# Can a non-regular language be made regular via concatenation when they don't share characters?

So this is a follow-on question to my other question (Can we make a non-regular language regular via concatentation?).

Given the following,

$L = \{0^n1^m2^m \mid n>1, m>1\}$

$A = \{0^n \mid n>1\}$

$B = \{1^m2^m \mid m>1\}$

Is $L$ in fact just $A$ and $B$ concatenated (I believe it is, but I want to verify that)?

Further, does proving $B$ non-regular prove that $L$ is non-regular since they don't share characters?

• If you want to prove that L is not regular, just use the Pumping Lemma. – Banach Tarski Mar 31 '16 at 10:25
• As I mentioned before, we'd prefer that you ask only one question per question. Thank you. – D.W. Mar 31 '16 at 22:36

First, the equality $L = AB$ is correct. Next, if you know that $B$ is non-regular, you can prove that $L$ is non-regular as follows.
Suppose that $L$ is regular. Then $(00)^{-1}L = \{0^n1^m2^m \mid n \geqslant 0, m>1\} = 0^*B$ is regular. It follows that $0^*B \cap \{1,2\}^* = B$ is regular. Contradiction.