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Unformatted text preview: Logic I  Session 11 Plan for today Damien’s comments on quiz My comments on teaching feedback A bit more on the TFcompleteness of SL Recap of proof of soundness of SD: If Γ ⊢ P in SD, then Γ ⊨ P Begin to prove completeness of SD: If Γ ⊨ P , then Γ ⊢ P in SD TFcompleteness We can express any truthfunction in SL. Find a sentence that expresses the TF for this TT schema: T T T T F F F T T F F F A&B A& ∼ B ∼ A&B ∼ A& ∼ B We want an iterated disjunction of CSs for the T rows: 1 and 3. (A&B) v ( ∼ A&B). TFcompleteness Strictly, we haven’t yet proven that SL is TFcomplete. We’d need to show that our algorithm always yields a sentence that expresses the truth function we want. See 6.1E (1d) and 6.2E (1). Not only is SL truthfunctionally complete, but so is any language that contains formulae TFequivalent to every sentence of SL. E.g. {&,v, ∼ }. (After all, that’s all we use in our algorithm!) In fact, we can achieve TFcompleteness with a single binary connective, ‘’. P Q P  Q T T F T F T F T T F F T TFcompleteness with ‘’ To see this, just add a step to our algorithm: translate the old sentence into one that only contains ‘’. The new one will be equivalent, so it will have the same TT, so it will expresses the same truthfunction. In our example, our algorithm generated (A&B) v ( ∼ A&B). To ¡nd an equivalent sentence, make replacements in stages. TFcompleteness with ‘’ We start with (A&B) v ( ∼ A&B) , which is of the form PvQ. Now, PvQ iff (PP)  (QQ). Substitute (A&B) and ( ∼ A&B) for P and Q (A&B) v ( ∼ A&B) ( (A&B)  (A&B) )  ( ( ∼ A&B)  ( ∼ A&B) ) Now replace the remaining subsentences. (A&B) iff (AB)(AB) . And ( ∼ A&B) iff ((AA)B)((AA)B)....
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 Fall '10
 DerekAllen
 Philosophy, Logic, Assumption of Mary, TVA, Γ*

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