the proof of the hard direction of the completeness theorem. The hard direction, recall, is: if Γ |= ϕ then Γ | ± ϕ The proof proceeds by showing the contrapositive: If not Γ | ± ϕ then not Γ |= ϕ We know that if not Γ | ± ϕ then Γ U { ¬ ϕ } is consistent (Lemma 1.5.5). So we are done if we can show that for every consistent set of propositions there is a valuation satisfying it. This would mean that if not Γ | ± ϕ then Γ U { ¬ ϕ } has a valuation, which would in turn imply that not Γ |= ϕ . Lecture 9 Completeness of Propositional Logic—Part III

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The claim that every consistent set of propositions Γ has a valuation is lemma 1.5.11. There are two major steps in proving it: 1) Construct a maximally consistent set Γ ± from a consistent set of propositions Γ . 2) Show that for any maximally consistent set Γ ± there is a valuation satisfying it. Since the maximally consistent set Γ ± contains Γ , the valuation of 2) also satisfies Γ . Last class we did 1) (Lemma 1.5.7). We are now going to move on to 2). Proving lemma 1.5.11
Suppose that we have a maximally consistent set Γ ± . We need to define a valuation such that it satisfies Γ ± . As a valuation is defined if it is defined for each propositional symbol p i , a candidate definition presents itself right away: v (p i ) = 1 if p i is in Γ ± = 0 if p i is not in Γ ± By definition we have that [|p i |] v = 1 if and only if p i is in Γ ± . What we have to show is that [|

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Lecture9 - Lecture 9 Completeness of Propositional...

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