diff --git a/main.tex b/main.tex
index 438013c10aa8c653535eb56011b86d10ec44b2e9..82cc4e04920521810f3d8d73731453a4e9838366 100644
--- a/main.tex
+++ b/main.tex
@@ -256,7 +256,7 @@ where $C(f)\in\{0,1,1'\}^n$ denotes a configuration with slots on the sites $C$
 	\begin{center}
     	\includegraphics{diagram_gap.pdf}
     \end{center}
-    \caption{\label{fig:diametergap} A configuration $C=\{1,2,4,7,9\}\subseteq[n]$ consisting of 5 slots shown by the red dots. The dotted line at the top depicts the rest of the circle which may be much larger. The diameter of this configuration is $\diam{C}=9$ as shown and the largest gap of $C$ is $\mathrm{gap}(C)=2$. Note that we do not count the rest of the circle as a gap, we only consider gaps \emph{within} the diameter of $C$.}
+    \caption{\label{fig:diametergap} A configuration $C=\{1,2,4,7,9\}\subseteq[n]$ consisting of 5 slots shown by the red dots. The blue squares denote the set $C_{><}$ which is all elements of $[n]\setminus C$ that lie within the interval spanned by $C$. The dotted line at the top depicts the rest of the circle which may be much larger. The diameter of this configuration is $\diam{C} = |C| + |C_{><}| =9$ as shown. The largest gap of $C$ is $\mathrm{gap}(C)=2$ which is the largest connected component of $C_{><}$. Note that we do not count the rest of the circle as a gap, we only consider gaps \emph{within} the diameter of $C$.}
 \end{figure}
 
 \begin{claim}[Strong cancellation claim] \label{claim:strongcancel}
@@ -528,7 +528,7 @@ The intuition of the following lemma is that the far right can only affect the z
 	The proof uses induction on $|I|$. For $|I|=1$ the statement is easy, since every resample sequence that resamples the $0$ vertex must produce at least $I_{\max}$ number of $0$-s during the resamplings.
 	
 	Induction step: For an event $A$ and $k>0$ let us denote by $A_k=A\cap$``Each vertex in $0,1,2,\ldots, k-1$ gets $0$ before termination (either by resampling or initialisation), but not $k$''. Observe that $V^{(0)}=\dot{\bigcup}_{k=1}^{\infty}V^{(0)}_k$.
-	Let $I_{<k}:=I\cap[k-1]$ and similarly $I_{>k}:=I\setminus[k]$, finally let $I_{><}:=\{I_{\min}+1,I_{\max}-1]\}\setminus I$. Suppose we proved the claim up to $|I|-1$, then the induction step can be shown by
+    Let $I_{<k}:=I\cap[k-1]$ and similarly $I_{>k}:=I\setminus[k]$, finally let $I_{><}:=\{I_{\min}+1,I_{\max}-1]\}\setminus I$ as shown in Figure \ref{fig:diametergap}. Suppose we proved the claim up to $|I|-1$, then the induction step can be shown by
 	\begin{align*}
 		P_{I}(V^{(0)})
 		&=\sum_{k=1}^{\infty}P(V^{(0)}_k)
