# Simplifying boolean function using boolean algebra

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 .

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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

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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

$=AC + BCD + ABD$.

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It should get AC + BCD + ABD . My question is how to reach that form ? –  Dina Nov 19 '12 at 20:34
Sorry, I figured the very last part was clear. I edited to explain it above. –  William Macrae Nov 19 '12 at 20:42