Let $(N_t)_{t \geq 0}$ be a rate-$\lambda$ Poisson process, and let $(X_i)_{i \geq 0}$ be IID $\text{Bernoulli}(1/2)$. Denote by $(N_t^i)_{t \geq 0, i \in \{0, 1\}}$ the thinned Poission processes defined by \begin{equation} N_t^i \triangleq \sum_{k \geq 1} 1\{S_k \leq t, X_k = i\}, \end{equation} where $(S_k)_{k \geq 1}$ are the increment times of $(N_t)$. Thus, $(N_t^0)$ accumulates the increments of $(N_t)$ occurring at times $S_k$ for which $X_k = 0$, and $(N_t^1)$ accumulates the increments of $(N_t)$ occurring at times $S_k$ for which $X_k = 1$. It can be proved, for instance by computing the characteristic function of $N_t^0$ and $N_t^1$, that $(N_t^0)$ and $(N_t^1)$ are independent $\lambda/2$ Poisson processes. While I understand the proof, the independence seems counterintuitive to me, and I want to obtain a better understanding of it.
Let, for example, $\lambda = 1$ and suppose that $S_1^1 = 1$ (i.e., the first increment of the $1$-thinned Poisson process occurs at time $1$.) Unconditionally, \begin{equation} N_1^0 \sim \text{Poisson}(1/2). \end{equation} Independence of the thinned processes implies that also \begin{equation} N_1^0 \mid \{S_1^1 = 1\} \sim \text{Poisson}(1/2), \end{equation} but this seems odd to me - surely, the fact that an increment occurs at time $1$ will affect the distribution of the number of increments $N_1$ in the [0, 1) interval, and thereby also $N_1^0$? Consider for example the conditional density of $S_1^1$ evaluated at time $1$: \begin{equation} f_{S_1^1}(1 \mid N_1^0 = k), \end{equation} where $k$ is some positive integer. Independence of the processes would imply that \begin{equation} f_{S_1^1}(1 \mid N_1^0 = k) = f_{S_1^1}(1) = \sum_{k \geq 1} f_{S_1^1}(1 \mid N_1^0 = k)\Pr[N_1^0 = k]. \end{equation} I suppose that I could verify the above equation, but why would I assume it to be true? It seems - considering the independence proof via the characteristic function - to be intimately related to the fact that the underlying process is Poisson.
I would much appreciate any explanation as to why the underlying process being Poisson suggests that its thinned sub-processes are independent.