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Simplify the following Boolean expression :(i) AB + AB’+ A’C + A’C’(ii) XY + XYZ’ + XYZ’ + XZY(iii) XY(X’YZ’+ XY’Z’+ XY’Z’) |
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Answer» (i) AB + AB’ + A’C + A’C’ = A(B + B’) + A’(C + C’) (B + B’ =1, C + C’ = 1) = A + A’ (A + A’ = 1) = 1 (ii) XY + XYZ’ + XYZ’ + XZY = XY(Z’) + XY(Z’ + Z) (Z + Z’ =1) = XY(Z’) + XY = XY(Z’ + 1) (Z’ + 1 = 1) = XY (iii) XY(X’YZ’ + XY’Z’ + XY’Z’) = XY[Z’(X’Y + XY’ + XY’)] = XY[Z’(X’Y + XY’(1 + 1)] = XY[Z’(X’Y + XY’)] = XYZ’(X’Y + X Y’) |
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