Question

show that language L4 = { wabw : w ∈ {a,b}* } is not regular, use pumping lemma

Answer 1

Pumping lemma for regular languages

** pumping lemma is a tool used to prove non regular languages as non regular .
** Pumping Lemma is based on the proof by contradiction .
** pumping Lemma uses pegion hole principal .
** to prove that a language is non - regular by using Pumping Lemma
steps
1. Assume that the given language is regular .
2. If a language is regular there exist a finite automata for that language . Assume that the finite automata has n- states .

3. Select some string Z from given language whose length is >= n . According to pegion howl principal whenever string length
is >= Number of states of finite automata, then there exist a loop or cycle in that automata .
4. divide the string Z into 3 parts u,v,x where v is the loop or the cycle string

Pumping Lemma condition
if a language is reguar for every i >= 1 in the string u vⁱ x which belongs to language L ,
for atleast one i if uvⁱ x does not belongs to L then given language is non - regular

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now the given language is L = { wabw : w ∈ {a,b}* }

ut be a st So, 2 a bb ab abb abb ab, abb 从 1c2 we have now do

ab bab bab abb 2 So, 0 yto