Answer :

Answer:

To show that the language L = {a

nb

kcnd

k : n ≥ 0,k > n} is not regular, we can use the pumping lemma for regular languages. Here is a proof by contradiction:

Assume that L is regular. Then, by the pumping lemma, there exists a positive integer p such that any string w in L of length at least p can be divided into three parts: w = xyz, such that |yx| < p, |y| > 0, and for all i > 0, w

(xy^i)z is also in L.