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 .
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)
A'BCD + 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):
$A'BCD + AB'CD' + AB'CD + ABC'D + ABCD' + ABCD = AC(B'D'+B'D+BD'+BD) + A'BCD + ABC'D$
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
$=AC + BCD + ABD$.