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I'm trying to solve the following Boolean expression:

A'B'C + A'BC' + A'BC + AB'C' + AB'C + ABC' + ABC = A + B + C

I have tried using the distributive law and trying to cancel out some of the letters, and I keep getting something along the lines of A'B + C. I'm not sure how to start this problem, because trying to take out a common factor and trying to cancel doesn't seem to workout for me. I basically need to show the steps on how it equals A + B + C.

Any suggestions? I've been stuck at this for a while.

Also a note, I haven't learned K-maps yet or any super advanced techniques yet.

bunkered_reageant
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1 Answers1

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The main laws used here are $x+x'=1$ and $x+x'y=x+y$. If you haven't learned them yet, they can be easily verified with a truth table.

\begin{align*} &A'B'C + A'BC' + A'BC + AB'C' + AB'C + ABC' + ABC\\ &=A'(B'C+BC'+BC)+A(B'C'+B'C+BC'+BC)\\ &=A'(B'C+B(C'+C))+A(B'(C'+C)+B(C'+C))\\ &=A'(B'C+B)+A(B'+B)\\ &=A'(B'C+B)+A\\ &=A'(B+C)+A\\ &=A+B+C \end{align*}

Riley
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