I'm trying to show that the language in the title is not regular but I don't know how to chose a decomposition $x = abc$ to express the word in the language because the $w \in \{a,b,c\}^*$ is kinda tricky for me, I know it means $ \{ a,b,c\} \cup\{ a,b,c\}^2 \cup ... \cup \{ a,b,c\}^k $ but I don't know how to chose the decompositons for a word in the language and proove that is not in the language by pumping lemma. Any tips ?
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After $n$ is fixed, you could choose $x = b^na^nb^n$. This string is clearly in $L$. By the Pumping lemma $x$ can be written as $uvw$, where $|uv| \leq n$ and $|v| \geq 1$. So for any choice of $u$ and $v$, $uv$ is a string consisting of only $b$s, and so is $v$. Thus $uv^iw = b...b(b...b)^ib...ba^nb^n$ which is not in $L$.
On @DavidRicherby comment:
The Pumping Lemma:
Let $L$ be a regular set. Then there is a constant $n$ such that if $z$ is any word in $L$, and $|z| \geq n$ we may write $z = uvw$ in such a way that $|uv| \leq n$, $|v| \geq 1$, and for all $i \geq 0$, $uv^iw \in L$.
How to use
1) Select the language you wish to to prove irregular.
2) The "adversary" picks $n$, the constant mentioned in the Pumping lemma. Once it has been picked, it may not be changed (fix it).
3) Select a string $z$ in $L$. Your choice may depend implicitly on the value of $n$.
We choose $z = b^na^nb^n$.
4) The "adversary" breaks $z$ into $u$, $v$, and $w$, subject to the constraints that $|uv| \leq n$ and $|v|$.
In my proof the prefix of $z$ consists of $n$ $b$'s, so any choice of $u$, $v$, and $w$ implies that $v$ consists solely of $b$s.
5) You achieve a contradiction to the pumping lemma by showing that there exists $i$ such that $uv^iw \notin L$.
Just choose $i=2n$ and we have unequal number of $b$s on the left and right sides of the substring $a^n$: $b^{2n + d}a^nb^n \notin L$.
This is how we prove.
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