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\end{center}

 Hence the contribution of all $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ paths together to $R_{b_1\land b_2}$ is given by

\[

\mathbb{P}[\xi_1]\cdot\mathbb{P}[\xi_2]\cdot(|\xi_1|+|\xi_2|) = \mathbb{P}[\xi_2]\cdot\mathbb{P}[\xi_1]\cdot|\xi_1| \;\; + \;\; \mathbb{P}[\xi_1]\cdot\mathbb{P}[\xi_2]\cdot|\xi_2|.

\]

Ideally we would now like to sum this expression over all possible paths $\xi_1,\xi_2$ and use $p_\mathrm{tot}:=\sum_{\xi\in\paths{b_i}} \mathbb{P}[\xi] = 1$ (which also holds up to arbitrary order in $p$). The above expression would then become $R_{b_1} + R_{b_2}$. However, not all paths in the sum would satisfy the independence condition so it seems we can't do this. We now argue that it works up to order $p^{k-1}$.

For all $\xi\in\paths{b_1\land b_2}$ we have that \emph{either} $\xi$ splits into two independent paths $\xi_1,\xi_2$ as above, \emph{or} it does not. In the latter case, when $\xi$ can not be split like that, we know $\mathbb{P}[\xi]$ contains a power $p^k$ or higher because there is a gap of size $k$  and the paths must have moved at least $k$ times `towards each other' (for example one path moves $m$ times to the right and the other path moves $k-m$ times to the left). So the total weight of such a combined path is at least order $p^k$. Therefore we have

\[

	R_{b_1\land b_2} = \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1| + \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_1]\mathbb{P}[\xi_2]|\xi_2| + \sum_{\mathclap{\xi\;\mathrm{dependent}}} \mathbb{P}[\xi]|\xi|.

\]

where last sum only contains only terms of order $p^{k}$ or higher. Now for the first sum, note that

\[

	\sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1|

    = \sum_{\xi_1\in\paths{b_1}} \sum_{\substack{\xi_2\in\paths{b_2}\\ \text{independent of }\xi_1}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1|

\]

where the sum over independent paths could be empty for certain $\xi_1$. Now we replace this last sum by a sum over \emph{all} paths $\xi_2\in\paths{b_2}$. This will change the sum but only for terms where $\xi_1,\xi_2$ are dependent. For those terms we already know that $\mathbb{P}[\xi_1]\mathbb{P}[\xi_2]$ contains a factor $p^k$ and hence we have 

\begin{align*}

    \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1|

    &= \sum_{\xi_1\in\paths{b_1}} \sum_{\xi_2\in\paths{b_2}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1| + \mathcal{O}(p^k) \\

    &= \sum_{\xi_1\in\paths{b_1}} \mathbb{P}[\xi_1]|\xi_1| + \mathcal{O}(p^k) \\

    &= R_{b_1} + \mathcal{O}(p^k)

\end{align*}

we can do the same with the second term and this proves the claim.

\end{proof}

~\\

\textbf{Proof of claim \ref{claim:weakcancel}}: We can assume $C$ consists of a group on the left with $l$ slots and a group on the right with $r$ slots (so $r+l=|C|$), with a gap of size $k=\mathrm{gap}(C)$ between these groups. Then on the left we have strings in $\{0,1'\}^l$ as possibilities and on the right we have strings in $\{0,1'\}^r$. The combined configuration can be described by strings $f=(a,b)\in\{0,1'\}^{l+r}$. The initial probability of such a state $C(a,b)$ is $\rho_{C(a,b)} = (-1)^{|a|+|b|} p^{r+l}$ and by claim \ref{claim:expectationsum} we know $R_{C(a,b)} = R_{C(a)} + R_{C(b)} + \mathcal{O}(p^k)$ where $C(a)$ indicates that only the left slots have been filled by $a$ and the other slots are filled with $1$s. The total contribution of these configurations is therefore

\begin{align*}

    \sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R_{C(f)}

    &= \sum_{a\in\{0,1'\}^l} \sum_{b\in\{0,1'\}^r} (-1)^{|a|+|b|}p^{r+l} \left( R_{C(a)} + R_{C(b)} + \mathcal{O}(p^k) \right) \\

    &=\;\;\; p^{r+l}\sum_{a\in\{0,1'\}^l} (-1)^{|a|} R_{C(a)} \sum_{b\in\{0,1'\}^r} (-1)^{|b|} \\

    &\quad + p^{r+l}\sum_{b\in\{0,1'\}^r} (-1)^{|b|} R_{C(b)} \sum_{a\in\{0,1'\}^l} (-1)^{|a|}

        + \mathcal{O}(p^{r+l+k})\\

    &= 0 + \mathcal{O}(p^{|C|+k})

\end{align*}

where we used the identity $\sum_{a\in\{0,1\}^l} (-1)^{|a|} = 0$.

