diff --git a/main.tex b/main.tex
index af2bd5aa5e7d5df7f31a82e70ea28f3e6fe66aa6..efc84dbb86206c78086ffdc2cbde7b352cda1299 100644
--- a/main.tex
+++ b/main.tex
@@ -618,16 +618,15 @@ Here, I (Tom) tried to set do the same Lemma but for the circle instead of the i
     Without loss of generality, we can assume that $0=j < i_* < s < n$. We will now consider intervals around vertex 0.
     For $l,r\geq 1$ and $l+r\leq n$, define the event ``zeroes patch'' $\mathrm{ZP}^{[-l,r]_0}$ as the event of getting zeroes inside the interval $[-l,r]_0$ but not on the boundary, i.e.
     $$\mathrm{ZP}^{[-l,r]_0} = \NZ{-l} \cap \Z{-l+1} \cap \cdots \cap \Z{0} \cap \cdots \cap \Z{r-1} \cap \NZ{r}$$
-    Note that there are $r+l-1$ `zeroes' in this event, so $\P_{J}(\mathrm{ZP}^{[-l,r]_0}) = \mathcal{O}(p^{r+l-1-|J|})$ for $J\subseteq[-l,r]_0$.
+    Note that there are $r+l-1$ `zeroes' in this event, so $\P_{J}(\mathrm{ZP}^{[-l,r]_0}) = \mathcal{O}(p^{r+l-1-|J|})$ for $J\subseteq[-l,r]_0$ is a lower bound on the order of $p$.\\
     Claim:
     \begin{align*}
         \P_{I}(\mathrm{ZP}^{[-l,r]_0}) &= \P_{I'}(\mathrm{ZP}^{[-l,r]_0})
-        + \mathcal{O}(p^{\min\left( \dist{r}(i_*,-l) , d(i_*,r)\right)+l+r-|I|})
+        + \mathcal{O}(p^{d(i_*,0)+1-|I|})
     \end{align*}
-    \todo{These special cases.}
-    If $r\geq i_*$ or $l\geq n-i_*$ then $\P_{I}(\mathrm{ZP}^{[-l,r]_0}) = \mathcal{O}(p^{d(i_*,0) + 1 - |I|})$.
-    If $-l\in I$ or $r\in I$ then the left hand side is zero so the claim holds.
-    If $[-l,r]_0$ has no overlap with $I$ then both sides of the above expression are zero so it also holds. We are left with the case where, $-l,r,\notin I$ and $[-l,r]_0 \cap I \neq \emptyset$ and $i_*\notin[-l,r]_0$.
+    If $r\geq i_*$ or $l\geq n-i_*$ then $\P_{I}(\mathrm{ZP}^{[-l,r]_0}) = \mathcal{O}(p^{d(i_*,0) + 1 - |I|})$ and also $\P_{I'}(\mathrm{ZP}^{[-l,r]_0}) = \mathcal{O}(p^{d(i_*,0) + 1 - |I|})$ so then the claim holds.
+    If $-l\in I$ or $r\in I$ (and $-l,r$ are both not $i_*$ because of the previous point) then the probability of $\mathrm{ZP}^{[-l,r]_0}$ is zero for both $I$ and $I'$ so the claim holds.
+    If $[-l,r]_0$ has no overlap with $I$ then both sides are also zero so it also holds. We are left with the case where: $-l,r,\notin I$ and $[-l,r]_0 \cap I \neq \emptyset$ and $i_*\notin[-l,r]_0$.
     The following diagram illustrates the situation
     \begin{center}
         \includegraphics{diagram_circle_lemma.pdf}
@@ -664,9 +663,12 @@ Here, I (Tom) tried to set do the same Lemma but for the circle instead of the i
         &= \P_{I'}(\mathrm{ZP}^{[-l,r]_0})
         + \mathcal{O}(p^{\min\left( \dist{r}(i_*,-l) , d(i_*,r)\right)+l+r-|I|})
     \end{align*}
-    Case separation shows that $\min\left( \dist{r}(i_*,-l) , d(i_*,r)\right) \geq d(i_*,0) + 1$ \todo{prove.}
+    Where we used Claim~\ref{claim:eventindependence} again.
+    Case separation shows that
+    $$\min\left( \dist{r}(i_*,-l) , d(i_*,r)\right) + l +r \geq d(i_*,0) + 1$$
+    which proves the claim.
