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')

=C(D'+A') + E[AA'+AD'+CA'+CD']

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

=C(D'+ED') + C(A'+EA') + EAD'

=CD' + CA' + EAD'

How can I turn this result into a product of sums ? Could anyone just give me an idea or a hint?
 A: 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 $AD(C'+E'+1) = AD$.
