prof. busch - lsu1 more applications of the pumping lemma

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  • Slide 1
  • Prof. Busch - LSU1 More Applications of the Pumping Lemma
  • Slide 2
  • Prof. Busch - LSU2 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
  • Prof. Busch - LSU3 Regular languages Non-regular languages
  • Slide 4
  • Prof. Busch - LSU4 Theorem: The language is not regular Proof: Use the Pumping Lemma
  • Slide 5
  • Prof. Busch - LSU5 Assume for contradiction that is a regular language Since is infinite we can apply the Pumping Lemma
  • Slide 6
  • Prof. Busch - LSU6 We pick Let be the critical length for Pick a string such that: lengthand
  • Slide 7
  • Prof. Busch - LSU7 we can write: with lengths: From the Pumping Lemma: Thus:
  • Slide 8
  • Prof. Busch - LSU8 From the Pumping Lemma: Thus:
  • Slide 9
  • Prof. Busch - LSU9 From the Pumping Lemma: Thus:
  • Slide 10
  • Prof. Busch - LSU10 BUT: CONTRADICTION!!!
  • Slide 11
  • Prof. Busch - LSU11 Our assumption that is a regular language is not true Conclusion: is not a regular language Therefore: END OF PROOF
  • Slide 12
  • Prof. Busch - LSU12 Regular languages Non-regular languages
  • Slide 13
  • Prof. Busch - LSU13 Theorem: The language is not regular Proof: Use the Pumping Lemma
  • Slide 14
  • Prof. Busch - LSU14 Assume for contradiction that is a regular language Since is infinite we can apply the Pumping Lemma
  • Slide 15
  • Prof. Busch - LSU15 We pick Let be the critical length of Pick a string such that: length and
  • Slide 16
  • Prof. Busch - LSU16 We can write With lengths From the Pumping Lemma: Thus:
  • Slide 17
  • Prof. Busch - LSU17 From the Pumping Lemma: Thus:
  • Slide 18
  • Prof. Busch - LSU18 From the Pumping Lemma: Thus:
  • Slide 19
  • Prof. Busch - LSU19 BUT: CONTRADICTION!!!
  • Slide 20
  • Prof. Busch - LSU20 Our assumption that is a regular language is not true Conclusion: is not a regular language Therefore: END OF PROOF
  • Slide 21
  • Prof. Busch - LSU21 Regular languages Non-regular languages
  • Slide 22
  • Prof. Busch - LSU22 Theorem: The language is not regular Proof: Use the Pumping Lemma
  • Slide 23
  • Prof. Busch - LSU23 Assume for contradiction that is a regular language Since is infinite we can apply the Pumping Lemma
  • Slide 24
  • Prof. Busch - LSU24 We pick Let be the critical length of Pick a string such that: length
  • Slide 25
  • Prof. Busch - LSU25 We can write With lengths From the Pumping Lemma: Thus:
  • Slide 26
  • Prof. Busch - LSU26 From the Pumping Lemma: Thus:
  • Slide 27
  • Prof. Busch - LSU27 From the Pumping Lemma: Thus:
  • Slide 28
  • Prof. Busch - LSU28 Since: There must exist such that:
  • Slide 29
  • Prof. Busch - LSU29 However:for for any
  • Slide 30
  • Prof. Busch - LSU30 BUT: CONTRADICTION!!!
  • Slide 31
  • Prof. Busch - LSU31 Our assumption that is a regular language is not true Conclusion: is not a regular language Therefore: END OF PROOF