Let $(B_t)$ a Brownian motion. I wrote the following at an exam : $$\mathbb P(\forall u\in [0,t], B_{u+s}\neq 0\mid B_s=x)=\mathbb P(\forall u\in [0,t], B_{u+s}-B_s\neq -x),$$ and I justified as follow : $$\mathbb P(\forall u\in [0,t], B_{u+s}\neq 0\mid B_s=x)$$ \begin{align*} &=\mathbb P(\forall u\in [0,t], B_{u+s}-B_s\neq -B_s\mid B_s=x)\\ &=\mathbb P(\forall u\in [0,t], B_{u+s}-B_s\neq -x\mid B_s=x)\\ &=\mathbb P(\forall u\in [0,t], B_{u+s}-B_s\neq -x), \end{align*}
where the last equality come from the fact that $B_{u+s}-B_t$ is independent of $B_s$. My correcteur says that argument doesn't hold, but I don't understand why. Why is this wrong ?