Simplify A'B+A'C'+A'b'C+ABC
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Answer:Simplify A'B+A'C'+A'b'C+ABC
Explanation:Assuming this to be a well defined Boolean expression, we have
ABC+AB′C+ABC′
=AC(B+B′)+ABC′=AC.I+ABC′=AC+ABC′ …since X.X′=I
=A(C+BC′)=A(C+B)(C+C′)=A(C+B)I=A(B+C) .. ….since both operations are distributive over one another in Boolean algebra.
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Answer:
Explanation:
AB+A′C+BC=AB+A′C+BC(A+A′)(A+A′=1, Complementarity law)=AB+A′C+ABC+A′BC=AB+ABC+A′C+ABC(Associative law)=AB+A′C(Absorption law)
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