# Conjunctive/disjunctive normal form

XodoX
Let's say I have p<-> q.. how do I get the disjunctive and conjunctive normal form ? I think I'd have to apply DeMorgan to get one of them. I don't know. Can anybody please explain this to me?

p<->q = (p $$\wedge$$ q) $$\vee$$ (~p $$\wedge$$ ~q)
=(p $$\wedge$$ q) $$\vee$$ (~p $$\wedge$$ ~q) $$\vee$$ (p $$\wedge$$ ~p) $$\vee$$ (q $$\wedge$$ ~q)
=(~p $$\vee$$ q) $$\wedge$$ (~q $$\vee$$ p)