diff --git a/main.tex b/main.tex
index 34ab2a1ac65ed8a4ba5806b5772e6df6fc8bcf9a..1d2be1de58b3a207d92ede0bc7dbf5de5131acd5 100644
--- a/main.tex
+++ b/main.tex
@@ -614,7 +614,7 @@ Consider the chain (instead of the cycle) for simplicity with vertices identifie
 New:
 
 \begin{lemma}[Conditional independence] \label{lemma:eventindependenceNew}
-	Let $i\neq j\in [n]$, and let $A_1$ be any event that depends only on the sites $[i,j]$ and similarly $A_2$ an event that depends only on the sites $[j,i]$. (For example $\mathrm{Z}^{(s)}$ or ``$s$ has been resampled $k$ times before termination'' for an $s$ on the correct side). Then we have
+	Let $i\neq j\in [n]$, and let $A_1$ be any event that depends only on the sites $[i,j]$ (meaning the initialization and resamples) and similarly $A_2$ an event that depends only on the sites $[j,i]$. (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}^{(i,j)}\cap A_1\cap A_2)
 	&=
@@ -632,31 +632,26 @@ New:
 
 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}
-	$\P^{[n]}(\Z{1})-\P^{[n-1]}(\Z{1}) = O(p^{n})$.
+	$\forall n\in \mathbb{N}_+:\P^{[n]}(\Z{1})-\P^{[n+1]}(\Z{1}) = O(p^{n})$. (Should be true with $O(p^{n+1})$ as well.)
 \end{lemma}
 \begin{proof}
-	\begin{figure}
-		\begin{center}
-			\includegraphics{diagram_proborders.pdf}
-		\end{center}
-		\caption{\label{fig:lemmaillustrationNew} Illustration of setup of Lemma \ref{lemma:probIndep}.}
-	\end{figure}
-	The proof uses induction on $n$. For $n=1$ the statement is easy, since $\P^{\emptyset}(\Z{1})=0$ and $\P^{[1]}(\Z{1})=p$.
+	The proof uses induction on $n$. For $n=1$ the statement is easy, since $\P^{[1]}(\Z{1})=p$ and $\P^{[2]}(\Z{1})=p+p^2+O(p^{3})$.
 	
-	Induction step: suppose we proved the claim for $n-1$
+	Induction step: suppose we proved the claim for $n-1$, then
 	\begin{align*}
-	\P^{[n]}(\Z{1})
-	&=\sum_{k=1}^{n}\P^{[n]}([k]\in\mathcal{P}) \tag{the events are a partition}\\
+	\P^{[n+1]}(\Z{1})
+	&=\sum_{k=1}^{n+1}\P^{[n]}([k]\in\mathcal{P}) \tag{the events are a partition}\\
 	&=\sum_{k=1}^{n-1}\P^{[n]}([k]\in\mathcal{P}) + O(p^{n})\\
-	&=\sum_{k=1}^{n-1}\P^{[k+1]}([k]\in\mathcal{P})\cdot \P^{[n-k]}(\NZ{1})/(1-p)+ O(p^{n}) \tag{by Claim~\ref{lemma:eventindependenceNew}}\\
-	&=\sum_{k=1}^{n-1}\P^{[k+1]}([k]\in\mathcal{P})\cdot \left(\P^{[n-k-1]}(\NZ{1})+O(p^{n-k})\right)/(1-p)+ O(p^{n}) \tag{by induction} \\	
-	&\overset{?}{=}\sum_{k=1}^{n-1}\P^{[k+1]}([k]\in\mathcal{P})\cdot \P^{[n-k-1]}(\NZ{1})/(1-p)+ O(p^{n}) \\	
-	&\overset{?}{=}\sum_{k=1}^{n-1}\P^{[n-1]}([k]\in\mathcal{P})+ O(p^{n}) \tag{by Claim~\ref{lemma:eventindependenceNew}}\\
-	&=\P^{[n-1]}(\Z{1})	+ O(p^{n}) 
+	&=\sum_{k=1}^{n-1}\P^{[k+1]}([k]\in\mathcal{P})\cdot \P^{[n-k+1]}(\NZ{1})/(1-p)+ O(p^{n}) \tag{by Claim~\ref{lemma:eventindependenceNew}}\\
+	&=\sum_{k=1}^{n-1}\P^{[k+1]}([k]\in\mathcal{P})\cdot \left(\P^{[n-k]}(\NZ{1})+O(p^{n-k})\right)/(1-p)+ O(p^{n}) \tag{by induction} \\	
+	&=\sum_{k=1}^{n-1}\P^{[k+1]}([k]\in\mathcal{P})\cdot \P^{[n-k]}(\NZ{1})/(1-p)+ O(p^{n}) \\	
+	&=\sum_{k=1}^{n-1}\P^{[n]}([k]\in\mathcal{P})+ O(p^{n}) \tag{by Claim~\ref{lemma:eventindependenceNew}}\\
+	&=\sum_{k=1}^{n}\P^{[n]}([k]\in\mathcal{P})+ O(p^{n}) \\
+	&=\P^{[n]}(\Z{1})	+ O(p^{n}) 
 	\end{align*}
 \end{proof}
 \begin{corollary}\label{cor:probIndepNew}
-	$\P^{[n]}(\Z{1})-\P^{[m]}(\Z{1}) = O(p^{\min(n,m)+1})$.
+	$\P^{[n]}(\Z{1})-\P^{[m]}(\Z{1}) = O(p^{\min(n,m)})$. (Should be true with $O(p^{\min(n,m)+1})$ as well.)
 \end{corollary}
 
