How to simplify this Boolean expression?$$A'BC + AB'C' + A'B'C' + AB'C + AB$$
I have solved it using Kmap and found the answer to be$$A + BC + B'C'$$
I tried simplifying it using the rules but only got to$$B'C' + C (A'B + AB') + AB$$
Seeing that I already have $B'C'$ in my answer, I assume I only need to simplify $C (A'B + AB') + AB$ to $A + BC$, but I don't know how.
I hope you can help me.
$\endgroup$ 12 Answers
$\begingroup$Write $AB=ABC+ABC’$ and go from there.
$\endgroup$ $\begingroup$\begin{align}A'BC+AB'C'+A'B'C'+AB'C+AB&=A'(BC+B'C')+A(B'C'+B'C+B)\\ &=A'(BC+B'C')+A(B'(C'+C)+B)\\ &=A'(BC+B'C')+A(B'+B)\\ &=A'(BC+B'C')+A(1)\\ &=A'(BC+B'C')+A(1+BC+B'C')\\ &=A'(BC+B'C')+A+A(BC+B'C')\\ &=(A'+A)(BC+B'C')+A\\ &=A+BC+B'C'\\ \end{align}
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