Does $a^n b^m$ with $n, m \ge 0$ and $m \ne n$ fail the pumping lemma?

Asked Sep 06 '16 at 02:21

Active Sep 06 '16 at 02:27

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I know it is not regular and that this can be demonstrated by using closure properties, but does the language pass the pumping lemma?

It seems to me that for any p I can pump either a or b (depending on which has a greater repetition count) without making n = m. But in an exam I am taking the answer key says it does not satisfy the pumping lemma.

edited Sep 06 '16 at 02:27

asked Sep 06 '16 at 02:21

Rodrigo Stv

What are your thoughts on the question? – Yuval Filmus Sep 06 '16 at 02:22
It seems to me that for any p I can pump either a or b (depending on which has a greater repetition count) without making n = m. – Rodrigo Stv Sep 06 '16 at 02:25
Have you tried proving your suspicion? That's how we find out things in math. – Yuval Filmus Sep 06 '16 at 02:27
Yes, I am scribbling a lot of stuff, but in the answer key from an exam it says this language does not pass the pumping lemma. So I am suspicious that the answer key might be incorrect. – Rodrigo Stv Sep 06 '16 at 02:28
The answer key is correct. – Yuval Filmus Sep 06 '16 at 02:29
So my proof is wrong. I am thinking of a string that can not be pumped anyhow. Do you have any hint on it? – Rodrigo Stv Sep 06 '16 at 02:30
Yes. That method solves my question. But the other question's title is hard to search for. – Rodrigo Stv Sep 06 '16 at 02:37

Does $a^n b^m$ with $n, m \ge 0$ and $m \ne n$ fail the pumping lemma?

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