Answered step by step
Verified Expert Solution
Question
1 Approved Answer
3. Using Pumping Lemma (slides 39-46 of the notes 'Regular Languages & Finite AutomataIV') one can show that the language L={anbnnN} is not regular (We
3. Using Pumping Lemma (slides 39-46 of the notes 'Regular Languages \& Finite AutomataIV') one can show that the language L={anbnnN} is not regular (We need this property in the notes 'Context-free Languages and Pushdown Automata I). This is done by way of contradiction. We assume L is regular. Since L is infinite, Pumping Lemma applies. We then consider the string s=ambm where m is the number of states in the DFA that recognizes L. Since the length of s is bigger than m, by Pumping Lemma, there exists strings x,y and z such that s=xyz,y=,xy2m and xykzL for all kN. If xy
Step by Step Solution
There are 3 Steps involved in it
Step: 1
Get Instant Access to Expert-Tailored Solutions
See step-by-step solutions with expert insights and AI powered tools for academic success
Step: 2
Step: 3
Ace Your Homework with AI
Get the answers you need in no time with our AI-driven, step-by-step assistance
Get Started