# Formal Languages and Automata Theory [duplicate]

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.

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

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.