Let $(\Omega, \mathcal{F},P, \mathbb{F}=(\mathcal{F}_{t})_{t \geq 0})$ be a filtered probability space and $(X_{t})_{t \geq 0}$ be a Markov process taking values in a polish space $(E,\mathcal{E})$, i.e $X$ is $\mathbb{F}$-adapted and $P[X_{t} \in A \mid \mathcal{F}_{s}]=P[X_{t} \in A \mid X_{s}]$ holds for $0 \leq s \leq t$ and $A \in \mathcal{E}$.
How can one prove that $$ P[X_{t_{n}} \in A_{n} \mid X_{t_{n-1}} \in A_{n-1},...,X_{t_{0}} \in A_{0}]=P[X_{t_{n}} \in A_{n} \mid X_{t_{n-1}} \in A_{n-1}] $$ holds for $0 \leq t_{0}< ... < t_{n}$ and $A_{0},...,A_{n} \in \mathcal{E}$ s.t. $P[X_{t_{n-1}} \in A_{n-1},...,X_{t_{0}} \in A_{0}] > 0$?
This sort of equality isn't true in general. Here's an example in discrete time, but it won't be hard to emulate it in continuous time with Brownian motion.
Example: $X$ is a discrete time symmetric random walk on the integers, with $P[X_0=0]=P[X_0=1]=1/2$. Let $A=\{1,2\}$. Then $$ P[X_2=3\mid X_0=0, X_1\in A]=0, $$ but $$ P[X_2=3\mid X_1\in A]={P[X_2=3,X_1=2]\over P[X_1\in A]}= {P[X_0=1,X_1=2,X_2=3]\over P[X_1\in A]}>0. $$