Let $I\subset\mathbb{Z}$ be a finite set of vertices.

    Let $b_I$ be the state where everything is $1$, apart from the vertices corresponding to $I$, which are set $0$. Define $\P^{(n)}_I(A)=\P^{(n)}_{b_I}(A)$ which is defined in Definition \ref{def:conditionedevents}.

\end{definition}

\begin{lemma}[Conditional independence] \label{lemma:eventindependenceNew}

	Let $v,w \in [n]$, and let $A$ be any event that depends only on the sites $[v,w]$ (meaning the initialization and resamples) and similarly $B$ an event that depends only on the sites $[w,v]$. (For example $\mathrm{Z}^{(s)}$ or ``$s$ has been resampled at least $k$ times'' for an $s$ on the correct interval). Then we have

	\begin{align*}

		\P^{(n)}(\mathrm{NZ}^{(v,w)}\cap A\cap B)

		=

		\P_{b_v=b_w=1}^{[v,w]}(\mathrm{NZ}^{(v,w)}\cap A)

		\; \cdot \;

		\P^{[w,v]}(\mathrm{NZ}^{(v,w)}\cap B),

	\end{align*}

	and similarly

	\begin{align*}

		\P^{[n]}(\mathrm{NZ}^{(v)}\cap A\cap B)

		=

		\P_{b_v=1}^{[v]}(\mathrm{NZ}^{(v)}\cap A)

		\; \cdot \;

		\P^{[v,n]}(\mathrm{NZ}^{(v)}\cap B)

	\end{align*}

	where there is no longer a condition on the starting state.

\end{lemma}

The intuition of the following lemma is that the far right can only affect the zero vertex if there is an interaction chain forming, which means that every vertex should get resampled to $0$ at least once.

\begin{lemma}\label{lemma:probIndepNew}

	$\forall n\in \mathbb{N}_+:\P^{[n]}(\Z{1})-\P^{[n+1]}(\Z{1}) = \bigO{p^{n}}$. (Should be true with $\bigO{p^{n+1}}$ as well.)

\end{lemma}

\begin{proof}

	The proof uses induction on $n$. For $n=1$ the statement is easy, since $\P^{[1]}(\Z{1})=p$ and $\P^{[2]}(\Z{1})=p+p^2+\bigO{p^{3}}$.

	Induction step: suppose we proved the claim for $n-1$, then

	\begin{align*}

	\P^{[n+1]}(\Z{1})

	&=\sum_{k=1}^{n+1}\P^{[n+1]}([k]\in\mathcal{P}) \tag{the events are a partition}\\

	&=\sum_{k=1}^{n-1}\P^{[n+1]}([k]\in\mathcal{P}) + \bigO{p^{n}}\tag*{$\left(\P^{[n+1]}([k]\in\mathcal{P})=O(p^{k})\right)$}\\	

	&=\sum_{k=1}^{n-1}\P^{[k+1]}_{b_{k+1}=1}([k]\in\mathcal{P})\cdot \P^{[n-k+1]}(\NZ{1})+ \bigO{p^{n}} \tag{by Claim~\ref{lemma:eventindependenceNew}}\\

	&=\sum_{k=1}^{n-1}\P^{[k+1]}_{b_{k+1}=1}([k]\in\mathcal{P})\cdot \left(\P^{[n-k]}(\NZ{1})+\bigO{p^{n-k}}\right)+ \bigO{p^{n}} \tag{by induction} \\	

	&=\sum_{k=1}^{n-1}\P^{[k+1]}_{b_{k+1}=1}([k]\in\mathcal{P})\cdot \P^{[n-k]}(\NZ{1})+ \bigO{p^{n}} \tag*{$\left(\P^{[k+1]}_{b_{k+1}=1}([k]\in\mathcal{P})=\bigO{p^{k}}\right)$}\\	

	&=\sum_{k=1}^{n-1}\P^{[n]}([k]\in\mathcal{P})+ \bigO{p^{n}} \tag{by Claim~\ref{lemma:eventindependenceNew}}\\

	&=\sum_{k=1}^{n}\P^{[n]}([k]\in\mathcal{P})+ \bigO{p^{n}} \tag*{$\left(\P^{[n]}([n]\in\mathcal{P})=\bigO{p^{n}}\right)$}\\	

	&=\P^{[n]}(\Z{1})	+ \bigO{p^{n}} 

	\end{align*}

\end{proof}

\begin{corollary}\label{cor:probIndepNew}

	$\P^{[n]}(\Z{1})-\P^{[m]}(\Z{1}) = \bigO{p^{\min(n,m)}}$. (Should be true with $\bigO{p^{\min(n,m)+1}}$ too.)

