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I have the following language i want to prove is not regular by using the pumping lemma but not sure on what to use as the xyz and the pumping length. can i get a point in the right direction for how to prove it is not regular. {z^7n c^5 c^n+7 c^m | n,m >= 0} i have tried condensing it down to z^7n c^n+m + 12 but dont know what word to pump down, as pumping up will just keep increasing m value.

Joe Hicks
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  • To be clear, this is $$\left{z^{7n}c^5c^{n+7}c^m:n,m\ge 0\right},?$$ If so, take $p$ as the pumping length, and start with $w=z^{7p}c^{p+12}$, i.e., with $n=p$ and $m=0$; all of the pumping will take place in the $z$ part. – Brian M. Scott Nov 04 '20 at 21:13
  • yes this the correct equation, still unsure how to pump after this. – Joe Hicks Nov 04 '20 at 21:21
  • more specifically how to split into xyz – Joe Hicks Nov 04 '20 at 21:27
  • You don’t get to say how it splits into $x,y$, and $u$. (I’m using $u$ instead of $z$, because $z$ is one of the letters in the alphabet for this language.) The only things that you can assume about the split are that $|xy|\le p$ and $|y|\ge 1$. From that information alone you have to show that there is some $k\ge 0$, such that $xy^ku$ is not in the language. And in this case that’s easy, because $k=1$ is the only value of $k$ for which $xy^ku$ is in the language: changing $k$ by $1$ changes the number of $z$s without changing the number of $c$s. – Brian M. Scott Nov 04 '20 at 21:31
  • sorry to keep questioning , but how would i translate this into a proof – Joe Hicks Nov 04 '20 at 21:40
  • something like this https://math.stackexchange.com/questions/2069389/prove-that-the-language-is-not-regular-using-pumping-lemma?rq=1 explains it in an easier format :) – Joe Hicks Nov 04 '20 at 21:41
  • Here and here are two pumping lemma arguments worked out in full detail. – Brian M. Scott Nov 04 '20 at 21:53

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