In my notes, one definition of poisson process is: $N_t$ is a poisson process of rate $\lambda$ if i) $N_0=0$; ii) for disjoint intervals $(s_i,t_i)$, $N_{(s_i,t_i]}$ are independent; iii) $N_{(s,t]} $ ~ $po(\lambda(t-s))$.
I understand that i) and iii) are true for superposition of poisson process. I think ii) is true intuitively (memoryless property?) but don't know how to formalize the argument.
Clarification. For two independent Poisson processes $M,N$ with same rate $\lambda$ their sum $M+N$ is a Poisson process with rate $2\lambda$.
Proof. Use \begin{align} &2^a\lambda^a(t-s)^a=\Big(\lambda(t-s)+\lambda(t-s)\Big)^a=\sum_{n=0}^a{a\choose n}\lambda^n(t-s)^n\lambda^{a-n}(t-s)^{a-n}\\ &=a!\sum_{n=0}^a\frac{\lambda^n(t-s)^n}{n!}\frac{\lambda^{a-n}(t-s)^{a-n}}{(a-n)!}\,. \end{align} Now let $M$ and $N$ be two independent Poisson processes with the same rate $\lambda.$ Then \begin{align} &\mathbb P\Big\{M_t+N_t-M_s-N_s=a\Big\}=\sum_{n=0}^a\mathbb P\Big\{M_t-M_s=a-n\Big\}\,\mathbb P\Big\{N_t-N_s=n\Big\}\\ &=e^{-2\lambda(t-s)}\sum_{n=0}^a\frac{\lambda^{a-n}(t-s)^{a-n}}{(a-n)!}\frac{\lambda^n(t-s)^n}{n!} =e^{-2\lambda(t-s)}\frac{(2\lambda)^a(t-s)^a}{a!}\,. \end{align}
About property (ii). Since $M$ and $N$ are independent their sum trivially has independent increments.