Is $S = $ {$∨, ∧,→,↔$} complete ? Can negation represented with the help of connectived from $S$? Inductively we can proof that negation operator cannot be replaced by any one of the connective belonging to $S$ . How can i prove that even the combination of connective from $S$ cannot represent negation ?
Or my assumption is wrong i.e $S$ is complete ?
 A: Your guess is right - that set of connectives is not functionally complete (unless you also include a primitive symbol $\perp$ for falsity, but that's cheating: that's really an additional nullary connective which combined with $\rightarrow$ gives you negation).
As often happens, in order to prove this you're going to have to prove something stronger. Intuitively, it should be impossible to express negation in terms of the connectives given, so let's make that precise:

If $\varphi$ is built from propositional atoms and the connectives $\wedge,\vee,\rightarrow,\leftrightarrow$, then the valuation making all propositional atoms true makes $\varphi$ true.

This immediately shows that this set of connectives cannot express negation (think about the formula $\neg p$). The proof meanwhile uses one of the standard techniques in this context, namely structural induction. The base case consists of just the propositional atoms on their own, for which the claim is trivial; the induction steps (of which there are four, one for each connective in the set) essentially rely on the fact that each of the relevant truth tables sends (True, True) to True. (Can you see how to make this precise?)
