Proving $\{ll^{R}l|l\in\{a,b\}^{*}\}$ is not context free using the pumping lemma How can I prove, using the pumping lemma for context free languages,
that $\{ll^{R}l|l\in\{a,b\}^{*}\}$is not a context free language
?
I tried to put $n$ as the pumping lemma constant and chose $a^{n}a^{n}a^{n}$
but this does seem to have a factorization since all I need is that
the amount of $a$ in the word is divisible by $3$.
I then tried to take $l':=a^{n}b^{n}b^{n}a^{n}a^{n}b^{n}$ ($l=a^{n}b^{n})$,
I am having a hard time even figuring out if this will lead to a contradiction:
My thoughts are that if $l'=uvwxy$ then since $|vwx|\leq n$ it must
be contained in the $a^{n}$ or $b^{n}$ or be something like $a^{k_{1}}b^{k2}$
or $b^{k_{1}}a^{k2}$. but I'm having a difficult time continue from
here, it seems that now I should go over each option and devide to
options of how to factor it to $v,w,x$ which gives many cases.
Am I on the right path or can I do something easier/smarter ?
Please note: Since this is important to me, as this is how I prepare
for my exam tomorrow (thus can't offer a bounty right now), I will
give $50$ points (+upvote+accept) for a somewhat full solution of
this question

Try $l = b a^n b$, for some cutting $l l^R l = u v x y w z$ with $\lvert v x w \rvert \le n$ you have $u v^k x y^k z$ in your language. This requires that $bb$ is in $x$ (can't repeat it), and the $b$ starting the last $l$ falls outside the repeat anyway. Repeating $x$ and $y$ gives inflated $l$'s in the first part that are too long for the last untouched $l$.
This is just a rough outline, need to fill in many details.
