Let $\{B(t):t \geq 0\}$ be a standard one-dimensional Brownian motion. How do we prove that $B(12)-B(10)$ is independent of $(B(3),B(5))$?
I can see that $B(12)-B(10)$ is independent of $B(5)$, and $B(12)-B(10)$ is independent of $B(3)$. But that doesn't imply that $B(12)-B(10)$ is independent of $(B(3),B(5))$.
Also, $B(3), B(5)$ aren't independent, so $B(3), B(5), B(12)-B(10)$ aren't mutually independent either.
Motivation: To prove that for a fixed $s>0$, the stochastic process $\{B(t+s)-B(s):t\geq 0\}$ is independent of the process $\{B(t): 0 \leq t \leq s\}$
For each $n\in N$ and any $r_1\le ...\le r_n\le s$ and any $a_i\le b_i$ the event $$\tag{1} \Big\{B(r_1)\in (a_1,b_1),B(r_2)-B(r_1)\in (a_1,b_1),...,B(r_n)-B(r_{n-1})\in (a_n,b_n)\Big\} $$ is independent of $B(t+s)-B(s)\,.$ The intersection of two events of the form (1) is again of that form. In other words, those events form a $\pi$-system. Consequently, the independence carries over to the entire $\sigma$-algebra that is generated by the events of the form (1).
The $\sigma$-algebra above is $\sigma(B(r);r\le s)\,.$
Proof. For each $n\in N$ and any $r_1\le ...\le r_n\le s$ we have obviously $$\tag{2} \sigma\Big(B(r_1),B(r_2)-B(r_1),...,B(r_n)-B(r_{n-1})\Big)=\sigma\Big( B(r_1),B(r_2),...,B(r_n)\Big)\,. $$ The $\sigma$-algebra generated by all events of the form (1) is the smallest $\sigma$-algebra that contains all $\sigma$-algebras on the LHS of (2) (when $n$ runs through $\mathbb N$ and $r_1\le...\le r_n\le s$ run through all those possible values). Therefore it is the smallest $\sigma$-algebra that contains all $\sigma$-algebras on the RHS of (2) which is nothing else than the definition of $\sigma(B(r);r\le s)\,.$ $\quad\quad\Box$