About $\sigma$-algebras of Brownian Motion

Question

Let $(B_t)_{t \geq 0}$ a standard brownian motion and $\tilde{B}_t = B_{t+s} - B_s$, for $t \geq 0$ and $s \geq 0$. And $\mathcal{F}^B$ = $(F_t^B)$ the natural filtration of $B$, where $F_t^B = \sigma(B_x: x \in [0,t])$. My question is, does $$F_\infty^B = F_\infty^\tilde{B}? $$ I have shown that $\tilde{B}_t$ is independent of $F_s^B$. My intuition says yes, since both $B_t$ and $\tilde{B}_t$ are standard brownian motions, but i dont think it's simple as that.