Let the process $(B_t)_{t\ge 0}$ with $B_t = (B_t^{(1)},B_t^{(2)})$ consist of two independent standard Brownian motions $B^{(1)}$ and $B^{(2)}$ both adapted to some filtration $(\mathcal F_t)_{t\ge 0}$ and for $t>s$ the increments $B_t^{(1)} - B_s^{(1)}$ and $B_t^{(2)} - B_s^{(2)}$ both independent of $\mathcal F_s$. I want to show that $B_t - B_s$ is independent of $\mathcal F_s$. I considered arbitrary sets $M\in \mathcal F_s$ and $N_1,N_2\in \mathcal B(\mathbb R)$ and want to proof the first equality in $$P(M\cap \lbrace B_t - B_s \in N_1 \times N_2 \rbrace ) = P(M) \cdot P(\lbrace B_t - B_s \in N_1 \times N_2 \rbrace ) = P(M) \cdot P (\lbrace B_t^{(1)} - B_s^{(1)} \in N_1 \rbrace) \cdot P (\lbrace B_t^{(2)} - B_s^{(2)} \in N_2 \rbrace).$$
So I'm wondering if $X$ and $Y$ are independent normally distributed random variables and $A$ some event independent of $X$ and independent of $Y$, is it then the case that $A$ is independent of $(X,Y)$?
I might have overlooked something very elementary and maybe this is an altogether trivial question.