For the Brownian motion $B$ on the probability space $(\Omega,\mathcal{F},\mathbb{P})$ with filtration $\mathcal{F}_n = \sigma(B_s,s\leq n)$ for $t\geq0$ and for any stopping time regarding $(\mathcal{F}_n)_{n\geq0}$ is $\tilde{B}:=B_{n+\tau}-B_{\tau}, t\geq 0$ a Brownian motion, which is independent of $\mathcal{F}_{\tau}$.
Let $(B_n)_{n\in\mathbb{N}}$ be a sequence of iid random variables with $\mathbb{P}(B_n = 1)=p$ and $\mathbb{P}(B_n=-1)=1-p$. We look at the simple random walk $S_n:=B_1+\dots+B_n$.
How can I use the strong Markov property to prove $(B_{n+\tau}-B_{\tau})_{n\in\mathbb{N}}$ is independent of $\mathcal{F}_{\tau}$?