Strong Markov property states that for bounded jointly measurable process $Y_s(\omega)$ on $ [0,\infty) \times \Omega$ and a stopping time $\tau$,
$$E^x(Y_{\tau} \circ \theta_{\tau})= \phi (B_{\tau}, \tau)$$ holds where $\phi(y,s) := E^y (Y_s)$ for $(y,s) \in \mathbb R \times [0, \infty)$ and $\theta_t (\omega):=\omega '$ with $\omega ' (s) := \omega(s+t)$. Note $E^y$ stands for the expectation w.r.t. Brwonian motion starting at $y \in \mathbb R$ and $B_s$ stands for the Brownian motion.
I want to show that, using the above Strong Markov property, $$Z_{t} := B_{t+\tau}-B_{\tau}$$ is a Brownian motion and independent with $\mathcal F_{\tau}$, the sigma-algebra associated with the stopping time $\tau$.
I have no idea where to start and only had some unsuccessful calculations by setting $Y_s(\omega)=\omega(t)-\omega(0)$ for a fixed $t\geq 0$ which is $B_t$ essentially.
Any help is appreciated.