Is <-> an Equivalence Relation on Propositions?

[soopo]

Homework Statement

We have an equivalence relation such that
A <-> B.

Prove that the equivalence relation is true.

The Attempt at a Solution

Let
P: A -> B
Q: B -> A

Let's prove the relation by contradiction.
Assume
\neg A -&gt; \neg B

The previous assumption is the same as Q. Thus, we have a contradiction, since
it is impossible that both of the following Q and \neg Q
are true at the same time, where

Q: B -> A and
\neg Q: \neg A -&gt; \neg B which is the same as B -> A.

Thus, the equivalence relation is true between A and B.

 

[qntty]

Homework Statement

We have an equivalence relation such that
A <-> B.

Prove that the equivalence relation is true.

Is A <-> B given or is that what you're trying to prove?

 

[AUMathTutor]

Are you trying to show that <-> is an equivalence relation on propositions?

If that is the case, you have to show the following three things:

A <-> A for all A.
if A <-> B then B <-> A.
if A <-> B AND B <-> C then A <-> C.

How should you show these? I'd probably use truth tables and the definition if <-> in terms of -> and "AND".