diff --git a/main.tex b/main.tex
index 21435d406f95bac71a58f6a45913ac8b6a7eca2a..de43fef3939747b2ed9a2620bf3de3a12122e15f 100644
--- a/main.tex
+++ b/main.tex
@@ -353,14 +353,20 @@ For all $\xi\in\paths{b_1\land b_2}$ we have that \emph{either} $\xi$ splits int
 \[
 	R_{b_1\land b_2} = \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1| + \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_1]\mathbb{P}[\xi_2]|\xi_2| + \sum_{\mathclap{\xi\;\mathrm{dependent}}} \mathbb{P}[\xi]|\xi|.
 \]
-The last sum only contains only terms of order $p^{k}$ or higher. Now for the first sum, note that
+where last sum only contains only terms of order $p^{k}$ or higher. Now for the first sum, note that
 \[
 	\sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1|
     = \sum_{\xi_1\in\paths{b_1}} \sum_{\substack{\xi_2\in\paths{b_2}\\ \text{independent of }\xi_1}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1|
 \]
-where the sum over independent paths could be empty.
+where the sum over independent paths could be empty for certain $\xi_1$. Now we replace this last sum by a sum over \emph{all} paths $\xi_2\in\paths{b_2}$. This will change the sum but only for terms where $\xi_1,\xi_2$ are dependent. For those terms we already know that $\mathbb{P}[\xi_1]\mathbb{P}[\xi_2]$ contains a factor $p^k$ and hence we have 
+\begin{align*}
+    \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1|
+    &= \sum_{\xi_1\in\paths{b_1}} \sum_{\xi_2\in\paths{b_2}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1| + \mathcal{O}(p^k) \\
+    &= \sum_{\xi_1\in\paths{b_1}} \mathbb{P}[\xi_1]|\xi_1| + \mathcal{O}(p^k) \\
+    &= R_{b_1} + \mathcal{O}(p^k)
+\end{align*}
+we can do the same with the second term and this proves the claim.
 \end{proof}
-
 \begin{center}
 \includegraphics{diagram_paths.pdf}
 \end{center}
@@ -566,4 +572,4 @@ where we used the identity $\sum_{a\in\{0,1\}^l} (-1)^{|a|} = 0$.
 	\bibliographystyle{alpha}
 	\bibliography{Resample.bib}
 	
-\end{document}
\ No newline at end of file
+\end{document}
