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

2. Courtesy Costas Busch - RPI2 The Pumping Lemma: Given a infinite regular language there exists an integer for any string with length we can write with and such that:

3. Courtesy Costas Busch - RPI3 Regular languages Non-regular languages

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

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

6. Courtesy Costas Busch - RPI6 We pick Let be the integer in the Pumping Lemma Pick a string such that: length and

7. Courtesy Costas Busch - RPI7 Write it must be that length From the Pumping Lemma Thus:

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

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

10. Courtesy Costas Busch - RPI10 BUT: CONTRADICTION!!!

11. Courtesy Costas Busch - RPI11 Our assumption that is a regular language is not true Conclusion: is not a regular language Therefore:

12. Courtesy Costas Busch - RPI12 Regular languages Non-regular languages

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

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

15. Courtesy Costas Busch - RPI15 We pick Let be the integer in the Pumping Lemma Pick a string such that: length and

16. Courtesy Costas Busch - RPI16 Write it must be that length From the Pumping Lemma Thus:

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

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

19. Courtesy Costas Busch - RPI19 BUT: CONTRADICTION!!!

20. Courtesy Costas Busch - RPI20 Our assumption that is a regular language is not true Conclusion: is not a regular language Therefore:

21. Courtesy Costas Busch - RPI21 Regular languages Non-regular languages

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

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

24. Courtesy Costas Busch - RPI24 We pick Let be the integer in the Pumping Lemma Pick a string such that: length

25. Courtesy Costas Busch - RPI25 Write it must be that length From the Pumping Lemma Thus:

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

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

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

29. Courtesy Costas Busch - RPI29 However:for for any

30. Courtesy Costas Busch - RPI30 BUT: CONTRADICTION!!!

31. Courtesy Costas Busch - RPI31 Our assumption that is a regular language is not true Conclusion: is not a regular language Therefore: