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.
The trick is to expand only one of the two implications materially at first:

$$\begin{align*}
(p\to q)\land(q \to p) &amp;\equiv (\neg p \lor q)\land (q\to p)\\
&amp;\equiv \underbrace{(\neg p \land q \to p)}_{\downarrow} \lor \underbrace{(q \land q\to p)}_\downarrow&amp; &amp; \text{distribution}\\
&amp;\equiv \underbrace{(\neg p \land \neg q) \hspace{1.5em} \lor\quad(p \land q)}_\downarrow\\
&amp;\equiv \hspace{2em}(p \lor q) \to (p \land q)
\end{align*}$$

I'll leave it up to you to take care of the transitions represented by the arrows.



Additional details: For the first two, you can rewrite the conditionals, then use distribution of the $\land$. For e.g. the first, this will leave $(\neg p \land \neg q) \lor (\neg p \land p)$, the second term of which is always false. Thus the whole expression is equivalent to $\neg p \land \neg q$.

A similar approach works for $q \land q \to p$.
You could write equivalence as $(p \land q) \lor (\lnot p \land \lnot q)$, which gives us, using commutation and implication law, the result you desire.