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This is weird but by pumping lemma, say

Let L be a regular language. There exists a constant n such that for every string w in L such that |w| >= n, we can break w in to xyz such that xy*z is also in L.

This lemma is strong because it argues for all regular languages. But what if the regular language L = a? There is only one word (a) in it. How the pumping lemma works for this case?

yuanqili
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If n = 2 then it is vacuously true that any w in L with |w| >= n satisfies the conclusion of the pumping lemma. No words in L are long enough to serve as counterexamples. More generally, if L is any finite language then L satisfies the pumping lemma: just take n to be greater than the length of the longest word in L.

John Coleman
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