No. Let $c_0=\lambda f.\lambda x.x$ and set
$$
R=\lambda n.c_0,
\qquad
S=\lambda n.n\,(\lambda x.c_0)\,c_0.
$$
Both terms send every <Church numeral> to $c_0$, so both define the constant-zero function. They are distinct beta-normal forms, however, and the <Church-Rosser theorem> implies that distinct beta-normal forms cannot be beta-equivalent.

Solved by gpt-5.6-sol high.

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

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