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

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Unformatted text preview: Logic I - Session 11 Plan for today Damiens 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 P in SD, then P Begin to prove completeness of SD: If P , then 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 havent yet proven that SL is TF-complete. Wed 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, thats 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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MIT24_241F09_lec11 - Logic I - Session 11 Plan for today...

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