# Boolean Algebra-Simplification Assistance Needed

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.

-
The sedond line: how did $(P' · Q' + R)$ turn into $(P' + Q' + R)$? –  user87690 Aug 28 '13 at 15:12
De Morgan's Law !(P.Q) = !P + !Q, did I do it wrong? –  Dromad Aug 28 '13 at 15:21
But there is $P' · Q'$ and not $(P · Q)'$. –  user87690 Aug 28 '13 at 15:39
Yeah sorry about that I wasn't sure how to type it out. It's P.Q Negated with the line above it, so I think it should be !(P.Q) I'll have to edit the question if this is right, –  Dromad Aug 28 '13 at 15:44
Actually since $Q'$ is in both conjunts of the second line expression, you can factor it out immediately: $(P' + Q' + R)(Q' + PR') = Q' + (P' + R)(PR')$ which implies $Q'$ –  user87690 Aug 28 '13 at 15:49

\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}
$\Large{+}^\infty$ –  Software Aug 28 '13 at 20:21
Yes your simplification is right, but can be done shorter way: $((PQ)' + R)(Q' + PR') = (P' + Q' + R)(Q' + PR') = Q' + (P' + R)(PR') = Q' + P'PR' + RPR' = Q' + 0 + 0 = Q'$.