# Determining class of language with pumping lemma?

I have the language $$L = \{ 0^{2l} 1^m | l,m >= 0 \} \ where \ \Sigma= \{0,1\}$$ which I am trying to find the class of language for, e.g. not context-free, context-free, regular.

By this notion I assume $$L$$ is context-free and has some pumping length $$p$$, and I will use the pumping lemma.

$$w = 0^{2p}1^p$$ and can be split into $$w = uv^ixy^iz, i >= 0.$$

$$|w| > p$$

I know that $$vxy$$ can only be in 3 places as it is of length at most $$p$$, these scenarios are:

-all 0's

-A mix of 0's and 1's.

-All 1's.

My thinking is that scenario 1 is invalid as it must always be an even length, and pumping up may break this. Scenarios 2 I am unsure although I think it follows the same logic as the first scenario, and scenario 3 could be pumped and still be in the language.

Does this mean $$L$$ is context-free? Can I also infer if $$L$$ is regular?

• As a reminder: the pumping lemma for regular languages (resp. for context-free languages) can be used to prove that a language is not regular (resp. context-free). Oct 27, 2022 at 9:11

Your language is regular. A regular expression is e.g. $$(00)^\ast1^\ast$$. The numbers $$l$$ and $$m$$ are not in any relation to each other. You just have an even number of $$0$$s followed by an arbitrary number of $$1$$s.