diff --git a/thesis/Abschlussarbeit.tex b/thesis/Abschlussarbeit.tex
index a3e10d2..47eb640 100644
--- a/thesis/Abschlussarbeit.tex
+++ b/thesis/Abschlussarbeit.tex
@@ -342,11 +342,12 @@ The \sdlog game is a variant of the discrete logarithm game which represents the
 The adversary has to call the \ioracle oracle with a commitment $\groupelement{R}$ to get a challenge from the challenger. Due to the nature of Algebraic Group Model the adversary also has to provide a representation of the group element $\groupelement{R}$, as the linear combination of all known group elements. Since only the generator of the group and the public key are known to the adversary the representation looks like this $\groupelement{R} = r_1 \groupelement{B} + r_2 \groupelement{A}$. Upon providing a valid solution for the game \igame the reduction also gains following equation $\groupelement{R} = 2^c s \groupelement{B} - 2^c c \groupelement{A}$. Both equations yield:
 
 % TODO: collision von cofactor c und challenge c
-\begin{align*}
+\begin{align}
 		r_1 \groupelement{B} + r_2 \groupelement{A} &= 2^c s \groupelement{B} - 2^c c \groupelement{A} \\
 		(r_2 + 2^c c)A &= (2^c s - r_1)B \\
 		A &= (2^c s - r_1)(r_2 + 2^c c)^{-1} B
-\end{align*}
+		\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.
 
@@ -418,9 +419,49 @@ TODO: vielleicht doch eher $\oraclequeries$  durch min entropy von $\{0,1\}^{2b}
 	
 	\item Finally, Game $G_2$ is well prepared to show that there exists an adversary $\adversary{B}$ satisfying
 	
-	\[ \Pr[G_2^{\adversary{A}} \Rightarrow 1] = \advantage{\sdlog}{\adversary{B}} \]
+	\begin{align}
+		\Pr[G_2^{\adversary{A}} \Rightarrow 1] = \advantage{\sdlog}{\adversary{B}} \label{eq:advbsdlog}
+	\end{align}
 	
-	\item This proofs Theorem \ref{theorem:advgamez}.
+	\begin{figure}
+		\hrule
+		\begin{multicols}{2}
+			\large
+			\begin{algorithmic}
+				\State \underline{Adversary $\adversary{B}(\groupelement{A})$}
+				\State $(s^*) \randomassign \adversary{A}^{\ioracle(\inp)}(\groupelement{A})$
+				\State $\exists \agmgroupelement{R^*}{r^*}, c^*: \groupelement{R}^* = 2^c (s^* \groupelement{B} - c^* \groupelement{A}) \wedge (\agmgroupelement{R^*}{r^*}, c^*) \in Q$
+				\State $R^* = r_1 \groupelement{B} + r_2 \groupelement{A}$
+				\State \Return $(2^c s^* - r_1)(r_2 + 2^c c^*)^{-1}$
+			\end{algorithmic}
+			\columnbreak
+			\begin{algorithmic}[1] 
+				\Procedure{\ioracle}{$\agmgroupelement{R_i}{r_i} \in \group{G}$}
+				\State $\groupelement{R}_i = r_1 \groupelement{B} + r_2 \groupelement{A}$
+				\State $c_i \randomsample \{0,1\}^{2b}$
+				\BeginBox[draw=blue]
+				\State \textbf{If} $2^c c_i = -r_2$ \textbf{then}
+				\State \text{	} $bad \assign true$
+				\BeginBox[draw=red,dashed]
+				\State \text{	} $abort$
+				\EndBox
+				\State \textbf{endIf}
+				\EndBox
+				\State $Q \assign Q \cup \{ (\agmgroupelement{R_i}{r_i}, c_i) \}$
+				\State \Return $c_i$
+				\EndProcedure
+			\end{algorithmic}
+		\end{multicols}
+		\hrule
+		\caption{Adversary $\adversary{B}$ breaking \sdlog}
+		\label{fig:adversarybsdlog}
+	\end{figure}
+	
+	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}).
+	
+	\item This proves Theorem \ref{theorem:advgamez}.
 \end{proof}
 
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