# fall 2006costas busch - rpi1 more applications of the pumping lemma

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• Slide 1
• Fall 2006Costas Busch - RPI1 More Applications of the Pumping Lemma
• Slide 2
• Fall 2006Costas Busch - RPI2 The Pumping Lemma: Given a infinite regular language there exists an integer (critical length) for any string with length we can write with and such that:
• Slide 3
• Fall 2006Costas Busch - RPI3 Regular languages Non-regular languages
• Slide 4
• Fall 2006Costas Busch - RPI4 Theorem: The language is not regular Proof: Use the Pumping Lemma
• Slide 5
• Fall 2006Costas Busch - RPI5 Assume for contradiction that is a regular language Since is infinite we can apply the Pumping Lemma
• Slide 6
• Fall 2006Costas Busch - RPI6 We pick Let be the critical length for Pick a string such that: lengthand
• Slide 7
• Fall 2006Costas Busch - RPI7 we can write: with lengths: From the Pumping Lemma: Thus:
• Slide 8
• Fall 2006Costas Busch - RPI8 From the Pumping Lemma: Thus:
• Slide 9
• Fall 2006Costas Busch - RPI9 From the Pumping Lemma: Thus:
• Slide 10
• Fall 2006Costas Busch - RPI10 BUT: CONTRADICTION!!!
• Slide 11
• Fall 2006Costas Busch - RPI11 Our assumption that is a regular language is not true Conclusion: is not a regular language Therefore: END OF PROOF
• Slide 12
• Fall 2006Costas Busch - RPI12 Regular languages Non-regular languages
• Slide 13
• Fall 2006Costas Busch - RPI13 Theorem: The language is not regular Proof: Use the Pumping Lemma
• Slide 14
• Fall 2006Costas Busch - RPI14 Assume for contradiction that is a regular language Since is infinite we can apply the Pumping Lemma
• Slide 15
• Fall 2006Costas Busch - RPI15 We pick Let be the critical length of Pick a string such that: length and
• Slide 16
• Fall 2006Costas Busch - RPI16 We can write With lengths From the Pumping Lemma: Thus:
• Slide 17
• Fall 2006Costas Busch - RPI17 From the Pumping Lemma: Thus:
• Slide 18
• Fall 2006Costas Busch - RPI18 From the Pumping Lemma: Thus:
• Slide 19
• Fall 2006Costas Busch - RPI19 BUT: CONTRADICTION!!!
• Slide 20
• Fall 2006Costas Busch - RPI20 Our assumption that is a regular language is not true Conclusion: is not a regular language Therefore: END OF PROOF
• Slide 21
• Fall 2006Costas Busch - RPI21 Regular languages Non-regular languages
• Slide 22
• Fall 2006Costas Busch - RPI22 Theorem: The language is not regular Proof: Use the Pumping Lemma
• Slide 23
• Fall 2006Costas Busch - RPI23 Assume for contradiction that is a regular language Since is infinite we can apply the Pumping Lemma
• Slide 24
• Fall 2006Costas Busch - RPI24 We pick Let be the critical length of Pick a string such that: length
• Slide 25
• Fall 2006Costas Busch - RPI25 We can write With lengths From the Pumping Lemma: Thus:
• Slide 26
• Fall 2006Costas Busch - RPI26 From the Pumping Lemma: Thus:
• Slide 27
• Fall 2006Costas Busch - RPI27 From the Pumping Lemma: Thus:
• Slide 28
• Fall 2006Costas Busch - RPI28 Since: There must exist such that:
• Slide 29
• Fall 2006Costas Busch - RPI29 However:for for any
• Slide 30
• Fall 2006Costas Busch - RPI30 BUT: CONTRADICTION!!!
• Slide 31
• Fall 2006Costas Busch - RPI31 Our assumption that is a regular language is not true Conclusion: is not a regular language Therefore: END OF PROOF

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