-1
$\begingroup$

This question already has an answer here:

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?

$\endgroup$

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I tried to improve your question, but there are still parts that do not make sense for me. $\endgroup$ – babou May 7 '15 at 22:20
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ yes i do agree but what if i want to use the pumping lemma? $\endgroup$ – user3841581 May 9 '15 at 5:35
  • $\begingroup$ Then keep working on it. $\endgroup$ – Yuval Filmus May 9 '15 at 5:42
  • $\begingroup$ 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 $\endgroup$ – user3841581 May 9 '15 at 7:43

Not the answer you're looking for? Browse other questions tagged or ask your own question.