diff --git a/ACNS/presentation/main.tex b/ACNS/presentation/main.tex
index eca90406eb915b0fbfeb31b54e5068fea7c7383b..dabf6c180129f01a21af508f654e4a392513638d 100644
--- a/ACNS/presentation/main.tex
+++ b/ACNS/presentation/main.tex
@@ -8,12 +8,12 @@
 \title{On Actively Secure Fine-Grained Access Structures from Isogeny Assumptions}
 \author{Fabio Campos\inst{1,2} \and \underline{Philipp Muth}\inst{3}}
 \institute{\relax
-  RheinMain University of Applied Sciences, Wiesbaden, Germany
+  $^1$RheinMain University of Applied Sciences, Wiesbaden, Germany
   \and
-  Radboud University, Nijmegen, The Netherlands %\\
+  $^2$Radboud University, Nijmegen, The Netherlands %\\
   %\email{campos@sopmac.de}
   \and
-  Technische Universität Darmstadt, Germany%\\
+  $^3$Technische Universität Darmstadt, Germany%\\
   %\email{philipp.muth@tu-darmstadt.de}
 }
 
diff --git a/ACNS/presentation/motivation.tex b/ACNS/presentation/motivation.tex
index 0eaf1219e439dc8a65495f1f20c1da71d79e002a..11cb711251338178c16b540ebde5bfa233b54bb8 100644
--- a/ACNS/presentation/motivation.tex
+++ b/ACNS/presentation/motivation.tex
@@ -27,10 +27,10 @@
 	\end{block}
 	\begin{remark}
 		For \(s,s'\in \Z_p\) and \(E\in\mathcal E\), we have
-		\[[s] \left(\left[s'\right] E\right) = \left[s+s'\right] E.\]
+		\[\left[s\right] \left(\left[s'\right] E\right) = \left[s+s'\right] E.\]
 	\end{remark}
 
-	\begin{block}{The Group Action Inverse Problem}
+	\begin{block}{The Group Action Inverse Problem (GAIP)}
 		Given two elements \(E,E' \in \mathcal E\), find \(g\in \mathcal G\) with
 		\[g\ast E = E'.\]
 	\end{block}
@@ -108,10 +108,10 @@
 		\draw [->] (cipher) -- (sh1);
 
 		\pause
-		\draw [->, >=Stealth, bend right] (sh1) edge node [midway, above] {$ E^1 = \left[L_{1} s_1\right] E_0$} (sh2) ;
-		\draw [->, >=Stealth, bend right] (sh2) edge node [midway, left] {$ E^2 = \left[L_{2} s_2\right] E_1$} (sh3) ;
-		\draw [->, >=Stealth, bend right] (sh3) edge node [midway, below] {$ E^3 = \left[L_{3} s_3\right] E_2$} (sh4) ;
-		\node [right = of sh4] (key) {$\key = \left[L_4 s_4\right] E^3$};
+		\draw [->, >=Stealth, bend right] (sh1) edge node [midway, above] {$ E^1 = \left[L_{1,S'} s_1\right] E_0$} (sh2) ;
+		\draw [->, >=Stealth, bend right] (sh2) edge node [midway, left] {$ E^2 = \left[L_{2,S'} s_2\right] E_1$} (sh3) ;
+		\draw [->, >=Stealth, bend right] (sh3) edge node [midway, below] {$ E^3 = \left[L_{3,S'} s_3\right] E_2$} (sh4) ;
+		\node [right = of sh4] (key) {$\key = \left[L_{4,S'} s_4\right] E^3$};
 		\draw [->] (sh4) -- (key);
 
 	\end{tikzpicture}
@@ -154,10 +154,10 @@
 		\draw [->] (cipher) -- (sh1);
 
 		%\pause
-		\draw [->, >=Stealth, bend right] (sh1) edge node [midway, above] {$ E^1 = \left[L_{1} s_1\right] E_0$} (sh2) ;
-		\draw [->, >=Stealth, bend right] (sh2) edge node [midway, left, color = red] {$ {E^2} \neq \left[L_{2} s_2\right] E_1$} (sh3) ;
-		\draw [->, >=Stealth, bend right] (sh3) edge node [midway, below] {$ E^3 = \left[L_{3} s_3\right] E_2$} (sh4) ;
-		\node [right = of sh4] (key) {$\key = \left[L_4 s_4\right] E^3$};
+		\draw [->, >=Stealth, bend right] (sh1) edge node [midway, above] {$ E^1 = \left[L_{1,S'} s_1\right] E_0$} (sh2) ;
+		\draw [->, >=Stealth, bend right] (sh2) edge node [midway, left, color = red] {$ {E^2} \neq \left[L_{2,S'} s_2\right] E_1$} (sh3) ;
+		\draw [->, >=Stealth, bend right] (sh3) edge node [midway, below] {$ E^3 = \left[L_{3,S'} s_3\right] E_2$} (sh4) ;
+		\node [right = of sh4] (key) {$\key = \left[L_{4,S'} s_4\right] E^3$};
 		\draw [->] (sh4) -- (key);
 
 	\end{tikzpicture}