Question

A'B'CD+A'B'C'D+ABC'D'+ABCD'
Reduce the expression
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Answer 1

Answer:

= AC + BCD + ABD.

How to simplify the following expression :

A'BCD + AB'CD' + AB'CD + ABC'D + ABCD' + ABCD ?

It should get AC + BCD + ABD using Kmap but using boolean algebra i am stuck no matter how i try

Triplicate ABCD. then reorder:

A'BCD + AB'CD' + AB'CD + ABC'D + ABCD' + ABCD = (A'BCD + ABCD) + (AB'CD' + =

AB'CD + ABCD' + ABCD) + (ABC'D + ABCD)

Now use A+A'=1 (etc)

ABCD + ABCD = (A'+A)BCD = BCD =

AB'CD' + AB'CD + ABCD' + ABCD = AC (as explained by William)

ABC'D + ABCD = ABD

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);

ABCD + AB'CD' + AB'CD

+ ABC'D + ABCD' + ABCD = AC(B'D' + B'D + BD' + BD)

+ ABCD + ABC'D

Then replace

1 = (B' + B)(D' + D)

= (B'D' + B'D+BD' + BD)

= AC + A' BCD + ABC'D.

Finally, observe that

AC + ABCD = AC + BCD,

because if C = B = D = 1, then either A=1 so AC =1 or A = 0 so

ABCD = 1. Similarly

AC + ABC'D= AC + ABD

because if A = B = D= 1, then either

C = 1 soC =1 or C = 0 so =

ABCD = 1. So make two final

substitutions to get

= AC + BCD + ABD.