I am having trouble proving two definition of brownian motions are equivalent. Let $(\Omega, F, (F_t), P)$ be a filtered probability space satisying the usual conditions. Let $(X_t)$ be a continuos adapted process valued in $\mathbb{R}$. Then the following are equivalent.
1) $(X_t)_t$ satisfies $\forall s \geq 0, t>0$ $X_{t+s}-X_s \sim N(0,t)$ and is independent of $F_S$.
2) $\forall s \geq 0, t>0$ and $\forall f:\mathbb{R}\rightarrow \mathbb{R}$ bounded and measurable $$ \mathbb{E}\left[f(X_{s+t})|F_s\right] = P_tf(X_s)$$ where $P_t$ is the heat semigroup, i.e. $$P_tf(x) = \int_{\mathbb{R}} \frac{1}{\sqrt{2 \pi t}} f(y) e^{\frac{-(x-y)^2}{2t} }$$.
In particular I am trying to prove that 2 implies 1 (I think I can prove 1 implies 2 using a monotone class argument). Any help/reference is really appreciated.
Let us assume that 2) holds, i.e. $X_t$ is a process adapted to $F_t$ and s.t. for any bounded, measurable, real function $f$ and for any $0\leq s<t$ $$ \mathbb E[f(X_{s+t} )| F_s] = P_t f(X_s) $$ First let us show that for any bounded, measurable function $f$ $$ \mathbb E[f(X_{t+s}-X_s)|F_s] = P_tf(0) $$ holds. It is known that for random variables $X,Y$ with finite expectations, and a $\sigma$-subalgebra $G \subset F$, where $Y$ is $G$ measurable, one has $$ \mathbb E[f(X+Y)| G] = \mathbb E[f(X+y)| G]\big|_{y=Y} .$$ Therefore $$ \mathbb E[f(X_{t+s}-X_s)|F_s] = \mathbb E[f(X_{t+s} - y)|F_s]\big|_{y=X_s} = P_tf(X_s-y)|_{y=X_s} = P_tf(0) . $$ Now let $f(x)=\mathbf 1(x<z)$ for some $z\in \mathbb R$. Then $$ \mathbb E[f(X_{t+s}-X_s)|F_s] = \mathbb P(X_t-X_s<z|F_s) $$ furthermore $$ \mathbb E[f(X_{t+s}-X_s)|F_s] = P_tf(0) = \frac1{\sqrt{2\pi t}}\int_{\mathbb R} f(y) e^{-y^2/2}dy = \frac{1}{\sqrt{2\pi t}}\int_{-\infty}^z e^{-y^2/2t} $$ and so $$ \mathbb P(X_{t+s}-X_s<z|F_s) = \Phi\bigg(\frac{z}{\sqrt t}\bigg) $$ meaning that $X_{t+s}-X_s |F_s \sim N(0,t)$ . This also means that $X_{t+s}-X_s$ is independent of $F_s$, but in order to show it in a more direct way, it is enough to show that $$ \mathbb E[f(X_{t+s}-X_s)|F_s] = \mathbb E[f(X_{t+s}-X_s)] $$ for any bounded, measurable $f$. This follows directly from 2) $$ \mathbb E[f(X_{t+s}-X_s)|F_s] = \int f(x) \gamma_t(x)dx = \mathbb E[f(X_t-X_s)] $$ where $\gamma_t(x)=\frac1{\sqrt{2\pi t}} e^{-x^2/2t}$.