Logical simplification clarification

Question

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

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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

Answer 1

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.

Answer 2

You cannot just remove brackets unless the operations are associative.   That's not the case here.

The rule you want is "absorption": $$(\psi\vee\neg\phi)\wedge\phi~=~\psi\wedge\phi$$

Answer 3

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!