regular language - Using condition 3 of the pumping lemma to prove irregularity -


so, i've been going through sipser's book on theory of computation, , 1 of exercises is:

let b language {0^n1^n | n≥0}.

prove b not regular.

the book continues give proof, using pumping lemma, letting s=0^p 1^p, s=xyz , testing 3 cases; when y 0's, 1's, 0's , 1's. can't understand how latter 2 options possible condition 3 of pumping lemma |xy|≤p. condition not imply y can 0's?

does condition not imply y can 0's?

no, not necessarily. fix p, , let string be

w = xyz = 01/2 p11/2 p

what prevents x being first character, z being last, , y being in between?


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