\subsection{Bounds on \somdl} \label{sec:somdl}

This section provides a lower bound on the hardness of the modified version of the one-more discrete logarithm problem in the generic group model. The variant of the one-more discrete logarithm problem was introduced in the definition \ref{def:somdl}. \somdl differs from the original one-more discrete logarithm problem by only allowing the adversary to query the discrete logarithm of all challenges but one. Also the discrete logarithms of the group elements in the challenge to the adversary are chosen from a predefined set that is the result of the special key generation algorithm used in EdDSA. The following proof uses the generic group model for twisted Edwards curves. There already exists a proof for the one-more discrete logarithm problem in the generic group model \cite{EPRINT:BauFucPlo21}. This proof provides a lower bound on the original definition of the one-more discrete logarithm problem. This proof is not directly applicable to this definition of \somdl, since the secret scalars are not chosen uniformly at random from $\field{L}$ and the group structure is not just a prime order group. Since a more restricted version of the one-more discrete logarithm problem is used a simpler proof than that in \cite{EPRINT:BauFucPlo21} can be used, providing a better bound on \somdl.

\begin{theorem}
	\label{theorem:somdl_ggm}
	Let $n$, $N$, $c$ be positive integers. Consider a twisted Edwards curve $\curve$ with a cofactor of $2^c$ and a generating set consisting of $(\groupelement{B}, \groupelement{E_2}, ..., \groupelement{E_m})$. Among these, let $\groupelement{B}$ be the generator of the largest prime order subgroup with an order of $L$. Let $\adversary{A}$ be a generic adversary against \somdl receiving $N$ group elements as challenge and making at most $\groupqueries$ group operations queries. Then,
	
	\[ \advantage{\curve, n, c, L, \adversary{A}}{\somdl} \leq \frac{2(\groupqueries + N + 2)^2 + 1}{2^{n-1-c}}. \]
\end{theorem}

\paragraph{\underline{Proof Overview}} This proof uses the same approach as the discrete logarithm proof in the generic group model by replacing the group elements with polynomials and choosing the challenge after the adversary provided its solution. The tricky part is that the adversary is able to query the discrete logarithms of all but one of the $N$ group elements, provided to it as a challenge. The proof starts by replacing all group elements with multivariate polynomials representing their discrete logarithms. The indeterminants of those polynomials are the discrete logarithms of each group element, provided to the adversary as challenges. Once the adversary requests the discrete logarithms for all but one group element of the challenge those discrete logarithms are chosen uniformly at random and all polynomials are partially evaluated. This leaves polynomials with just one indeterminate, representing the discrete logarithm of the last challenge. This challenge is then chosen after the adversary provided its solution, leaving the adversay no option but to guess the remaining discrete logarithm. The \somdl game in the generic group model is depicted in figure \ref{fig:somdl_ggm}.

\begin{figure}[h]
	\hrule
	\vspace{2mm}
	\begin{algorithmic}
		\Statex \underline{\game \somdl}
		\State \textbf{for} $i \in \{1,2,...,N\}$
		\State \quad $a_i \randomsample \{ 2^{n-1}, 2^{n-1} + 2^c, ..., 2^{n} - 2^c \}$
		\State \quad $\groupelement{A_i} \assign a_i \groupelement{B}$
		\State $(a'_1, a'_2, ..., a'_N) \randomassign \adversary{A}^{GOp(\inp, \inp, \inp), \textit{Reveal}(\inp)}(Enc(\groupelement{B}), Enc(\groupelement{E_2}), ..., Enc(\groupelement{E_m}), Enc(\groupelement{A_1}), ..., Enc(\groupelement{A_N}))$
		\State \Return $(a_1, a_2, ..., a_N) \test (a'_1, a'_2, ..., a'_N)$
	\end{algorithmic}
	\vspace{1mm}
	\begin{algorithmic}
		\Statex \underline{\oracle \textit{Reveal}($j \in \{1,2,...,N\}$)}
		\Comment{max. one query}
		\State \Return $\{a_i | i \in \{1,2,...,N\} \backslash \{j\}\}$
	\end{algorithmic}
	\vspace{1mm}
	\begin{algorithmic}
		\Statex \underline{\oracle GOp($x, y \in \mathbf{S}, b \in \{0,1\}$)}
		\State \quad \Return $Enc(\sum^{-1}[x] + (-1)^b \sum^{-1}[y])$
	\end{algorithmic}
	\vspace{1mm}
	\begin{algorithmic}
		\Statex \underline{\textbf{Procedure} Enc($\groupelement{X} \in \curve$)}
		\vspace{1mm}
		\State \textbf{If } $\sum[\groupelement{X}] = \bot$ \textbf{ then}
		\State \quad $\sum[\groupelement{X}] \randomsample \{0,1\}^{\lceil log_2(|\curve|) \rceil} \backslash \pset{S}$
		\State \quad $\mathbf{S} \assign \pset{S} \cup \{\sum[X]\}$
		\State \Return $\sum[\groupelement{X}]$
	\end{algorithmic}
	\hrule
	\caption{\somdl in the generic group model}
	\label{fig:somdl_ggm}
\end{figure}

