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?
-1
$\begingroup$
$\endgroup$
-
$\begingroup$ 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
$\begingroup$
$\endgroup$
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.
answered May 7 '15 at 22:51
-
$\begingroup$ yes i do agree but what if i want to use the pumping lemma? $\endgroup$ – user3841581 May 9 '15 at 5:35
-
-
$\begingroup$ 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
