Take two worlds $r<s$ and let $p$ be forced only at $s$. Neither world forces $\neg p$: at $s$ this follows from $s\Vdash p$, while at $r$ the extension $s$ forces $p$. Consequently every extension of $r$ that forces $\neg p$ also forces $p$ vacuously, so
$$
r\Vdash\neg p\to p.
$$
But $r\nVdash p$. The implication clause therefore gives
$$
r\nVdash(\neg p\to p)\to p,
$$
which is a finite <Kripke countermodel> and proves that the formula is not intuitionistically valid.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 120 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-120/1/c/solution