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Given the language $L:= { \{ c^{2k} w \ \vert \ k \ge 1, \ w \in \{a,b,c\}^* \ and \ \vert w\vert_a \ = \ \vert w\vert_b \} \ \cup \ \{ a,b \}^* }$

I'm really unsure how to even start because of the union.

I tried it with $w=a^nb^n$ but my correction said that it's pumpable because it is in $\{ a,b \}^*$ which makes sense.

What would be a good word to start with? I guess there are several cases i need to show?

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If $L$ was regular, so would be $L \cap c^+\Sigma^* = \{c^{2k}w\mid k\geq 1, w\in\{a,b,c\}^* \text{ and }|w|_a = |w|_b\}$.

That means that if you prove that this language is not regular, then $L$ cannot be regular too (that way, we got rid of the union).

Now you can take it from here with pumping lemma, starting from $c^2a^nb^n$!

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  • $\begingroup$ Thanks for the answer. I tried to make a little "sketch". So i take the word c^2a^nb^n as you said. Because |xy| <= n and |y| >= 1, y has to be y = a^k and 1<=k<=n. With that i can pump xyz to xyyz and get c^2a^(n+k)b^n which is it not regular because the amount of as and bs is not the same. Would this make sense or did i miss something (I feel like i did). $\endgroup$
    – momos
    May 21 at 17:52
  • $\begingroup$ Technically, $y$ could contain $c$'s, but overall you have the idea. $\endgroup$
    – Nathaniel
    May 21 at 18:07
  • $\begingroup$ So i would have to make several cases? $\endgroup$
    – momos
    May 21 at 18:10
  • $\begingroup$ Yes, and find a contradiction in each case. $\endgroup$
    – Nathaniel
    May 21 at 18:11
  • $\begingroup$ Thank you very much! $\endgroup$
    – momos
    May 21 at 18:12

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