prove the following using the identities

(P-Q) ^ (Q-P) equivalent to (P v Q)-(P ^ Q)

Question

prove the following using the identities

(P->Q) ^ (Q->P) equivalent to (P v Q)->(P ^ Q)

walters · Answer

@terenzreignz HELP

anonymous · Answer

$$(p\to q)\wedge(q\to p)\equiv(p\vee q)\to(p\wedge q)~~~?$$

walters · Answer

YES

anonymous · Answer

Are you supposed to do this with truth tables, or just apply some other known identities?

walters · Answer

apply known identities

anonymous · Answer

Hmm, one that comes to mind is $p\to q\equiv \neg ~p\vee q$.

So the left side is the same as
$$(\neg~p\vee q)\wedge(\neg ~ q \vee p)$$
Does this resemble any other identities you know?

anonymous · Answer

You may also wish to use the same identity to express the right side differently:
$$(\neg~p\vee q)\wedge(\neg ~ q \vee p)\stackrel{?}\equiv\neg(p\vee q)\vee(p\wedge q)\
(\neg~p\vee q)\wedge(\neg ~ q \vee p)\stackrel{?}\equiv(\neg ~p\wedge \neg~q)\vee(p\wedge q)$$
I think one of DeMorgan's laws might work here, if I recall them correctly...

anonymous · Answer

Not DeMorgan's, sorry. The distributive ones. Sorry, I haven't studied this stuff in a while :P

terenzreignz · Answer

Sorry... I must have dozed off :(