diff --git a/main.tex b/main.tex
index b826b535b105828b8fda669d5ff0bf40f744870d..6119dd84d3ca3a2ae4fca124879e10d5963820f6 100644
--- a/main.tex
+++ b/main.tex
@@ -402,24 +402,87 @@ where we used the identity $\sum_{a\in\{0,1\}^l} (-1)^{|a|} = 0$.
 
 ~
 
+It is useful to introduce some new notation: for any event $A$ (where an event is a set of paths), define
+\begin{align*}
+    \mathbb{P}_b(A) &= \mathbb{P}(A \;|\; \text{start in }b) \\
+    R_{b,A} &= \mathbb{E}( \#resamples \;|\; A\;,\; \text{start in }b)
+\end{align*}
+Denote by $\mathrm{Z}_j$ the event that site $j$ becomes zero at any point in time before the Markov Chain terminates. Denote the complement by $\mathrm{NZ}_j$, i.e. the event that site $j$ does \emph{not} become zero before it terminates.
+
 The proof of claim \ref{claim:expectationsum} also proves the following claim
-\begin{claim}[Probability independence] \label{claim:pathindependence}
-    As in \ref{claim:expectationsum}, let $b=b_1\land b_2\in\{0,1\}^n$ be a state with two groups ($b_1\lor b_2 = 1^n$) of zeroes. Let $j_1$, $j_2$ be indices `inbetween' the groups (or only one index in case of the infinite line). Denote by $\mathrm{NZ}_j$ the event that site $j$ does not become zero before the Markov Chain terminates. Then we have
+\begin{claim}[Conditional independence] \label{claim:eventindependence}
+    As in \ref{claim:expectationsum}, let $b=b_1\land b_2\in\{0,1\}^n$ be a state with two groups ($b_1\lor b_2 = 1^n$) of zeroes. Let $j_1$, $j_2$ be indices `inbetween' the groups (or only one index in case of the infinite line). Then we have
     \begin{align*}
-        \mathbb{P}[\mathrm{NZ}_{j_1} , \mathrm{NZ}_{j_2} |\;\text{start in }b]
-        =
-        \mathbb{P}[\mathrm{NZ}_{j_1} , \mathrm{NZ}_{j_2} \;|\;\text{start in }b_1]
+        \mathbb{P}_b(\mathrm{NZ}_j)
+        &=
+        \mathbb{P}_{b_1}(\mathrm{NZ}_j)
         \; \cdot \;
-        \mathbb{P}[\mathrm{NZ}_{j_1} , \mathrm{NZ}_{j_2} \;|\;\text{start in }b_2]
+        \mathbb{P}_{b_2}(\mathrm{NZ}_j) \\
+        R_{b,\mathrm{NZ}_j}
+        &=
+        R_{b_1,\mathrm{NZ}_j}
+        \; + \;
+        R_{b_2,\mathrm{NZ}_j}
     \end{align*}
-up to any order in $p$.
+    up to any order in $p$. Furthermore the equalities also hold when $\mathrm{NZ}_j$ is replaced by any subset $A\subseteq\mathrm{NZ}_j$.
 \end{claim}
-Since the left hand side is defined as
+\begin{proof}
+    Note that any path $\xi\in\paths{b} \cap \mathrm{NZ}_j$ can be split into paths $\xi_1\in\paths{b_1}\cap \mathrm{NZ}_j$ and $\xi_2\in\paths{b_2}\cap\mathrm{NZ}_j$ and by the same reasoning as in the proof of claim \ref{claim:expectationsum}, we obtain
+    \begin{align*}
+        \mathbb{P}_b(\mathrm{NZ}_j)
+        = \sum_{\substack{\xi\in\paths{b}\\\xi \in \mathrm{NZ}_j}} \mathbb{P}[\xi]
+        &= \sum_{\substack{\xi_1\in\paths{b_1}\\\xi_1 \in \mathrm{NZ}_j}}
+          \sum_{\substack{\xi_2\in\paths{b_1}\\\xi_2 \in \mathrm{NZ}_j}}
+        \mathbb{P}[\xi_1]\cdot\mathbb{P}[\xi_2] \\
+        &=
+        \mathbb{P}_{b_1}(\mathrm{NZ}_j)
+        \; \cdot \;
+        \mathbb{P}_{b_2}(\mathrm{NZ}_j).
+    \end{align*}
+    For the second equality, note that again by the same reasoning as in the proof of claim \ref{claim:expectationsum} we have
+    \begin{align*}
+        \mathbb{P}_b(\mathrm{NZ}_j) R_{b,\mathrm{NZ}_j}
+        := \sum_{\substack{\xi\in\paths{b}\\\xi \in \mathrm{NZ}_j}} \mathbb{P}[\xi] |\xi| 
+        &= \sum_{\substack{\xi_1\in\paths{b_1}\\\xi_1 \in \mathrm{NZ}_j}}
