diff --git a/main.tex b/main.tex
index abd3cde6df99068eb8c35021caca298e0c8427c9..b59a0a477f53431cf926c5f24cf4caeb78da314d 100644
--- a/main.tex
+++ b/main.tex
@@ -580,6 +580,49 @@ The intuition of the following lemma is that the far right can only affect the z
 	
 	Also this claim finally ``sees'' how many empty places are between slots. These properties make it possible to use this lemma to prove the sought linear bound. We show it for the infinite chain, but with a little care it should also translate to the circle.
 
+~
+
+Here, I (Tom) tried to set up the same Lemma but for the circle instead of the infinite chain.
+This time, it is no longer $I_\mathrm{max}$ but any vertex $i_* \in I$, and $I' = I \setminus \{i_*\}$. Without loss of generality, we can assume that $i_* \leq n/2$ (because if not then we can relabel the vertices and count the other way around so that $i_* \to n-i_*$). The goal is now to prove:
+\begin{align*}
+    P_I(Z^{(0)}) = P_{I'}(Z^{(0)}) + \mathcal{O}(p^{i_* + 1 - |I|})
+\end{align*}
+Note that when we refer to an interval $[a,b]$ on the circle we could be referring to two possible intervals because of the periodicity of the circle. In the following, whenever we refer to an interval $[a,b]$ we refer to the interval with vertex 0 on the \emph{inside}.
+
+For $a,b\in[n]$, define the event ``zeroes patch'' as the event of getting zeroes inside the interval $[a,b]$ but not on the boundary, i.e.  $\mathrm{ZP}^{[a,b]} = \mathrm{NZ}^{(a)} \cap \mathrm{Z}^{(a+1)} \cap \mathrm{Z}^{(a+2)} \cap \cdots \cap \mathrm{Z}^{(b-1)} \cap \mathrm{NZ}^{(b)}$ (where we assume that $\mathrm{Z}^{(0)}$ is part of this intersection).
+
+Furthermore, define the `inside' and `outside' of $I$ as $I_{\mathrm{in}(a,b)} = I\cap[a,b]$ and $I_{\mathrm{out}(a,b)} = I \setminus [a,b]$.
+The following diagram illustrates these definitions.
+\begin{center}
+    \includegraphics{diagram_circle_lemma.pdf}
+\end{center}
+\begin{align*}
+    P_{I}(\mathrm{Z}^{(0)})
+    &=\sum_{\substack{l,k=1\\k+l<n}}
+    P_I(\mathrm{ZP}^{[n-l,k]}) \tag{the events are a partition}\\
+    &=\sum_{\substack{l,k=1\\k+l<n\\k,n-l\notin I}}
+    P_I(\mathrm{ZP}^{[n-l,k]}) \tag{$\mathbb{P}(\mathrm{ZP}^{[a,b]})=0$ for $a\in I$ or $b\in I$}
+\end{align*}
+Note that if $[-l,k]$ does not `touch' $I$ then $P_I(\mathrm{ZP}^{[-l,k]}) = 0$.
+Furthermore, we have $P_I(\mathrm{ZP}^{[n-l,k]}) = \mathcal{O}(p^{k+l-1-|I_{\mathrm{in}(n-l,k)}|})$. If $k\geq i_*$ or $l\geq i_*$ then this gives $P_I(\mathrm{ZP}^{[n-l,k]}) = \mathcal{O}(p^{i_* - 1 - |I|})$ since $|I_\mathrm{in}| \leq |I|$. Therefore we have
+\begin{align*}
+    P_I(\mathrm{Z}^{(0)})
+    &=\sum_{\substack{l,k=1\\k,n-l\notin I}}^{i_*-1}
+    P_I(\mathrm{ZP}^{[n-l,k]})
+    + \mathcal{O}(p^{i_* - 1 - |I|}) \\
+    &=\sum_{\substack{l,k=1\\k,n-l\notin I}}^{i_*-1}
+    P_{I_{\mathrm{in}(n-l,k)}}(\mathrm{ZP}^{[n-l,k]}) \cdot
+    P_{I_{\mathrm{out}(n-l,k)}}(\mathrm{NZ}^{(n-l,k)})
+    + \mathcal{O}(p^{i_* - 1 - |I|}) \\
+    \tag{by Claim~\ref{claim:eventindependence} for $n-l,k\notin I$} \\
+    &=\sum_{\substack{l,k=1\\k,n-l\notin I}}^{i_*-1}
+    P_{I'_{\mathrm{in}(n-l,k)}}(\mathrm{ZP}^{[n-l,k]}) \cdot
+    P_{I_{\mathrm{out}(n-l,k)}}(\mathrm{NZ}^{(n-l,k)})
+    + \mathcal{O}(p^{i_* - 1 - |I|})
+\end{align*}
+Now we are supposed to use the induction step, but this is where I got stuck.
+
+
 \begin{definition}[Connected patches]
 	Let $\mathcal{P}\subset 2^{\mathbb{Z}}$ be a finite system of finite subsets of $\mathbb{Z}$. We say that the patch set of a resample sequence is $\mathcal{P}$,
 	if the connected components of the vertices that have ever become $0$ are exactly the elements of $\mathcal{P}$. We denote by $A^{(\mathcal{P})}$ the event that the set of patches is $\mathcal{P}$. For a patch $P$ let $A^{(P)}=\bigcup_{\mathcal{P}:P\in \mathcal{P}}A^{(\mathcal{P})}$.