Below is the proof on the semigroup property of the Brownian motion given in Schilling's Brownian Motion.
Let $B_t$ be a $d$- dimensional Brownian motion, and $u$ is a bounded Borel measurable function on $\mathbb{R}^d$. I'm having difficulty understanding the second equality below. The author uses $(6.4)$ for the step, however, I don't understand how we can get it from this. Namely, in the (6.4) provided below, we have the expectation taken over the probability space $P^0$, that is, where the Brownian motion starts at $0$. However, since we have $E^x$ in the setting of this problem, wouldn't we get $E^{B_s + x} [u(B_t)]$ from the (6.4)? How can I rigorously understand this step? I would greatly appreciate any help.
Call $W_t = B_t - x$ the translated copy of $B$ that starts at $0$, and $v(y) = u(x+y)$. Then we have $$ \newcommand{\F}{\mathcal F} \newcommand{\E}{\mathbb E} \E^{B_0 =x}(u(B_{t+s}) | \F_s) = \E^{B_0=x}(u(x+W_{t+s}) | \F_s)) = \E^{W_0 = 0}(v(W_{t+s}) | \mathcal F_s).$$ Let $X_t$ be another brownian motion starting at $0$, and let $C_t = x+X_t$ Under the expectation, we can replace things with other things that are equal in distribution. Application of (6.4) thus gives: $$ \E^{X_0 = W_s}(v(X_{t})) = \E^{X_0 = W_s}(v(x+X_{t} - x) ) = \E^{X_0 =W_s}(v(C_t-x)) = \E^{C_0= B_s}(u(C_{t})) .$$
After relabelling, this is the required line.