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• If a language L is regular there is a DFA M that recognizes it. Of course, when applying the pumping lemma to prove that a language is not regular, you don't actually play this game with another person. You get to do the roles of yourself and of your opponent. You can think of it like you're having identity disorders (here we laugh) and the two personalities are your opponent and yourself.

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A: We assume that the language IS REGULAR, and then prove a contradiction. Q: Okay, where does the PL come in? A: We prove that the PL is violated. 2.4 The Pumping Lemma for Context-Free Languages.

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You can think of it like you're having identity disorders (here we laugh) and the two personalities are your opponent and yourself. CANNOT use pumping lemma to prove regular language BUT we can prove it is NOT regular What is pumping lemma? Describes the nec For any regular language A, there is a constant p, or pumping length, that is equal to the amount of states in the DFA that accepts this language For any string,s, in a whose length is greater than p, we can break s into 3 substrings like so: S = xyz Length of y >0 $\begingroup$ You may take into consideration the 3rd property of the pumping lemma, that is $$\left | xy \right| \leq p$$ where p is the pumping length. $\endgroup$ – Rrjrjtlokrthjji Oct 8 '14 at 16:38

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Beispiele mit Beweis ✓ mit  Pumping lemma for regular languages.

2017-09-14 Pumping Lemma . Let L be a regular language. Then there exists a number n such that all w ∈ L where |w| ≥ n, there exists a prefix of w whose length is less than n which contains a pump.
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Proof.

regularity. Pumping Lemma for Regular Languages - Automata - Tutorial Pumping lemma for regular set h1.
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An example L = {0n1n: n ≥ 0} is not regular.