\newpage

\section{Proving the strong cancellation claim}

It is useful to introduce some new notation. We will consider variations of the Markov Chains:

\begin{itemize}

    \item $\P^{(n)}$ refers to the original process on the length-$n$ cycle.

    \item $\P^{[a,b]}$ or $\P^{[n]}$ refers to a similar Markov Chain but on a finite chain ($[a,b]$ or $[1,n]$).

\end{itemize}

The process on the finite chain has the following modification at the boundary: if a boundary site is resampled, it can not resample one of its neighbors so it ignores it and only draws two new bits.

%Note that an \emph{event} is a subset of all possible paths of the Markov Chain.

\begin{definition}[Events conditioned on starting state] \label{def:conditionedevents}

    \begin{align*}

    \end{align*}

    Furthermore, for the Markov Chain on the finite chain, define

    \begin{align*}

        \P^{[n]}_{\partial=1}(A) &= \P^{[n]}(A \;|\; \text{boundary is initialized to }1)

    \end{align*}

    where the boundary of $[n]$ is site $1$ and site $n$, and the boundary of $[a,b]$ are $a$ and $b$.

\end{definition}

\begin{definition}[Vertex visiting event] \label{def:visitingResamplings}

    Denote by $\mathrm{Z}^{(v)}$ the event that site $v$ becomes zero at any point in time before the Markov Chain terminates. Denote the complement by $\mathrm{NZ}^{(v)}$, i.e. the event that site $v$ does \emph{not} become zero before it terminates. Furthermore define $\mathrm{NZ}^{(v,w)} := \mathrm{NZ}^{(v)} \cap \mathrm{NZ}^{(w)}$, i.e. the event that \emph{both} $v$ and $w$ do not become zero before termination.

\end{definition}

    Let $b=b_1\land b_2\in\{0,1\}^n$ be a state with two groups of zeroes that are separated by at least one site inbetween, as in Figure \ref{fig:separatedgroups}. Let $v$, $w$ be any indices inbetween the groups, such that $b_1$ lies on one side of them and $b_2$ on the other, as shown in the figure. Furthermore, let $A_1$ be any event that depends only on the sites ``on the $b_1$ side of $v,w$'', and similar for $A_2$ (for example $\mathrm{Z}^{(i)}$ for an $i$ on the correct side). Then we have

    \begin{align*}

        \P^{(n)}_b(\mathrm{NZ}^{(v,w)}, A_1, A_2)

        &=

        \P^{(n)}_{b_1}(\mathrm{NZ}^{(v,w)}, A_1)

        \; \cdot \;

        \P^{(n)}_{b_2}(\mathrm{NZ}^{(v,w)}, A_2) \\

        \P^{(n)}_b(A_1, A_2 \mid \mathrm{NZ}^{(v,w)})

        &=

        \P^{(n)}_{b_1}(A_1 \mid \mathrm{NZ}^{(v,w)})

        \; \cdot \;

        \P^{(n)}_{b_2}(A_2 \mid \mathrm{NZ}^{(v,w)}) .%\\

        %R_{b,\mathrm{NZ}^{(v,w)},A_1,A_2}

        %&=

        %R_{b_1,\mathrm{NZ}^{(v,w)},A_1}

        %\; + \;

        %R_{b_2,\mathrm{NZ}^{(v,w)},A_2}

    \end{align*}

    %up to any order in $p$.

