# Simplifying Boolean Algebra

I am trying to prove that BC + !A!B + !A!C = ABC +!A

I have attempted using De Morgan laws, and substituting X for !A!B and Y for !A!C, however I made no headway in this.

I've also tried grouping the A's like so, !A(!B+!C), however again I couldn't get anywhere. If anyone could point me in the right direction, help me solve, show me a tool that can do it, etc. I'd be grateful.

If everything else fails, rewrite each term on both sides as a sum of terms where each contains every variable either direct or negated. For example, $BC = (A+\bar A)BC = \bar ABC + ABC$. Then check whether you get the same terms on each side of the equation.
In standard logical notation, with which I'm much more familiar, you want to prove $$(B \land C) \lor (\lnot A \land \lnot B) \lor (\lnot A \land \lnot C) \;\equiv\; (A \land B \land C) \lor \lnot A$$ We can simplify the right hand side as follows: \begin{align} & (A \land B \land C) \lor \lnot A \\ \equiv & \;\;\;\;\;\text{"use negation of \lnot A in other side of \lor"} \\ & (\textrm{true} \land B \land C) \lor \lnot A \\ \equiv & \;\;\;\;\;\text{"simplify"} \\ & (B \land C) \lor \lnot A \\ \end{align} and the left hand side like this: \begin{align} & (B \land C) \lor (\lnot A \land \lnot B) \lor (\lnot A \land \lnot C) \\ \equiv & \;\;\;\;\;\text{"factor out \lnot A"} \\ & (B \land C) \lor (\lnot A \land (\lnot B \lor \lnot C)) \\ \equiv & \;\;\;\;\;\text{"use De Morgan"} \\ & (B \land C) \lor \lnot(A \lor (B \land C)) \\ \equiv & \;\;\;\;\;\text{"use negation of B \land C in other side of \lor"} \\ & (B \land C) \lor \lnot(A \lor \textrm{false}) \\ \equiv & \;\;\;\;\;\text{"simplify"} \\ & (B \land C) \lor \lnot A \\ \end{align} which shows that both sides are logically equivalent.