Boolean algebra: simplify (A+B)$'C'$(C+D)

Question

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.

Answer 1

$C'C=0$ 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.

Answer 2

The output for the above expressing is 1 in only 1 out of the 16 possible cases, I guess you must have missed that one, let's have a look,

if $C = 1$, $$C = 1 \implies C' = 0 \implies (A+B)′′(C+D) = 0$$

Now, if $C = 0$, $$(A+B)′′(C+D) \equiv (A+B)′D \equiv A'B'D$$

Clearly, $A'B'D$ can be equal to 1 only when $A=0,B=0,D=1$

Therefore, $(A+B)′′(C+D) = 1$ only when $A=0,B=0,C=0,D=1$, otherwise, $(A+B)′′(C+D) = 0$

Answer 3

Here's a handy equivalence:

Reduction

$P(P'+Q)=PQ$

To see why this is true: note that $P$ has to be true, and hence for $P'+Q$ to be true as well, it is clear that (since $P'$ is now false), $Q$ has to be true. We thus say that in the context of $P$, the term $P'+Q$ 'reduces' to $Q$.

You can of course derive this from more basic principles:

$P(P'+Q)=PP'+PQ=0+PQ=PQ$

but I find that having Reduction as a 1 step equivalence really makes simplification a lot faster in practice. Indeed, for your problem, we get:

$(A+B)'C'(C+D)\overset{DeMorgan, Reduction}{=}A'B'C'D$

Done!

So yes, definitely put Reduction in your handy-dandy-toolbag for Boolean logic!