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

The following is a logical derivation using only A1, A2 and the modus ponens rule. The first two lines are assumptions:Thus $\boxed{\{\alpha \to \beta ,\beta \to \gamma \}⊢\alpha \to \gamma }$.

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 $\boxed{\Gamma ⊢\alpha }$.