# Prove derivability without using the deduction theorem. [closed]

It is necessary to prove the derivability: $$\vdash((\lnot B \to B) \to \lnot \lnot B)$$.
My steps:
$$\lnot B \to B \vdash \lnot \lnot B$$
1)$$\lnot B \to B$$ - hypothesis
2)$$(((\lnot B) \to B)) \to (((\lnot B) \to B) \to \lnot \lnot B)$$ (A3)
3)$$(((\lnot B) \to \lnot \lnot B) \to \lnot \lnot B)$$ (1,2 MP)
But then I don't understand what to do.

• Can you list the precise axioms you're using? It seems it's a Hilbert calculus, but different textbooks may use different names for the axioms... Oct 27 at 21:51

Your application of A3 seems problematic, you can do a more careful inspection to see is your step 2) really correct by invoking A3 of your system, which seems Hilbert $$H_2$$ axiom system:
A3: $$(\neg A\to \neg B)\to ((\neg A\to B)\to A)$$
One way to proceed is to use proof by negation, assume $$\lnot B$$, then using MP you can easily get $$B$$ and clearly you arrive at $$\bot$$, a contradiction. So you then can safely conclude $$(\lnot B \to B) \to \lnot \lnot B$$.