Is it possible to apply De-Morgan's laws in the expression: $\rm [(A' + C) (A + B)']'$ and obtain $\rm (A'C)' + (AB)''$? Is it possible to apply De-Morgan's laws in the expression: $\rm [(A' + C) (A + B)']'$ and obtain $\rm (A'C)' + (AB)''$?
 A: Your solution is a bit off. I'm getting this:
$$\begin{align*}
[ (\neg A + C) (A + B)'] = &(A' + C)' + (A + B)''] &\text{(de Morgan)} &\\
= &(A' + C)' + A + B &\text{(double-negative cancellation)} &\\
= &A C' + A + B &\text{(de Morgan)} \\
\end{align*}$$
whereas your proposed simplification, after a de Morgan application on the first term and a cancellation of negatives on the second, is equivalent to $A + C' + AB$. In particular, $(A', B, C)$ satisfies the original expression but not your proposed simplification.
(Edit: The ending expression even simplifies to $A + B$, because $A C'$ implies $A$.)
A: It is possible but your solution is not quite correct.
$$\begin{align}\rm [(A' + C) (A + B)']' = &(A'+C)' + (A+B) \tag 1 &\\= &AC' + A + B  \tag 2&\\= &A(C' + 1) + B \tag 3&\\= &A+ B \end{align}$$
$(1)$ : $(A\cdot B)' = A' + B'$
$(2)$ : $(A + B)' = A' \cdot B'$
$(3)$ : $A + 1 = 1$
A: In order to find out whether two expressions are equivalent, 
you can construct their truth tables.
Let $X=((A'+C)(A+B)')'$ and $Y=(A'C)'+(AB)''$.


*

*If A=0; B=0; C=0, then X=0 and Y=1.

*If A=0; B=0; C=1, then X=0 and Y=0.

*If A=0; B=1; C=0, then X=1 and Y=1.

*If A=0; B=1; C=1, then X=1 and Y=0.

*If A=1; B=0; C=0, then X=1 and Y=1.

*If A=1; B=0; C=1, then X=1 and Y=1.

*If A=1; B=1; C=0, then X=1 and Y=1.

*If A=1; B=1; C=1, then X=1 and Y=1.


Since there exist values of A,B and C such that $X \neq Y$,
e.g. A=B=C=0,
conclude that you cannot apply de Morgan's law (or any other law)
to show that they are the same.
A: The first expressions simplifies as follows:
$$[(A' + C) (A + B)']' = \text{ (DeMorgan)}$$
$$(A' + C)' +(A + B)'' = \text{ (DeMorgan)}$$
$$A''C' +(A + B)'' = \text{ (Double Negation)}$$
$$AC' +(A + B) = \text{ (Association)}$$
$$(AC' +A) + B = \text{ (Distribution)}$$
$$A(C' +1) + B = \text{ (Annihilation)}$$
$$A1 + B = \text{ (Identity)}$$
$$A + B$$
The second expressions simplifies as follows:
$$(A'C)' + (AB)'' = \text{ (DeMorgan)}$$
$$(A''+C') + (AB)'' = \text{ (Double Negation)}$$
$$(A+C') + AB = \text{ (Commutation)}$$
$$(C'+A) + AB = \text{ (Association)}$$
$$C'+(A + AB) = \text{ (Distribution)}$$
$$C'+A(1 + B) = \text{ (Annihilation)}$$
$$C'+A1 = \text{ (Identity)}$$
$$C'+A$$
It should be pretty clear that these are not equivalent. Indeed, with $A$ being False, and $B$ and $C$ being True, $A + B$ is True, but $C'+A$ is False.
