position of atomic propositons in bi-conditionals In implication position of $p$ and $q$ is important and can't be interchanged but I guess in case of bi-conditionals these two can be interchanged freely.
I mean to say $p\to q$ and $q\to p$ will not be equivilent necesserily but $p\leftrightarrow q$ and $q\leftrightarrow p$ convey the same meaning.
For example I have a statement 
If you watch television your mind will decay and conversely.

so expressing it by using if and only if 
(1) Your mind will decay if and only if you watch TV.
(2) you watch TV if and only if you mind decays
So are statements 1 and 2 equivalent?
 A: Yes, they are equivalent. The biconditional connective $(\leftrightarrow)$ is commutative. This can easily be seen when we expand $$p \leftrightarrow q \equiv (p\rightarrow q) \land (q \rightarrow p)$$
Since the connective $\land$, as we know, is commutative, we have have $$\begin{align} p \leftrightarrow q & \equiv (p\rightarrow q) \land (q \rightarrow p)\\ \\ & \equiv (q\rightarrow p) \land (p \rightarrow q) \\ \\ &\equiv (q \leftrightarrow p)\end{align}$$
The truth of a biconditional statement  depends only on the agreement in the truth value of the propositions it connects.
So $$p \leftrightarrow q \equiv q \leftrightarrow p$$
and is true if and only if (both $p$ and $q$ are true) or (both $p$ and $q$ are false). Since this says the same as being true if and only if (both $q$ and $p$ are true) or (both $q$ and $p$ are false), which "side" of the biconditional connective each of $p, q$ appear really doesn't matter.
We can easily confirm this all by constructing a truth-table for each, and seeing their line-by-line agreement. Note the "quirky" symbol Wolfram Alpha uses to denote the connective $\iff$!

A: Yes, (1) and (2) are equivalent under propositional calculus.
You can see the difference with the method of analytic tableaux. You start with the negation of your original formula then apply a series of contradiction hunting rules. If each of the paths of the tableau is closed (i.e. contains a contradiction), then the original formula is a tautology. If some paths are open (i.e. do not contain contradictions), then each open path provides a counterexample to the original formula being a tautology.
Observe:

This next one, when completed, is closed, but I will leave it to you to finish. Here:

Do you see it now?

 The second tableau continues from "?" in almost exactly the same way as it does on the left.

