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 ) $


For every proposition $P$, the deduction that $P\land\lnot P\to\bot$, that is, that $\lnot(P\land\lnot P)$, is simply applying modus ponens, therefore intuitionistically valid; this is because $\lnot P$ just means $P\to\bot$. Apply this for $P=\lnot\alpha$.
