Problem Detail: 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?
Asked By : Simon S
Answered By : Gilles
$L = {a^n b^n mid ninmathbb{N}}$ is context-free but not regular (classical example). So is $L’ = {a^n b^n mid ninmathbb{N}} cup {a,b}$. $L’^ast = {a,b}^ast$ is regular.
Best Answer from StackOverflow
Question Source : http://cs.stackexchange.com/questions/2081
