6 added 752 characters in body edited Aug 30 '17 at 21:08 fade2black 8,74711 gold badge1919 silver badges3232 bronze badges You interpret the lemma incorrectly. The lemma says that "there is $$n$$ such that any string $$y \in L$$, with $$|y| \geq n$$...". So the lemma does not provide a particular $$n$$, it only says "there is", not "for all". If the language is finite then, the pumping length is greater than the longest string in the language. And hence the three conditions of the lemma fails for all strings in the language, meaning there is nothing to "pump". Nevertheless, the lemma holds. Just recall the implication rule $$p \implies q$$: if $$p$$ is false then it does not matter if $$q$$ is true or false, the whole implication is true. In your example with a single string $$a$$ take $$n$$, the pumping length to be $$2$$. So there is no string of the length greater than or equal to $$2$$. But it does not mean that the lemma fails. Update:: the following predicate formula expresses the Pumping lemma for regular languages $$[ L \text{ is regular }] \implies [\exists n\forall x[ \left((x \in L) \land (|x| \geq n) \right) \implies (\exists u \exists uv \exists w(x=uvw \land |uv| \leq n \land |v| > 0 \land \forall m[m>0 \implies uv^mw \in L))]]]$$ and this is antecedent (conclusion) part of the lemma: $$\exists n \forall x[ \left((x \in L) \land (|x| \geq n) \right) \implies (\exists u \exists uv \exists w(x=uvw \land |uv| \leq n \land |v| > 0 \land \forall m[m>0 \implies uv^mw \in L))]]$$ This premise of the antecedent $$(x \in L) \land (|x| \geq n)$$ is false since $$|x|$$ is less than $$n$$ (for each $$x \in L$$) and hence the whole antecedent is true. You interpret the lemma incorrectly. The lemma says that "there is $$n$$ such that any string $$y \in L$$, with $$|y| \geq n$$...". So the lemma does not provide a particular $$n$$, it only says "there is", not "for all". If the language is finite then, the pumping length is greater than the longest string in the language. And hence the three conditions of the lemma fails for all strings in the language, meaning there is nothing to "pump". Nevertheless, the lemma holds. Just recall the implication rule $$p \implies q$$: if $$p$$ is false then it does not matter if $$q$$ is true or false, the whole implication is true. In your example with a single string $$a$$ take $$n$$, the pumping length to be $$2$$. So there is no string of the length greater than or equal to $$2$$. But it does not mean that the lemma fails. You interpret the lemma incorrectly. The lemma says that "there is $$n$$ such that any string $$y \in L$$, with $$|y| \geq n$$...". So the lemma does not provide a particular $$n$$, it only says "there is", not "for all". If the language is finite then, the pumping length is greater than the longest string in the language. And hence the three conditions of the lemma fails for all strings in the language, meaning there is nothing to "pump". Nevertheless, the lemma holds. Just recall the implication rule $$p \implies q$$: if $$p$$ is false then it does not matter if $$q$$ is true or false, the whole implication is true. In your example with a single string $$a$$ take $$n$$, the pumping length to be $$2$$. So there is no string of the length greater than or equal to $$2$$. But it does not mean that the lemma fails. Update:: the following predicate formula expresses the Pumping lemma for regular languages $$[ L \text{ is regular }] \implies [\exists n\forall x[ \left((x \in L) \land (|x| \geq n) \right) \implies (\exists u \exists uv \exists w(x=uvw \land |uv| \leq n \land |v| > 0 \land \forall m[m>0 \implies uv^mw \in L))]]]$$ and this is antecedent (conclusion) part of the lemma: $$\exists n \forall x[ \left((x \in L) \land (|x| \geq n) \right) \implies (\exists u \exists uv \exists w(x=uvw \land |uv| \leq n \land |v| > 0 \land \forall m[m>0 \implies uv^mw \in L))]]$$ This premise of the antecedent $$(x \in L) \land (|x| \geq n)$$ is false since $$|x|$$ is less than $$n$$ (for each $$x \in L$$) and hence the whole antecedent is true. 5 added 3 characters in body edited Aug 30 '17 at 4:33 fade2black 8,74711 gold badge1919 silver badges3232 bronze badges You interpret the lemma incorrectly. The lemma says that "there is $$n$$ such that any string $$y \in L$$, with $$|y| \geq n$$...". So the lemma does not provide a particular $$n$$, it only says "there is", not "for all". If the language is finite then, the pumping length is greater than the longest string in the language. And hence the three conditions of the lemma fails for all strings in the language, meaning there is nothing to "pump". Nevertheless, the lemma holds. Just recall the implication rule $$p \implies q$$: if $$p$$ is false then it does not matter if $$q$$ is true or false, the