After starting with something significantly larger, asked in another question, I found myself stuck with essentially this proof. What law can I use here? I've tried to look for distribution, but I don't see how it would work. Any explanation would be helpful.
$$(p \land q) \land (¬q \lor r) \iff p \land q\land r$$
$$ ( p \wedge q ) \wedge ( \neg q \vee r ) \equiv ( p \wedge q \wedge \neg q) \vee (p \wedge q \wedge r) \equiv (p \wedge q \wedge r) $$ While $( p \wedge q \wedge \neg q)$ is a contradiction, of course.
The easiest way is to simply make a truth table/work by cases. The RHS is true precisely when all of the variables are true, by definition. It's easy to check that that assignment gives the LHS as true as well. Then you just need to check that no other assignment satisfies the LHS.
Here is a handy equivalence:
Reduction
$P \land (P \lor Q) \Leftrightarrow P$
This equivalence can of course be derived from more basic principles, but it happens so often when doing Boolean algebra, that you would do well to put it in your toolbox.
Indeed, with this equivalence:
$(p \land q) \land (\neg q\land r) \Leftrightarrow $ (Association)
$ p \land (q \land (\neg q \land r)) \Leftrightarrow $ (Reduction)
$p \land (q \land r)$
In fact, if you allow for generalized conjunctions like $p \land q \land r$, you basically make any Associations implicit, so in that case it is just one step:
$(p \land q) \land (\neg q \lor r) \Leftrightarrow $ (Reduction)
$p \land q \land r$
Done!