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l5_6soln

# l5_6soln - CS 245 Winter 2009 Solution Set for Lectures 5 6...

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CS 245 Winter 2009 Solution Set for Lectures 5 & 6 Nissanke Ch. 5 Exercise 5.1 (p. 65) 1. p q , ¬ p Therefore, ¬ q Counterexample: v ( p ) = F ,v ( q ) = T Premises: v ( p q ) = v ( p ) OR v ( q ) = F OR T = T v ( ¬ p ) = NOT v ( p ) = NOT F = T Conclusion: v ( ¬ q ) = NOT v ( q ) = NOT T = F 2. p q , p r , r Therefore, p Counterexample: v ( p ) = F ,v ( q ) = F ,v ( r ) = T Premises: v ( p q ) = v ( p ) IFF v ( q ) = F IFF F = T v ( p r ) = v ( p ) IMP v ( r ) = F IMP T = T v ( r ) = T Conclusion: v ( p ) = F 1

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3. p q , q Therefore, p Counterexample: v ( p ) = F ,v ( q ) = T Premises: v ( p q ) = v ( p ) OR v ( q ) = F OR T = T v ( q ) = T Conclusion: v ( p ) = F 4. p q , q p Therefore, p q Counterexample: v ( p ) = F ,v ( q ) = F Premises: v ( p q ) = v ( p ) IMP v ( q ) = F IMP F = T v ( q p ) = v ( q ) IMP v ( p ) = F IMP F = T Conclusion: v ( p q ) = v ( p ) AND v ( q ) = F AND F = F 2
Exercise 5.3 (p. 76) A B,C D,B D E, ¬ E | = ND ¬ A ∧ ¬ C 1 A B premise 2 C D premise 3 B D E premise 4 ¬ E premise 5 A assumption 6 B E 1 , 5 7 B D I 6 8 E E 3 , 7 9 false ¬ E 4 , 8 10 ¬ A RAA 5 - 9 11 C assumption 12 D E 2 , 11 13 B D I 12 14 E E 3 , 13 15 false ¬ E 4 , 14 16 ¬ C RAA 11 - 15 17 ¬ A ∧ ¬ C I 10 , 16 Exercise 5.4 (p. 76) 1. A ⇒ ¬ B, ( B C ) D, ¬ C D A, ¬ C | = ND D 1 A ⇒ ¬ B premise 2 ( B C ) D premise 3 ¬ C D A premise

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l5_6soln - CS 245 Winter 2009 Solution Set for Lectures 5 6...

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