Take
$$
OR=\lambda a:Bool_\sigma.\lambda b:Bool_\sigma.
\lambda x:\sigma.\lambda y:\sigma.a\,x\,(b\,x\,y).
$$
It has type $Bool_\sigma\to Bool_\sigma\to Bool_\sigma$. If $a\equiv_\beta\top$, the body selects $x$. If $a\equiv_\beta\bot$, it reduces to $bxy$, which selects $x$ exactly when $b\equiv_\beta\top$. Thus it has the stated truth table.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 120 / 3 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-120/3/e/solution