00:01
Hello students, so here the proof by using the pumping lemma.
00:04
We aim to prove that the language n, l1, o1, i, j, ok.
00:10
So here i is greater than j and k is less than i.
00:14
So here k is greater than i is not a regular using the pumping lemma.
00:18
So we prove that assume that sake of the contradiction l1 is a regular language.
00:24
According to the pumping lemma there exist a pumping length p such that any string s in l1 with length at least p can be split into the three part x s is equal to x y z satisfying the following condition.
00:42
Y is greater than 0 x y is less than or equal to p for i is greater than or equal to 0 x y z belongs to l1.
00:50
Let's choose the string as 0 p 1 p 0 p plus 1 it clearly satisfies s greater than or equal to p and can be split into the s equal to x y z and x y is less than or equal to p and y is greater than 0.
01:08
So here now by pumping on pumping up y that is considering x x x k or z we also get a string that violate the condition of i greater than j k greater than i since the number 0 and 1 have no longer satisfy this condition...