diff --git a/main.tex b/main.tex
index dc9ce03d46a66d3759fb20823dcd29614deadf4d..773be6389995515275ebe45a210cca969b0de548 100644
--- a/main.tex
+++ b/main.tex
@@ -438,10 +438,10 @@ The process on the finite chain has the following modification at the boundary:
 
 %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}
-    For any state $b\in\{0,1\}^n$, define $\textsc{start}(b)$ as the event that the starting state of the chain is the state $b$. For any event $A$, define
+    For any state $b\in\{0,1\}^n$, define $\start{b}$ as the event that the starting state of the chain is the state $b$. For any event $A$, define
     \begin{align*}
-        \P^{(n)}_b(A) &= \P^{(n)}(A \;|\; \textsc{start}(b)) %\\
-        %R_{b,A} &= \mathbb{E}( \#resamples \;|\; A \; , \; \textsc{start}(b))
+        \P^{(n)}_b(A) &= \P^{(n)}(A \;|\; \start{b}) %\\
+        %R_{b,A} &= \mathbb{E}( \#resamples \;|\; A \; , \; \start{b})
     \end{align*}
     Furthermore, for the Markov Chain on the finite chain, define
     \begin{align*}
@@ -449,7 +449,7 @@ The process on the finite chain has the following modification at the boundary:
     \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}
-%Note that we have $\P^{(n)}(\textsc{start}(b)) = (1-p)^{|b|}p^{n-|b|}$ by definition of our Markov Chain.
+%Note that we have $\P^{(n)}(\start{b}) = (1-p)^{|b|}p^{n-|b|}$ by definition of our Markov Chain.
 \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}
@@ -576,7 +576,7 @@ The lemma says that conditioned on $v$ and $w$ not being crossed, the two halves
     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}
+\begin{lemma}[Conditional independence 2] \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)
@@ -595,6 +595,49 @@ The lemma says that conditioned on $v$ and $w$ not being crossed, the two halves
 	\end{align*}
 	where there is no longer a condition on the starting state.
 \end{lemma}
+\begin{proof}
+    We start by relating the different Markov Chains.
+    If $b$ is a starting state where all the zeroes are inside an interval $[v,w]$ (not on the boundary) then we can switch between the cycle and the finite chain:
+    \begin{align*}
+        \P^{(n)}_{b} (\NZ{v,w} \cap A) = \P^{[v,w]}_b (\NZ{v,w}\cap A) .
+    \end{align*}
+    If vertex $v$ and $w$ never become zero, then the zeroes never get outside of the interval $[v,w]$ and we can ignore the entire circle and only focus on the process within $[v,w]$.
+    We can apply this to the result of Lemma \ref{lemma:eventindependence}, to get
+    \begin{align*}
+        \P^{(n)}_b(\mathrm{NZ}^{(v,w)} \cap A \cap B)
+        &=
+        \P^{[v,w]}_{b|_{[v,w]}}(\mathrm{NZ}^{(v,w)} \cap A)
+        \; \cdot \;
+        \P^{[v,w]}_{b|_{[w,v]}}(\mathrm{NZ}^{(v,w)} \cap B)
+    \end{align*}
+    Note that this also holds if $b$ has zeroes on the boundary (i.e. $b_v=0$ or $b_w=0$), because then both sides of the equations are zero.
+    For the starting state we have the expression $\P^{(n)}(\start{b}) = (1-p)^{|b|} p^{n-|b|}$ so it splits into a product
+    \begin{align*}
+        \P^{(n)}(\start{b}) = \P^{[v,w]}(\start{b|_{[v+1,w-1]}}) \;\; \P^{[w,v]}(\start{b|_{[w,v]}})
+    \end{align*}
+    where we have to be careful to count the boudary only once.
+    We now have
+    \begin{align*}
+		\P^{(n)}(\mathrm{NZ}^{(v,w)}\cap A\cap B)
+        &= \sum_{b\in\{0,1\}^n} \P^{(n)}_b(\mathrm{NZ}^{(v,w)}\cap A\cap B) \; \P^{(n)}(\start{b}) \\
+        &= \sum_{b\in\{0,1\}^n}
+            \P^{[v,w]}_{b|_{[v,w]}}(\mathrm{NZ}^{(v,w)}\cap A)
+            \P^{[v,w]}(\start{b|_{[v+1,w-1]}})
+            \\ &\qquad\qquad\quad
+            \P^{[w,v]}_{b|_{[w,v]}}(\mathrm{NZ}^{(v,w)}\cap B)
+            \P^{[w,v]}(\start{b|_{[w,v]}}) \\
+        &= \left( \sum_{\substack{b_1\in\{0,1\}^{[v,w]}\\ b_v=b_w=1}}
+            \P^{[v,w]}_{b_1}(\mathrm{NZ}^{(v,w)}\cap A)
+            \P^{[v,w]}(\start{b_1}) \right)
+            \\ &\qquad \cdot
+           \left( \sum_{b_2\in\{0,1\}^{[w,v]}}
+            \P^{[w,v]}_{b_2}(\mathrm{NZ}^{(v,w)}\cap B)
+            \P^{[w,v]}(\start{b_2}) \right) \\
+        &=  \P^{[v,w]}_{b_v=b_w=1}(\mathrm{NZ}^{(v,w)}\cap A) \cdot
+            \P^{[w,v]}(\mathrm{NZ}^{(v,w)}\cap B)
+    \end{align*}
+    The second equality follows in a similar way.
+\end{proof}
 
 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}
