# Simplification of a Boolean Expression

I want to simplify this expression: ACD' + E(A+C)(A'+D') + A'C .

The result must be a product of sums, where every sum should be consisted of just two variables. For example (A+B)(C+A)(Z+Y) ...

This is how I did it so far:

C(AD' + A') + E(A+C)(A'+D')

=CD' + CA' + EAD' + ECA' +ECD'



How can I turn this result into a product of sums ? Could anyone just give me an idea or a hint?

-

In the second line of your work, you lost an $A$. So you should have $C(AD'+A')+E[AA'+AD'+CA'+CD']$, which does change the final result. I would suggest a different path. Instead of trying to simplify first, I would take advantage of the fact that $(A')' = A$ to change what you are simplifying. The idea being that if you need $(A+B)(C+A)$ that can be gotten from $(A'B'+C'A')'$. We will also be using DeMorgan's Laws $(AB)' = A'+B'$ and $(A+B)' = A'B'$.

So if we do a double negation at the beginning and use the first negation to rewrite and simplify the expression, then when we take the final negation we will end up in the correct form. In general, there is no guarantee that you will have exactly two variables per expression.

$ACD'+E(A+C)(A'+D')+A'C \\ \Longleftrightarrow \left[(ACD'+E(A+C)(A'+D')+A'C)'\right]' \\ \Longleftrightarrow \left[(ACD')'\left(E(A+C)(A'+D')\right)'(A'C)' \right]' \\ \Longleftrightarrow \left[(A'+C'+D)(E'+A'C'+AD)(A+C')\right]' \\ \Longleftrightarrow \left[\left((A'+C'+D)(A+C')\right)(E'+A'C'+AD)\right]' \\ \Longleftrightarrow \left[\left(C'+(A'+D)(A)\right)(E'+A'C'+AD)\right]' \\ \Longleftrightarrow \left[(C'+AD)(E'+A'C'+AD)\right]' \\ \Longleftrightarrow \left[\left((C')(E'+A'C')\right)+AD\right]' \\ \Longleftrightarrow \left[C'E'+A'C'+AD\right]' \\ \Longleftrightarrow (C+E)(A+C)(A'+D')$

I took advantage of the fact that in boolean algebra each operation distributes over the other to make the simplification a little faster. If that causes trouble, just do it like you were doing above and you'll end at the same result.

Edit I'm adding the details in order to get to the final result I have from the point OP got stuck in the comments.

$\left[(C'+AD)(E'+A'C'+AD)\right]' \\ \Longleftrightarrow \left[C'E'+A'C'C'+AC'D+ADE'+AA'C'D+ADAD\right]' \\ \Longleftrightarrow \left[C'E'+A'C'+AC'D+ADE'+AD\right]' \\ \Longleftrightarrow \left[C'E'+A'C'+AD\right]' \Longleftrightarrow (C+E)(A+C)(A'+D')$

I suspect the issue might be this last step of $C'E'+A'C'+AC'D+ADE'+AD\Longleftrightarrow C'E'+A'C'+AD$. If you factor out the common $AD$ you get $AD(C'+E'+1) = AD$.

-
Nice work, John! (+1) (You'll soon find that boolean-algebra-tagged questions earn little in the way of reputation (for some unfortunate reason). So don't take it personally if you find you earn little in the way of reputation on answers to boolean-algebra questions. – amWhy Jan 29 '14 at 13:57
@amWhy Hello amWhy, John claimed that I lost an A. In fact I didn't. I found a rule on the internet, according to this rule: x + x'y = x + y . What do you think ? – Mcrp Jan 29 '14 at 14:06
@John Habert Hey, I followed your example but for some reason my result is [C'D + C'A' + C'E' +DA]' which equals to your result according to wolfram alpha CNF form. link here: wolframalpha.com/input/… I just don't get it! – Mcrp Jan 29 '14 at 14:15
You are absolutely right, Mcrp, that $C(AD' + A') = C(D' + A') = CD'+CA'$. That is a valid move you made. – amWhy Jan 29 '14 at 14:23
@amWhy Cheers for your input. But for some reason I cant reach John's result. Look I followed his example with De Morgans etc. I reached this point: [(C' + DA)(E' + A'C' +AD)]' = ... = [C'E' + A'C' +C'AD + DAE' + DA]' =...= [C'E' + C'D + C'A' +DA]' But despite it is correct (according to wolfram), it isn't the same, and I don't get it all DAMN. – Mcrp Jan 29 '14 at 14:38