Boolean Algebra-Simplification Assistance Needed

Question

I have to show that (!(P.Q) + R)(!Q + P.!R) => !Q by simplifying it using De Morgan's Laws. Here is what I did but I'm not sure it's right.

(!(P.Q) + R)(!Q + P.!R) => !Q

(!P + !Q + R)(!Q + P.!R)

!(P.Q) + !P.P.!R + !Q + !Q.P.!R + R.!Q + R.P.!R

!(P.Q) + 0 + !Q + !Q.P.!Q + R.!Q + 0

!(P.Q) + !Q + !Q(1 + P.!R + R)

!(P.Q) + !Q + !Q

!Q + !(P.Q)

!Q(1 + !P)

!Q

Hope that's clear enough.

Answer 1

Yes your simplification is right, but can be done shorter way: $((PQ)' + R)(Q' + PR') = (P' + Q' + R)(Q' + PR') = Q' + (P' + R)(PR') = Q' + P'PR' + RPR' = Q' + 0 + 0 = Q'$.

Answer 2

Indeed, your work is correct. Let's shorten things up, though, using the Distributive Law (D.L.) twice immediately following the application of DeMorgan's:

$\begin{align}(\overline{P\cdot Q} + R)\cdot (\overline Q + P\cdot \overline R) &= (\overline P + \overline Q + R)\cdot(\overline Q + P\cdot \overline R) \tag{DeMorgan's} \\ \\ & = \overline Q + (\overline P + R)(P\cdot \overline R) \tag{D.L.} \\ \\ & = \overline Q + (\color{blue}{\overline P\cdot P}\cdot \overline R) + (\color{red}{R}\cdot P\cdot \color{red}{ \overline R}) \tag{D.L.}\\ \\ & = \overline Q + \color{blue}{ 0} + \color{red}{\bf 0} \tag{$A \cdot \overline A = 0$}\\ \\ &= Q'\end{align}$