Strong Markov property, independence, regular conditional probability

I have a question on an argument appearing in this paper P.

Setting

Let $S=(1,\infty) \times (-1,1) \subset \mathbb{R}^2$ be a split, and let $X=(\{X_t\},\{P_x\}_{x \in S})$ be a Brownian motion in $S$ conditioned to hit $\{1 \} \times (-1,1)$.

We denote by $r(t)$, $y(t)$ the first coordinate process of $X$ and the second coordinate process of $X$, respectively. Let $\tau_r=\inf\{t>0 \mid r(t)=r\}$, $r \ge 1$.

Question

We consider random variables:\begin{align*}R_{k}=\int_{\tau_{k}}^{\tau_{k-1}}\frac{1}{r(s)^2}\,ds,\quad 2 \le k \le n. \end{align*}

I would like to ask why $\{R_k\}_{k=2}^{n}$ are independent under $P_{n,y}(\cdot \mid y(\tau_j)=y_j,\quad j=1,\cdots,n-1)$ for each $n \ge 2$ and $y \in (-1,1)$.

The author of the above paper claims that strong Markov property yields the independence. However, I couldn't know how to use it. The event $\{y(\tau_j) \in B_j,\quad j=1,\cdots, n-1\}$, $B_j \in \mathcal{B}((-1,1))$ seems like future events...

ADD

As RaphaelB4 said, under $P_{n,y}(\cdot \mid y(\tau_j)=y_j,\quad j=1,\cdots,n-1)$, $X_t$ on $[\tau_i,\tau_{i-1}]$ and $X_{t}$ on $[\tau_{j},\tau_{j-1}]$$(i \neq j)$ would be independent. But, I don't know how to prove this statement.