Let $$L=\left\{0^n 1^{2n} \mid n > 1\right\}.$$ Show that $L$ is not regular.
Attempt:
If the language is regular, it must satisfy the Pumping lemma. $P$ will be our pumping length and our string $w = 0^p 1^p 1^p$. $x$ and $y$ must only contain zeroes because the length of $xy$ is less than or equal to $p$. Since the length of $y$ is greater than $0$, it must also contain some zeroes. However, if the string $xyz$ was balanced in the langauge, $xyyz$ will unbalance it by introducing additonal zeroes. Pumping lemma is a necessary property of regular languages, and one of its conditions state that for $w = xyz$, $xyyz$ must hold. Because $xyyz$ does not satisfy the language’s definition, this language is not regular.
Is this solution correct?