$p \leftrightarrow q \equiv (p \to q) \wedge (q \to p)$ Prove using conditional identities

Question

Q. $p \longleftrightarrow q \equiv (p \to q) \wedge (q \to p)$

R.H.S:

$(p \to q) \wedge (q \to p)= (\neg p \vee q) \wedge (\neg q \vee p)$

Let $m = \neg p \vee q$

$= m \wedge (\neg q \vee p)$

$= (m \wedge\neg q) \vee (m \wedge p)$ ......using Distributive law

(-> Put value of $m$ back)

= $((\neg p \vee q) \wedge \neg q) \vee (\neg p \vee q) \wedge p)$

= ?

I can't like this:
$(\neg q \wedge \ q) \neg p) \vee (\neg p \wedge p) \vee q)$ Right?

Note: I was not able to write in correct mathematical form as I don't know.

Answer 1

You are correct that you can't do what you propose. Associativity works only within the same connective.

From $$ ((-p \lor q) \land -q) \lor ((-p \lor q) \land p)$$

We get, using the distributive property twice, $$(\lnot p \land \lnot q)\lor(q\land \lnot q) \lor (\lnot p \land p) \lor (q \land p)$$

$$\equiv (\lnot p \land \lnot q) \lor (p \land q)$$

This is exactly how the biconditional $p\longleftrightarrow q$ is defined. It means either $(p\land q) \lor (\lnot p \land \lnot q)$