# 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

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