Proving $ \neg ( \neg \alpha \wedge \neg \neg \alpha )$ I'm training to prove this statement , but first I need to know if this statement can be proved in : 
1 -  both in classical and Intuitionistic logic ( in this case i need to provide demonstration in Intuitionistic logic )
2 -  classical logic but not Intuitionistic logic ( in this case i need to provide a Kripke Counter-Models ) 
3 - not provable in either classic and Intuitionistic logic ( in this case i need to provide a classic Counter-Models )
My question is how to distinguish if a statement is provable in one of this cases ? 
PS :  I know the Intuitionistic logic doesn't allow the elimination of double negation 

$ \neg ( \neg \alpha \wedge \neg \neg \alpha ) $


This statement can be proved in minimal logic. When you rewrite the negations as implications in the usual way, the statement is
$$
((\alpha \to \bot) \land ((\alpha \to \bot) \to \bot) \to \bot
$$
which is really of the form
$$
(X \land X \to Y) \to Y
$$
which is just a form of modus ponens. The provability of the statement has nothing to do with negation, really, apart from rewriting the negations as implications in the usual way. 
