diff --git a/main.tex b/main.tex
index b59a0a477f53431cf926c5f24cf4caeb78da314d..baddb12b70fe3995a2936fa00b739de83a4b6297 100644
--- a/main.tex
+++ b/main.tex
@@ -443,6 +443,11 @@ It is useful to introduce some new notation:
         \mathbb{P}_{b_1}(\mathrm{NZ}^{(j_1,j_2)}, A_1)
         \; \cdot \;
         \mathbb{P}_{b_2}(\mathrm{NZ}^{(j_1,j_2)}, A_2) \\
+        \mathbb{P}_b(A_1, A_2 \mid \mathrm{NZ}^{(j_1,j_2)})
+        &=
+        \mathbb{P}_{b_1}(A_1 \mid \mathrm{NZ}^{(j_1,j_2)})
+        \; \cdot \;
+        \mathbb{P}_{b_2}(A_2 \mid \mathrm{NZ}^{(j_1,j_2)}) \\
         R_{b,\mathrm{NZ}^{(j_1,j_2)},A_1,A_2}
         &=
         R_{b_1,\mathrm{NZ}^{(j_1,j_2)},A_1}
@@ -466,7 +471,8 @@ The lemma says that conditioned on $j_1$ and $j_2$ not being crossed, the two ha
         \; \cdot \;
         \mathbb{P}_{b_2}(\mathrm{NZ}^{(j_1,j_2)},A_2).
     \end{align*}
-    For the second equality, note that again by the same reasoning as in the proof of claim \ref{claim:expectationsum} we have
+    The second equality follows directly from Bayes rule and removing $A_1,A_2$.
+    For the third 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_1,j_2)},A_1,A_2) R_{b,\mathrm{NZ}^{(j_1,j_2)},A_1,A_2}
         &:= \sum_{\substack{\xi\in\paths{b}\\\xi \in \mathrm{NZ}^{(j_1,j_2)}\cap A_1\cap A_2}} \mathbb{P}[\xi] |\xi| \\
@@ -478,7 +484,7 @@ The lemma says that conditioned on $j_1$ and $j_2$ not being crossed, the two ha
         &\quad +
         \mathbb{P}_{b_1}(\mathrm{NZ}^{(j_1,j_2)},A_1) \mathbb{P}_{b_2}(\mathrm{NZ}^{(j_1,j_2)},A_2) R_{b_2,\mathrm{NZ}^{(j_1,j_2)},A_2} .
     \end{align*}
-    Dividing by $\mathbb{P}_b(\mathrm{NZ}_{j_1}\cap\mathrm{NZ}_{j_2},A_1,A_2)$ and using the first equality gives the desired result.
+    Dividing by $\mathbb{P}_b(\mathrm{NZ}_{(j_1,j_2)},A_1,A_2)$ and using the first equality gives the desired result.
 \end{proof}
 
 \begin{comment}
@@ -583,9 +589,9 @@ The intuition of the following lemma is that the far right can only affect the z
 ~
 
 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:
+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 $i_* \leq n/2$ so that the distance to $0$ is simply $d(i_*,0)=i_*$ (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|})
+    P_I(Z^{(0)}) = P_{I'}(Z^{(0)}) + \mathcal{O}(p^{\mathrm{d}(i_*,0) + 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}.
 
@@ -604,21 +610,21 @@ The following diagram illustrates these definitions.
     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
+Furthermore, we have $P_I(\mathrm{ZP}^{[n-l,k]}) = \mathcal{O}(p^{k+l-1-|I_{\mathrm{in}(n-l,k)}|})$. If $k > \mathrm{d}(i_*,0)$ or $l > \mathrm{d}(i_*,0)$ then this gives $P_I(\mathrm{ZP}^{[n-l,k]}) = \mathcal{O}(p^{\mathrm{d}(i_*,0) + 1 - |I|})$ 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|}) \\
+    + \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|}) \\
+    + \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|})
+    + \mathcal{O}(p^{i_* + 1 - |I|})
 \end{align*}
 Now we are supposed to use the induction step, but this is where I got stuck.
 
