derive the boolean equation for A=B when A and B are 4- bit numbers Solution Dis

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Distributing (ie. multiplying): x = a+((a'b+bc)+c) + abd'+ abe' x = a + a'b+ bc + c + abd'+ abe' Group, factor, and reduce: x = a + a'b+ (bc + c) + abd'+ abe' x = a + a'b+ c(b + 1) + abd'+ abe' x = a + a'b+ c(1) + abd'+ abe' x = a + a'b+ c + abd'+ abe' Reorder, Group, factor, and reduce: x = a + a'b+ c + abd'+ abe' x = a'b+ c + a + abd'+ abe' x = a'b+ c + (a + abd'+ abe') x = a'b+ c + a(1 + bd'+ be') x = a'b+ c + a(1) x = a'b+ c + a Reorder: x = a + a'b+ c Expand "a": x = a(1) + a'b + c x = a(1 +b) + a'b + c x = a + ab + a'b + c Group, factor, reduce: x = a + (ab + a'b) + c x = a + b(a + a') + c x = a + b(1) + c x = a + b + c The truth tables for these are: a b c d e ==> x x(reduced) 0 0 0 0 0 ==> 0 0 0 0 0 0 1 ==> 0 0 0 0 0 1 0 ==> 0 0 0 0 0 1 1 ==> 0 0 0 0 1 0 0 ==> 1 1 0 0 1 0 1 ==> 1 1 0 0 1 1 0 ==> 1 1 0 0 1 1 1 ==> 1 1 0 1 0 0 0 ==> 1 1 0 1 0 0 1 ==> 1 1 0 1 0 1 0 ==> 1 1 0 1 0 1 1 ==> 1 1 0 1 1 0 0 ==> 1 1 0 1 1 0 1 ==> 1 1 0 1 1 1 0 ==> 1 1 0 1 1 1 1 ==> 1 1 1 0 0 0 0 ==> 1 1 1 0 0 0 1 ==> 1 1 1 0 0 1 0 ==> 1 1 1 0 0 1 1 ==> 1 1 1 0 1 0 0 ==> 1 1 1 0 1 0 1 ==> 1 1 1 0 1 1 0 ==> 1 1 1 0 1 1 1 ==> 1 1 1 1 0 0 0 ==> 1 1 1 1 0 0 1 ==> 1 1 1 1 0 1 0 ==> 1 1 1 1 0 1 1 ==> 1 1 1 1 1 0 0 ==> 1 1 1 1 1 0 1 ==> 1 1 1 1 1 1 0 ==> 1 1 1 1 1 1 1 ==> 1 1 Finally, to verify another way, compare to the output of an automated system which can reduce booleans. Using wolframalpha.com, the input expression

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