Simplify the expression : A'B'C' + A'B'C + A'B C + A'B C' +AB C'
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Explanation:
From your last line, as you "prematurely grouped" the A′BC term into A′B, the absorption law can "recreate" an A′BC out of A′B:
p=A′B+C(A′B′+AB′+AB)=(A′B+A′BC)+C(A′B′+AB′+AB)=A′B+C(A′B+A′B′+AB′+AB)=A′B+C(A′+A)(B′+B)=A′B+C
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