Proof that {w ∈ {a, b} ∗ | |w|a = |w|b} is not a regular language

Asked May 12 '17 at 00:08

Active May 12 '17 at 04:14

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I know that the language {w ∈ {a, b}∗| |w|a = |w|b} is not regular since somehow, you can't store all the information needed in a DFA. I've seen that normally it's done with reduction to absurdity or pumping lemma and also having into consideration that the language a^nb^n is not regular But I would like to see a formal approach of it to fully understand why and not only because of "intuition"

Could someone please help me? Most of examples just always show the same and I don't see how to apply them to my case.

edited May 12 '17 at 00:58

asked May 12 '17 at 00:08

Meyer

1

https://cs.stackexchange.com/q/1031/66948 – Aditya May 12 '17 at 00:29
It's not exactly the same since I was looking for an exact explanation. The answer is that if you use the pumping lemma you can demonstrate that for a given w = xyz, you have a y that makes the condition goes false. In this case, aaabbb or any a^nb^n makes the case. – Meyer May 15 '17 at 10:03

Proof that {w ∈ {a, b} ∗ | |w|a = |w|b} is not a regular language

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