diff --git a/main.tex b/main.tex
index 7d637a34353ca9dfcacb90c355094351323afd4a..d1cad055bc49ef6190a224e51eb23566459ed372 100644
--- a/main.tex
+++ b/main.tex
@@ -244,7 +244,7 @@
     We observe that for the initial $\rho$ we have that $\text{rank}(\rho)=n$, therefore $\text{rank}(\rho*(M_{(n)}^k))\geq n+k$, and so $\rho*(M_{(n)}^k)*\mathbbm{1}$ is obviously divisible by $p^{k}$. This implies that $a_k^{(n)}$ can be calculated by only looking at $\rho*(M_{(n)}^1)*\mathbbm{1}, \ldots, \rho*(M_{(n)}^k)*\mathbbm{1}$.
     
 \newpage
-\section{Proving the strong cancellation claim}
+\section{Proving that $a_k^{(k+1)}=a_k^{(n)}$ for all $n>k$}
 It is useful to introduce some new notation. We will consider variations of the Markov Chains:
 \begin{itemize}
     \item $\P^{(n)}$ refers to the original process on the length-$n$ cycle.