Q. Prove -- 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 a regular language with m state DFA.
Now we take a string of this given language and check pumping lemma.
Let (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:
Now check xz ∊ L satisfy because it belongs to .
Also xyz ∊ L because also belongs to .
Another case ?
From the above string, it is clear that it is violating the pumping lemma.
So, 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, violet pumping lemma so, it is not a Regular Language.