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 .

Answer 1

[Matt Fig]

>> syms A B C D
>> simplify((1-A)*B*C*D + A*(1-B)*C*(1-D) + A*(1-B)*C*D +...
            A*B*(1-C)*D + A*B*C*(1-D) + A*B*C*D)
ans =

A*C + A*B*D + B*C*D - 2*A*B*C*D

Now look closely at the last term. The only case where it is nonzero is when all A,B,C,D are nonzero. If any non-zero value for the overall expression counts as true, then the last term is redundant.