Let $B$ be a standard Brownian motion in $1$ dimension. Define \begin{equation} \tau = \inf \bigg\{ t \geq 0 : B_t = \max_{0 \leq s \leq 1} B_s \bigg\}. \end{equation} We want to show that $(B_{t+ \tau} - B_{\tau} )_{t \geq 0}$ is not a Brownian motion.
My ideas:
Let $\tilde{B}_t= B_{t+ \tau} - B_{\tau} $. Suppose on the contrary that $\{\tilde{B}_t\}$ is a Brownian motion.
I can only deduce by the definition that $$\mathbb{P} \bigg( \bigcup_{q \in \mathbb{Q}} \bigcap_{t \in [0,q]} \{\tilde{B}_t \leq 0 \} \bigg) =1.$$ Any ideas of how that leads to a contradiction?
(Should I somehow use the fact that $\tau <1$ a.s.?)
Every Brownian motion starting from $0$ enters $(0,+\infty)$ instantaneously, thus, $$\mathbb{P} \bigg( \bigcup_{q \in \mathbb{Q}} \bigcap_{t \in [0,q]} \{B_t \leqslant 0 \} \bigg) =0.$$