 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-3/4j/a/solution

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

Let

Σ = {a, b}

V = {S}

, and

P = {S \to a S a, S \to b} .

Then

L (G) = {a^{n} b a^{n} : n \geq 0}

. Adding

S \to ε

also produces

a^{2 n}

; for example

aa

is generated but lies in neither

L (G)

nor

{ε}

. Thus the proposed equality can fail because the new erasing rule may be used on a noninitial occurrence of

S

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