\paragraph{\underline{Formal Proof}}

\begin{figure}[H]
	\hrule
	\vspace{2mm}
	\begin{algorithmic}
		\Statex \underline{\game \textcolor{black}{$G_0$} / \textcolor{blue}{$G_1$} /\textcolor{red}{$G_2$} / \textcolor{green}{$G_3$} / \textcolor{orange}{$G_4$}}
		\State \textbf{for} $i \in \{1,2,...,N\}$
		\BeginBox[draw=black]
		\State \quad $a_i \randomsample \{ 2^{n-1}, 2^{n-1} + 2^c, ..., 2^{n} - 2^c \}$
		\Comment{$G_0 - G_4$}
		\EndBox
		\BeginBox[draw=black]
		\State \quad $\groupelement{A_i} \assign a_i \groupelement{B}$
		\Comment{$G_0$}
		\EndBox
		\BeginBox[draw=blue]
		\State \quad $\groupelement{A_i} \assign (a_i, 0, ..., 0)$
		\Comment{$G_1$}
		\EndBox
		\BeginBox[draw=red]
		\State \quad $P_i \assign Z_i$
		\Comment{$G_2 - G_4$}
		\State \quad $\groupelement{A_i} \assign (P_i, 0, ..., 0)$
		\EndBox
		\State $(a'_1, a'_2, ..., a'_N) \randomassign \adversary{A}^{GOp(\inp, \inp, \inp), \textit{Reveal}(\inp)}(Enc(\groupelement{B}), Enc(\groupelement{E_2}), ..., Enc(\groupelement{E_m}), Enc(\groupelement{A_1}), ..., Enc(\groupelement{A_N}))$
		\State \Return $(a_1, a_2, ..., a_N) \test (a'_1, a'_2, ..., a'_N)$
	\end{algorithmic}
	\vspace{1mm}
	\begin{algorithmic}
		\Statex \underline{\oracle \textit{Reveal}($j \in \{1,2,...,N\}$)}
		\BeginBox[draw=green]
		\State \textbf{for } $P_i \in \pset{P}$
		\Comment{$G_3 - G_4$}
		\State \quad Let $P_i = R_i + S_i, R_i \in \field{L}[Z_1,...,Z_{j-1},Z_{j+1},...,Z_N], S_i \in \field{L}[Z_j]$
		\State \quad $\pset{R} \assign \pset{R} \cup \{R_i\}$
		\State \textbf{if } $\exists R_i, R_j \in \pset{R}: R_i(\overset{\rightharpoonup}{a}) = R_j(\overset{\rightharpoonup}{a}) \wedge R_i \neq R_j$
		\State \quad $bad_1 \assign true$
		\BeginBox[draw=orange,dashed]
		\State \quad abort
		\Comment{$G_4$}
		\EndBox
		\State \textbf{for } $P_i \in \pset{P}$
		\State \quad $\sum[R_i(\overset{\rightharpoonup}{a}) + S_i] = \sum[P_i]$
		\State \quad $P_i \assign R_i(\overset{\rightharpoonup}{a}) + S_i$
		\EndBox
		\State \Return $\{a_i | i \in \{1,2,...,N\} \backslash \{j\}\}$
	\end{algorithmic}
	\vspace{1mm}
	\begin{algorithmic}
		\Statex \underline{\oracle GOp($x, y \in \mathbf{S}, b \in \{0,1\}$)}
		\State \quad \Return $Enc(\sum^{-1}[x] + (-1)^b \sum^{-1}[y])$
	\end{algorithmic}
	\vspace{1mm}
	\begin{algorithmic}
		\BeginBox[draw=black]
		\State \underline{\textbf{Procedure} Enc($\groupelement{X} \in \curve$)}
		\Comment{$G_0$}
		\EndBox
		\BeginBox[draw=blue]
		\State \underline{\textbf{Procedure} Enc($\groupelement{X} \in \field{L} \times \field{ord(E_2)} \times ... \times \field{ord(E_n)}$)}
		\Comment{$G_1$}
		\EndBox
		\BeginBox[draw=red]
		\State \underline{\textbf{Procedure} Enc($\groupelement{X} \in \field{L}[Z_1,...,Z_N] \times \field{ord(E_2)} \times ... \times \field{ord(E_n)}$)}
		\Comment{$G_2 - G_4$}
		\State Let $X = (P, x_2, ..., x_n)$
		\State $\pset{P} = \pset{P} \cup \{P\}$
		\State $X \assign (P(\overset{\rightharpoonup}{a}), x_2, ..., x_n)$
		\EndBox
		\State \textbf{If } $\sum[\groupelement{X}] = \bot$ \textbf{ then}
		\State \quad $\sum[\groupelement{X}] \randomsample \{0,1\}^{\lceil log_2(|\curve|) \rceil} \backslash \pset{S}$
		\State \quad $\mathbf{S} \assign \pset{S} \cup \{\sum[X]\}$
		\State \Return $\sum[\groupelement{X}]$
	\end{algorithmic}
	\hrule
	\caption{$G_0 - G_4$}
	\label{fig:somdl_games_ggm_1}
\end{figure}

