From 91ec89c0400d5e5d14d0251675b5efc114d34882 Mon Sep 17 00:00:00 2001
From: Aaron Kaiser <a_kaiser@posteo.de>
Date: Thu, 2 Mar 2023 10:46:40 +0100
Subject: [PATCH] fixed typo

---
 thesis/Abschlussarbeit.tex | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/thesis/Abschlussarbeit.tex b/thesis/Abschlussarbeit.tex
index 57dc9c1..9c7aea4 100644
--- a/thesis/Abschlussarbeit.tex
+++ b/thesis/Abschlussarbeit.tex
@@ -349,7 +349,7 @@ The adversary has to call the \ioracle oracle with a commitment $\groupelement{R
 		\label{eq:Adlog}
 \end{align}
 
-Assuming that $r_2 + 2^c c$ is invertable in $\field{L}$ (not equal to $0$) we can use both equations to calculate the discrete logarithm of $\groupelement{A}$. To ensure that $r_2 + 2^c c$ is invertable the reduction has to abort if $-r_2$ equals $2^c c$ with $c$ being randomly choosen in the \ioracle oracle.
+Assuming that $r_2 + 2^c c$ is invertible in $\field{L}$ (not equal to $0$) we can use both equations to calculate the discrete logarithm of $\groupelement{A}$. To ensure that $r_2 + 2^c c$ is invertible the reduction has to abort if $-r_2$ equals $2^c c$ with $c$ being randomly chosen in the \ioracle oracle.
 
 \begin{figure}
 	\hrule
@@ -452,7 +452,7 @@ Game $G_0$ is defined in Figure \ref{fig:igamewithabort} by ignoring all boxes.
 	
 	To prove (\ref{eq:advbsdlog}), we define an adversary $\adversary{B}$ attacking \sdlog that simulates $\adversary{A}$'s view on $G_2$. Adversary $\adversary{B}$ formally defined in figure \ref{fig:adversarybsdlog} is run in the \sdlog game and has access to \ioracle. \ioracle is perfectly simulated.
 	
-	Finally, consider $\adversary{A}$ output $s^*$. If bad is not set $r_2 + 2^c c^*$ is invertable in $\field{L}$ and $(2^c s^* - r_1)(r_2 + 2^c c^*)^{-1}$ is the discrete logarithm of $\adversary{A}$ to the basis of $\adversary{B}$ as shown in (\ref{eq:Adlog}).
+	Finally, consider $\adversary{A}$ output $s^*$. If bad is not set $r_2 + 2^c c^*$ is invertible in $\field{L}$ and $(2^c s^* - r_1)(r_2 + 2^c c^*)^{-1}$ is the discrete logarithm of $\adversary{A}$ to the basis of $\adversary{B}$ as shown in (\ref{eq:Adlog}).
 	
 	\item This proves Theorem \ref{theorem:advgamez}.
 \end{proof}