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

Question

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 ) $

Answer 1

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.

Answer 2

Hint: Assume $\neg\alpha \wedge \neg\neg \alpha$, derive a contradiction. Does the resulting proof requires some use of the double negation elimation rule?

Answer 3

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 $P=\lnot\alpha$.