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

Published: 01 January 1970
on channel: Easy Theory

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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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