 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/4f/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Choose a new start variable

S_{0} \in / V \cup V^{'}

H = (Σ, V \cup V^{'} \cup {S_{0}}, P \cup P^{'} \cup {S_{0} \to S S^{'}}, S_{0}) .

Indeed, its derivations first produce a word from

G

and a word from

G^{'}

in their original order, so

L (H) = L (G) L (G^{'})

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