Prove with pumping lemma that the language { $a^n b^n b^m a^m | n ≠ m $ } is not context free

Asked May 08 '21 at 07:16

Active May 08 '21 at 20:59

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I'm having a trouble proving it to be non-context-free.

For example, if I take w = $a^k b^k b^{k+1} a^{k+1}$, it would be problematic if the partition of $vxy$ with $|v| = |y|$ was in the $ b^{k+1} a^{k+1} $ part as I can't get it to be equal with $k$ or make the amount of $a$s and $b$s different.

We are also allowed to use Ogden's Lemma.

edited May 08 '21 at 20:59

John L.

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asked May 08 '21 at 07:16

user10131421414

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Are you familiar with the pumping lemma proof that ${a^nb^m \mid n \neq m}$ isn't regular? – Yuval Filmus May 08 '21 at 13:08
The trick is explained here: Prove that the following language is not regular: ${ 0^i1^j \mid i\neq j }$. – Hendrik Jan May 08 '21 at 15:36
Consider $ab b^{p!+1} a^{p!+1}$. If $n,m$ can be zero, we can also consider $b^{p!} a^{p!}$. Here $p$ is a pumping length. – John L. May 08 '21 at 21:02

Prove with pumping lemma that the language { $a^n b^n b^m a^m | n ≠ m $ } is not context free

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