Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space and $(B_t)$ a brownian motion. We denote $\mathbb P^x(B_t\in U):=\mathbb P(\{B_t\in U\}-x)$ where for $B\in \mathcal F$, we define $$B-x=\{\omega \in \Omega \mid \exists \omega '\in B : \omega =\omega '-x \}$$ and for $C\in \mathcal B(\mathbb R)$, $$C-x:=\{c-x\mid c\in C\}.$$ Let $\alpha $ be a probability measure, and define $$\mathbb P^\alpha (E)=\int_{\mathbb R}\mathbb P^x(E)\alpha (dx)$$ for all $E\in \mathcal F$. I want to prove that for all $U,V\in \mathcal B(\mathbb R)$, $0<t<s$, $$\mathbb P^\alpha (B_t-B_s\in U,B_s-B_0\in V)=\mathbb P^\alpha (B_t-B_s\in U)\mathbb P^\alpha (B_s-B_0 \in V).$$
Attempts
\begin{align*} &\mathbb P^\alpha (B_t-B_s\in U,B_s-B_0\in V)\\ &=\int_{\mathbb R}\mathbb P(\{B_t-B_s\in U,B_s\in V\}-x)\alpha (dx)\\ &=\int_{\mathbb R}\mathbb P(B_t-B_s\in U-x,B_s\in V-x)\alpha (dx)\\ &=\int_{\mathbb R}\mathbb P(B_t-B_s\in U-x)\mathbb P(B_s-B_0\in V-x)\alpha (dx)\\ &=\int_{\mathbb R}\mathbb P(\{B_t-B_s\in U\}-x)\mathbb P(\{B_s-B_0\in V\}-x)\alpha (dx)\\ &=\int_{\mathbb R}\mathbb P^x(B_t-B_s\in U)\mathbb P^x(B_s-B_0\in V)\alpha (dx), \end{align*} but unfortunately, I can't see how getting $$\int_{\mathbb R}\mathbb P^x(B_t-B_s\in U)P^y(B_s-B_0\in V) \alpha (dy)\alpha (dx)=\mathbb P^\alpha (B_t-B_s\in U)\mathbb P^\alpha (B_s-B_0\in V).$$
Any idea or hint would be appreciated.