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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?

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  • $\begingroup$ 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). $\endgroup$
    – Nathaniel
    Commented Oct 27, 2022 at 9:11

1 Answer 1

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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.

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