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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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This homework help was uploaded on 01/30/2008 for the course ICS 6D taught by Professor Jarecki during the Fall '07 term at UC Irvine.
 Fall '07
 Jarecki

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