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Assume that we want to use the Pumping Lemma to prove the language L is not a regular language. Here are the steps we follow: 1. We assume that L is a regular laguage with a pumping length value of P. 2. We define a string in the language L, called S such that |S| 3. We consider all different cases that string S can be divided into three segments: S=x.y.z for all these cases, the following conditions can not be satisfied at the same time: |y|>0; │x.y| ; x.yi.ze for all i=0,1,2,3,... therefore L cannot be a regular language.

Assume that we want to use the Pumping Lemma to prove the language L is not a regular language. Here are the steps we follow: 1. We assume that L is a regular laguage with a pumping length value of P. 2. We define a string in the language L, called S such that |S| 3. We consider all different cases that string S can be divided into three segments: S=x.y.z for all these cases, the following conditions can not be satisfied at the same time: |y|>0; │x.y| ; x.yi.ze for all i=0,1,2,3,... therefore L cannot be a regular language.

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Assume that we want to use the Pumping Lemma to prove the language L is not a regular language. Here are the steps we follow: 1. We assume that L is a regular laguage with a pumping length value of P. 2. We define a string in the language L, called S such that |S| 3. We consider all different cases that string S can be divided into three segments: S=x.y.z for all these cases, the following conditions can not be satisfied at the same time: |y|>0; │x.y| ; x.yi.ze for all i=0,1,2,3,... therefore L cannot be a regular language.
Transcribed Image Text:Assume that we want to use the Pumping Lemma to prove the language L is not a regular language. Here are the steps we follow: 1. We assume that L is a regular laguage with a pumping length value of P. 2. We define a string in the language L, called S such that |S| 3. We consider all different cases that string S can be divided into three segments: S=x.y.z for all these cases, the following conditions can not be satisfied at the same time: |y|>0; │x.y| ; x.yi.ze for all i=0,1,2,3,... therefore L cannot be a regular language.
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