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Unformatted text preview: ICS 6D ( Discrete Mathematic): Solution to Homework #2 Section 1.5 4. (a) simplification (b) disjunctive syllogism (c) modus ponens (d) addition (e) hypothetical syllogism 6. Let p be It rains, q be It is foggy, r be Sailing race is held, s be Lifesaving demonstration goes on and t be Trophy is awarded. Our premises are ( p q ) ( r s ), r t and t . And the conclusion is p . The following argument shows how to derive the truth of p . r t t : r r : r s ( p q ) ( r s ) ( r s ) : ( p q ) p q : p (Above we also used the DeMorgan Law twice: First that r s is equivalent to ( r s ), and then the same for p,q instead of r,s , which implies in particular that ( p q ) is equivalent to p q . 24. The error occurs in step 2. We cannot say that p ( c ) is true when p ( c ) q ( c ) is true. In fact this might look like simplification rule but note that simplification only applies for conjunction. Section 2.1 4. There are 4 4 different combination for subsets from which the following statements are true. A A , B A , B B , C A , C C , C D , D D . 8. (a) True, (b) True, (c) False, (d) True, (e) True, (f) True, (g) True. 22. (a) It is not power set, (b) Power set of { a } , (c) It is not power set, (d) Power set of { a,b } 34. (a) There exists a real number x , such that x 3 = 1. True (b) There exists an integer x , such that x + 1 > x . True (c) For every integer x , x 1 is an integer. True (d) For every integer x , x 2 is an integer. True Section 2.2Section 2....
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 Fall '07
 Jarecki

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