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I know that there are non-regular languages, so that $L^*$ is regular, but all examples I can find are context-sensitive but not context free.

In case there are none how do you prove it?

Raphael
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Simon S
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1 Answers1

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$L = \{a^n b^n \mid n\in\mathbb{N}\}$ is context-free but not regular (classical example). So is $L' = \{a^n b^n \mid n\in\mathbb{N}\} \cup \{a,b\}$.

$L'^\ast = \{a,b\}^\ast$ is regular.

Gilles 'SO- stop being evil'
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