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I've been working a Boolean algebra problem for probably 2 hours at this point, and while I arrive at a much simplified equivalent expression, there's a simpler one yet.

Basically, I start out with a truth table and am asked to get it to a simplified SOP form. So far, so good. Then I'm asked to convert it to POS form, which I successfully do, but I can't seem to get it to its simplest form.

I used DeMorgan's theorem twice for the conversion, and almost at the end, I bump into this wall.

I arrive at $AB'+ C'A' + C'B'$ as my answer before applying DeMorgan's theorem for the second time. I know that $AB' + C'A'$ is equivalent, but for the life of me, I cannot derive it.

Can anyone help me understand the steps used? It seems like it should be straight forward, but I just can't get there. If needed, the SOP I simplified to was $A'C + AB$.

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You can use $1=A+A'$ to get

$AB'+C'A'+C'B'=AB'+C'A'+C'B'(A'+A)=AB'+C'A'+A'C'B'+AC'B'=A(B'+C'B')+A'(C'+C'B')=AB'(1+C')+A'C'(1+B')=AB'+A'C'$

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  • $\begingroup$ Thank you so much! I can get to sleep now :) How long did it take you to see it? It seems a rather unintuitive thing to do. $\endgroup$ – orn Jan 25 '15 at 23:31
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    $\begingroup$ Well, it is pretty obvious that the missing part was $C'B' => AB' + C'A'$. To see how that can be shown i used a venn diagram and transformed the "geometrical solution" into a formula. You'll get used to it :) $\endgroup$ – Gregor de Cillia Jan 25 '15 at 23:37

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