Past exam of the logic and set theory course of the University of Oxford / 2010 / A / 2 / a / iii

The old system does derive the replacement axiom. Here is a syntactic proof using only its axioms, modus ponens rule, and the permitted deduction theorem for propositional logic.

First derive explosion. From $U$ and $\neg U$, the first axiom gives $\neg V\to \neg U$. The old third axiom then gives $U\to V$, and modus ponens rule with $U$ gives any desired $V$.

Next derive the classical reductio rule. Suppose $\Gamma ,\neg B$ derives both $C$ and $\neg C$. Let $T$ be the theorem $B\to B$, obtainable by applying the deduction theorem for propositional logic to the identity derivation from assumption $B$. Explosion under $\Gamma ,\neg B$ gives $\neg T$, so the deduction theorem for propositional logic gives $\Gamma ⊢\neg B\to \neg T$. The old third axiom with $\alpha =T$, $\beta =B$ yields $\Gamma ⊢T\to B$. Since $T$ is a theorem, $\Gamma ⊢B$. Thus a contradiction under $\neg B$ permits deriving $B$ with that temporary assumption discharged.

Now assume $h=\neg \beta \to \alpha$ and $\neg \alpha$, and temporarily also assume $\neg \beta$. Modus ponens rule gives $\alpha$, contradicting $\neg \alpha$. The just-proved reductio rule discharges $\neg \beta$ and gives $\beta$. Finally discharge $\neg \alpha$ and $h$ by the deduction theorem for propositional logic:No propositional completeness theorem is needed in this derivation.