whole implication is true. In your example with a single string $$a$$ take $$n$$, the pumping length to be $$2$$. So there is no string of the length greater than or equal to $$2$$. But it does not mean that the lemma fails. You interpret the lemma incorrectly. The lemma says that "there is $$n$$ such that any string $$y \in L$$, with $$|y| \geq n$$...". So the lemma does not provide a particular $$n$$, it only says "there is", not "for all". If the language is finite then, the pumping length is greater than the longest string in the language. And hence the three conditions of the lemma fails for all strings in the language, meaning there is nothing to "pump". Nevertheless, the lemma holds. Just recall the implication rule $$p \implies q$$: if $$p$$ is false then it does not matter if $$q$$ is true or false, the whole implication is true. In your example with a single string $$a$$ take $$n$$, the pumping length to be $$2$$. So there is no string of the length greater than or equal to $$2$$. But does not mean that the lemma fails. You interpret the lemma incorrectly. The lemma says that "there is $$n$$ such that any string $$y \in L$$, with $$|y| \geq n$$...". So the lemma does not provide a particular $$n$$, it only says "there is", not "for all". If the language is finite then, the pumping length is greater than the longest string in the language. And hence the three conditions of the lemma fails for all strings in the language, meaning there is nothing to "pump". Nevertheless, the lemma holds. Just recall the implication rule $$p \implies q$$: if $$p$$ is false then it does not matter if $$q$$ is true or false, the whole implication is true. In your example with a single string $$a$$ take $$n$$, the pumping length to be $$2$$. So there is no string of the length greater than or equal to $$2$$. But it does not mean that the lemma fails. 4 edited body edited Aug 30 '17 at 3:42 fade2black 8,74711 gold badge1919 silver badges3232 bronze badges You interpret the lemma incorrectly. The lemma says that "there is $$n$$ such that any string $$y \in L$$, with $$|y| \geq n$$...". So the lemma does not provide a particular $$n$$, it only says "there is", not "for all". If the language is finite then, the pumping length is greater than the longest string in the language. And hence the three conditions of the lemma fails for all strings ofin the language, meaning there is nothing to "pump". Nevertheless, the lemma holds. Just recall the implication rule $$p \implies q$$: if $$p$$ is false then it does not matter if $$q$$ is true or false, the whole implication is true. In your example with a single string $$a$$ take $$n$$, the pumping length to be $$2$$. So there is no string of the length greater than or equal to $$2$$. But does not mean that the lemma fails. You interpret the lemma incorrectly. The lemma says that "there is $$n$$ such that any string $$y \in L$$, with $$|y| \geq n$$...". So the lemma does not provide a particular $$n$$, it only says "there is", not "for all". If the language is finite then, the pumping length is greater than the longest string in the language. And hence the three conditions of the lemma fails for all strings of the language, meaning there is nothing to "pump". Nevertheless, the lemma holds. Just recall the implication rule $$p \implies q$$: if $$p$$ is false then it does not matter if $$q$$ is true or false, the whole implication is true. In your example with a single string $$a$$ take $$n$$, the pumping length to be $$2$$. So there is no string of the length greater than or equal to $$2$$. But does not mean that the lemma fails. You interpret the lemma incorrectly. The lemma says that "there is $$n$$ such that any string $$y \in L$$, with $$|y| \geq n$$...". So the lemma does not provide a particular $$n$$, it only says "there is", not "for all". If the language is finite then, the pumping length is greater than the longest string in the language. And hence the three conditions of the lemma fails for all strings in the language, meaning there is nothing to "pump". Nevertheless, the lemma holds. Just recall the implication rule $$p \implies q$$: if $$p$$ is false then it does not matter if $$q$$ is true or false, the whole implication is true. In your example with a single string $$a$$ take $$n$$, the pumping length to be $$2$$. So there is no string of the length greater than or equal to $$2$$. But does not mean that the lemma fails. 3 added 15 characters in body edited Aug 30 '17 at 3:28 fade2black 8,74711 gold badge1919 silver badges3232 bronze badges 2 added 178 characters in body edited Aug 30 '17 at 3:17 fade2black 8,74711 gold badge1919 silver badges3232 bronze badges 1 answered Aug 30 '17 at 3:09 fade2black 8,74711 gold badge1919 silver badges3232 bronze badges