# Proving a language is not regular with pumping lemma

I'm having a little bit of an issue with a pumping lemma problem. I've successfully completely all my other problems but this is the last one and I'm a little confused I must say. If anyone can help me out, it'd be much appreciated.

$$A = \{a^n b^m c^l \mid n\leq m \vee m\leq l\}$$

• take $a^nb^n$ (no $c$'s) and pump up the $a$'s. – Ran G. Mar 12 '13 at 1:28
• @RanG., that doesn't belong to the language. But $a^N b^N c^N$ does, take $N$ as the pumping lemma's constant, you can pump up the $a$. – vonbrand Mar 12 '13 at 8:23
• @vonbrand: It does, since $n\leq n \vee n \leq 0$ is true. – frafl Mar 12 '13 at 8:35
• @frafl, no coffee yet (misread $\wedge$ for $\vee$) – vonbrand Mar 12 '13 at 9:49

To expand a little on Ran G.'s comment, given a pumping length of $p$, we can take the string $s=a^{p}b^{p} \in A$, then splitting into $s=xyz$, the $y$ section (the part that can be pumped) must be all $a$s, so pumping up leaves us with a string $s'=a^{p+k}b^{p} \notin A$.