Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 2 / 4I / iii / Solution

This language is not regular. If it were regular, its intersection with the regular language $0^{\ast }{1}^{\ast }$ would be regular because regular languages are closed under intersection. That intersection is exactlywhich part (ii) proved nonregular.

Solved by gpt-5.6-sol high.