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I'm reading Billingsley and getting tripped up on a few details in his explanation on how to see Brownian Motion has the Markov Property:

Fix $t_0 \geq 0$ and put $W'_t = W_{t_0+t}-W_{t_0}$. Let $\mathcal{F}_t=\sigma[W_s:s\leq t]$. The random variables $W'_t$ are independent of $\mathcal{F}_{t_0}$. To see this, suppose $0 \leq s_1 \leq s_2 ... \leq s_j \leq t_0$ and $0 \leq t_1 \leq t_2 ... \leq t_k$. Put $u_i = t_0 + t_i$. Since the increments are independent, $(W'_{t_1}, W'_{t_2}-W'_{t_1}, ... W'_{t_k}-W'_{t_{k-1}}) = (W_{u_1}-W_{t_0}, W_{u_2}-W_{u_1},...,W_{u_k}-W_{u_k})$ is independent of $(W_{s_1}, W_{s_2}-W_{s_1}, ... W_{t_j}-W_{t_{j-1}})$. But then $(W'_{t_1},W'_{t_2},...W'_{t_k})$ is independent of $(W_{s_1},W_{s_2},...W_{s_j})$

Why is the bolded statement true?

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Let $F_k(x_1,\ldots,x_k):=(x_1,x_2+x_1,\ldots,x_k+x_{k-1})$. Then $F_k$ is Borel measurable and for any Borel sets $B_1\subset \mathbb{R}^k$ and $B_2\subset \mathbb{R}^j$, \begin{align} &\mathsf{P}((W_{t_1}',\ldots,W_{t_k}')\in B_1,(W_{s_1},\ldots,W_{s_j})\in B_2) \\ &\qquad=\mathsf{P}((W_{t_1}',W_{t_2}'-W_{t_1}',\ldots,W_{t_k}'-W_{t_{k-1}}')\in F_k^{-1}(B_1)) \\ &\qquad\quad\times\mathsf{P}((W_{s_1},W_{s_2}-W_{s_1},\ldots,W_{s_j}-W_{s_{j-1}})\in F_j^{-1}(B_2)) \\ &\qquad =\ldots \end{align}

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