Boolean algebra: simplify (A+B)$'C'$(C+D) I need to solve / simplify the output of a logic circuit. The output is $(A+B)'C'(C+D)$
I wrote the truth table and it returns 0 in all outputs. I tried to solve it using laws of boolean algebra:
$(A+B)' = A'B'$
And $C′(C+D) = (C′C + C′D)$
$C′C = 0$ [Boolean law $A.A′ = 0$]
Does that mean $(C′C + C′D) = 0$ as well?
If that's the case, the result of the entire output would be zero. Am I right or is there any mistakes? Is there any other law applied here?
Update: I made a mistake in the truth table. It doesn't return zero in all outputs.

$C'C=0$ will not imply $C'C + C'D=0$, and $C'C + C'D=1$ when $C=0,D=1.$ 
Here is how to simplify the expression:
\begin{align}
&(A+B)'C'(C+D)\\
&\equiv(A'B')C'(C+D)\tag*{de Morgan’s Theorem}\\
&\equiv(A'B')(C'C+C'D)\tag*{Distributive law}\\
&\equiv(A'B')(0+C'D)\tag*{Complement}\\
&\equiv(A'B')(C'D)\tag*{Identity}\\
\end{align}
Hence it has A'B'C'D as it's minimal CNF&DNF form.
