Let $(B^1,B^2)$ be a two-dimensional Brownian motion.
Let $t>s$. Is it true that $$ E[B^1(t) \lvert B^2(t),B^2(s) ] = E[B^1(t) \lvert B^2(t),B^2(t)-B^2(s) ] = E[B^1(t) \lvert B^2(t)] $$ since $(x,y) \mapsto (x,x-y)$ is measurable and bijective and $B^1(t)$ and $B^2(t)$ are independent of $B^2(t)-B^2(s)$?
Intuitively I would say it should be the case, since $B^2(s)$ would only contain information of $B^1(t)$ through the correlation between $B^1(t)$ and $B^2(t)$ and the autocorrelation in $B^1$.