Logical simplification clarification My question is "Using Logical simplification, prove that the following is a tautology, contradiction or neither."
  $p \land ((\lnot p \lor q)\land \lnot q)$
My logic is probably wrong but this is what I attempted:
Removing the brackets to get to this
  $(p \land \lnot p) \lor (q\land \lnot q)$
then using negation laws this evaluates to:
  $(F) \lor (F) $
proving that this is a contradiction as every outcome is false.
Is there anything stopping me from applying the logic I used in the first step? If not what is the correct approach to doing this?
EDIT 1

so using absorption as mentioned below I got to this:
  $ p \land ((\lnot p \lor q)\land \lnot q)$
  $ p \land (\lnot p \land \lnot q)$
then using associative getting this
  $(p \land \lnot p)\land \lnot q$
  $(F) \land \lnot q$
which doesn't prove if it is a tautology, contradiction or neither
 A: Just use the distributive properties.  Or distributes over And :
$$(a \land b) \lor c \equiv (a \lor c) \land (b \lor c)$$
and And distributes over Or :
$$(a \lor b) \land c \equiv (a \land c) \lor (b \land c)$$
So:
$$P \land (\underbrace{(\lnot P \lor Q) \land \lnot Q}_\text{Distr. And})$$
$$\underbrace{P \land (\overbrace{(\lnot P \land Q) \lor (Q \land \lnot Q)}^\text{To this})}_\text{Commute And}$$
$$\underbrace{((\lnot P \land Q) \lor (Q \land \lnot Q)) \land P}_\text{Distr. And}$$
$$((\lnot P \land Q) \land P) \lor ((Q \land \lnot Q) \land P)$$
Then just finish it up.  This is btw just a symbolic way of doing truth tables.
A: You cannot just remove brackets unless the operations are associative.   That's not the case here.
The rule you want is "absorption": $$(\psi\vee\neg\phi)\wedge\phi~=~\psi\wedge\phi$$
A: Here's a very handy equivalence rule that unfortunately many texts don't provide you outright:
Reduction
$$\neg P \land (P \lor Q) \Leftrightarrow \neg P \land Q$$
The intuitive idea here is that (from left to right) given $\neg P$, the term $P \lor Q$ can be 'reduced' to just $Q$.  But obviously you can go from right to left as well, so this is an equivalence. And of course you also have its dual:
$$\neg P \lor (P \land Q) \Leftrightarrow \neg P \lor Q$$
Applied to your statement:
$$P \land ((\neg P \lor Q) \land \neg Q) \Leftrightarrow P \land (\neg P \land \neg Q) \Leftrightarrow (P \land \neg P) \land \neg Q \Leftrightarrow \bot \land \neg Q \Leftrightarrow \bot$$
So ... it's a contradiction!
