Notes

Prove that the language L={w | w is a palindrome} is not regular.
We choose s = 101 => 1^p 0^p 1^p
p should have a value at least equal to |s| which is 3 so 111000111.
with the third condition we know that xy <=p
that means y should be in the first section of the string with all 1s,
if we pump y we get 111....111000111, which is not palindrome,
so we can state that the language is not regular

Prove that Language L2 = {w | w has even length and the first half of w has more 0s than the second half of w} is not regular.
We choose s = 01 => 0^p 1^p => 0011
with third condition we know that xy <=p that means y must be in the first half of the string, more precisily it must be x = 0, y = 0 and z = 11
with the first condition - for each i ≥ 0, xy^i z ∈ A
if we choose i = 0 or i odd there's no way to divide the string in xyz, because y^i will always get an odd number which is not in the language,
so we can state that the language is not regular