Logic1012 - all5 and some 1 2 3 HW#4 Puzzles involving all...

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all and some 1. HW #4 2. Puzzles involving all and some 3. The Barber Paradox
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Midterm Monday 19th 9pm Covers Smullyan, pp.1 – 119 (up to Wednesday’s lecture) HW #5 due Wednesday 1. 11.1(j) on p.90 2. 11.3(b) on p.94 3. For synthetic tableaux, show: if X=>Y and ~X=>Y are provable, so is Y.
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HW #4, Q3 There are quite a few errors in the solu- tion (p.82). Case 1: “T” is E -related to X and X. Case 3: X’ is E -related to X and T’. The following is not covered by Cases 1- 4. X and X’: T’ is E -related to X and X’ (if you discovered this, that’s extra 2 points).
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HW #4, Q2. is not definable from =>. From the truth table, the # of F’s in p q p q p=>q T T T T T F F F F T F T F F F T
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Claim: The # of F’s in a formula con- sisting of p, q, => is 0-2. Case 1: the # of F’s in p and q is 2, re- spectively. Case 2: Assume that the # of F’s in X and Y is 0-2. Then, the # of F’s in X=>Y the # of F’s in Y = 0-2. By Cases 1 & 2, the claim is proved.
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Students’ solutions that don’t work Every possible combination of p, q, => has one of these truth tables. [that’s begging the question . Since this claim is exactly what has to be shown.] p q p=> p p=> q q=> p (q=>p) =>q … or some more T T T T T T T F T F T T F T T T F T F F T T T F
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Begging the question #2: Assume
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Logic1012 - all5 and some 1 2 3 HW#4 Puzzles involving all...

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