+          \sum_{\substack{\xi_2\in\paths{b_2}\\\xi_2 \in \mathrm{NZ}_j}}
+        \mathbb{P}[\xi_1]\mathbb{P}[\xi_2] (|\xi_1| + |\xi_2|) \\
+        &=
+        \mathbb{P}_{b_2}(\mathrm{NZ}_j) \mathbb{P}_{b_1}(\mathrm{NZ}_j) R_{b_1,\mathrm{NZ}_j}
+        \; + \;
+        \mathbb{P}_{b_1}(\mathrm{NZ}_j) \mathbb{P}_{b_2}(\mathrm{NZ}_j) R_{b_2,\mathrm{NZ}_j} .
+    \end{align*}
+    Dividing by $\mathbb{P}_b(\mathrm{NZ}_j)$ and using the first equality gives the desired result.
+\end{proof}
+
+~
+
+TEST: Although a proof of claim \ref{claim:expectationsum} was already given, I'm trying to prove it in an alternate way using claim \ref{claim:eventindependence}.\\
+Assume that $b_1$ ranges up to site $0$, the gap ranges from sites $1,...,k$ and $b_2$ ranges from site $k+1$ and onwards. For $j=1,...,k$ define the partial-zero event $\mathrm{PZ}_j = \mathrm{Z}_1 \cap \mathrm{Z}_2 \cap ... \cap \mathrm{Z}_{j-1} \cap \mathrm{NZ}_j$ i.e. the first $j-1$ sites of the gap become zero and site $j$ does not become zero. Also define the all-zero event $\mathrm{AZ} = \mathrm{Z}_1 \cap ... \cap \mathrm{Z}_k$, where all sites of the gap become zero. Note that these events partition the space, so we have for all $b$ that $\sum_{j=1}^k \mathbb{P}_b(\mathrm{PZ}_j) = 1 - \mathbb{P}_b(\mathrm{AZ}) = 1 - \mathcal{O}(p^k)$.
+
+Furthermore, if site $j$ becomes zero from $b_1$ it means all sites to the left of $j$ become zero as well. Similarly, from $b_2$ it implies all the sites to the right of $j$ become zero.
+Because of that, we have
+\begin{align*}
+    \mathbb{P}_{b_1}(\mathrm{PZ}_j) &= \mathbb{P}_{b_1}(\mathrm{Z}_{j-1} \cap \mathrm{NZ}_j) = \mathcal{O}(p^{j-1}) \\
+    \mathbb{P}_{b_2}(\mathrm{PZ}_j) &= \mathbb{P}_{b_2}(\mathrm{NZ}_j) = 1 - \mathbb{P}_{b_2}(\mathrm{Z}_j) = 1 - \mathcal{O}(p^{k-j+1})
+\end{align*}
+Now observe that
 \begin{align*}
-    \mathbb{P}[\mathrm{NZ}_{j_1} , \mathrm{NZ}_{j_2} |\;\text{start in }b]
-    = \sum_{\substack{\xi\in\paths{b}\\j_1,j_2 \text{ not 0 in } \xi}} \mathbb{P}[\xi]
+    R_b &= \sum_{j=1}^k \mathbb{P}_b(\mathrm{PZ}_j) R_{b,\mathrm{PZ}_j} + \mathbb{P}_b(\mathrm{AZ}) R_{b,\mathrm{AZ}} \\
+        &= \sum_{j=1}^k \mathbb{P}_{b_2}(\mathrm{PZ}_j)\mathbb{P}_{b_{1}}(\mathrm{PZ}_j) R_{b_1,\mathrm{PZ}_j}
+        + \sum_{j=1}^k \mathbb{P}_{b_1}(\mathrm{PZ}_j)\mathbb{P}_{b_{2}}(\mathrm{PZ}_j) R_{b_2,\mathrm{PZ}_j}
+        + \mathcal{O}(p^k) \\
+        &= \sum_{j=1}^k \mathbb{P}_{b_{1}}(\mathrm{PZ}_j) R_{b_1,\mathrm{PZ}_j}
+        - \sum_{j=1}^k \mathbb{P}_{b_2}(\mathrm{Z}_j)\mathbb{P}_{b_{1}}(\mathrm{PZ}_j) R_{b_1,\mathrm{PZ}_j}
+        + \sum_{j=1}^k \mathbb{P}_{b_1}(\mathrm{PZ}_j)\mathbb{P}_{b_{2}}(\mathrm{PZ}_j) R_{b_2,\mathrm{PZ}_j}
+        + \mathcal{O}(p^k) \\
+        &= \sum_{j=1}^k \mathbb{P}_{b_{1}}(\mathrm{PZ}_j) R_{b_1,\mathrm{PZ}_j}
+        + \sum_{j=1}^k \mathbb{P}_{b_1}(\mathrm{PZ}_j)\mathbb{P}_{b_{2}}(\mathrm{PZ}_j) R_{b_2,\mathrm{PZ}_j}
+        + \mathcal{O}(p^k) \\
+        &= R_{b_1}
+        + \sum_{j=1}^k \mathbb{P}_{b_1}(\mathrm{PZ}_j)\mathbb{P}_{b_{2}}(\mathrm{PZ}_j) R_{b_2,\mathrm{PZ}_j}
+        + \mathcal{O}(p^k) \\
+        &= R_{b_1} + R_{b_2} + \mathcal{O}(p^k)
 \end{align*}
-we see that all such paths $\xi$ can be split into paths $\xi_1\in\paths{b_1}$ and $\xi_2\in\paths{b_2}$ and by the same reasoning as in the proof of claim \ref{claim:expectationsum}, we obtain the right hand side.
 
 \newpage
     \subsection{Sketch of the (false) proof of the linear bound \ref{it:const}}
