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.

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