# independent increments property implies Markov property

Let $\{X_t\}_{t\in\mathbb R^+}$ be a stochastic process with values in $\mathbb R$. Suppose that $\{X_t\}$ has independent increments, namely for every $t_1<t_2<\ldots<t_k$ the random variables $X_{t_2}-X_{t_1}$, $X_{t_3}-X_{t_2}$, $\ldots,X_{t_k}-X_{t_{k-1}}$ are independent. I have to prove the Markov property, that is $$P(X_t\in B\,|\, \mathcal F_s)=P(X_t\in B\,|\, X_s)$$ for $s<t$, $B$ measurable and $\mathcal F_s=\sigma(X_s:s\le t)$. Can you help me in order to formalize the details of this proof?

As $X_t-X_s$ is independent of $\mathcal{F}_s$ and $X_s$ is $\mathcal{F}_s$-measurable, it follows that
$$\mathbb{P}(X_t \in B \mid \mathcal{F}_s) = \mathbb{E}(1_B((X_t-X_s)+X_s) \mid \mathcal{F}_s) = \mathbb{E}(1_B(X_t-X_s+x)) \bigg|_{x=X_s}. \tag{1}$$
Obviously, the right-hand side is $\sigma(X_s)$-measurable and therefore the tower property implies
$$\mathbb{P}(X_t \in B \mid X_s) =\mathbb{E}(1_B(X_t-X_s+x)) \bigg|_{x=X_s} \tag{2}.$$
Combining $(1)$ and $(2)$ finishes the proof.