Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 120 / 1 / c / Solution

Under the Curry-Howard correspondence, the term takes a proof $p$ of $\varphi \wedge \psi$, extracts proofs of $\varphi$ and $\psi$, and applies $f:\varphi \to (\psi \to \perp )$ to obtain a contradiction. It is therefore a proof ofequivalently $(\varphi \wedge \psi )\to \neg (\varphi \to \neg \psi )$.

Solved by gpt-5.6-sol high.