How do you use pumping lemma to prove that languages are non regular?

1. The Pumping Lemma is used for proving that a language is not regular. Here is the Pumping Lemma.
2. Let L = {0k1k : k ∈ N}. We prove that L is not regular.
3. Let L = {(10)p1q : p, q ∈ N, p ≥ q}. We prove that L is not regular.
4. There are 3 cases to consider: (a) v starts with 0 and ends with 0.

Can a non regular language satisfying pumping lemma?

Note: The converse of the pumping lemma is not true! That is, a language satisfying the lemma may still be non-regular. Since s ∈ C and |s| ≥ p, the pumping lemma implies that s can be divided into three pieces s = xyz, such that xyiz ∈ D for all i ≥ 0.

Does DFA accept non regular language?

The DFA will either accept a string not in the language or reject a string in the language, which it shouldn’t be able to do. Can’t place all these strings into different states; there are only finitely many states! Theorem: The language L = { anbn | n ∈ ℕ } is not regular.

What are the applications of pumping lemma for regular languages?

Pumping Lemma is to be applied to show that certain languages are not regular. It should never be used to show a language is regular. If L is regular, it satisfies Pumping Lemma. If L does not satisfy Pumping Lemma, it is non-regular.

How to prove that language L is not regular?

Method to prove that a language L is not regular 1 At first, we have to assume that L is regular. 2 So, the pumping lemma should hold for L. 3 Use the pumping lemma to obtain a contradiction − Select w such that |w| ≥ c Select y such that |y| ≥ 1 Select x such that |xy| ≤ c

How is the pumping lemma used in CFL?

Pumping Lemma for Context-free Languages (CFL) Pumping Lemma for CFL states that for any Context Free Language L, it is possible to find two substrings that can be ‘pumped’ any number of times and still be in the same language. For any language L, we break its strings into five parts and pump second and fourth substring.

How is pumping lemma used in theory of computation?

Pumping Lemma, here also, is used as a tool to prove that a language is not CFL. Because, if any one string does not satisfy its conditions, then the language is not CFL. For above example, 0 n 1 n is CFL, as any string can be the result of pumping at two places, one for 0 and other for 1.