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Simplify the following Boolean expression. Your final equation should be in minimized sum-of-products form.

W = B' (A + BC + B'CD + A'D)

So far I have:

Use distributing:

AB' + B'BC + B'B'CD + A'B'D

Complementary:

AB' + 0 + B'B'CD + A'B'D

Is this right so far? What's next?

user9419099
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1 Answers1

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So far so good.

Next,
\begin{align} AB' + B'CD + A'B'D &= B'(A + CD + A'D)\\ &= B'(A + (A+A')CD + A'D)\\ &= B'((A+ACD) + (A'CD + A'D))\\ &= B'(A + A'D)\tag{Absortion laws}\\ &= B'((A+A')(A+D))\tag{Distributivity}\\ &= B'(A+D). \end{align} I suppose that the equalities that I didn't justify are quite immediate.

Edit. In sum of products, just distribute again, to get $AB' + B'D$.

amrsa
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