Discrete Math Logical Equivalences

Question

I've been having trouble figuring on this logical equivalent problem. Any help would be appreciated.

(p ^ q) v (~p ^ ~q) = p <-> q

My work so far:

p <-> q

=(p->q) ^ (q->p)

Using conditional equivalence

=(~p v q) ^ (~q v p)

At this point I got stuck and I don't know what to do. I assume I use demorgans law now so it would be

=(p ^ ~q) v (q v~p)

I feel that that's wrong though. Thank you for any help.

Answer 1

Use Distribution:

$$(\neg p \lor q) \land (\neg q \lor p) \overset{Distribution}{=}$$

$$(\neg p \land \neg q) \lor (\neg p \land p) \lor (q \land \neg q) \lor (q \land p) \overset{Complement}{=}$$

$$(\neg p \land \neg q) \lor \bot \lor \bot \lor (q \land p) \overset{Identity}{=}$$

$$(\neg p \land \neg q) \lor (q \land p) \overset{Commutation}{=}$$

$$(p \land q) \lor (\neg p \lor \neg q)$$

Answer 2

=(~p v q) ^ (~q v p)

It is okay up to here.   The next step is distribution.

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

Then contradiction and identity and you are done.