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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