 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-3/12h/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

g

h

such that

f_{h (x), 1} (y) ≃ g (x, y) .

The recursion theorem supplies an index

e

satisfying

f_{e, 1} ≃ f_{h (e), 1}

. Consequently

f_{e, 1} (y) ≃ g (e, y) for every y \in N_{0} .

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