Take
$$
\delta=\lambda x.\lambda f.f(xf).
$$
If $L$ is a <fixed-point combinator>, then $Lf\equiv_\beta f(Lf)$, so eta-conversion gives
$$
L\equiv_\eta\lambda f.Lf\equiv_\beta\lambda f.f(Lf)=\delta L.
$$
Conversely, if $L\equiv_{\beta\eta}\delta L$, application to an arbitrary $f$ gives $Lf\equiv_{\beta\eta}f(Lf)$, which is precisely the fixed-point-combinator property.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 120 / 2 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-120/2/e/solution