I want to prove that the complement of $\{0^n1^n \mid n \geq{} 0\}$ is not regular using closure properties.

I understand pumping lemma can be used to prove that $\{0^n1^n \mid n \geq{} 0\}$ is not a regular language. I also understand regular languages are closed under complement operation. However, does that also imply that a non-regular language's complement is also non-regular?


1 Answer 1


Yes. Since the complement of a regular language is also a regular language, then it follows that the complement of a non-regular language must also be non-regular. Strictly speaking, this works since the complement is its own inverse.

  • 3
    $\begingroup$ Just to explicitly state it, the proof would run along these lines: Towards a contradiction, let L be the complement of the given language. If L were regular then the complement of L, which is the given language: $\{0^n 1^n \mid n \geq 0\}$, would be regular as well by the property of regular languages being closed over the compliment operation. Therefore L is not regular. $\endgroup$ Sep 27, 2012 at 2:39

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