Let $B=(B_t,t\ge 0)$ be a Brownian motion and $$B'_t:=B_{T+t}-B_T\;\;\;\text{for }t\ge 0$$ for some $T\ge 0$. Especially, $B$ has independent increments, i.e. $$\left(B_{t_i}-B_{t_{i-1}}\right)_{i=1,\ldots,n}$$ is independent, for all $0\le t_0<\ldots<t_n$.
I want to prove, that $B'$ and $(B_s)_{0\le s\le T}$ are independent, i.e. for all $0\le s_0<\ldots<s_m\le T$ and $0\le t_0<\ldots<t_n<\infty$ $$\left(B_{s_0},\ldots,B_{s_m}\right)\;\;\;\text{and}\;\;\;\left(B'_{t_0},\ldots,B'_{t_n}\right)\tag{1}$$ are independent.
Since $$B'_{t_i}-B'_{t_{i-1}}=B_{T+t_i}-B_{T+t_{i-1}}$$ and $$0\le s_i<T+t_0<\ldots<T+t_n$$ it's easy to see (since $B$ has independent increments) that $$B_{T+t_n}-B_{T+t_{n-1}},\ldots,B_{T+t_1}-B_{T+t_0},B_{T+t_0}-B_{s_i},B_{s_i}-B_0=B_{s_i}$$ are independent. Thus, $$B_{s_i}\;\;\;\text{and}\;\;\;\left(B'_{t_j}-B'_{t_{j-1}}\right)_{j=1,\ldots,n}\tag{2}$$ are independent, for all $1\le i\le m$. Can we conclude $(1)$ from $(2)$?
For any $0 \leq s_0 < \ldots < s_m \leq T =T + t_0 < \ldots < T+t_n$ the random variables $$\underbrace{B_{t_n+T}-B_{t_{n-1}+T}}_{B'_{t_n}-B'_{t_{n-1}}},\ldots,\underbrace{B_{t_1+T}-B_{t_{0}+T}}_{B'_{t_1}-B'_{t_{0}}}, B_{s_m}-B_{s_{m-1}},\ldots,B_{s_1}-B_{s_0} \tag{1}$$
are independent. Recall that for any two measurable mappings $f: \mathbb{R}^m \to \mathbb{R}^m$, $g: \mathbb{R}^n \to \mathbb{R}^n$ the implication $$X_1,\ldots,X_m, Y_1,\ldots,Y_n \, \text{independent} \Rightarrow f(X_1,\ldots,X_m) \, \text{and} \, g(Y_1,\ldots,Y_n) \, \text{independent}$$
holds true. In particular for $$f(x_1,\ldots,x_m) := \left(x_1,x_1+x_2,\ldots, \sum_{i=1}^m x_i \right)$$ and $$g(y_1,\ldots,y_n) := \left(y_1,y_1+y_2,\ldots, \sum_{i=1}^n y_i \right)$$ we obtain (using (1)) that
$$f(B_{s_0},B_{s_1}-B_{s_0},\ldots, B_{s_m}-B_{s_{m-1}}) = (B_{s_0},B_{s_1},\ldots,B_{s_m})$$
and
$$g(B'_{t_0},B'_{t_1}-B'_{t_0},\ldots, B'_{t_n}-B'_{t_{n-1}}) = (B'_{t_0},B'_{t_1},\ldots,B'_{t_n})$$
are independent.