How to prove $\mathbb P [X_{n} = x_{n} |X_m = x_m, \ldots, X_0=x_0] = \mathbb P [X_{n} = x_{n} |X_m = x_m]$? Let $S$ be a countable set and $(X_n)$ an $S$-valued discrete Markov chain. Then
$$
\mathbb P [X_{n+1} = x_{n+1} |X_n = x_n, \ldots, X_0=x_0] = \mathbb P [X_{n+1} = x_{n+1} |X_n = x_n]
$$
for all $x_0, \ldots, x_{n+1} \in S$. This is called the Markov property. Previously, I proved that

Theorem 1
$$
\mathbb P [X_{n+1} = x_{n+1} |X_n = x_n, \ldots, X_m=x_m] = \mathbb P [X_{n+1} = x_{n+1} |X_n = x_n]
$$
for all $m \le n$ and $x_0, \ldots, x_{n+1} \in S$.

Could you provide some hints on how to prove below result?

Theorem 2
$$
\mathbb P [X_{n} = x_{n} |X_m = x_m, \ldots, X_0=x_0] = \mathbb P [X_{n} = x_{n} |X_m = x_m]
$$
for all $m \le n$ and $x_0, \ldots, x_{n} \in S$.

 A: Taking for example $m = n-2$, we have
\begin{align}
&P(X_n = x_n, X_{n-2}=x_{n-2}, \ldots, X_0=x_0)
\\
&= \sum_{x_{n-1}} P(X_n = x_n, X_{n-1}=x_{n-1}, X_{n-2}=x_{n-2}, \ldots, X_0=x_0)
\\
&= \sum_{x_{n-1}} P(X_n = x_n \mid X_{n-1} = x_{n-1}, \ldots, X_0 = x_0)
P(X_{n-1} = x_{n-1} \mid X_{n-2} = x_{n-2},\ldots, X_0=x_0)
P(X_{n-2}=x_{n-2},\ldots, X_0=x_0)
\\
&= P(X_{n-2}=x_{n-2},\ldots, X_0=x_0)
\sum_{x_{n-1}} P(X_n=x_n \mid X_{n-1} = x_{n-1})P(X_{n-1}=x_{n-1} \mid X_{n-2} = x_{n-2}),
\end{align}
where the last step is by your definition of Markov chain.
Dividing both sides by $P(X_{n-2}=x_{n-2}, \ldots, X_0=x_0)$ yields
\begin{align}
&P(X_n = x_n \mid X_{n-2}=x_{n-2}, \ldots, X_0=x_0)
\\
&= \sum_{x_{n-1}} P(X_n=x_n \mid X_{n-1} = x_{n-1})P(X_{n-1}=x_{n-1} \mid X_{n-2} = x_{n-2})
\\
&= \sum_{x_{n-1}} P(X_n=x_n \mid X_{n-1} = x_{n-1}, X_{n-2} = x_{n-2})P(X_{n-1}=x_{n-1} \mid X_{n-2} = x_{n-2})
\\
&= \sum_{x_{n-1}} P(X_n=x_n, X_{n-1}=x_{n-1} \mid X_{n-2}=x_{n-2})
\\
&= P(X_n=x_n \mid X_{n-2} = x_{n-2})
\end{align}
where the second equality is due to Theorem 1.
The proof for general $m \le n$ should be similar.
