Logical Equivalence with → I am given the problem of proving:
$p → (q\land r) \equiv (p→q) \land(p→r)$
Using known logical equivalences. I'm not well practiced in transforming logical statements that contain →'s in them into other forms, and i'm at a bit of a loss, so:
What logical equivalences are available the best to use when presented with →'s, as i'm certainly going to see them again soon and which would be best suited to solving my problem.
Thank you!
 A: you simply have 
$$p → (q\land r) \equiv p'\lor (q\land r)\\
\equiv (p'\lor q) \land(p'\land r)\\ \equiv (p→q) \land(p→r)
$$
A: You can either work out the truth table for each of these propositions, fill in the assignments and you will see that the truth tables are identical, which means that these propositions are indeed equivalent.
Of you can use other equivalences and just unwind and rewind the proposition from one end to the other, for example: $$p\rightarrow (q\land r)\equiv \lnot p\lor(q\land r)\equiv \ldots\equiv(p\rightarrow q)\land(p\rightarrow r).$$
(And I leave you with the task of filling in those $\ldots$ there with the needed equivalences.)
A: For anybody else from the future.
Just remember that to get rid of the right arrow, this is an easy equivalence to remember: $p \rightarrow q \equiv \neg p \vee q$
You basically just use a negation operator on the entire left side, and just use an or operator in place of the arrow. The right side stays the same.
Same goes for parenthesis. $(p \wedge r) \rightarrow q \equiv \neg(p \wedge r) \vee q$. From there, just apply De Morgan's Law.
