# pumping lemma for $L=\{a^n b^m c^k \mid n = m \vee m\neq k\}$ [duplicate]

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Using pumping lemma, how can I prove that $L=\{a^n b^m c^k \mid n = m \vee m\neq k\}$ is not regular?.

If I choose $w= a^m b^m c^m$ and pump up with $i=2$, if have $a^m=1 b^m c^m$ but the string is still in the language. Any hint?

## marked as duplicate by D.W.♦, Nicholas Mancuso, Luke Mathieson, David Richerby, JuhoMay 8 '15 at 9:07

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• I tried to improve your question, but there are still parts that do not make sense for me. – babou May 7 '15 at 22:20

## 1 Answer

The easiest way to show that $L$ isn't regular is by noticing that $$L \cap b^+c^+ = \{ b^m c^k : m,k \geq 1, m \neq k \}.$$ This should look familiar.

• yes i do agree but what if i want to use the pumping lemma? – user3841581 May 9 '15 at 5:35
• Then keep working on it. – Yuval Filmus May 9 '15 at 5:42
• i think i got an idea; what if for a given m, we choose w=a^m b^m a^m; clearly this w is in L; and we can easily find the contradiction of the pumping lemma here – user3841581 May 9 '15 at 7:43