I am reading the proof of the reflection principle of Brownian motion from René Schilling's Brownian motion and stochastic Calculus. There are two parts in the below identities that I cannot understand. In the below proof, it says that $B_{\tau_b + (t-\tau_b)} - B_{\tau_b} \in \mathscr{F}_\infty^W \coprod \mathscr{F}_{\tau_b}^B$ and $\sim W_{t-\tau_b}$. This follows from the SMP or Theorem 6.5 below. However, for $W_{t-\tau_b} := B_{\tau_b + (t-\tau_b)} - B_{\tau_b}$ to make sense, we need $t-\tau_b \ge 0$, i.e. condition on $\{\tau_b \le t\}$. Hence all this only holds when we assume $\tau_b \le t$, which is an event in $\mathscr{F}_{\tau_b}^B$. So how do we ensure that $1_{\tau \le b}B_{\tau_b + (t-\tau_b)} - B_{\tau_b}$ is independent of $\mathscr{F}_{\tau_b}$ and distributed according to $W_{t-\tau_b}$ from the Strong Markov Property in Theorem 6.5?
Also, where is independence of $\{ B_{{\tau_b}+(t-\tau_b)} - B_{\tau_b} <0\}$ and $\mathscr{F}_{\tau_b}^B $ under the RHS of the first equality actually used here? From what I see, all we need is that $B_{{\tau_b}+(t-\tau_b)} - B_{\tau_b} $ is distributed as $W_{t-\tau_b}$, which is symmetric about the origin, so we get the next equality, and the final equality follows from $B_{\tau_b}=b$. But again, I don't know how to interpret $W_{t-\tau_b}$ herebe cause $t-\tau_b$ is random, whereas $t$ in $W_t$ of 6.5 is not.
I couldn't figure out these questions on my own. I would greatly appreciate any help.
The Strong Markov Property, Theorem 6.5, is the following from the book.
Can I recommend Jean-Francois Le Gall's book, Brownian Motion, Martingales and Stochastic Calculus, in particular, Section 2.4 titled the Strong Markov Property of Brownian Motion. His proof goes into a bit more detail in handling the particular steps you mentioned.
I do have the proof typed up in a fair amount of details since this bothered me a lot as well but it is rather messy.
Letting $(\Omega,\mathcal{F}, \mathbb{P})$ be the probability space, the idea is to note that $\{\omega \in \Omega : \tau_b \leq t \} \cap \{\omega \in \Omega : W_{t - \tau_b} < 0\} = \{ \omega \in \Omega : (\tau_b, W_t) \in A\} $ where $A := \{ (s,w) \in \mathbb{R}_+ \times C(\mathbb{R}_+, \mathbb{R}^d) : s \leq t, w(t - s) < 0\}$. On $\mathbb{R}_+ \times C(\mathbb{R}_+, \mathbb{R}^d)$, we can declare the product $\sigma$-algebra generated by the Borel $\sigma$-algebra on $\mathbb{R}_+$ and the canonical one on $C(\mathbb{R}_+, \mathbb{R}^d)$. R. Schilling mentions it in his text when he introduces the Wiener measure at the start of Chapter $4$. We have to check also that $A$ is measurable with respect to the product $\sigma$-algebra.
Note that since $\tau_b$ and $W$ are independent, then for the product mapping $(\tau_b, W_t): \Omega \rightarrow \mathbb{R}_+ \times C(\mathbb{R}_+, \mathbb{R}^d)$ given by $\omega \mapsto (\tau_b(\omega),W_t(\omega))$, we actually have the the joint distribution is just the product measure, that is $(\tau_b, W_t)_\#\mathbb{P} = (\tau_b)_\#\mathbb{P} \otimes (W_t)_\# \mathbb{P}$. But note that the law of the Brownian motion, $(W_t)_\#\mathbb{P}$ could be replaced with any other Brownian motion (See R. Schilling's Remark $4.4$ in Chapter $4$), in particular, we replace it with $(-W_t)_\# \mathbb{P}$, since $- W_t$ is also a Brownian motion. Also, note that since $W_t$ is independent of $\tau_b$, then $- W_t$ is also independent of $\tau_b$. A similar line of reasoning gives that $(\tau_b, -W_t)_\# \mathbb{P} = (\tau_b)_\#\mathbb{P} \otimes (-W_t)_\# \mathbb{P}$.
Putting everything together, we have that $(\tau_b, W_t)_\# \mathbb{P} = (\tau_b, -W_t)_\# \mathbb{P}$ so that $$ \mathbb{P}(\{\omega \in \Omega : \tau_b \leq t \} \cap \{\omega \in \Omega : W_{t - \tau_b} < 0\}) = \mathbb{P}(\{ \omega \in \Omega : (\tau_b, W_t) \in A\}) = (\tau_b, W_t)_\# \mathbb{P}(A) = (\tau_b, -W_t)_\# \mathbb{P}(A) = \mathbb{P}(\{ \omega \in \Omega : (\tau_b, -W_t) \in A\}) = \mathbb{P}(\{\omega \in \Omega : \tau_b \leq t \} \cap \{\omega \in \Omega : -W_{t - \tau_b} < 0\}) $$ from which the equality you were after follows.
I hope this helps!
Just in case: $f_\# \mathbb{P}$ is the pushforward measure, given by $f_\# \mathbb{P}(A) = \mathbb{P}(\{ f \in A\})$.