Question

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2. Prove that {a"b"c" | m,n 20}is not a regular language. Answer:

Answer 1

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We will solve this using pumping lemma.

Let L= where m,n>=0.

a 772 197 "C

Lets define constant of Pumping lemma as N.

So, some word=$a^{1}b^{N}c^{N}$ and this is a part of L. If we assume Language L as regular, we can say the following.

word=xyz such that |xy|<=N such that $xy^{k}z$ for all k>=0. Now if x=$\epsilon$, y=b,z=c . This means for k=0, we have :-

$xy^{0}z=b^{N}c^{N}$which does not belong to language L as a is missing. So, this statement is contracdicted.

So, Language L is not Regular as per pumping lemma.