\begin{figure}[H]
	\hrule
	\vspace{2mm}
	\begin{algorithmic}
		\Statex \underline{\game \textcolor{black}{$G_4$} / \textcolor{blue}{$G_5$} /\textcolor{red}{$G_6$} / \textcolor{green}{$G_7$} / \textcolor{orange}{$G_8$}}
		\State \textbf{for} $i \in \{1,2,...,N\}$
		\BeginBox[draw=black]
		\State \quad $a_i \randomsample \{ 2^{n-1}, 2^{n-1} + 2^c, ..., 2^{n} - 2^c \}$
		\Comment{$G_4 - G_7$}
		\EndBox
		\State \quad $P_i \assign Z_i$
		\State \quad $\groupelement{A_i} \assign (P_i, 0, ..., 0)$
		\State $(a'_1, a'_2, ..., a'_N) \randomassign \adversary{A}^{GOp(\inp, \inp, \inp), \textit{Reveal}(\inp)}(Enc(\groupelement{B}), Enc(\groupelement{E_2}), ..., Enc(\groupelement{E_m}), Enc(\groupelement{A_1}), ..., Enc(\groupelement{A_N}))$
		\BeginBox[draw=orange]
		\State \textbf{for } $i \in \{1,2,...,N\}$
		\Comment{$G_8$}
		\State \quad \textbf{if } $a_i = \bot$
		\State \qquad $a_i \randomsample \{ 2^{n-1}, 2^{n-1} + 2^c, ..., 2^{n} - 2^c \}$
		\EndBox
		\BeginBox[draw=blue]
		\State \textbf{if } $\exists P_i, P_j \in \pset{P}: P_i(\overset{\rightharpoonup}{a}) = P_j(\overset{\rightharpoonup}{a}) \wedge P_i \neq P_j$
		\Comment{$G_5 - G_8$}
		\State \quad $bad_2 \assign true$
		\BeginBox[draw=red,dashed]
		\State \quad abort
		\Comment{$G_6 - G_8$}
		\EndBox
		\EndBox
		\State \Return $(a_1, a_2, ..., a_N) \test (a'_1, a'_2, ..., a'_N)$
	\end{algorithmic}
	\vspace{1mm}
	\begin{algorithmic}
		\Statex \underline{\oracle \textit{Reveal}($j \in \{1,2,...,N\}$)}
		\BeginBox[draw=orange]
		\State \textbf{for } $i \in \{1,2,...,N\} \backslash \{j\}$
		\Comment{$G_8$}
		\State \quad $a_i \randomsample \{ 2^{n-1}, 2^{n-1} + 2^c, ..., 2^{n} - 2^c \}$
		\EndBox 
		\State \textbf{for } $P_i \in \pset{P}$
		\State \quad Let $P_i = R_i + S_i, R_i \in \field{L}[Z_1,...,Z_{j-1},Z_{j+1},...,Z_N], S_i \in \field{L}[Z_j]$
		\State \quad $\pset{R} \assign \pset{R} \cup \{R_i\}$
		\State \textbf{if } $\exists R_i, R_j \in \pset{R}: R_i(\overset{\rightharpoonup}{a}) = R_j(\overset{\rightharpoonup}{a}) \wedge R_i \neq R_j$
		\State \quad $bad_1 \assign true$
		\State \quad abort
		\State \textbf{for } $P_i \in \pset{P}$
		\State \quad $\sum[R_i(\overset{\rightharpoonup}{a}) + S_i] = \sum[P_i]$
		\State \quad $P_i \assign R_i(\overset{\rightharpoonup}{a}) + S_i$
		\State \Return $\{a_i | i \in \{1,2,...,N\} \backslash \{j\}\}$
	\end{algorithmic}
	\vspace{1mm}
	\begin{algorithmic}
		\Statex \underline{\oracle GOp($x, y \in \mathbf{S}, b \in \{0,1\}$)}
		\State \quad \Return $Enc(\sum^{-1}[x] + (-1)^b \sum^{-1}[y])$
	\end{algorithmic}
	\vspace{1mm}
	\begin{algorithmic}
		\State \underline{\textbf{Procedure} Enc($\groupelement{X} \in \field{L}[Z_1,...,Z_N] \times \field{ord(E_2)} \times ... \times \field{ord(E_n)}$)}
		\State Let $X = (P, x_2, ..., x_n)$
		\State $\pset{P} = \pset{P} \cup \{P\}$
		\BeginBox[draw=black]
		\State $X \assign (P(\overset{\rightharpoonup}{a}), x_2, ..., x_n)$
		\Comment{$G_4 - G_6$}
		\EndBox
		\State \textbf{If } $\sum[\groupelement{X}] = \bot$ \textbf{ then}
		\State \quad $\sum[\groupelement{X}] \randomsample \{0,1\}^{\lceil log_2(|\curve|) \rceil} \backslash \pset{S}$
		\State \quad $\mathbf{S} \assign \pset{S} \cup \{\sum[X]\}$
		\State \Return $\sum[\groupelement{X}]$
	\end{algorithmic}
	\hrule
	\caption{$G_4 - G_8$}
	\label{fig:somdl_games_ggm_2}
\end{figure}

