How to prove that this language is not regular?
Now if we have $UVVW$ that means $V$ is repeating some of the characters in $x$ and therefore it must repeat some $x_i$. Therefore we have reached a contradiction, which shows that $L$ is not a regular language.


This is not necessarily true. Suppose $x=[0^p, 0^{p-1}, \dots, 00, 0]$. If $V$ is of length $k&lt;p$ and entirely in $0^p$, we get $UVVW = [0^{p+k}, 0^{p-1}, \dots, 00, 0]$ which is still in $L$.

Normally when doing a proof using pumping lemma, we try to use a more specific $x$ such that the pumping lemma fails. Like this:

Suppose $L$ is regular and let $p$ be its pumping length. We choose $x=[0^p, 0^{p-1}, \dots, 00, 0]\in L$. By the pumping lemma, we can split it into 3 parts $UVW$. Noticed that $V$ cannot include the symbol [ otherwise $UVVW$ contains two [ which is not in $L$. Then, the choices for $V$ is either $0, 00, \dots, 0^{p-1}$. Note that it cannot be $0^p$ because it is subject to the constraint $|UV|\leq p$. No matter what we choose for $V=0^k$, the string $UW=[0^{p-k}, 0^{p-1}, \dots, 0]$ is not in $L$ because $0^{p-k}$ must be one of $0, 00, \dots, 0^{p-1}$. Hence $L$ is not regular.