logic 3-13 - Def. is formally complete just in case for all...

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- Midterm 3/27 - Problem set 5 due 3/25 - Review problems on the web by Sunday CT0: CT1: Def. A set of sentences Γ is formally consistent just in case Γ ├/-- ┴ (just in case you can’t derive falsum from gamma.) Otherwise, we call Γ formally inconsistent . Lemma: Gamma proves S ( Γ├ S) iff Γv (~S) is formally inconsistent. Cor 1: S is not derived from gamma (S is not a logical consequence of gamma) iff Γ v ~S is formally consistent. CT2: Γ v ~S is formally consistent then Γ╞/= S iff there is a valuation v such that v*(Γ) = T and v*(S)=F. PLAN TO PROVE COMPLETENESS THEOREM: 1. Suppose we have gamma and S such that S is a logical consequence of gamma (Γ╞ S). 2. Suppose for a contradiction that you couldn’t prove S from gamma. Hence Γ v ~S is formally consistent. 3. Extend Γ v ~S to a set Γ* which has as a subset, Γ v ~S such that Γ* is not complete. 4. Use Γ* to define a valuation v such that v*(Γ*) = T. 5. Observe that v*(S)=T and v*(~S)=T ==> v*(S)=F.
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Unformatted text preview: Def. is formally complete just in case for all S Sent PL ,(a sentence in propositional logic) either S or /-- S. Lemma 2: Suppose is formally complete and formally consistent. Then: 1. Gamma proves P and Q iff gamma proves P and gamma proves Q. 2. Gamma proves P or Q iff gamma proves P or gamma proves Q. 3. Gamma proves not Q if gamma does not prove Q. 4. Gamma proves P Q iff does prove P or proves Q. 5. proves PQ iff either gamma proves P and gamma proves Q or gamma does not prove P and gamma does not prove Q. Lemma 3: Suppose gamma is formally complete and formally consistent. Then there exists valuation v such that v*() = T. Lemma 4: If gamma is formally complete iff A or gamma does not prove A for all i N. Lemma 5: Every formally consistent set gamma can be extended to some set * and is a subset of * and * is formally complete. FINALLY result is the COMPLETENESS THEOREM...
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logic 3-13 - Def. is formally complete just in case for all...

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