\end{lemma}

\begin{proof}

    From any path $\xi\in\start{b} \cap \mathrm{NZ}^{(v,w)}$ we can construct paths $\xi_1\in\start{b_1}\cap \mathrm{NZ}^{(v,w)}$ and $\xi_2\in\start{b_2}\cap\mathrm{NZ}^{(v,w)}$ as follows. Let us write the path $\xi$ as

    $$\xi=\left( (\text{initialize }b), (z_1, s_1, r_1), (z_2, s_2, r_2), ..., (z_{|\xi|}, s_{|\xi|}, r_{|\xi|}) \right)$$

    where $z_i\in[n]$ denotes the number of zeroes in the state before the $i$th step, $s_i\in [n]$ denotes the site that was resampled and $r_i\in \{0,1\}^3$ is the result of the three resampled bits. We have

    \begin{align*}

        \P^{(n)}_b[\xi] &= \P(\text{pick }s_1 | z_1) \P(r_1) \P(\text{pick }s_2 | z_2) \P(r_2) \cdots \P(\text{pick }s_{|\xi|} | z_{|\xi|}) \P(r_{|\xi|}) \\

                &= \frac{1}{z_1} \P(r_1) \frac{1}{z_2} \P(r_2) \cdots \frac{1}{z_{|\xi|}} \P(r_{|\xi|}) .

    \end{align*}

    To construct $\xi_1$ and $\xi_2$, start with $\xi_1 = \left( (\text{initialize }b_1) \right)$ and $\xi_2 = \left( (\text{initialize }b_2) \right)$. For each step $(z_i,s_i,r_i)$ in $\xi$ do the following: if $s_i$ is ``on the $b_1$ side of $v,w$'' then append $(z^{(1)}_i,s_i,r_i)$ to $\xi_1$ and if its ``on the $b_2$ side of $v,w$'' then append $(z^{(2)}_i,s_i,r_i)$ to $\xi_2$. Here $z^{(1)}_i$ is the number of zeroes that were on the $b_1$ side and $z^{(2)}_i$ is the number of zeroes on the $b_2$ side so we have $z_i = z^{(1)}_i + z^{(2)}_i$.

    %Let the resulting paths be

    %\begin{align*}

    %    \xi_1 &= \left( (z^{(1)}_{a_1}, s_{a_1}, r_{a_1}), (z^{(1)}_{a_2}, s_{a_2}, r_{a_2}), ..., (z^{(1)}_{a_{|\xi_1|}}, s_{a_{|\xi_1|}}, r_{a_{|\xi_1|}}) \right) \\

    %    \xi_2 &= \left( (z^{(2)}_{b_1}, s_{b_1}, r_{b_1}), (z^{(2)}_{b_2}, s_{b_2}, r_{b_2}), ..., (z^{(2)}_{b_{|\xi_1|}}, s_{b_{|\xi_1|}}, r_{b_{|\xi_1|}}) \right)

    %\end{align*}

    Now $\xi_1$ is a valid (terminating) path from $b_1$ to $\mathbf{1}$, because in the original path $\xi$, all zeroes ``on the $b_1$ side'' have been resampled by resamplings ``on the $b_1$ side''. Since the sites $v,w$ inbetween never become zero, there can not be any zero ``on the $b_1$ side'' that was resampled by a resampling ``on the $b_2$ side''.

    Vice versa, any two paths $\xi_1\in\start{b_1}\cap \mathrm{NZ}^{(v,w)}$ and $\xi_2\in\start{b_2}\cap\mathrm{NZ}^{(v,w)}$ also induce a path $\xi\in\start{b} \cap \mathrm{NZ}^{(v,w)}$ by simply interleaving the resampling positions. Note that $\xi_1,\xi_2$ actually induce $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ paths $\xi$ because of the possible orderings of interleaving the resamplings in $\xi_1$ and $\xi_2$.

    For a fixed $\xi_1,\xi_2$ we will now show the following:

    \begin{align*}

        \sum_{\substack{\xi\in\start{b} \cap \mathrm{NZ}^{(v,w)} \text{ s.t.}\\ \xi \text{ decomposes into } \xi_1,\xi_2 }} \P^{(n)}_b[\xi] &=

        \sum_{\text{interleavings of }\xi_1,\xi_2} \P(\text{interleaving}) \cdot \P^{(n)}_{b_1}[\xi_1] \cdot \P^{(n)}_{b_2}[\xi_2] \\

        &= \P^{(n)}_{b_1}[\xi_1] \cdot \P^{(n)}_{b_2}[\xi_2]

    \end{align*}

    where both sums are over $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ terms.

