Prove using laws and axiom of logic So I have been looking at this question all day and have made a few attempts but can't seem to get any further.
Can someone please help with me proving this algebra equation and what laws I would need to use?
Forgive me for not knowing how to use the proper symbols on this. first time here, Didn't even know there was a forum for this.
I have also looked through already answered questions, and could not find anything similar.
This is the equation I need to prove

$p \leftrightarrow q \equiv (p \lor q) \rightarrow (p \land q)$

So far I have

$ p \leftrightarrow q \equiv (p \to q) \land (q \to p) $       The Equivalence Law
$(p \to q) \land (q \to p) \equiv (\lnot p \lor q) \land (\lnot q \lor p)  $  Implication Law
$(\lnot p \lor q) \land (\lnot q \lor p) \equiv (p \land q) \lor (\lnot p \land \lnot q) $    not sure what law(I used a truth table)

I am not sure if I am even on the right track but if I am, when does it become proved? I think I may need like 3 or 4 more laws to be used.
 A: That last equivalence, while true, is not in the 'standard' list of equivalences. Below is how you prove it using 'standard' equivalences:
$$p \leftrightarrow q \equiv \text{ Equivalence}$$
$$(p \rightarrow q) \land (q \rightarrow p) \text{ Implication}$$
$$(\neg p \lor q) \land (\neg q \lor p) \text{ Distribution}$$
$$((\neg p \lor q) \land \neg q)) \lor ((\neg p \lor q) \land p) \text{ Distribution x 2}$$ 
$$(\neg p \land \neg q) \lor (q \land \neg q) \lor (\neg p \land p) \lor (q \land p) \text{ Complement x 2}$$ 
$$(\neg p \land \neg q) \lor \bot \lor \bot \lor (q \land p) \text{ Identity x 2}$$ 
$$(\neg p \land \neg q)\lor (q \land p) \text{ DeMorgan}$$ 
$$\neg (p \lor q) \lor (q \land p) \text{ Implication}$$ 
$$(p \lor q) \rightarrow (q \land p)$$ 
So here I used the following Equivalences in addition to Equivalence and Implication:
Distribution
$p \land (q \lor r) \equiv (p \land q) \lor (p \land r)$
Complement
$p \land \neg p \equiv \bot$
Identity
$p \lor \bot \equiv p$
DeMorgan
$\neg(p \lor q) \equiv \neg p \land \neg q$
