A language is defined as $w = \{ 0^m1^n \mid m, n \in \Bbb N \}$. Is this a regular language? I have seen people proving for both the sides.

Thread saying it is regular

Proof for it being non-regular


closed as off-topic by eranreches, Namaste, Shailesh, J.-E. Pin, Sahiba Arora Jan 6 '18 at 12:43

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    $\begingroup$ The second link is for the different language where $m$ and $n$ are restricted to be unequal, so it doesn't apply here. $\endgroup$ – Daniel Schepler Jan 5 '18 at 23:33
  • $\begingroup$ Could you also help me see why adding that restriction stops it from being a regular language, I have intuition, that if you take y = 0, and pump that you might have a case where the number of 0s and 1s will be same in the string. Thanks. $\endgroup$ – Swapnil Raj Jan 5 '18 at 23:37
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    $\begingroup$ Isn't $0^*1^*$ trivially a regular expression? $\endgroup$ – Jack D'Aurizio Jan 5 '18 at 23:43

The language $L = \{ 0^{m}1^{n} : m, n \in \mathbb{N} \}$ is given by the regular expression $0^{*}1^{*}$, which Jack D'Aurizio pointed out.

Now the language $L^{\prime} = \{0^{m}1^{n} : m \neq n \}$ is not regular. To see this, we use the fact that regular languages are closed under the set difference operation. Suppose to the contrary that $L^{\prime}$ is regular. Then: $$0^*1^* \setminus L^{\prime} = \{ 0^{n}1^{n} : n \in \mathbb{N} \}$$

Is also regular. However, we know that $\{0^{n}1^{n} : n \in \mathbb{N} \}$ is not regular, a contradiction. So $L^{\prime}$ is not regular.


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