    This is best explained by an example. Lets consider the following fixed $\xi_1,\xi_2$ and an example interleaving where we choose steps from $\xi_2,\xi_1,\xi_1,\xi_2,\cdots$:

    \begin{align*}

        \xi_1 &= \left( (z_1, s_1, r_1), (z_2, s_2, r_2), (z_3, s_3, r_3), (z_4, s_4, r_4),\cdots  \right) \\

        \xi_2 &= \left( (z_1', s_1', r_1'), (z_2', s_2', r_2'), (z_3', s_3', r_3'), (z_4', s_4', r_4'),\cdots  \right) \\

        \xi   &= \left( (z_1 + z_1', s_1', r_1'), (z_1+z_2', s_1, r_1), (z_2+z_2', s_2, r_2), (z_3+z_2', s_2', r_2'), \cdots \right)

    \end{align*}

    The probability of $\xi_1$, started from $b_1$, is given by

    \begin{align*}

        \P^{(n)}_{b_1}[\xi_1] &= \P(\text{pick }s_1|z_1) \P(r_1) \P(\text{pick }s_2|z_2) \P(r_2) \cdots \P(\text{pick }s_{|\xi_1|}|z_{|\xi_1|}) \P(r_{|\xi_1|}) \\

                &= \frac{1}{z_1} \P(r_1) \frac{1}{z_2} \P(r_2) \cdots \frac{1}{z_{|\xi_1|}} \P(r_{|\xi_1|}) .

    \end{align*}

    and similar for $\xi_2$ but with primes.

    The following diagram illustrates all possible interleavings, and the red line corresponds to the particular interleaving $\xi$ in the example above.

    \begin{center}

        \includegraphics{diagram_paths2.pdf}

    \end{center}

    For the labels shown within the grid, define $p_{ij} = \frac{z_i}{z_i + z_j'}$.

    The probability of $\xi$ is given by

    \begin{align*}

        \P^{(n)}_b[\xi] &= \frac{1}{z_1+z_1'} \P(r_1') \frac{1}{z_1+z_2'} \P(r_1) \frac{1}{z_2+z_2'} \P(r_2) \frac{1}{z_3+z_2'} \P(r_2') \cdots \tag{by definition}\\

        &=

        \frac{z_1'}{z_1+z_1'} \frac{1}{z_1'} \P(r_1') \;

        \frac{z_1 }{z_1+z_2'} \frac{1}{z_1 } \P(r_1 ) \;

        \frac{z_2 }{z_2+z_2'} \frac{1}{z_2 } \P(r_2 ) \;

        \frac{z_2'}{z_3+z_2'} \frac{1}{z_2'} \P(r_2')

        \cdots \tag{rewrite fractions}\\

        &=

        \frac{z_1'}{z_1+z_1'} \;

        \frac{z_1 }{z_1+z_2'} \;

        \frac{z_2 }{z_2+z_2'} \;

        \frac{z_2'}{z_3+z_2'}

        \cdots

        \P^{(n)}_{b_1}[\xi_1] \; \P^{(n)}_{b_2}[\xi_2] \tag{definition of $\P^{(n)}_{b_i}[\xi_i]$} \\

        &= (1-p_{1,1}) \; p_{1,2} \; p_{2,2} \; (1-p_{3,2}) \; \P^{(n)}_{b_1}[\xi_1] \; \P^{(n)}_{b_2}[\xi_2] \tag{definition of $p_{i,j}$} \\

        &= \P(\text{path in grid}) \; \P^{(n)}_{b_1}[\xi_1] \; \P^{(n)}_{b_2}[\xi_2]

    \end{align*}

    In the grid we see that at every point the probabilities sum to 1, and we always reach the end, so we know the sum of all paths in the grid is 1. This proves the required equality.

    We obtain

    \begin{align*}

        \P^{(n)}_b(\mathrm{NZ}^{(v,w)},A_1,A_2)

        &= \sum_{\substack{\xi\in\start{b} \cap \\ \mathrm{NZ}^{(v,w)}\cap A_1\cap A_2}} \P^{(n)}_b(\xi) \\

        &= \sum_{\substack{\xi_1\in\start{b_1} \cap \\ \mathrm{NZ}^{(v,w)}\cap A_1}} \;\;

          \sum_{\substack{\xi_2\in\start{b_1} \cap \\ \mathrm{NZ}^{(v,w)}\cap A_2}}

        \P^{(n)}_{b_1}(\xi_1)\cdot\P^{(n)}_{b_2}(\xi_2) \\

        &=

        \P^{(n)}_{b_1}(\mathrm{NZ}^{(v,w)},A_1)

        \; \cdot \;

        \P^{(n)}_{b_2}(\mathrm{NZ}^{(v,w)},A_2).

