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More Applications of the Pumping Lemma. 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:. Non-regular languages.

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the pumping lemma
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:

Prof. Busch - LSU

slide3

Non-regular languages

Regular languages

Prof. Busch - LSU

slide4

Theorem:

The language

is not regular

Proof:

Use the Pumping Lemma

Prof. Busch - LSU

slide5

Assume for contradiction

that is a regular language

Since is infinite

we can apply the Pumping Lemma

Prof. Busch - LSU

slide6

Let be the critical length for

Pick a string such that:

and

length

We pick

Prof. Busch - LSU

slide7

From the Pumping Lemma:

we can write:

with lengths:

Thus:

Prof. Busch - LSU

slide8

From the Pumping Lemma:

Thus:

Prof. Busch - LSU

slide9

From the Pumping Lemma:

Thus:

Prof. Busch - LSU

slide10

BUT:

CONTRADICTION!!!

Prof. Busch - LSU

slide11

Therefore:

Our assumption that

is a regular language is not true

Conclusion:

is not a regular language

END OF PROOF

Prof. Busch - LSU

slide12

Non-regular languages

Regular languages

Prof. Busch - LSU

slide13

Theorem:

The language

is not regular

Proof:

Use the Pumping Lemma

Prof. Busch - LSU

slide14

Assume for contradiction

that is a regular language

Since is infinite

we can apply the Pumping Lemma

Prof. Busch - LSU

slide15

Let be the critical length of

Pick a string such that:

and

length

We pick

Prof. Busch - LSU

slide16

From the Pumping Lemma:

We can write

With lengths

Thus:

Prof. Busch - LSU

slide17

From the Pumping Lemma:

Thus:

Prof. Busch - LSU

slide18

From the Pumping Lemma:

Thus:

Prof. Busch - LSU

slide19

BUT:

CONTRADICTION!!!

Prof. Busch - LSU

slide20

Therefore:

Our assumption that

is a regular language is not true

Conclusion:

is not a regular language

END OF PROOF

Prof. Busch - LSU

slide21

Non-regular languages

Regular languages

Prof. Busch - LSU

slide22

Theorem:

The language

is not regular

Proof:

Use the Pumping Lemma

Prof. Busch - LSU

slide23

Assume for contradiction

that is a regular language

Since is infinite

we can apply the Pumping Lemma

Prof. Busch - LSU

slide24

Let be the critical length of

Pick a string such that:

length

We pick

Prof. Busch - LSU

slide25

From the Pumping Lemma:

We can write

With lengths

Thus:

Prof. Busch - LSU

slide26

From the Pumping Lemma:

Thus:

Prof. Busch - LSU

slide27

From the Pumping Lemma:

Thus:

Prof. Busch - LSU

slide28

Since:

There must exist such that:

Prof. Busch - LSU

slide29

However:

for

for any

Prof. Busch - LSU

slide30

BUT:

CONTRADICTION!!!

Prof. Busch - LSU

slide31

Therefore:

Our assumption that

is a regular language is not true

Conclusion:

is not a regular language

END OF PROOF

Prof. Busch - LSU