# Simplify Boolean expression

Solve 4th problem.

Reposting the problem with the correct orientation

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Option (c) is the correct option
below is the solution

Using Boolean Algebra won’t give you the answer directly, I will show the result obtained using the Boolean Algebra

Z = ABCD + AB’CD + A’CB’D
Z = ACD(B+B’) + A’CB’D {taking ACD common from the first two terms}
Z = ACD + A’CB’D {X+X’ = 1}
Z = CD(A + A’B’) {taking CD common}
Z = CD(A+B’) {X + X’Y = X+Y, known as distributive law of Boolean Algebra}
Z = ACD + B’CD {Distributive law}

And, if you try to get the minimized result using the K’Map you will get the same result i.e. Z = ACD + B’CD

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Thanks
But how can i recognize such type of problem where Boolean alzebra not gives correct answer?

@chaituvarma there is mistake in your calculations, A + A’CB’D {=}\llap{/\,} CB’D
you said A + A’ is 1, which is true but you have A + A’CB’D

use distributive law
A + A’X = (A + A’)(A + X) { where X = CB’D}
A + A’X = 1.(A + X) {since A + A’ = 1}
A + A’CB’D = A + CB’D {substituting X = CB’D}

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