Proof in logic that $P \Leftrightarrow Q$ is the same as $ (\lnot P \lor Q) \land (\lnot Q \lor P)$

Question

How is it possible to prove that $P \Leftrightarrow Q$ is the same as $ (\lnot P \lor Q) \land (\lnot Q \lor P)$ using logic laws?

Answer 1

$$\begin{align} p \iff q & \equiv (p \rightarrow q) \land (q\rightarrow p)\\ \\ & \equiv (\lnot p \lor q) \land (\lnot q \lor p)\end{align}$$

The first step is definitional. In the second step, I use, twice, the fact that $$a \rightarrow b \equiv \lnot a \lor b$$

Answer 2

Use truth tables to prove that

(!p OR q) <-> p -> q

and

(!q OR p) <-> q -> p.

Recall that p -> q is defined as such:

p q p -> q

T T T

T F F

F T T

F F T

So prove that ~p or q is equivalent to the above

Hmm...is p <-> equivalent to (p -> q) and (q -> p) by definition?

If so, QED.

If not, ...