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Unformatted text preview: ECE 190 Fall, 2006 Assignment 3 Solutions Section 1.3 36 See text. 42 from (b) and (d) and modus tollens, we can conclude q (1) From (1) and (a) and elimination, we can conclude p (2) From (e) and (1) and modus ponens, we can conclude u s (3) From (30 and specialization, we can conclude s (4) From (2) and (4) and conjunction, we can conclude p s (5) From (c) and (5) and modus ponens, we can conclude t , as required. Section 2.1 7 (b) Truth set is { 1 , 2 , 3 , 6 } (d) Truth set is { 2 , 1 , 1 , 2 } 10 Both a = 1, since (1 1) / 1 = 0 Z and a = 1, since ( 1 1) / ( 1) = 2 Z provide counterexamples 12 As one counterexample, we can take x = y = 1 giving x + y = 2 while x + y = 2 14 See text. 18 (a) See text. (b) See text. (c) s ( C ( s ) E ( s )) (d) s ( C ( s ) M ( s )) (e) See text. 19 See text. 28 (a) There is an integer that is prime and not odd. This is true: 2 is such a value....
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This homework help was uploaded on 04/19/2008 for the course ECE 190 taught by Professor Carter during the Fall '06 term at University of Toronto Toronto.
 Fall '06
 Carter

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