Pumping lemma occurrence of c > d I'm trying to prove a language is not regular through using pumping lemma, but can't seem to come up with any way of doing it. 

The alphabet is:

$$ \Sigma = \{c, d\} $$

The language is:

$$ A = \{z \in \Sigma^* \mid c(z) &gt; d(z)\} , \text{where $c(w)$ and $d(w)$ means occurrences of $c$ or $d\in z$  }$$

I tried to define a string like so:

Pumping length = P

$$ s = c^{(P+1)}d^P $$

Here is how I tried to solve it:

If we select P = 5. Then s = ccccccddddd and split it in 3 parts. Where x = ccccccd, y = ddd and z = d. (Is that even allowed though as |xy| &lt;= P, but now |xy|=10 and 10>5). If I then do y^2. Then I get s = ccccccdddddddd, so now d>c which proves the string is not in the language. Does this then prove that the language is not regular?
You need to show that with that arbitrary pumping length $P$, for any partition $c^{(P+1)}d^P = xyz$ such that $|y|&gt;0$ and $|xy|\leq P$, there is some $i$ such that $xy^iz \not\in A$.

Since for any partition we require that $|xy|\leq P$, then $x$ and $y$ necessarily consist only of $c$'s. Also, you know that $|y| &gt;0$ (it's also a requirement). Can you find an $i\in\mathbb{N}$ such that the amount of $c$'s in $xy^iz$ is less than or equal to the amount of $d$'s?