How to prove that if a language A is not regular then A* isn't regular either?
I have tried the usual methods with no result.
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possible duplicate of How to prove that a language is not regular? – Juho Nov 04 '13 at 10:00
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What are the "usual methods" you tried? – Juho Nov 04 '13 at 10:01
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I tried everything in that link, and couldn't work it out. That works as reply to both. – MattGenius Nov 04 '13 at 10:14
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4What makes you think that this claim is correct? – Shaull Nov 04 '13 at 11:01
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3If you don't find a proof, look for a counter-example... – Denis Nov 04 '13 at 11:09
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It is just an intuition. I unsuccessfully looked for counter-examples too. – MattGenius Nov 04 '13 at 11:15
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1Hint: try a non-regular language over a unary alphabet ${a}$. See where that gets you. – Shaull Nov 04 '13 at 11:30
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Turns out that my intuition was wrong. I did not think of non-regular languages over unary alphabets! – MattGenius Nov 04 '13 at 11:34
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2For Kleene star for a language over unary alphabet, see here. – Hendrik Jan Nov 04 '13 at 13:38
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2Actually, you don't even need to consider languages over a unary alphabet. Consider, for instance, non-regular languages containing at least all strings of length one over any alphabet. – Patrick87 Nov 04 '13 at 14:45
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Hint: Suppose $L$ is any language over the alphabet $\Sigma$. If $L$ is not regular then so is $L+\Sigma$, yet $(L+\Sigma)^* = \Sigma^*$ is regular.
Yuval Filmus
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1Excellent answer; although I fail to see how this is a hint, so much as a complete proof that the claim is false. – Patrick87 Nov 04 '13 at 19:31
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