Let $X_t,t\ge 0$ be continuous, $n$-dimensional process and filtration $\mathcal{F}_t=\sigma(X_s,0\le s \le t)$. We want to show the equivalence of the two definition in the following: $$ E[f(X_t)|X_{t_n},\cdots,X_{t_1}] = E[f(X_t)|X_{t_n}] $$ for all $t>t_n>\cdots>t_1\ge 0$ and $f$ is continous, bounded and measurable.
Another definition is, $$E[f(X_t)|\mathcal{F}_s]=E[f(X_t)|X_s]$$ for all $t>s\ge 0$ and $f$ is continous, bounded and measurable.
I can show from second definition implies first one, but not sure how to show from first to second. I tried the following: Given that $E[f(X_t)|X_{t_n},\cdots,X_{t_1}] = E[f(X_t)|X_{t_n}]$. Since $E[f(X_t)|X_{t_n}]=E[E[f(X_t)|X_{t_n}]|X_{t_n},\cdots,X_{t_1}]$ and $E[f(X_t)|X_{t_n},\cdots,X_{t_1}]=E[E[f(X_t)|\mathcal{F}_{t_n}]|X_{t_n},\cdots,X_{t_1}]$. Therefore, $$E[E[f(X_t)|X_{t_n}]|X_{t_n},\cdots,X_{t_1}] = E[E[f(X_t)|\mathcal{F}_{t_n}]|X_{t_n},\cdots,X_{t_1}]$$.
From here, I am not sure if I can conclude that $E[f(X_t)|X_{t_n}] = E[f(X_t)|\mathcal{F}_{t_n}]$?
Assume for simplicity that $X_t\in\mathbb{R}$. Note that $\mathcal{F}_t=\sigma(X_s:s\le t)$ is generated by the collection of cylinder sets $$ \mathcal{C}_t=\{\{X_{t_1}\in B_1,\ldots X_{t_n}\in B_n\}: n\in \mathbb{N}, B_i\in \mathcal{B}(\mathbb{R}), t_i\le t\}, $$ and $\mathcal{C}$ is a $\pi$-system ($\because$ it is closed under finite intersections). For $s<t$, let $$ \mathcal{L}_s=\{A\in \mathcal{F}_s:\mathsf{E}[f(X_t)1_A]=\mathsf{E}[\mathsf{E}[f(X_t)\mid X_s]1_A]\}. $$ It is clear that $\mathcal{C}_s\subset\mathcal{L}_s$. Thus, by the $\pi$-$\lambda$ theorem $\mathcal{F}_s=\sigma(\mathcal{C}_s)\subseteq\mathcal{L}_s$. (It remains to show that $\mathcal{L}$ is a Dynkin system.)