Proof there was no tautologically equivalent wff to $(A\leftrightarrow B)$, constructed only by conditional symbol . The way I did was to simply construct a truth table about $\rightarrow$, and treat the 2,3 line as ordered pairs in vertical direction. (Given $A: T T FF$ $B:TFTF$.)
I got $(T,F)$ from $A$ and $(F,T)$ from $B$. Thus all the possible in the 2,3 line were $(T,F),(F,T),(T,T)$ by operation $\rightarrow$. Thus the pair $(F,F)$ in $(A\leftrightarrow B)$ was not possible by $\rightarrow$. 
However, is there any other way to prove it without listing all the posibilities?
 A: Consider any expression using only the implication $\to$. Possibly it is very long, such as $(((A\to B)\to ((A\to A)\to(B\to A)))\to (A\to B))$ or what have you. Perhaps it has millions of symbols, or just a few. 
In any case, consider the very last propositional variable that appears in the statement, such as the variable $B$ in the example above, which appears at the extreme right. Let's imagine a case where that variable is true. It follows that the implication it appears in is also true, and so the next level implication that appears in is true, and so on all the way out to the top-level implication. So the whole expression will be true, provided that we specify merely that that variable at the end is true. But $A\leftrightarrow B$ cannot be guaranteed true by knowing only that one of the variables is true. Therefore, we cannot express $A\leftrightarrow B$ in terms of the conditional $\to$.   
A: One way of doing it: Prove by (strong) induction on the number of $\implies$ in your expression that if either of $(A,B)=(T,F)$ or $(A,B)=(F,T)$ in a statement that uses only $\implies$ gives false (which it should if we're trying to make $\iff$), then the other one gives true.
