midterm-2008-sol - York University Department of...

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York University Department of Mathematics and Statistics Faculty of Science and Engineering MATH1090 A. Mid Term Test, October 28, 2008 Solutions Professor George Tourlakis Question 1. (a) (2 MARKS) Through truth tables or related short cuts show that | = taut ± A B C D ² ± A ( B ( C D )) ² ( i ) Answer. Take a state v . We want the formula truth-value to be computed as t . There are two cases: (1) v ± A B C D ² = f . Then v ( D ) = f ( ii ) and v ( A B C ) = t hence v ( A ) = v ( B ) = v ( C ) = t ( iii ) The truth-value of the right-hand side of ( i ) is computed immediately below. The truth- values of the subformulae are computed in the order (1)–(3) based on ( ii ) and ( iii ) and are written —as is customary— under the corresponding “ ”. A ( B ( C D )) f f f (3) (2) (1) Hence both sides of ( i ) are f . OK! (2) v ± A B C D ² = t . There are two sub-cases, (I) and (II): (I) v ( D ) = t . From the table for “ ” I obtain immediately A ( B ( C D )) t t t (3) (2) (1) Hence both sides of ( i ) are t . OK! (II) v ( D ) = f . Then v ( A B C ) = f hence at least one of A,B,C is f . If it is C , then C D is t . But then, as in sub-case (I), B ( C D ) is t
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This note was uploaded on 05/09/2010 for the course LAPS CSE 1090 taught by Professor Peter during the Winter '10 term at York University.

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midterm-2008-sol - York University Department of...

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