TITLE: boolean algebra simplification
QUESTION [5 upvotes]: a) $(\lnot(P \land Q)) \lor (Q \land R)$
b) $(P \lor Q) \land \lnot(Q)$ 
How do I simplify these 2 expression?

REPLY [5 votes]: a) 
1. $(\lnot(P \land Q)) \lor (Q \land R)$
2. $(\lnot P \lor \lnot Q) \lor (Q\land R)$
3. $((\lnot P \lor \lnot Q) \lor Q) \land ((\lnot P \lor \lnot Q) \lor R)$
4. $(\lnot P \lor (\lnot Q \lor Q)) \land (\lnot P \lor \lnot Q \lor R)$
5. $(\lnot P \lor True) \land (\lnot P \lor \lnot Q \lor R)$
6. $(True ) \land (\lnot P \lor \lnot Q \lor R)$
7. $\lnot P \lor \lnot Q \lor R\qquad$ (Simplified normal disjunctive form)
Note: By DeMorgan's, (7) is equivalent to $\lnot(P \land Q) \lor R$
b) 
1.  $(P \lor Q) \land \lnot(Q)$ 
2.  $(P \land \lnot Q) \lor (Q \land \lnot Q)$
3.  $(P \land \lnot Q) \lor (False)$ 
4.  $(P \land \lnot Q)\qquad$ (simplified in normal disjunctive form)
Can you understand how to move from one step to the next? Essentially, I am simply applying "distribution":
e.g. $(a \lor b) \land c \equiv (a \land c) \lor (b \land c)$ 
also $a \lor (b \land c) \equiv (a \lor b) \land (a \lor c)$ 
and I've used DeMorgan's in the first simplifications.