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Sagot :
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.
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