General Markov property Let $X_0,X_1,\ldots$ be a Markov chain with state space $\mathbb{Z}$. Now I found in a textbook that we have
$$
\mathbb{P}(X_{n+1} \in A \mid X_n = i, (X_{0},X_1,\ldots,X_{n-1}) \in B) = \mathbb{P}(X_{n+1} \in A \mid X_n = i)
$$
for all $n \in \mathbb{N}$, $i \in \mathbb{Z}$, $A \subset \mathbb{Z}$, $B \subset \mathbb{Z}^n$. This was given as a Proposition without a proof? How can we prove it and what is the difference to the original Markov property for Markov chain?
 A: A similar proposition is proved in chapter 1 of “Introduction to Stochastic Processes” by Hoel (page 12), using the solution of exercise 4 of the very same chapter. I will solve the exercise and show how it can be used to prove your proposition.
Let's consider a probability space $(\Omega,F,\mathbb{P})$, and assume the sets mentioned belong to $F$.
We want to prove that if $D_k$ are disjoint and $\mathbb{P}(C \mid D_k) = p$ independently of k, then $\mathbb{P}(C \mid \cup_{k}D_k) = p$.
Note that: $$\mathbb{P}(C \mid \cup_{k}D_k) = \frac{\mathbb{P}(C \cap (\cup_{k}D_k))}{\mathbb{P}(\cup_{k}D_k)}=
\frac{\sum_{k}{\mathbb{P}(C \cap D_k)}}{\sum_{k}{\mathbb{P}(D_k)}}=
\frac{\sum_{k}{\mathbb{P}(C \mid D_k)\mathbb{P}(D_k)}}{\sum_{k}{\mathbb{P}(D_k)}}
= \frac{p\sum_{k}{\mathbb{P}(D_k)}}{\sum_{k}{\mathbb{P}(D_k)}} =p
$$
Now that we proved that, let's go back to your initial proposition.
Let $B_k = (X_o=i^{(k)}_0,...,X_{n-1} =i^{(k)}_{n-1})$, such that $B = \cup_{k}{B_k}$, and $B_k$ are disjoint. Since we have a Markov chain:
$$ \mathbb{P}(X_{n+1} \in A \mid \{{ X_n =i\}} \cap B_k)=\mathbb{P}(X_{n+1} \in A \mid X_o=i^{(k)}_0,...,X_n=i) =\mathbb{P}(X_{n+1} \in A \mid X_n = i) = p~,$$
independently of $k$.
So what we have to show is that:
\begin{align}
\mathbb{P}(X_{n+1} \in A \mid 
\{{ X_n =i\}} \cap (\cup_{k}{B_k})) &= \mathbb{P}(X_{n+1} \in A \mid X_n=i,(X_o,...,X_{n-1}) \in B) \\
&= \mathbb{P}(X_{n+1} \in A \mid X_n=i)
\end{align}
Therefore, the proof to your proposition is achieved by making 
$C = \{{X_{n+1}\in A\}}$ and $D_k = \{{\{{ X_n =i\}} \cap {B_k}\}}$, and applying the solution previously demonstrated.
