There are two statements about Markov property: $B_t $ is Brownian motion and $\mathcal{F}$ is generated by $B$
If $s>0$ and $Y$ is bounded and measuable, then $$\mathbb{E}(Y\circ\theta_s|\mathcal{F_s^+})=\mathbb{E}_{B_s}Y.$$
and
$B_{t+s}-B_t$ is independent of $\mathcal{F_s^+}.$
They are both called Markov in different books, but I don't know how to prove the equivalance.
These two are not equivalent.
The first one, is the "real" Markov property which is used in order to define so-called Markov processes. It basically says that the future of the process does not depend on the past, but only on the present. An equivalent formulation is the following: $$\mathbb{P}(B_{t+s} \in B \mid \mathcal{F}_s) = \mathbb{P}(B_{t+s} \in B \mid B_s)$$ for any $t \geq 0$.
The second one says that a Brownian motion has independent increments, i.e. that $B_{t_n}-B_{t_{n-1}},\ldots,B_{t_1}-B_{t_0}$ are independent for any $0 = t_0 \leq t_1 \leq \ldots \leq t_n$. This implies in particular the Markov property (i.e. the first statement).
Reference: Brownian Motion - An Introduction to Stochastic Processes by René L. Schilling & Lothar Partzsch, Chapter 6.
