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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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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