This document presents proofs that three languages are not regular using the pumping lemma. It first proves that the language {a^n b^n | n >= 0} is not regular by picking the string "abba" and showing it cannot be pumped. It then proves that the language {w | the number of a's is not equal to the number of b's} is not regular, again by picking a string and showing it cannot be pumped. Finally, it proves that the language of strings with an equal number of 1's and 0's is not regular using a similar pumping argument.

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More Applications of the Pumping Lemma
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3. © 2020 Wael Badawy

4. © 2020 Wael Badawy
The Pumping Lemma:
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• Given a infinite regular language L
• there exists an integer (critical length)p
• for any string with lengthLwÎ pw ³||
• we can write zyxw =
• with andpyx £|| 1|| ³y
• such that: Lzyx i
Î ...,2,1,0=i

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Regular languages
Non-regular languages *}:{ SÎ= vvvL R

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Theorem: The language
is not regular
Proof: Use the Pumping Lemma
*}:{ SÎ= vvvL R
},{ ba=S

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Assume for contradiction
that is a regular languageL
Since is infinite
we can apply the Pumping Lemma
L
*}:{ SÎ= vvvL R

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pppp
abbaw =We pick
Let be the critical length for
Pick a string such that:w Lw Î
pw ³||length
p
and
*}:{ SÎ= vvvL R
L

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we can write: zyxabbaw pppp
==
with lengths:
From the Pumping Lemma:
ababbabaaaaxyz ..................=
x y z
p p p p
1||,|| ³£ ypyx
pkay k
££= 1,Thus:
=w

10. © 2020 Wael Badawy
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From the Pumping Lemma: Lzyx i
Î
...,2,1,0=i
Thus: Lzyx Î2
pppp
abbazyx = pkay k
££= 1,

11. © 2020 Wael Badawy
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From the Pumping Lemma:
Lababbabaaaaaazxy ∈.....................=2
x y z
kp + p p p
y
Lzyx Î2
Thus: Labba pppkp
Î+
pppp
abbazyx = pkay k
££= 1,

12. © 2020 Wael Badawy
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Labba pppkp
Ï+
BUT:
CONTRADICTION!!!
1³k
*}:{ SÎ= vvvL R
Labba pppkp
Î+

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Our assumption that
is a regular language is not true
L
Conclusion: L is not a regular language
Therefore:
END OF PROOF

14. © 2020 Wael Badawy
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Regular languages
Non-regular languages
}0,:{ ³= +
lncbaL lnln

15. © 2020 Wael Badawy
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Theorem: The language
is not regular
Proof: Use the Pumping Lemma
}0,:{ ³= +
lncbaL lnln

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Assume for contradiction
that is a regular languageL
Since is infinite
we can apply the Pumping Lemma
L
}0,:{ ³= +
lncbaL lnln

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ppp
cbaw 2
=We pick
Let be the critical length of
Pick a string such that:w Lw Î
pw ³||length
p
}0,:{ ³= +
lncbaL lnln
and
L

18. © 2020 Wael Badawy
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We can write zyxcbaw ppp
== 2
With lengths
From the Pumping Lemma:
cccbcabaaaaaxyz ..................=
x y z
p p p2
1||,|| ³£ ypyx
Thus:
=w
pkay k
££= 1,

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From the Pumping Lemma: Lzyx i
Î
...,2,1,0=i
Thus:
ppp
cbazyx 2
=
Lxzzyx ∈=0
pkay k
££= 1,

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From the Pumping Lemma:
Lcccbcabaaaxz Î= ...............
x z
kp - p p2
Lxz Î
Thus: Lcba ppkp
Î- 2
ppp
cbazyx 2
= pkay k
££= 1,

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Lcba ppkp
Ï- 2
BUT:
CONTRADICTION!!!
}0,:{ ³= +
lncbaL lnln
1³kLcba ppkp
Î- 2

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Our assumption that
is a regular language is not true
L
Conclusion: L is not a regular language
Therefore:
END OF PROOF

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Regular languages
Non-regular languages }0:{ !
³= naL n

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Theorem: The language
is not regular
Proof: Use the Pumping Lemma
}0:{ !
³= naL n
nnn ×-×= )1(21! !

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Assume for contradiction
that is a regular languageL
Since is infinite
we can apply the Pumping Lemma
L
}0:{ !
³= naL n

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!p
aw =We pick
Let be the critical length of
Pick a string such that:w Lw Î
pw ³||length
p
}0:{ !
³= naL n
L

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We can write zyxaw p
== !
With lengths
From the Pumping Lemma:
aaaaaaaaaaaxyz p
...............!
==
x y z
p pp -!
1||,|| ³£ ypyx
pkay k
££= 1,Thus:
=w

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From the Pumping Lemma: Lzyx i
Î
...,2,1,0=i
Thus:
!p
azyx =
Lzyx Î2
pkay k
££= 1,

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From the Pumping Lemma:
Laaaaaaaaaaaazxy Î= ..................2
x y z
kp + pp -!
Thus:
!p
azyx = pkay k
££= 1,
Lzyx Î2
y
La kp
Î+!

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La kp
Î+!
!! zkp =+
}0:{ !
³= naL nSince:
pk ££1
There must exist such that:z

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However:
)!1(
)1(!
!!
!!
!
+=
+=
+<
+£
+£
p
pp
ppp
pp
ppkp +! for 1>p
)!1(! +<+ pkp
!! zkp ¹+ for any z

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for 1=p
we could pick string
!p
aw ¢
=
where pp >¢
and we would obtain the same conclusion:
!! zkp ¹+¢ for any z

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La kp
Î+!
La kp
Ï+!
BUT:
CONTRADICTION!!!
}0:{ !
³= naL n
pk ££1

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Our assumption that
is a regular language is not true
L
Conclusion: L is not a regular language
Therefore:
END OF PROOF