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

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Courtesy Costas Busch - RPI1 More Applications of the Pumping Lemma

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:

Courtesy Costas Busch - RPI3 Regular languages Non-regular languages

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

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

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

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

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

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

Courtesy Costas Busch - RPI10 BUT: CONTRADICTION!!!

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

Courtesy Costas Busch - RPI12 Regular languages Non-regular languages

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

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

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

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

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

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

Courtesy Costas Busch - RPI19 BUT: CONTRADICTION!!!

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

Courtesy Costas Busch - RPI21 Regular languages Non-regular languages

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

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

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

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

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

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

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

Courtesy Costas Busch - RPI29 However:for for any

Courtesy Costas Busch - RPI30 BUT: CONTRADICTION!!!

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

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