Definitions: A Poisson Process is a counting process that satisfies $N(t)\sim\mathcal{P}oi(\lambda t)$ and has the independent and stationary increment properties.
The independent increment property is where the number of arrivals of a phenomenon in disjoint intervals (i.e. increments) are independent. That is, if $(t_1, t_2] $ and $ (t_3, t_4 ] $ are disjoint, then $N(t_2) - N(t_1) $ and $N(t_4) - N(t_3)$ are independent.
I've seen this reasoning many, many times:
(Source: An Introduction to Probability Models by Sheldon Ross)
Question: I don't understand why the highlighted condition disappears because of "independent increments." The condition is written as the first arrival time. How can you write this as an equivalent event with increment(s) disjoint from $(s, s+t]$? I tried $\{ T_1 = s \} = \{ N(s) = 1 \cap N(t) < 1 \quad \forall t \in (0,s)\} $, but I can't see how I can fit this into the definition of independent increments.
If someone can formally show this, I would really appreciate it.
The relevant event is the conjunction. $$\{T_2{>}t~,T_1{=}s\} = \{N(t){-}N(s){=}0,\forall \tau~(\tau{<}s\leftrightarrow N(\tau){=}0)\}$$ Now, while the LHS is not, the RHS is an intersection of independent events. Since the count of arrivals in $(s..t]$ is independent of the count of arrivals in $(0..s]$, so therefore the count of arrivals in $(s..t]$ is independent of the first arrival occurring exactly at $s$ (a point within $(0..s]$). $$\begin{align}\mathsf P(T_2>t\mid T_1=s)&=\mathsf P(N(t){-}N(s){=}0\mid T_1=s)\\[1ex]&=\mathsf P(N(t){-}N(s){=}0)\end{align}$$
Alternatively, if we use the notation $N_\mathcal I$ to represent the count for arrivals within interval $\mathcal I$, then
$$\{T_2{>}t~,T_1{=}s\} = {\{N_{(s..t]}{=}0, N_{\{s\}}=1, N_{(0..s)}=0\}}$$
So $$\begin{align}\mathsf P(T_2{>}t\mid T_1{=}s)&=\mathsf P(N_{(s..t]}{=}0 \mid N_{\{s\}}=1, N_{(0..s)}=0)\\[1ex]&=\mathsf P(N_{(s..t]}{=}0)\end{align}$$