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I need to prove that the following language is not regular

$\{c^mb^na^n \mid n>0,m\geq0\}$

But I am not sure how to do that for this particular one. I vaguely understand pumping lemma, but every example I see is in the form

$\{a^nb^n \mid n>0\}$

Or something similiar. This does not help me with my problem, because I dont fully understand pumping lemma and am not sure how to do it for two variables that have no dependency on each other. I know to take a length P from L, and I also need a string s>=P. Which then becomes P=xyz, but it is this step that confuses me.


marked as duplicate by Raphael Nov 19 '14 at 17:27

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  • $\begingroup$ Hm. You're planning to show that something is not regular, but you don't understand the pumping lemma? That's not going to work. You could try Myhill-Nerode, but most find that more difficult. $\endgroup$ – john_leo Nov 19 '14 at 16:15
  • 1
    $\begingroup$ The language of words $c^mb^n$ would be regular. It's the $ab$ part you need to think about. The $m$ is largely irrelevant to the proof. In particular, the word $c^0b^na^n$ is in the language. $\endgroup$ – Karolis Juodelė Nov 19 '14 at 16:22
  • $\begingroup$ Okay, so I could finish the pumping lemma like |xy|<=n, y=a^k where 0<k<=n, x=a^q where 0<=q<m, z=a^(n-k-q)b^n. then go say xyyz, which would leave a^(k+n)b^(n), which is not a regular language as (k+n)!=n? $\endgroup$ – tyuip Nov 19 '14 at 16:27

There are many ways to solve this. Here are some:

  1. If $L = \{c^m b^n a^n : n>0, m\geq 0\}$ were regular then so would $L \cap b^*a^*$ be.

  2. The words $\{ b^n : n > 0 \}$ are pairwise inequivalent modulo $L$, so $L$ is irregular according to Myhill-Nerode.

  3. The pumping lemma works if you ignore the $c$ part.


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