Pumping Lemma Proof (Type of wcw language) I have the language $L = \{ dkd\space \mid d \in \{a,b\}^*, k \in \{a,b\} \}$ and i have to show that it's non-regular using the pumping lemma.
The structure of the language i think can be explained as (based on definition above) $\{a^{n}b^{n}ca^{n}b^{n} \mid c \in \{a,b\}\}$
Based on pumping lemma, it must have this 3 properties:

$\lvert y \rvert \gt 0$
$\lvert xy \rvert \le p$
$xy^iz\in L$ for all $i \ge 0$.

Based on that, i have $a^{p}b^{p}ca^{p}b^{p}$ and $ 4p+1 \ge p $
Based on 2nd property there must be 9 scenarios (i guessed that from the theory):
One of them should be this one:

x = $\{a^{i}$} $\space $ y = $a^{p-i}$ $\space$ z= $b^{p}ca^{p}b^{p}$

Based on the 3rd property i can use $xy^{2}z$ = $a^{i}$$a^{p-i}$$a^{p-i}$$b^{p}ca^{p}b^{p}$ eventually giving me $a^{p}$$a^{p-i}$$b^{p}ca^{p}b^{p}$, the explaination is: it fails since $a$'s of the start arent the same (in number) with the $a$'s of the end.
If the above is correct how i find the other 8 scenarios and it does make any difference that my $c$ is ($a$ or $b$) instead of a single constant character?
The structure of the language i think can be explained as (based on definition above) $\{a^n b^n ~c~ a^n b^n |c \in \{a,b\} \}$

As explained in the comments, this is a wrong inference. If
$$
L = \left \{ d k d ~|~ d \in \{ a, b \}^{*}, k \in \{a, b\} \right \}
$$
then $d$ can be any finite sequence (of any length) of $a$ or $b$ terminals. Recall that, for an alphabet $\Sigma$, Kleene star is defined inductively as
\begin{align}
\Sigma^0  &amp; = \{ \epsilon \} \\
\Sigma^n  &amp; = \{ \omega a ~|~ \omega \in \Sigma^{n - 1}, a \in \Sigma \} \\
\Sigma^{*}  &amp; = \bigcup_{n \ge 0}{\Sigma^n}
\end{align}
So for example
$$
\{ \epsilon, a, b, ab, ba, abab, abbb, baba \} \subset \{ a, b \} ^{*}
$$
while instead
$$
\{ a^n b^n | n \ge 0 \} = \{ \epsilon, ab, aabb, aaabbb, \ldots \}
$$
To tackle your problem, I can recommend negating the Pumping lemma and then

Fix an arbitrary $N &gt; 0$
Take $\omega = a^N b a^N$ with $d = a^N$ and $k = b$
Equate $\omega = xyz$ and note that it must be $|xy| \le N$ and $|y| &gt; 0$

Can you conclude your proof?

 We just chose our $\omega \in L$. We must show that, whatever our decomposition, $\exists k \ge 0. ~ x y^k z \not \in L$. Let $x, y, z$ be such that \begin{cases} \omega &amp; = xyz \\ |xy| &amp; \le N \\ |y| &amp; &gt; 0 \end{cases} then, whatever $y$, for $k = 0$ you always have \begin{equation} x y^0 z = a^M b a^N \end{equation} with $M \neq N$.  Thanks @plop for strenghening the proof.