Prove that Language is not Regular using closure(s) I'm having a really tough time trying to comprehend proofs via closure and pumping lemma.
None of the examples I can find are similar to the one's im looking to prove.  Can anyone explain if I'm totally missing the mark here or is this valid?

Question:
Prove that
$$ 
L = \{0^x 1^y 2^{x+y} \mid x \ge 0 \text{ and } y \ge1\}\qquad\text{is not Regular.}
$$ 

Must use Closure(s).
Your "destination" should be
$$
\{ a^z b^z \mid z \ge 1\}, \qquad\text{ which we know is not Regular.}
$$ 

Start with $L$

(show a sequence of operations and results)

End with $\{a^z b^z\}$ which is not regular

Answer:

If $L$ is regular then $L'$ must be regular
$$
L' = \{(0^x)' (1^y)' (2^{x+y})' \mid x \ge 0, y \ge1 \}
$$
Let $E$ be an element of $L'$ where $E = \{a^z b^z \mid z\ge 1\}$. Since we know $\{a^z b^z \mid z\ge 1\}$ is not regular from using pumping lemma, etc.
$L$ is not regular
What is your " ' " operator? Could it perhaps be the homomorphism $0\rightarrow a, 1\rightarrow a, 2\rightarrow b$? If not, consider this as a hint, since regular languages are closed under homomorphisms.