~ABC + ~BCD + ~A~BC~D + ABCD = ?
The solution book gives this answer:
~AC + CD
I arrived this one:
ACD
My question is if the method used to simplify can affect the result or if should it be equal?
Note: ~ stands for not ( ~A = Not A)
My steps:
~ABC + ~BCD + ~A~BC~D + ABCD =
=CD(~B+AB) + ~AC(B+~B~D)=
=CD(~B+A) + ~AC(B+~D)=
=~BCD + ACD + ~ABC + ~AC~D=
=C(~BD + AD + ~AB + ~A~D)=
=C[D(~B + A + ~AB + ~A~D)]=
=C[D(~B + A + B + ~A~D)]=
=C[D(A + ~A~D)]=
=C[D(A + ~D)]=
=CDA + CD~D=
=ACD
It's possible to get an answer that looks different, but it should always be logically equivalent, no matter what route you take.
Unfortunately, your answer is not equivalent to the one provided by the book: If $A$ is False, and $C$ is True, then your expression is False, but the one from the book is True.
Try this: take the second and third term: $B'CD$ and $A'B'CD'$. Notice that they both have $B'C$ in common, so that definitely has to be true in order for one of those terms to be true. However, for the first term to be true, you also need $D$, while for the second you need $A'D'$, and so for one of those to hold you either need $D$ or $A'$ (because you already have $D'$ if not $D$). So, these two terms can be simplified to $B'CD$ and $A'B'C$
Formally:
$B'CD+A'B'CD'= B'C(D + A'D')= B'C(D+A')(D+D')=B'C(D+A')1=B'C(D+A')=B'CD+B'CA'$
