The

"a^n b^n" language isn't a

regular language; this can be easily proved with the

pumping lemma.

Suppose the language were regular. Since it's infinite, for large enough n the pumping lemma guarantees the decomposition `a^n b^n = x y z`, with nonempty `y`, such that `x y^k z` is always in the language.

This is clearly absurd: either `y` consists wholly of `a`'s, and then `x y^2 z` contains more a's than b's, or it consists wholly of b's, and then that word contains more b's than a's, or else it contains a's and then b's - but then `x y^2 z` is of the form `a^n_1 b^n_2 a^n_3 b^n_4`, which clearly is not in the language.