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
Your second line, where you have $A'B'C$, should read $A'B'C'$. Once that's fixed, your steps from there should get you what you need, with minor corrections (and removal of a redundant line). In particular,
$$\begin{align}F_0 &= A'B'C' + A B' C' D' + A B' D\\ &= A'B'C' + AB'(C'D'+D)\\ &= A'B'C' + AB'(C'+D)\\ &= B'((A'C')+A(C'+D))\\ &= B'(A'C'+AC'+AD)\\ &= B'((A'+A)C'+AD)\\ &= B'(C'+AD)\end{align}$$
More generally, I'd say your steps look fine. Just be careful not to drop any symbols.