I am trying to prove with the pumping lemma if the expression described in the title is a regular language or not, this is what I have tried:
∑={a,b}
∃(w)∈L y w = a^i b^j a , siendo i,j>=0
|w| = 2n+1 >=n , cumple la condicion de |w| >= n
We know that w = xyz
with which:
|xy| <= n ^ 1 <= |y| <= n
Being X=a^p
, y=a^q
and that p+q<=n
I would stay in the endz=a^(n-p-q) b^n a
This way it would stay|w|= p+q+n-p-q+n+1
checking the property of: x y^j z ∈ L
for all j >=0, we have to create a new chain with j=2: w' = x y^2 z
with which it would be:
p + 2q + n - p - q + n + 1 = 2n + 1
In the end it is simplified to: q+2n+1 = 2n+1
that is only true if q=0
which cannot be since the property of the pumping lemma itself says1<=|y|<=n
What I don't understand is that if I apply this with a string, for example:
w = aaaba
With n=2
we can divide w
into:
x=a
y=a
z=aba
With which for any y^i
that occurs to us as long as it is >=0, it should be accepted, since it ends in a
no ??