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Unformatted text preview: S. Therefore E i E j = for i j and S E n i i = = U 1 . We need to show that ( E i B ) ( E j B ) = for i j and U n i i B E 1 ) ( = = B . Note that ( E i B ) ( E j B ) = E i E j B = B since E i E j = for i j = for i j and U n i i B E 1 ) ( = = B E n i i = U 1 by the distributive law = S B since S E n i i = = U 1 = B . Problem 2 Problem 3 Problem 4 Important: Let S = S c , G = G c , F = F c , ( S G F ) = ( S G F ) c ....
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This note was uploaded on 10/25/2010 for the course ISE 220 taught by Professor Ross during the Fall '05 term at USC.
 Fall '05
 Ross

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