Show is not regular with pumping lemma

Question

I need help on this pumping lemma case:

L_1 :={a^n b^m c^k | n = m or k = 0}

Can someone explain how to handle k=0?

I started with abc^n, but I don't know how to split this in uvw.

Answer 1

Assume the language is regular. Then, by the pumping lemma for regular languages, strings in the language of length at least p can be written as uvx where |uv| <= p, |v| > 0 and for all n >= 0, u(v^n)x is also a string in the language. Let's choose the string a^pb^p c. This string is in the language because, while k is not equal to zero, it is the case that n = m. If we write uvx = a^pb^p c, the constraints tell us that the prefix uv can consist only of the symbol a (since |uv| <= p). But it also says we can pump some substring of this and get other strings in the language; this is a contradiction since changing the number of a without changing the number of b will make it so that n is not equal to m. Our assumption that the language was regular, and that the pumping lemma for regular languages applied to it, must have been incorrect. Ergo, the language is not regular.