Considering $W$ the canonical process on $C([0,1],\mathbb{R})$ and the row filtration generated by the coordinate process of $W$, I want to prove that $\mathbb{P}_x[\tau_0>t]=\mathbb{P}_0[\tau_x>t]$ where $\tau_0=\inf\{r>0:W_r=0\}$ and $\tau_x=\inf\{r>0:W_r=x\}$.
I tried the following way.
$\mathbb{P}_x[\tau_0>t]=\mathbb{E}_{W_s}[I_{\tau_0>t}]\mid_{x=W_s}=\mathbb{E}_{0}[I_{\tau_0>t}\circ \theta_s\mid \mathcal{H}_s]\mid_{x=W_s}$ but here I am lost and I don't know how to continue. Thanks for any help
Without loss of generality, we may assume $x>0$. By the very definition of the stopping time, we have
$$\begin{align*} \mathbb{P}_x(\tau_0>t) &= \mathbb{P}_x \left( \left\{w \in C[0,1]; \inf_{s \in [0,t]} (w(s)-x) +x >0 \right\} \right) \end{align*}$$
Since $\mathbb{P}_x$ is defined by translational invariance (it is the image measure of the mapping $C[0,1] \ni w \mapsto w(\cdot)-x \in (C[0,1],\mathcal{B}(C[0,1]),\mathbb{P}_0)$, we get
$$\mathbb{P}_x(\tau_0>t) = \mathbb{P}_0 \left( \left\{w \in C[0,1]; \inf_{s \in [0,t]} w(s)>-x \right\} \right) = \mathbb{P}_0(\tau_{-x}>t).$$
Finally, using symmetry, we find
$$\mathbb{P}_x(\tau_0>t) = \mathbb{P}_0(\tau_{x}>t).$$