Discrete Event Systems Solution to Exercise Sheet 3 1 Pumping Lemma [ Exam ]
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چکیده
a) Language L1 can be shown to be non-regular using the pumping lemma. Assume for contradiction that L1 is regular and let p be the corresponding pumping length. Choose w to be the word 01101. Because w is an element of L1 and has length more than p, the pumping lemma guarantees that w can be split into three parts, w = xyz, where |xy| ≤ p and for any i ≥ 0, we have xyz ∈ L1. In order to obtain the contradiction, we must prove that for every possible partition into three parts w = xyz where |xy| ≤ p, the word w cannot be pumped. We therefore consider the various cases.
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تاریخ انتشار 2013