Suppose $\mathbf{X} = \{X_t\}_{t \in T}$ is a random process, with $T$ being totally ordered. Let $\{\mathcal{F}_t\}_{t \in T}$ be the natural filtration generated by $\mathbf{X}$.
I have seen the following definitions of the Markov property.
We say $\mathbf{X}$ is Markov if, for every bounded Borel function $h\colon \mathbb{R} \to \mathbb{R}$, $$\mathbf{E}[h(X_t) \vert \mathcal{F}_s] = \mathbf{E}[h(X_t) \vert X_s] \quad \forall s < t, \, s, t \in T.$$
If $T$ is discrete (say, $T = \mathbb{Z}_{\ge 0}$), and $X_t$ takes values in a common countable set $\mathcal{S}$ for every $t \in T$, then we say $\mathbf{X}$ is Markov if $$\mathbf{P}(X_{n + 1} = x_{n + 1} \vert X_n = x_n, \, X_{n - 1} = x_{n - 1}, \, \dots, \, X_0 = x_0) = \mathbf{P}(X_{n + 1} = x_{n + 1} \vert X_n = x_n)$$ for every nonnegative integer $n$ and every sequence $\{x_k\}_{k = 0}^{\infty} \subseteq \mathcal{S}$.
If $T$ is continuous (say, $T = \mathbb{R}_{\ge 0}$), and $X_t$ takes values in a common countable set $\mathcal{S}$ for every $t \in T$, then we say $\mathbf{X}$ is Markov if $$\mathbf{P}(X_{t_{n + 1}} = x_{n + 1} \vert X_{t_n} = x_n, \, X_{t_{n - 1}} = x_{n - 1}, \, \dots, \, X_{t_0} = x_0) = \mathbf{P}(X_{t_{n + 1}} = x_{n + 1} \vert X_{t_n} = x_n)$$ for every nonnegative integer $n$, every increasing sequence $\{t_k\}_{k = 0}^{\infty} \subseteq T$ and every sequence $\{x_k\}_{k = 0}^{\infty} \subseteq \mathcal{S}$.
I believe the first is the most general, since there is no restriction on the type of time we are dealing with.
How do we show that the first is equivalent to the second and equivalent to the third in the respective time contexts?
I believe to handle equivalence to the second, we can use the fact that $\mathcal{F}_n$ is generated by the countable collection of events $$\{\{X_0 = x_0, \, X_1 = x_1, \, \dots, \, X_n = x_n\} \;\vert\; \{x_k\}_{k = 0}^{n} \subseteq \mathcal{S}\},$$ but I am not so sure about equivalence to the third.