Consider the language L = {omega element {a, b}^*| omega ends in a}. This is a r
ID: 3843740 • Letter: C
Question
Consider the language L = {omega element {a, b}^*| omega ends in a}. This is a regular language. However, in the Pumping Lemma "proof" below, it might appear at first glance that the author has managed to prove that L is non-regular. Explain in one or two sentences what is wrong with the "proof". The language L = {omega element {a, b}^*| omega ends in a} is non-regular. Suppose L is regular, and let p greaterthanorequalto 1 be the pumping length guaranteed by the Pumping Lemma. We choose a string s = b^p -1 a. This string is in L (because it ends in a) and |s| = p, which is long enough. Now, suppose the adversary chooses the decomposition x = b^p -1, y = a, z = element. This decomposition satisfies |y| greaterthanorequalto 1 and |xy| lessthanorequalto p. However, we can show that it does not satisfy the third condition of the Pumping Lemma becase xy^0 z = xz = (b^p -1)(element) = b^p -1. This string does not end in a, and so xy^0 z notelement L. This contradicts the Pumping Lemma, and so our assumption that L is regular must have been false. Therefore, L is non-regular.Explanation / Answer
We cant take z as null.Because only middle string pump(generate) strings. last charcater must me fixed as a.