Proving the Uncontrollable Fun in CFL Pumping Lemma!

Have you ever wondered how to prove that certain languages are not context-free?

Let's dive into the intriguing world of CFL pumping lemma to uncover the truth behind the context-free nature of languages!

Exploring the Magic of CFL Pumping Lemma

Using the CFL pumping lemma, we have proved that none of the languages L1, L2, and L3 are context-free.

In all cases, we reach a contradiction. Hence, L1 is not context-free.

In all cases, we reach a contradiction. Hence, L2 is not context-free.

In all cases, we reach a contradiction. Hence, L3 is not context-free.

Diving into the Details of CFL Pumping Lemma Proofs

(a) To prove that L1 is not context-free using the CFL pumping lemma, we assume that L1 is context-free and reach a contradiction.

Assume L1 is context-free. Let p be the pumping length given by the pumping lemma.

Consider the string s = a^pb^pb^pa^p. According to L1, s should be in the language. Since s is a sufficiently long string (|s| >= p), it can be divided into five parts: uvxyz.

By the pumping lemma, different cases are considered to reach a contradiction, ultimately proving that L1 is not context-free.

(b) To prove that L2 is not context-free using the CFL pumping lemma, we follow a similar approach of reaching a contradiction when assuming L2 is context-free.

Consider the string s = a^pb^(2p) for L2 and go through the process of contradiction to establish that L2 is not context-free.

(c) The proof for L3 being not context-free follows the same structure with unique string considerations and contradiction processes.