pumpig lemmahow to prove non rugalarity lemma : i'm minni how pumping lemma works 1) All strings in th elanguage can be repeated (pumped) if they are at least as long as a certain length (i.e. the pumping length).example: A={all strings that end in 11} ; pumping length = p = 30 1110 11011 11If A is a rugular language, then there is a number p (the pumping length) where if s is any string in A that is at least length p, then s may be divided into 3 pieces, s=xyz,satisfying the following conditions: 1) for each i>=0, x y^i z belongs to A2) |y| > 0 (y cannot be an empty string because we can not pumped an empty string ) (but x and z can be empty )3) |xy| <= p Summary:-1 -> All reegular languages satisfyy the pumping lemma.A={strings that end with 11}2 -> we used pumping lemma as a way to prove that some languages are not regular.regular language is one that can be expressed using finite automata or regular expressions and non-regular language on the other hand cannothere is the example of a non-regular language,B={ 0^n 1^n | n>=0 }if you try to construct a finite automata for this language, you will find that it is impossible and that is because NFAs and DFAs have a finite amount of