Contradiction: Formulas And Consequences Of L1 Is A Regular Language

1.(a)

1.(b)

1.(c) 2.(a)
Proof by contradiction: Assume for contradiction L1 is a regular language
As L1 is infinite we can apply Pumping Lemma
Let m be the critical length for L1
Pick a String w such that: w L1 and length |w|>=m
We pick w=
For the pumping lemma let’s write w= =xyz
With lengths |x y|<=m, |y|>=1
=>x= and y= and z= and a+b<=m
Consider i=3, we need to prove that xy³z = is not in L1 i.e, we need to show that m²+2b is not a square.
Since b>=1 we have m²+2b > m²
Since a+b<=m, we have m²+2b<= m²+2m<= (m+2)²
Therefore L1 is not a regular language

2.(b)
Proof by contradiction: Assume for contradiction L2 is a regular language
As L3 is infinite we can apply Pumping Lemma
Let m be the

And hence L4 is not a regular language

3.(a) Let’s use Pumping lemma to find whether L1 is regular or a non-regular language
Assume that L1 is a regular language
Let m be the critical length for L – Pumping length
Pick a string w such that : w L1 and length |w|>=m
We pick w= which is a string from L1
For the pumping lemma let’s write w= = xyz
With lengths |x y|<=m, |y|>=1
Consider x= , y=1, z=
For a regular language, for every i, x z should be in language L1 where i>=0
Lets consider the case where i=0 x z = xz= = As 2m-1 never equals 2m, our assumption was wrong
Therefore L1 is a non-regular language

3.(b)

If a single string w Є L2 is proven to contradict, then the whole language is Non-regular
Lets consider that L2 is regular language
As L2 can go to infinity , we can apply Pumping Lemma
Let m be the critical length for L2 – Pumping length
Pick a string w= such that: wЄL2
For the pumping lemma let’s write w = =xyz
With lengths |x y|<=m, |y|>=1
Consider x= , y=a, z=
For a regular language, for every i, x z should be in language L1 where i>=0
Lets consider the case where