How to prove the independence of increments of the summation of two independent Poisson processes? Suppose that $\{N_1(t)\,t \ge 0\}$ and $\{N_1(t)\,t \ge 0\}$ are two independent Poisson processes with rates of $\lambda_1$ and $\lambda_2$ respectively, then prove the sum process where $N(t) = N_1(t)+N_2(t)$ for all $t \ge 0$ is a Poisson process with rate of $\lambda_1+\lambda_2$.
The proof on my textbook includes the step of proving that $N(t)$ has independent increments, of which the reason is that $N_1(t)$ and $N_2(t)$ are independent and both possess stationary and independent increments properties. But I couldn't understand such an obvious step. 
Let $I_1$ and $I_2$ be two disjoint intervals, then our goal is to show that $N_1(I_1)+N_2(I_1)$ and $N_1(I_1)+N_2(I_1)$ are independent. Obviously, we have:


*

*$N_1(I_1)$, $N_1(I_2)$, $N_2(I_1)$, $N_2(I_1)$ are independent of each other;

*$(N_1(I_1),N_1(I_2))$ and $(N_2(I_1),N_2(I_2))$ are independent random vectors since $\{N_1(t)\,t \ge 0\}$ and $\{N_1(t)\,t \ge 0\}$ are two independent processes.


Next, I only need to prove that $(N_1(I_1),N_2(I_1))$ and $(N_1(I_2),N_2(I_2))$ are two independent vectors to achieve our goal. But I was stuck here. It seems that I'm complicating the question. Does someone have a better idea? Please help me out here.
 A: $$\begin{aligned} & P\left(N_{1}\left(I_{1}\right)+N_{2}\left(I_{1}\right)=r\wedge N_{1}\left(I_{2}\right)+N_{2}\left(I_{2}\right)=s\right)\\
 & =\sum_{r_{1}+r_{2}=r}\sum_{s_{1}+s_{2}=s}P\left(N_{1}\left(I_{1}\right)=r_{1}\wedge N_{2}\left(I_{1}\right)=r_{2}\wedge N_{1}\left(I_{2}\right)=s_{1}\wedge N_{2}\left(I_{2}\right)=s_{2}\right)\\
 & =\sum_{r_{1}+r_{2}=r}\sum_{s_{1}+s_{2}=s}P\left(N_{1}\left(I_{1}\right)=r_{1}\right)P\left(N_{2}\left(I_{1}\right)=r_{2}\right)P\left(N_{1}\left(I_{2}\right)=s_{1}\right)P\left(N_{2}\left(I_{2}\right)=s_{2}\right)\\
 & =\sum_{r_{1}+r_{2}=r}\left[P\left(N_{1}\left(I_{1}\right)=r_{1}\right)P\left(N_{2}\left(I_{1}\right)=r_{2}\right)\sum_{s_{1}+s_{2}=s}P\left(N_{1}\left(I_{2}\right)=s_{1}\right)P\left(N_{2}\left(I_{2}\right)=s_{2}\right)\right]\\
 & =\left[\sum_{s_{1}+s_{2}=s}P\left(N_{1}\left(I_{2}\right)=s_{1}\right)P\left(N_{2}\left(I_{2}\right)=s_{2}\right)\right]\times\left[\sum_{r_{1}+r_{2}=r}P\left(N_{1}\left(I_{1}\right)=r_{1}\right)P\left(N_{2}\left(I_{1}\right)=r_{2}\right)\right]\\
 & =P\left(N_{1}\left(I_{2}\right)+N_{2}\left(I_{2}\right)=s\right)\times P\left(N_{1}\left(I_{1}\right)+N_{2}\left(I_{1}\right)=r\right)
\end{aligned}
$$
showing that $N_{1}\left(I_{1}\right)+N_{2}\left(I_{1}\right)$ and
$N_{1}\left(I_{2}\right)+N_{2}\left(I_{2}\right)$ are independent.
