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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.

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2 Answers 2

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Write $AB=ABC+ABC’$ and go from there.

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\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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    $\begingroup$ In the 6th line, you do not need $AA’$ because $A’(BC+B’C’)+A(BC+B’C’)=BC+B’C’$ $\endgroup$ Aug 30, 2020 at 5:53
  • $\begingroup$ That is a nice simplification of the proof. Thank you @RezhaAdrianTanuharja. $\endgroup$
    – user400188
    Aug 30, 2020 at 5:56
  • $\begingroup$ Where did the (1 + BC + B'C') come from in the 5th line? $\endgroup$
    – pi3.14
    Aug 30, 2020 at 6:24
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    $\begingroup$ @pi3.14 $A$ is the same as $A$ AND $1$. Also; $1$ is the same as $1$ OR something else. So $A$ must be the same as $A$ AND ($1$ OR something else). Is that helpful? $\endgroup$
    – user400188
    Aug 30, 2020 at 7:13
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    $\begingroup$ @user400188 yes, it is! Thank you so much! $\endgroup$
    – pi3.14
    Aug 30, 2020 at 7:46

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