\end{corollary}

 		&=\!\!\!\sum_{\ell, r\in [k]: \ell<r-1}\!\!\!

 		\P^{[\ell+1]}_{b_{\ell+1}=1}([\ell]\in\mathcal{P})

 		\left(\P^{[\ell+1,k]}(\NZ{\ell+1})

 		\P^{[1,r-1]}_{b_{r-1}=1}(\NZ{r-1})\right)

 		\P^{[r-1,k]}([r,k]\in\mathcal{P})

 		+\bigO{p^{k}} \tag{by Corrolary~\ref{cor:probIndepNew}}\\

 		&=\!\!\!\sum_{\ell, r\in [k]: \ell<r-1}\!\!\!

 		\P^{[k]}([\ell]\in\mathcal{P})

 		\P^{[k]}([r,k]\in\mathcal{P})

 		+\bigO{p^{k}} \tag{by Lemma~\ref{lemma:eventindependenceNew}}\\

 		&=\left(\sum_{\ell\in [k]}\P^{[k]}([\ell]\in\mathcal{P})\right)

 		\left(\sum_{r\in [k]}\P^{[k]}([r,k]\in\mathcal{P})\right)

 		+\bigO{p^{k}} \tag*{$\left(\P^{[k]}([\ell]\in\mathcal{P})=\bigO{p^{\ell}}\right)$}\\	

 		&=\P^{[k]}(\Z{1})\P^{[k]}(\Z{k})

 		+\bigO{p^{k}}.	

 		\end{align*}

 	\end{proof}

	Again the intuition of the final theorem is simmilar to the previous lemmas. A site can only realise the length of the cycle after an interaction chain was formed around the cycle, implying that every vertex was resampled to $0$ at least once.

	\begin{theorem}

		$R^{(n)}-R^{(m)}=\bigO{p^{\min(n,m)}}$.

	\end{theorem}

	\begin{proof}

		Let $N\geq \max(2n,2m)$, then

		\begin{align*}

			R^{(n)}

	\end{proof} 	

Old:

The intuition of the following lemma is that the far right can only affect the zero vertex if there is an interaction chain forming, which means that every vertex should get resampled to $0$ at least once.

\begin{lemma}\label{lemma:probIndep}

	Suppose we have a finite set $I\subset\mathbb{N}_+$ of vertices.

    Let $I_{\max}:=\max(I)$ and $I':=I\setminus\{I_{\max}\}$, and similarly let $I_{\min}:=\min(I)$. These definitions are illustraded in Figure \ref{fig:lemmaillustration}.

	Then $\P^\infty_{I}(\Z{0})-\P^\infty_{I'}(\Z{0}) = O(p^{I_{\max}-|I'|})$.

\end{lemma}

\begin{proof}

\begin{figure}

	\begin{center}

    	\includegraphics{diagram_proborders.pdf}

    \end{center}

    \caption{\label{fig:lemmaillustration} Illustration of setup of Lemma \ref{lemma:probIndep}.}

\end{figure}

	The proof uses induction on $|I|$. For $|I|=1$ the statement is easy, since every resample sequence that resamples vertex $0$ to zero must produce at least $I_{\max}$ zeroes in-between.

    Induction step: For an event $A$ and $k>0$ let us denote $A_k = A\cap\left(\cap_{j=0}^{k-1} \mathrm{Z}^{(j)}\right)\cap \NZ{(k)}$, i.e. $A_k$ is the event $A$ \emph{and} ``Each vertex in $0,1,2,\ldots, k-1$ becomes $0$ at some point before termination (either by resampling or initialisation), but vertex $k$ does not''. Observe that these events form a partition, so $\Z{(0)}=\dot{\bigcup}_{k=1}^{\infty}\Z{(0)}_k$.

    Let $I_{<k}:=I\cap[1,k-1]$ and similarly $I_{>k}:=I\setminus[1,k]$, finally let $I_{><}:=\{I_{\min}+1,I_{\max}-1]\}\setminus I$ (note that $I_{><} = \gaps{I}$ as shown in Figure \ref{fig:diametergap}). Suppose we have proven the claim up to $|I|-1$, then the induction step can be shown by

	\begin{align*}

		\P^\infty_{I}(\Z{(0)})

		&=\sum_{k=1}^{\infty}\P^\infty(\Z{(0)}_k) \tag{the events are a partition}\\

        &=\sum_{k\in \mathbb{N}\setminus I}\P^\infty(\Z{(0)}_k) \tag{$\mathbb{\P^\infty}(A_k)=0$ for $k\in I$}\\