let $\mathbb B$ be a brownian motion, why is $B_{t}-B_{s}$ independent of $\mathcal{F}_{s}$

Question

let $\mathbb B$ be a brownian motion, and $t>s$, define $\mathcal{F}_{w}:=\sigma(B_{v}\vert v \leq w)$, i.e. the canonical filtration.

My question I have seen that $\mathbb E[ B_{t}-B_{s} \vert \mathcal{F}_{s}]=\mathbb E[ B_{t}-B_{s} ]$ which means that $B_{t}-B_{s}$ independent of $\mathcal{F}_{s}$.

Is it because $\sigma (B_{s})=\sigma (B_{s}-0)=\sigma (B_{s}-B_{0})$ and by definition of brownian motion $B_{s}-B_{0}$ and $B_{t}-B_{s}$ are independent. But then I still need to account for for $\sigma (B_{w})$ where $w < s$. Any explanations/clarifications would help.

We defined brownian motion as a process $\mathbb B$ so that

i) $B_{0}=0$-a.s. for all $t\geq 0$

ii) idependent increments where $B_{t}-B_{s}$~ $\mathcal{N}(0, t-s)$

iii) $t\to B_{t}$ is continuous almost surely

Answer 1

One proof has already been suggested in d.k.o 's comment above. Another proof can be constructed easily using the following:

if $X_1,X_2,...,X_n$ are jointly normal they they are independent iff the variance-covariance matrix is diagonal.