 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-2/4i/iii/solution

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

This language is not regular. If it were regular, its intersection with the regular language

0^{*} 1^{*}

would be regular because regular languages are closed under intersection. That intersection is exactly

{0^{n} 1^{n} : n \geq 0},

which part (ii) proved nonregular.

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