# Poisson- process inter arrival times

I don't understand the proof of inter arrival timesof a poisson process.

Consider the poisson process with rate $\lambda$ and let $\{N (t); t \geq 0\}$ and let $T_1$ be the time up to the first event and $T_n$ be the time between ($n-1$)th and $n$ the events.
Then distribution of $T_1$ is $F_{ T_1}(t)=\Pr (T_1 \leq t)$. Then I don't understand how it was written as $F_{ T_1}(t)=\Pr (T_1 \leq t)=1-\Pr (N (t)=0)$.
Also $F_{ T_2}(t)=\Pr (T_2 \leq t)$ is written as
$1- \int_{0}^{ \infty} \Pr (T_2>t|T_1=s) f_{T_1}(s) ds$. How can we write this?
Also I don't understand how $\Pr (T_2>t|T_1=s)= \Pr (\text{zero events between } s \text{ and } s+t)$.
Can someone please carrify this to me

1. $F_{T_1}(t) = \Pr(T_1 \leq t) = 1 - \Pr(N(t) = 0)$, since if by the time $t$ the first event hasn't occured ($N(t) = 0$), then it must occur after $t$, so $T_1 \geq t$.
2. $\Pr(T_2 \leq t) = 1 - \Pr(T_2 > t)$. As a result of the law of total expectation (applied to $\mathsf{E}\; 1_{T_2 > t} = \Pr(T_2 > t)$), $\Pr(T_2 > t) = \int_0^\infty ds \Pr(T_2 > t | T_1 = s) f_{T_1}(s)$.
3. If the first event occured at time $s$ and by the time $t + s$ no more events have occured, this means that the second event cannot have occured during that interval, thus, at least $t$ time has passed between the first event and the second event.