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Unformatted text preview: M325K HW 02 - Solutions to Graded Questions (Out of 25 points) Section 1.3 23. (4pts) Its form: q p ∨ q p → r q ~ ∨ ∴ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ → ← premises ⎯ ⎯ ⎯ → ← conclusion 28 . (2pts) Its form is: q p → Invalid, a converse error. q p ∴ 38. (5pts) c) Answer: There is only one knave and only one knight. There are 4 cases to consider, namely a) Both are knights c) E is a knight and F is a knave. b) Both are knaves d) E is a knave and F is s knight. Observe that (a) is not possible, because then both would also be telling the truth and both would also be knaves. Similarly, (b) is not possible because in this case both would be telling the truth which would make them knights, a contradiction. Hence there is only one knight and only one knave. Note that in case c and d, both E and F would have spoken as reported, without a contradiction. It is OK if you combine c and d and consider it as “if one is a knight and the other is a knave”. But your solution should clearly explain/mention why this case is possible (does not give a contradiction)...
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This note was uploaded on 04/29/2010 for the course MATH m325k taught by Professor Seckin during the Spring '10 term at University of Texas-Tyler.
- Spring '10