Need help simplifying this Boolean expression From my textbook, by using distributive law, its able to simplify:
$[(p \land \lnot q) \lor (p \land q)] \land q$ 
To:
$[p \land (\lnot q \lor q)] \land q $
I don't know how to get to this step, and here is how I've tried by distributing the first expression :
$[(p \lor p) \land (p \lor q) \land (\lnot q \lor p) \land (\lnot q \lor q)] \land q$
If I try continue to expand this, it will become very complex. I think I'm doing this in the wrong way and I will appreciate it very much if anyone can explain how the textbook got the simplified answer.
Thanks
 A: $$[(\color{blue}{p} \land \lnot q) \lor (\color{blue}{p} \land q)] \land q \equiv [\color{blue}{p} \land (\color{red}{{\lnot q \lor q}})] \land q\tag{distributivity}$$
$$ \equiv (\color{blue}{p} \land \color{red} \top) \land q$$
$$\equiv \color{blue}{p} \land q$$
That is, the distributive law declares that $$[p\land (\lnot q \lor q)] \equiv [(p \land \lnot q) \lor (p \land q)]$$  That means that it is the same law when we write it: $$[(p \land \lnot q) \lor (p \land q)] \equiv [p\land (\lnot q \lor q)]$$

Addendum



A: The Distributive law goes both ways, because it is an equivalence.
That is, using the Distributive Law you can go from $p \land (\neg q \lor q)$ to $(p \land \neg q) \lor (p \land q)$, but you can use that same Distributive Law to go from the latter to the former, and that is what they did here.
I believe the name is somewhat to blame for the fact that so many students, like you, only think of the Distributive Law going one way, because going from $(p \land \neg q) \lor (p \land q)$ to $p \land (\neg q \lor q)$ does not feel like Distribution, but rather as a kind of 'Un-Distribution' or 'Reverse Distribution', or maybe even 'Taking out a common Factor' ... but it is still called Distribution in logic.
