Lecture8

Lecture8 - Lecture 8 Completeness of Propositional...

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The completeness theorem Γ | ± ϕ if and only if Γ |= ϕ Last class we proved the direction: if Γ | ± ϕ then Γ |= ϕ . In other words, we proved that the formal system of natural deduction with the rules ˭ I, ˭ E, I, E, , and RAA is sound. Today we are going to start the proof of the hard direction: if Γ |= ϕ then Γ | ± ϕ Lecture 8 Completeness of Propositional Logic—Part II

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The hard direction: if Γ |= ϕ then Γ | ± ϕ The proof does not proceed by assuming Γ |= ϕ and then showing that there is a derivation witnessing Γ | ± ϕ . Instead, the contrapositive of the if-then statement is proven. We assume that ϕ is not derivable from Γ and then show that there is valuation that satisfies Γ but does not satisfy ϕ . Before the plan for doing this can be sketched, we need one more lemma. Lemma 1.5.5 a) If Γ U { ϕ } is inconsistent, then Γ | ± ¬ ϕ . b) If Γ U { ¬ ϕ } is inconsistent, then Γ | ± ϕ . ϕ Proof D Statement a): If Γ U { ϕ } is inconsistent, there is a derivation whose hypotheses are contained in Γ U { ϕ }. Applying the rule of I to the derivation we obtain a derivation confirming Γ | ± ¬ ϕ . Statement b): Same reasoning, except that we apply the RAA rule to a given derivation to get ϕ .
The hard direction: if Γ |= ϕ then Γ | ± ϕ What this lemma tells us is that if not Γ | ± ϕ then the set Γ U { ¬ ϕ } is consistent (the contrapositive again). This fits into proving the hard direction as follows. We prove the statement if Γ |= ϕ then Γ | ± ϕ if we prove the statement if not Γ | ± ϕ then not Γ |= ϕ Now, by the lemma, we have proved this statement once we have proved if Γ U { ¬ ϕ } is consistent then not Γ |= ϕ

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if Γ |= ϕ then Γ | ± ϕ If not Γ | ± ϕ then the set
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Lecture8 - Lecture 8 Completeness of Propositional...

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