Let $(X_t)_{t \in \mathbb N}$ be a stochastic process on a probability space $(\Omega, \mathcal F, \mathbb P)$ with countable state space $S$.
$(A)$:
Let $n \ge 1, A_1,\ldots,A_{n+1}\subseteq S, P[X_n \in A_n, \ldots X_1 \in A_1]>0$
$\mathbb P[X_{n+1} \in A_{n+1}|X_n \in A_n, \ldots , X_1 \in A_1]=\mathbb P[X_{n+1} \in A_{n+1}|X_n \in A_n]$
$(B)$
Let $n \ge 1, t_1<t_2<\ldots <t_{n+1} \in \mathbb N, A_1,\ldots,A_{n+1}\subseteq S, P[X_{t_n} \in A_n, \ldots X_{t_1} \in A_1]>0$
$\mathbb P[X_{t_{n+1}} \in A_{n+1}|X_{t_n} \in A_n, \ldots , X_{t_1} \in A_1]=\mathbb P[X_{t_{n+1}} \in A_{n+1}|X_{t_n} \in A_n]$
Clearly, $B$ implies $A$.
Does $A$ imply $B$ as well? If yes, how can I show it?
Yes, A) does imply B). All you have to do is to take some of the sets $A_i$ to be $S$. (You can take some simple cases to see what is happening).
A typical example of how this works: $$P(X_5\in A_5|X_1\in A_1, X_4\in A_4)\equiv P(X_5\in A_5|X_1\in A_1,X_2 \in S,X_3 \in S, X_4\in A_4)$$ $$ =P(X_5\in A_5|X_4\in A_4)$$
by A).