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This question already has an answer here:

How can I show that $L = \{a^m b^n \mid (m > n \text{ or } m < n) \text{ and } m, n ≥ 1\}$ is not a regular language.

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marked as duplicate by David Richerby, Yuval Filmus, D.W., Ran G., Raphael Jun 5 '15 at 22:07

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  • $\begingroup$ What have you tried? What is stopping you? What techniques have you learned for doing such a proof? Regarding style, it is nicer to ask for help rather than using a sentence that looks like an order, for example "How do I show ...., I tried this and that but I am stuck because ...". Also, -please use LaTeX for writing math. $\endgroup$ – babou Jun 5 '15 at 21:41
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Use the fact that compliment of a regular grammar is regular. So assume this language is regular. Then its complement is also regular that is $$\{{a^{m}b^{m}}\}$$ Now using pumping lemma you can show that this is not a regular language.

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    $\begingroup$ Thanks for restricting to a hint, since this looks a lot like a homework question. However, you're missing a step: the complement of the language in the question isn't the one you give. $\endgroup$ – David Richerby Jun 5 '15 at 19:20
  • $\begingroup$ I deliberately left it incomplete and gave away the major step $\endgroup$ – iLoveCamelCase Jun 5 '15 at 19:27

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