MIT24_241F09_lec11

# MIT24_241F09_lec11 - Logic I Session 11 Plan for today...

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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 TF-completeness 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 TF-completeness We can express any truth-function 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). TF-completeness Strictly, we haven’t yet proven that SL is TF-complete. 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 truth-functionally complete, but so is any language that contains formulae TF-equivalent to every sentence of SL. E.g. {&,v, ∼ }. (After all, that’s all we use in our algorithm!) In fact, we can achieve TF-completeness with a single binary connective, ‘|’. P Q P | Q T T F T F T F T T F F T TF-completeness 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 truth-function. In our example, our algorithm generated (A&B) v ( ∼ A&B). To ¡nd an equivalent sentence, make replacements in stages. TF-completeness with ‘|’ We start with (A&B) v ( ∼ A&B) , which is of the form PvQ. Now, PvQ iff (P|P) | (Q|Q). 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 sub-sentences. (A&B) iff (A|B)|(A|B) . And ( ∼ A&B) iff ((A|A)|B)|((A|A)|B)....
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## This note was uploaded on 01/25/2012 for the course PHIL 201H1F taught by Professor Derekallen during the Fall '10 term at University of Toronto.

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MIT24_241F09_lec11 - Logic I Session 11 Plan for today...

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