show that p <--> q and ((p /\ q) \/ (~p /\ ~q)) are equal?

Question

Please help me out on this!

Answer 1

p -> q = q \/ ~p

q -> p = p \/ ~q

p <->q = (p -> q) /\ (q ->p) = (q \/ ~p) /\ (p \/ ~q) =
= (q \/ ~p) /\ p \/ (q \/ ~p) /\ q =
= (q /\ p) \/ (~p /\ p) \/ (q /\ p) \/ (~q /\ q) =
= (q /\ p) \/ (~q /\ p)

Answer 2

You can also do this using truth tables. The table for p<-->q will be the same, line for line, as that for the other expression.

p<-->q is true exactly when p and q have the same truth value, so it is true precisely when p and q are T and when p and q are F.
p ^ q is true exactly when p and q are T
~p ^ ~q is T exactly when p and q are F
(p ^ q) v (~p ^ ` q) is true exactly when one side is true or the other is true (or both - but that doesn;'t happen here), ie when p and q are both T or p and q are both F.

Answer 3

((p / q) / (~p / ~q) v is such as addition / is similiar to multiplication ~isn't or negation p-q-~p-~q-p/q-~p / ~q-((p / q) / (~p / ~q)) one million-one million-0-0-one million-0-one million one million-0-0-one million-0-0-0 0-one million-one million-0-0-0-0 0-0-one million-one million-0-one million-one million final colon is p <--> q for this reason p <--> q and ((p / q) / (~p / ~q)) are equivalent.

Answer 4

I do not understand your operands.

What does <--> and V mean?