Past exam of the logic and set theory course of the University of Oxford / 2011 / A / 1 / b / ii / Solution

An instance of A3 isThe second assumed theorem and the modus ponens rule therefore give $\Gamma ⊢\beta \to \alpha$. Apply the preceding implication-composition derivation to this theorem and $\Gamma ⊢\neg \alpha \to \beta$. We obtain $\Gamma ⊢\neg \alpha \to \alpha$. The permitted theorem $(\neg \alpha \to \alpha )\to \alpha$ and one more modus ponens rule application yield $\boxed{\Gamma ⊢\alpha }$.