Poisson process on nonintersecting sets I try to show that for a homogeneous Poisson process $N$ with intensity $\lambda$ and nonintersecting sets $A,B$ the amount of events happening in the two sets are independent. My idea is to prove the following 
$$P(N(A) = n, N(B) =z)=P(N(A\cup B)= n+z)=P(N(A) = n)P(N(B) =z)$$
the first equality comes from $A\cap B=\emptyset$ but I fail to show the next one because of the factorial
$$P(N(A\cup B)= n+z)=\frac{(\lambda|A\cup B|)^{n+z}}{(n+z)!}e^{-\lambda(|A\cup B|)}$$
I would appreciate some help, thanks
Edit:
My definition of a HPP 
$N$:


*

*For some $\lambda > 0$ and finite planar region $A$, we have a Poisson distribution, i.e. $$N(A)\sim \text{Poi}(\lambda \cdot  |A|)$$

*Given $N(A)=n$, the $n$ events in $A$ are an independent sample from the uniform distribution on $A$
 A: For simplicity of notation let $X=N(A)$, $Y=N(B)$ and $Z=X+Y=N(A\cup B)=N(A)+N(B)$, let $a = EX = \lambda|A|$ and $b=EY$.
The probability that $(X,Y)=(k,l)$ can be found by conditioning on $Z$ as follows:
$$\begin{align*}
P((X,Y)=(k,l)) &= \sum_{n\ge0}P(Z=n)\times P((X,Y)=(k,l)|Z=n)\\
&=P(Z=k+l)\times  P((X,Y)=(k,l)|Z=k+l)\tag{*}\\
&=\frac{(a+b)^{k+l}}{(k+l)!}e^{-(a+b)}\times P((X,Y)=(k,l)|Z=k+l)\\
&= \frac{(a+b)^{k+l}}{(k+l)!}e^{-(a+b)}\times  \binom{k+l}k 
\left(\frac a{a+b}\right)^k \left(\frac b{a+b}\right)^l\\
&= \frac{(a+b)^{k+l}}{(k+l)!}e^{-(a+b)}\times \frac{(k+l)!}{k!l!} 
\left(\frac a{a+b}\right)^k \left(\frac b{a+b}\right)^l\\
&= \frac{a^k}{k!}e^{-a}\,\frac {b^l}{l!} e^{-b}\\
&= P(X=k)P(Y=l).
\end{align*}$$
Here (*) holds because all terms $P((X,Y)=(k,l)|Z=n)$ vanish, except when $n=k+l$.
In words: the variables $X,Y,Z$ are definitely not independent. 
 But when $Z$ is Poisson and $X$ conditional on $Z$ is binomial and conditional on $Z$ we have $Y=Z-X$, magically the right factorials cancel to make $X$ and $Y$ independent.
