Does$\{w\in\Sigma^*=\{a,b\}^*: \nexists u\in \Sigma^*, w=uu\}$ uphold the pumping lemma

Question

I have the $L=\{w\in\Sigma^*=\{a,b\}^*: \nexists u\in \Sigma^*, w=uu\}$.

Does it uphold the pumping lemma (regardless of it being regular or otherwise)?

Answer 1

Apparently it does not. Proof:

For every $n>0$ we can choose the word $w=a^nba^mb$ where $m=n+n!$ ($n\neq m$). Which ever way we split $w$ into $w=xyz$ (such that $|xy|\leq n, |y|>0$) it must be that $y$ is a sequence of $k$ $a$'s where $k\leq n$, thereforewe can always choose to pump $y$ $i$ times, where $n-k+ik=m\Rightarrow i=(m-n+k)/k=n!/k+1$ and since $k\leq n$ we know this number is in $\mathbb N$, hence $xy^iz=a^mba^mb\notin L$.