# general strategies for maximizing boolean simplification from a specific example

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?

Joe

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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. 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’+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}