I am trying to prove the following: If $X =\{X_t : t\geq 0\}$ is a real-valued stochastic process with independent increments then it is also a Markov process: $$ p\{X_t \in A | \mathcal{F}_s\} = p\{X_t \in A | X_s\}, \quad \forall A \in \mathcal{F}. $$ where $\mathcal{F} = \sigma(X) = \{\mathcal{F}_s:s \geq 0\}$ denotes the natural filtration and $\mathcal{F}_s = \sigma(X_k: 0\leq k \leq s$).
This should be enough to prove but I can't really justify what I've done: Supose $f$ is bounded and measurable. For every $t\geq s \geq 0$,$X_s $ is $\mathcal{F}_s$-measurable and the increment $X_t - X_s$ is independent of $\mathcal{F}_s$, therefore \begin{align*} \mathbb{E}[f(X_t) | \mathcal{F}_s] &= E[f(X_t - X_s + X_s) | \mathcal{F}_s]\\ &= E[f(X_t - X_s + y)]\bigg\rvert_{y = X_s} \\ &= \mathbb{E}[f(X_t - X_s + y) | \sigma(X_s)] \bigg\rvert_{y = X_s}\\ &= \mathbb{E}[f(X_t) | \sigma(X_s)]. \end{align*}
Now is enough to consider $f = \mathbb{1}_A$ and this concludes. Is the third equality right? Am I missing some steps in between? Something tells me that the tower property should be applied in there but I just don't see it :/
I want to write a clear proof about this and want to justify every detail.
Thanks in advance.