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?

  • 1
    $\begingroup$ Can be answered with the same techniques as cs.stackexchange.com/questions/1549 $\endgroup$
    – sdcvvc
    May 25, 2012 at 20:54
  • 2
    $\begingroup$ Hint: all languages which contain the alphabet have a very simple Kleene closure. $\endgroup$
    – Raphael
    Dec 15, 2014 at 22:58

1 Answer 1


$L = \{a^n b^n \mid n\in\mathbb{N}\}$ 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.

  • 2
    $\begingroup$ Brute-force, but valid. $\endgroup$
    – Raphael
    May 25, 2012 at 22:51
  • $\begingroup$ $L' = L$, actually... $\endgroup$
    – vonbrand
    Nov 30, 2016 at 1:48
  • $\begingroup$ @vonbrand ... actually not. $\endgroup$ Oct 8, 2020 at 19:00

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