  	\begin{lemma}\label{lemma:independenetSidesNew}	
@@ -671,7 +666,7 @@ The intuition of the following lemma is that the far right can only affect the z
  		$$\P^{[k]}(\Z{1})=\sum_{P\text{ patch}\,:\,1\in P}\P^{[k]}(P\in\mathcal{P})$$
  		$$\P^{[k]}(\Z{k})=\sum_{P\text{ patch}\,:\,k\in P}\P^{[k]}(P\in\mathcal{P})$$
  		
- 		Suppose we proved the statement for all $\ell\leq k$, then we proceed using induction similarly to the above (let us use the notation $>I<:=I\cap \overline{P_l}\cap \overline{P_r}$ for simplicity)
+ 		Suppose we proved the statement up to $k-$, then we proceed using induction similarly to the above
  		\begin{align*}
  		&\P^{[k]}(\Z{1}\cap \Z{k})=\\
  		&=\sum_{\ell, r\in [k]: \ell<r-1}\P^{[k]}([\ell],[r,k]\in\mathcal{P})
@@ -719,8 +714,9 @@ The intuition of the following lemma is that the far right can only affect the z
 		&= \frac{1}{n}\sum_{P\text{ patch}}\E^{(n)}(\# \text{ resamples in }P|P\in \mathcal{P})\\			
 		&= \sum_{s=1}^{n-1}\E^{(n)}(\# \text{ resamples in }[s]|[s]\in \mathcal{P})+\mathcal{O}(p^{n})\tag{by translation symmetry}\\   		
 		&= \sum_{s=1}^{n-1}\E^{[0,s+1]}(\# \text{ resamples in }[s]|[s]\in \mathcal{P})\P^{[s+1,n]}(\NZ{s+1}\cap\NZ{n})/(1+p)^2+\mathcal{O}(p^{n}) \tag{by Lemma~\ref{lemma:eventindependenceNew}}\\   
-		&= \sum_{s=1}^{n-1}\E^{[0,s+1]}(\# \text{ resamples in }[s]|[s]\in \mathcal{P})\left(\P^{[s+1,n]}(\NZ{s+1})\right)^2/(1+p)^2+\mathcal{O}(p^{n}) \tag{by Lemma~\ref{lemma:independenetSidesNew}}\\   	
-		&= \sum_{s=1}^{n-1}\E^{[-N,N]}(\# \text{ resamples in }[s]|[s]\in \mathcal{P})+\mathcal{O}(p^{n}) \tag{by Lemma~\ref{lemma:eventindependenceNew},~\ref{lemma:independenetSidesNew}}\\   	
+		&= \sum_{s=1}^{n-1}\E^{[0,s+1]}(\# \text{ resamples in }[s]|[s]\in \mathcal{P})\left(\P^{[s+1,n]}(\NZ{s+1})\right)^2/(1+p)^2+\mathcal{O}(p^{n}) \tag{by Lemma~\ref{lemma:independenetSidesNew}}\\   
+		&= \sum_{s=1}^{n-1}\E^{[0,s+1]}(\# \text{ resamples in }[s]|[s]\in \mathcal{P})\left(\P^{[s+1,N]}(\NZ{s+1})\right)^2/(1+p)^2+\mathcal{O}(p^{n}) \tag{by Corollary~\ref{cor:probIndepNew}}\\   			
+		&= \sum_{s=1}^{n-1}\E^{[-N,N]}(\# \text{ resamples in }[s]|[s]\in \mathcal{P})+\mathcal{O}(p^{n}) \tag{by Lemma~\ref{lemma:eventindependenceNew}, Corollary~\ref{cor:probIndepNew}}\\   	
 		&= \sum_{s=1}^{N}\E^{[-N,N]}(\# \text{ resamples in }[s]|[s]\in \mathcal{P})+\mathcal{O}(p^{n}).
 		\end{align*}
 		Repeating the same calculation with $m$, and comparing the two expressions completes the proof.
