I'm trying to completely simplify
$F_0 =A' B' C' D' + A' B' C' D + A B' C' D' + A B' C' D + A B' C D$
I got as far as
$\begin{align} &= A’B’C’ + A B' C' D' + A B' D\\ &= A’B’C + AB’(C’D’+D)\\ &= A’B’C + AB’(C’D’+D)\\ &= A’B’C + AB’(C’+D)\\ &= B’((A’C)+A(C’+D))\\ &= B’(A’C+AC’+AD)\\ &= B'A'C+B'AC'+B'AD\end{align}$
Unfortunately two different pieces of software both returned
$B'C'+AB'D$
1) Can anyone prove (without benefit of a truth table) that the two statements are equivalent?
2) What are some general strategies in the early stages of boolean simplification that will help me avoid getting stuck?
Many thanks in advance
Joe