    \end{align*}

    The second equality follows directly from $\mathbb{P}(A\mid B)=\mathbb{P}(A,B)/\mathbb{P}(B)$ and setting $A_1,A_2$ to the always-true event.

    %For the third equality, by the same reasoning we can decompose the paths

    %\begin{align*}

    %    \P^{(n)}_b(\mathrm{NZ}^{(v,w)},A_1,A_2) R_{b,\mathrm{NZ}^{(v,w)},A_1,A_2}

    %    &\equiv \sum_{\substack{\xi\in\start{b}\\\xi \in \mathrm{NZ}^{(v,w)}\cap A_1\cap A_2}} \P^{(n)}[\xi] |\xi| \\

    %    &= \sum_{\substack{\xi_1\in\start{b_1}\\\xi_1 \in \mathrm{NZ}^{(v,w)}\cap A_1}}

    %      \sum_{\substack{\xi_2\in\start{b_2}\\\xi_2 \in \mathrm{NZ}^{(v,w)}\cap A_2}}

    %    \P^{(n)}[\xi_1]\P^{(n)}[\xi_2] (|\xi_1| + |\xi_2|) \\

    %    &=

    %    \P^{(n)}_{b_2}(\mathrm{NZ}^{(v,w)},A_2) \P^{(n)}_{b_1}(\mathrm{NZ}^{(v,w)},A_1) R_{b_1,\mathrm{NZ}^{(v,w)},A_1} \\

    %    &\quad +

    %    \P^{(n)}_{b_1}(\mathrm{NZ}^{(v,w)},A_1) \P^{(n)}_{b_2}(\mathrm{NZ}^{(v,w)},A_2) R_{b_2,\mathrm{NZ}^{(v,w)},A_2} .

    %\end{align*}

    %Dividing by $\P^{(n)}_b(\mathrm{NZ}_{(v,w)},A_1,A_2)$ and using the first equality gives the desired result.

\end{proof}

\begin{definition}[Starting state dependent probability distribution.]

	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}

	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}

    Some notation: let $P$ be an interval $[a,b]$. We say $P$ is a \emph{patch} when the $\Z{i}$ event holds for all $i \in [a,b]$ and $\NZ{a-1}$ and $\NZ{b+1}$ holds. We denote this event by $P\in\mathcal{P}$, so

	\begin{align*}

	P\in\mathcal{P} \equiv \NZ{a-1} \cap \Z{a} \cap \Z{a+1} \cap \cdots \cap \Z{b-1} \cap \Z{b} \cap \NZ{b+1} .

	\end{align*}

	Note that we have the following partition of the event $\Z{v}$ for any vertex $v\in[n]$:

	\begin{align*}

	\Z{v} = \dot\bigcup_{P : v\in P} (P\in\mathcal{P})

	\end{align*}

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}

	The intuition of the following lemma is simmilar to the previous. The events on the two sides should be independent unless an interaction chain is forming, implying that every vertex gets resampled to $0$ at least once.

 	\begin{lemma}\label{lemma:independenetSidesNew}	

 		$$\P^{[k]}(\Z{1}\cap \Z{k})=\P^{[k]}(\Z{1})\P^{[k]}(\Z{k})+\bigO{p^{k}}=\left(\P^{[k]}(\Z{1})\right)^2+\bigO{p^{k}}.$$

 	\end{lemma}   

 	Note that using De Morgan's law and the inclusion-exclusion formula we can see that this is equivalent to saying:

 	$$\P^{[k]}(\NZ{1}\cap \NZ{k})=\P^{[k]}(\NZ{1})\P^{[k]}(\NZ{k})+\bigO{p^{k}}.$$

 	\begin{proof}

 		We proceed by induction on $k$. For $k=1,2$ the statement is trivial.

 		Now observe that:

 		$$\P^{[k]}(\Z{1})=\sum_{P\text{ patch}\,:\,1\in P}\P^{[k]}(P\in\mathcal{P})$$

 		$$\P^{[k]}(\Z{k})=\sum_{P\text{ patch}\,:\,k\in P}\P^{[k]}(P\in\mathcal{P})$$