Using Logic Laws to prove $p \leftrightarrow q \equiv (p\lor q)\to(p \land q)$ I am trying to prove that

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

and am really lost in the steps to solve this.
So far I have:

$p \leftrightarrow q \equiv (p\to q)\land(q\to p) \qquad$|equivalence
   $p \leftrightarrow q\equiv (\neg p\lor q)\land(\neg q\lor p) \qquad$|implication

and I am not sure how to proceed from here. Any advice would be greatly appreciated!
edit:
Thanks to lord farin i now have 

≡ (~p ∨ q) ∧ (q → p)  Implication Law
  ≡ (q → p) ∧ (~p ∨ q)  Commutative Law
  ≡ (q → p ∧ ~p ) ∨ (q → p ∧ q) Distributive Law
  ≡ (~p ∧ q → p) ∨ (q ∧ q → p) Commutative Law

but i am unsure of how to get there still. 

\begin{align}
(p\lor q)\to(p \land q) & \equiv (p\lor q)'\lor(p \land q)\\
 & \equiv (p'\land q')\lor(p \land q)\\
 & \equiv (p'\lor(p \land q))\land (q'\lor(p \land q))\\
 & \equiv ((p'\lor p) \land (p'\lor q))\land ((q'\lor p) \land (q'\lor q))\\
 & \equiv (T \land (p'\lor q))\land ((q'\lor p) \land T)\\
 & \equiv (p'\lor q)\land (q'\lor p)\\
 & \equiv (p\rightarrow q)\land (q\rightarrow p)\\
 & \equiv p \leftrightarrow q 
\end{align}
