# Show $\{0^𝑚1^𝑛|𝑚≠𝑛\}$ is not regular

So I have the question: show "Show $$\{0^𝑚1^𝑛|𝑚≠𝑛\}$$ is not regular". I've already seen various proofs for this question, but they all have one step I don't get.

They all take: $$\bar{L}∩(0^∗1^∗)$$ ($$\bar{L}$$ is the complement of $$L$$) and show that it's not regular. I don't get why we can't just take $$\bar{L}$$. Because isn't $$\bar{L} = \{0^𝑚1^𝑛|m=n\}$$ which is the same as $$\{0^n1^n|n ≥ 0\}$$ which we know is not regular? What am I missing?

Try to express in natural language what $$\overline{L}$$ contains; that is, what words $$L$$ doesn't contain. Most obviously, it's "words of the form $$0^m0^n$$, with $$m = n$$." However, it also contains "words that are not of the form $$0^m1^n$$", such as "$$101010$$". That's why the intersection with $$0^*1^*$$ is employed, to not bother with these words.
The demonstration is then possible because intersection and complementation are closed properties on the regular language. So if $$\overline{L} \cap L(0^*1^*)$$ is not regular, we know that $$\overline{L}$$ must not be regular, so its complement, $$L$$, must not be regular.
Assume m ≠ m'. After processing $$0^m$$ and $$0^{m'}$$ you cant be in the same state, because in one state adding $$1^m$$ is not accepted and in the other state adding $$1^m$$ is accepted. Since is the case for all m ≠ m', the number of states is not finite.