 
-    Now we can prove the required equalities:
+    The first equality that we have to prove now follows from the fact that the ``zeroes patch'' events are a partition of $\Z{0}$:
     \begin{align*}
         \P_{I}(\Z{0})
         &=\sum_{\substack{l,r\geq 1\\l+r\leq n}}
@@ -674,7 +676,8 @@ Here, I (Tom) tried to set do the same Lemma but for the circle instead of the i
         \tag{the events are a partition of $\Z{0}$}\\
         &=\sum_{\substack{l,r\geq 1\\l+r\leq n}}
         \P_{I'}(\mathrm{ZP}^{[-l,r]_0})
-        + \mathcal{O}(p^{\min\left( \dist{r}(i_*,-l) , d(i_*,r)\right)+l+r-|I|}) \\
+        + \mathcal{O}(p^{d(i_*,0)+1-|I|})
+        \tag{by claim} \\
         &= \P_{I'}(\Z{0}) + \mathcal{O}(p^{d(i_*,0)+1-|I|})
     \end{align*}
     Similarly, we have
@@ -683,13 +686,35 @@ Here, I (Tom) tried to set do the same Lemma but for the circle instead of the i
         &=\sum_{l=1}^{n-s}\sum_{r=1}^{s}
         \P_{I}(\mathrm{ZP}^{[-l,r]_0},\NZ{s})
         \tag{partition of $\Z{0}$}\\
+        &=\sum_{l=1}^{n-s}\sum_{r=1}^{i_*-1}
+        \P_{I}(\mathrm{ZP}^{[-l,r]_0},\NZ{s})
+        +\mathcal{O}(p^{\dist{s}(i_*,0)+1-|I|}) \\
+        &=\sum_{l=1}^{n-s}\sum_{r=1}^{i_*-1}
+        \P_{I\cap[s,r]_0}(\mathrm{ZP}^{[-l,r]_0},\NZ{s}) \cdot
+        \P_{I\setminus [s,r]_0}(\NZ{r},\NZ{s})
+        +\mathcal{O}(p^{\dist{s}(i_*,0)+1-|I|})
+        \tag{Claim~\ref{claim:eventindependence}}\\
+        &=\sum_{l=1}^{n-s}\sum_{r=1}^{s}
+        \P_{I'\cap[s,r]_0}(\mathrm{ZP}^{[-l,r]_0},\NZ{s}) \cdot
+        \P_{I\setminus [s,r]_0}(\NZ{r},\NZ{s})
+        +\mathcal{O}(p^{\dist{s}(i_*,0)+1-|I|})
+        \tag{$i_*\in I \setminus[s,r]_0$}\\
+        &=\sum_{l=1}^{n-s}\sum_{r=1}^{s}
+        \P_{I'\cap[s,r]_0}(\mathrm{ZP}^{[-l,r]_0},\NZ{s}) \cdot
+        \P_{I'\setminus [s,r]_0}(\NZ{r},\NZ{s}) \\
+        &\qquad +\mathcal{O}(p^{\min\left( \dist{r}(i_*,s) , d(i_*,r)\right)+l+r-|I|})
+        +\mathcal{O}(p^{\dist{s}(i_*,0)+1-|I|})
+        \tag{same argument as before}\\
         &=\sum_{l=1}^{n-s}\sum_{r=1}^{s}
-        \P_{I'}(\mathrm{ZP}^{[-l,r]_0},\NZ{s})
-        + \mathcal{O}(p^{something}) \\
+        \P_{I'\cap[s,r]_0}(\mathrm{ZP}^{[-l,r]_0},\NZ{s}) \cdot
+        \P_{I'\setminus [s,r]_0}(\NZ{r},\NZ{s}) \\
+        &\qquad
+        +\mathcal{O}(p^{\dist{s}(i_*,0)+1-|I|})
+        \tag{case separation \todo{does not seem to work?}}\\
         &= \P_{I'}(\Z{0} , \NZ{s})
-        + \sum_{l=1}^{n-s}\sum_{r=1}^{s} + \mathcal{O}(p^{something})
+        +\mathcal{O}(p^{\dist{s}(i_*,0)+1-|I|})
     \end{align*}
-    \todo{finish details}
+    This finishes the proof.
 \end{proof}
 
 \begin{definition}[Connected patches]