\sum_{\xi^{G\setminus X} \in \NZ{S}\cap A^Y}

            \P^{G\setminus Y}_S(\xi^{G\setminus Y}) \cdot

            \P^{G\setminus X}_S(\xi^{G\setminus X}) \\

        &= \P^{G\setminus Y}_S(\NZ{S} \cap A^X) \; \cdot \; \P^{G\setminus X}_S(\NZ{S} \cap A^Y)

    \end{align*}

\end{proof}

The intuition of the following lemma is that if two sites have distance $d$ in the graph, then the only way they can affect each other is that an interaction chain is forming between them, meaning that every vertex should get resampled to $0$ at least once in between them.

\begin{lemma}\label{lemma:distancePower}

	Suppose $G=(V,E)$ is a graph, $X,Y\subseteq V$ and $A^X$ is a local event on $X$. Then

	$$\P^{G}(A^X)-\P^{G\setminus Y}(A^X)=\bigO{p^{d(X,Y)}}.$$

\end{lemma}

\begin{proof}

	The proof goes by induction on $d(X,Y)$. The statement is trivial for $d(X,Y)=0$, and is easy to check for $d(X,Y)=1$, by looking at resample sequences that reach the all $1$ state in at most $0$ step (which is simply the case when everything is sampled to $1$ initially).

	Now we show the inductive step, assuming we know the statement for $d$, and that $d(X,Y)=d+1$.

	First we assume, that $\NZ{X}\subseteq\overline{A^X}$, i.e., $A^X\subseteq \Z{X}$.

	For $i\in[d]$ we define $A_i^X:=A^X\cap{\NZ{\overline{\partial}(X,i)}}\cap\bigcap_{j\in[i-1]}\Z{\overline{\partial}(X,j)}$, 

	and define $A_{d+1}^X:=A^X\cap\bigcap_{j\in[d]}\Z{\overline{\partial}(X,j)}$,

	so that they form a partition $A^X=\dot\bigcup_{i\in [d+1]}A_i^X$. 

	It is easy to see that for all $i\in[d+1]$ we have $A_{i}^X\subseteq\Z{X}\cap\bigcap_{j\in[i-1]}\Z{\overline{\partial}(X,j)}$, and therefore 

	\begin{equation}\label{eq:AXorder}

		\P^G(A_{i}^X)=\bigO{p^{i}}.