2
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ Is this duplicate correct? this quiestion is asking for the formula ((¬A → A) → A) while the other question is asking for ((A → ¬A) → A). $\endgroup$ – Mauro curto Oct 2 '20 at 18:48