1. Fall 2006Costas Busch - RPI1 More Applications of the Pumping Lemma

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:

3. Fall 2006Costas Busch - RPI3 Regular languages Non-regular languages

4. Fall 2006Costas Busch - RPI4 Theorem: The language is not regular Proof: Use the Pumping Lemma

5. Fall 2006Costas Busch - RPI5 Assume for contradiction that is a regular language Since is infinite we can apply the Pumping Lemma

6. Fall 2006Costas Busch - RPI6 We pick Let be the critical length for Pick a string such that: lengthand

7. Fall 2006Costas Busch - RPI7 we can write: with lengths: From the Pumping Lemma: Thus:

8. Fall 2006Costas Busch - RPI8 From the Pumping Lemma: Thus:

9. Fall 2006Costas Busch - RPI9 From the Pumping Lemma: Thus:

10. Fall 2006Costas Busch - RPI10 BUT: CONTRADICTION!!!

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

12. Fall 2006Costas Busch - RPI12 Regular languages Non-regular languages

13. Fall 2006Costas Busch - RPI13 Theorem: The language is not regular Proof: Use the Pumping Lemma

14. Fall 2006Costas Busch - RPI14 Assume for contradiction that is a regular language Since is infinite we can apply the Pumping Lemma

15. Fall 2006Costas Busch - RPI15 We pick Let be the critical length of Pick a string such that: length and

16. Fall 2006Costas Busch - RPI16 We can write With lengths From the Pumping Lemma: Thus:

17. Fall 2006Costas Busch - RPI17 From the Pumping Lemma: Thus:

18. Fall 2006Costas Busch - RPI18 From the Pumping Lemma: Thus:

19. Fall 2006Costas Busch - RPI19 BUT: CONTRADICTION!!!

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

21. Fall 2006Costas Busch - RPI21 Regular languages Non-regular languages

22. Fall 2006Costas Busch - RPI22 Theorem: The language is not regular Proof: Use the Pumping Lemma

23. Fall 2006Costas Busch - RPI23 Assume for contradiction that is a regular language Since is infinite we can apply the Pumping Lemma

24. Fall 2006Costas Busch - RPI24 We pick Let be the critical length of Pick a string such that: length

25. Fall 2006Costas Busch - RPI25 We can write With lengths From the Pumping Lemma: Thus:

26. Fall 2006Costas Busch - RPI26 From the Pumping Lemma: Thus:

27. Fall 2006Costas Busch - RPI27 From the Pumping Lemma: Thus:

28. Fall 2006Costas Busch - RPI28 Since: There must exist such that:

29. Fall 2006Costas Busch - RPI29 However:for for any

30. Fall 2006Costas Busch - RPI30 BUT: CONTRADICTION!!!

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