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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.

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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.

(This is the algebraic equivalent of drawing Venn diagrams for the two sides and seeing whether they match).

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You could rewrite each side as @henning-makholm said. For another specific method apart from Venn diagrams, you could use a Karnaugh map, either by hand or using a tool like this one:

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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.

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