Q. Prove L = {anbn | n  0} -- it is not regular by pumping lemma.




Let us assume DFA of the above language has m state (still we know for this language DFA is not possible). 

Actually we assume  anbn | n  0 a regular language with m state DFA.

Now we take a string of this given language and check pumping lemma.


Let ω = a3b3 (aaabbb) |ω| = 6 so, |ω| ≥ m now we decompose it at xyz as 



Here ω ∊ L, |xy| = 4 so, |xy| ≤ m and y > 0 both are satisfied.


Here we want to prove it,



Now we check that all string that accepts pumping lemma is belongs to L or not:


[Satisfy all condition]


Now check xz ∊ L satisfy because aax bbz it belongs to anbn (L).


Also xyz ∊ L because  aax aby bbz also belongs to anbn (L).


Another case xy2z L?



From the above string, it is clear that it is violating the pumping lemma


So,  L = {anbn | n  0} violet pumping lemma, so it is not Regular Language.


We can also prove this language by general way :


Let n number of states.







When we put i = 0 then it will be,



 If we put i = 2 then it will be,



So, anbn violet pumping lemma so, it is not a Regular Language.