Let $(X_n,\F_n,\P^x)$ be a Markov chain in $S$, $A\in\F_n$, $B\in\F_n^\prime$. Prove that $\P^x(A\cap B|X_n)=\P^x(A|X_n)\P^x(B|X_n)$.
Since $B$ is in $\F_n^\prime$, there is some $C\in\F^X$ such that $B=[\Theta_n\in C]$ and therefore by the extended Markov property:
\begin{eqnarray*}
\P^x(A\cap B|X_n)
&=&\E^x(\E^x(I_AI_B|\F_n)|X_n)
=\E^x(I_A\E^x(I_B|\F_n)|X_n)\\
&=&\E^x(I_A\E^x(I_C\circ\Theta_n|\F_n)|X_n)\\
&=&\E^x(I_A\E^{X_n}I_C|X_n)
=\E^x(I_A|X_n)\E^{X_n}I_C
\end{eqnarray*}
and $\P^x(B|X_n)=\E^x(I_B|\F_n)=\E^{X_n}I_C$.