booleanboolean-logicboolean-expressionboolean-operationsboolean-algebra

Any one can solve this: Simplify the boolean expression Z=A+A'B + A'B'C+ A'B'C'D

Any one can solve this: Simplify the boolean expression Z=A+A'B + A'B'C+ A'B'C'D What will be the final answer of this question.

Solution 
  • Z = A + A'B + A'B'C + A'B'C'D <br>
Z = A + A'(B + B'C + B'C'D)        (distributivity) 
Z = A + A'(B + B'(C + C'D))       (distributivity) 
Z = A + A'(B + B'(C(D+1) + C'D))  (null law) 
Z = A + A'(B + B'(CD + C + C'D))  (distributivity) 
Z = A + A'(B + B'(C + D(C + C'))) (distributivity) 
Z = A + A'(B + B'(C + D))         (inverse law) 
Z = A + A'(B + C + D)             (same 4 steps applied above) 
Z = A + B + C + D                 (same as above)

    So, whole expression actually can be solved with this rule:
    A + A'B = A + B (absorption law)