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I know that this language is not regular L = {w | na(w) = nb(w)} where na(w) is the number of a's in w. But what if now the language changes to that the number of a's has to be greater than b's? I think not, because we do not have finite number of prefix equivalence class. And in that case, how can I prove it? Or if it was regular, how would it be?

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    – D.W.
    May 29, 2021 at 7:24

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Both mentioned languages are irregular

This can be easily shown for both languages using the pumping lemma

First consider L = {w | w has more as than bs} Let p be the pumping length then clearly b^p a^(p+1) is in L and it cannot be pumped

We know that |xy| ≤ p ,and thus x and y will have only bs , then pumping the string for i > 1 , xy^i z will give you a string with bs equal to (for i=2) or more than ( for i >2) as, since this string cannot be pumped L is irregular

In a similar manner the language with the number of as equal to bs can be shown to be non-regular , consider a^p b^p clearly it has equal number of as and bs , using the same argument as above x and y will have only as and thus pumping the string gives more as than bs and thus this language is also non-regular

Intuitively we know both are irregular since we need to keep count of the number of as and bs to compare them something , and the number of as and bs is not a finite specified number .

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