A'B'CD+A'B'C'D+ABC'D'+ABCD'
Reduce the expression
.
Answers
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.