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$L=\{a^{2k}b^nb^k\mid k\geq0, n\geq0\}$ over alphabet $\{a,b\}$

How do I prove that $L$ is not regular using Pumping Lemma? All the examples I've come across had same exponents all around, and I'm a bit confused how should I write this down.

Should I start with $w=\{a^pb^pb^p\mid p\leq k, p\leq n\}$ where $w$ is a subset of $L$, or how should this be tackled?

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    $\begingroup$ You cannot start with your $w$ since it's a set rather than a word. A good starting word would be $a^{2k} b^k$ for an appropriate $k$. $\endgroup$ – Yuval Filmus Oct 23 '14 at 4:24
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  • Choose for example the word $x=a^{2k}b^{k}\in L$ ( with $|x|=3k\ge k$)
  • Possible partitions of $x=uvw$ are $u=a^l,\;v=a^s,\;w=a^mb^{k}$ with $s\ge 1,\; l+s\le k$ and $l+s+m=2k$
  • Now lets look at $uv^2w=a^{2k+s}b^k$ which is clearly not in $L$. If $s$ is odd, than the number of $a$'s is odd and therefore $uv^2w\notin L$. If otherwise $s$ is even, then the number of $b$'s would have to be larger than $(2k+s)/2$, which it is not ($k<(2k+s)/2)$. Therefore $uv^2w\notin L$ holds for each $s\ge 1$.

$\rightarrow$ $L$ is not regular.

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  • $\begingroup$ Nice. You could even have eliminated everything but the first sentence of your third bulleted item. $\endgroup$ – Rick Decker Oct 23 '14 at 13:27
  • $\begingroup$ I wanted to avoid a question like "Why is $uv^2w$ not in $L$?" :) $\endgroup$ – Danny Oct 23 '14 at 13:32

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