# How do I show that $\vdash (\lnot \alpha \to \alpha) \to \alpha$? [duplicate]

I want to show that: $$\vdash (\lnot \alpha \to \alpha) \to \alpha$$, and here's what I've tried:

1. Using the deduction theorem, this is the same as proving $$\{(\lnot \alpha \to \alpha)\} \vdash \alpha$$
2. $$\alpha \to (\lnot \alpha \to \alpha)$$ (axiom)
3. $$(\alpha \to (\lnot \alpha \to \alpha)) \to ((\alpha \to \lnot \alpha) \to (\alpha \to \alpha))$$ (axiom)
4. $$(\alpha \to \lnot \alpha) \to (\alpha \to \alpha)$$ (Modus-Ponens on 2 and 3)

How do I go ahead from here? Thanks!

List of Axioms:

• $$\alpha \to (\beta \to \alpha)$$ (1)
• $$(\alpha \to (\beta \to \gamma)) \to ((\alpha \to \beta) \to (\alpha \to \gamma))$$ (2)
• $$(\lnot \beta \to \lnot \alpha) \to (\alpha \to \beta)$$ (3)

and Modus-Ponens is the sole rule of inference.

• Is this duplicate correct? this quiestion is asking for the formula ((¬A → A) → A) while the other question is asking for ((A → ¬A) → A). – Mauro curto Oct 2 '20 at 18:48