john doe claims that the language l of all strings over the alphabet a b that contain an even nu

1. The assumption that L is regular is correct.

Question

John Doe claims that the language L, of all strings over the alphabet Σ = {a, b} that contain an even number of occurrences of the letter 'a', is not a regular language. He offers the following "pumping lemma proof". Explain what is wrong with the "proof" given below. "Pumping Lemma Proof" We assume that L is regular. Then, according to the pumping lemma, every long string in L (of length m or more) must be "pumpable". We choose the specific string w = a^2m. Clearly, w is in L, and |w| >= m, so this w must be "pumpable". That means that the string w consists of 3 substrings, x, y, and z. In other words, w = xyz. We choose y = a. Then if we "pump up" to produce the string xy^2z, this string should also be in the language L, according to the Pumping Lemma. But xy^2z = a^(2m+1), and clearly xy^2z is NOT in L, since this new string has an odd number of a's. Therefore, our original assumption that L is regular was incorrect. Q.E.D.

          John Doe claims that the language L, of all strings over the alphabet Σ = {a, b} that contain an even number of occurrences of the letter 'a', is not a regular language. He offers the following "pumping lemma proof". Explain what is wrong with the "proof" given below.

"Pumping Lemma Proof"

We assume that L is regular. Then, according to the pumping lemma, every long string in L (of length m or more) must be "pumpable".

We choose the specific string w = a^2m. Clearly, w is in L, and |w| >= m, so this w must be "pumpable".

That means that the string w consists of 3 substrings, x, y, and z. In other words, w = xyz. We choose y = a.

Then if we "pump up" to produce the string xy^2z, this string should also be in the language L, according to the Pumping Lemma.

But xy^2z = a^(2m+1), and clearly xy^2z is NOT in L, since this new string has an odd number of a's.

Therefore, our original assumption that L is regular was incorrect. Q.E.D.

Added by Stephanie M.

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