Pumping Lemma for Regular Languages Example: 0ⁿ1ᵐ, n != m (HARD!)

Here we do a very hard pumping lemma example of all strings of the form 0^n 1^m where n != m. The reason it is hard is given by the n != m condition, because to find a string to get a contradiction, we need to have the string of the form 0^n 1^n (i.e., same number of 0s and 1s). It's very easy to "miss" the two if the decomposition of the original string can be anything. That's why we use the "p factorial" method.
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[@EasyTheory]

Thanks to Micah Wood (Platinum), and Josh Hibschman, Timmy Gy, Patrik Keinonen, and Travis Schnider (Silver) for helping support this video. If you want to contribute, links are in the video description.

[@galbalandroid]

Thank you for your videos! I've noticed you're uploading very often lately and hope you'll get the exposure you deserve! Universities should learn from you about how to teach this stuff.

[@EasyTheory]

Thanks very much!

[@miyamotomusashi4556]

What if we took 0^p 1^(p+1) as my string? wouldn't be easier to prove its not regular this way?

[@amitbarnahum4732]

That's a useful trick, thank you for that!

[@EasyTheory]

You're welcome!

[@danijose1100]

can i use the same methood if i have 0ⁿ1ᵐ, n > m? or would i have to do it another way?