3 ms·
The proof for a^n b^n, as I was taught, being that -> there exists a integer N such that pumping lemma holds for any w in L longer than L. (That is we can divi
by chakkepolja 5y ago
The proof for a^n b^n, as I was taught, being that
-> there exists a integer N such that pumping lemma holds for any w in L longer than L. (That is we can divide w into xyz and pump y, where |xy| ≤ n, and y is nonempty).
-> now consider a^N b^N, we cant divide it into xyz such that we can pump y and w still remains in a^n b^n.