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Unformatted text preview: p , in light of the requirements that | xy | p and y 6 = . Since weve supposed that L is regular, s must be pumpable, so it must be that xy 2 z L . But, wait: whats the length of xy 2 z ? Well, since | xyz | = p 2 , it must be that | xy 2 z | = p 2 + | y | p 2 + p , since, if | xy | p , it must be that y p . Now wait a minute. The length of s = a p 2 was p 2 . Shouldnt the length of the next string in L s quadratic sequence, a ( p +1) 2 , be ( p +1) 2 = p 2 +2 p +1? Indeed! But note that p 2 < p 2 + p < p 2 + 2 p + 1, the implication of which is that the length of xy 2 z is strictly between two consecutive squares! Hence, it cant be that xy 2 z = a k , where k = q 2 for some q 0, in which case xy 2 z isnt in L , which contradicts the pumping lemma. Since the pumping lemma is provably true, the aw in our argument must be our assumption that L is regular. Ergo, L is not regular. QED . 1...
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- Fall '09
- Computer Science