Question

(a) Simplify the Boolean function F= A'B'C'+B'CD' + A'BCD' + AB'C'
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Answer 1

Answer:

I'm assuming A′ is notation for "not A". Observe that you can factor out AC from several terms (the second, third, fifth, and sixth):

A′BCD+AB′CD′+AB′CD+ABC′D+ABCD′+ABCD=AC(B′D′+B′D+BD′+BD)+A′BCD+ABC′D

Then replace 1=(B′+B)(D′+D)=(B′D′+B′D+BD′+BD) to get

=AC+A′BCD+ABC′D.

Finally, observe that AC+A′BCD=AC+BCD, because if C=B=D=1, then either A=1 so AC=1 or A=0 so A′BCD=1. Similarly AC+ABC′D=AC+ABD because if A=B=D=1, then either C=1 so AC=1 or C=0 so ABC′D=1. So make two final substitutions to get

=AC+BCD+ABD.

Explanation:

PAWAN KULORA