\begin{proof}
	\item The proof starts by replacing group elements with polynomials. This happens in games $G_1$ and $G_2$. After that it is argued that the challenger makes a mistake in its simulation with only negligible probability by comparing polynomials directly instead of evaluating them. This is shown in $G_3 - G_6$. At last, since the polynomials are not evaluated during the simulation, one discrete logarithm is not used before the adversary provides its solution. Therefore, it can be chosen after the adversary provided its solution, which is shown in $G_7$ and $G_8$.
	
	\item \paragraph{\underline{$G_0:$}} Let $G_0$ be depicted in figure \ref{fig:somdl_games_ggm_1} by excluding all boxes but the black ones. Clearly, this is equivalent to the \somdl game in the generic group model. Therefore, 
	
	\[ \advantage{\curve, n, c, L, \adversary{A}}{\somdl} = \prone{G_0^{\adversary{A}}}. \]
	
	\item \paragraph{\underline{$G_1:$}} $G_1$ now replaces the group elements in the challenger with their discrete logarithms. This change is purely conceptual, since the adversary only sees the labels of the group elements, and each group element can be uniquely identified by its discrete logarithm. As in the \sdlog proof, the discrete logarithm of a group element is denoted by an integer vector, where each element in the vector represents the discrete logarithm with respect to a generator from the generating set. For this reason,
	
	\[ \prone{G_0^{\adversary{A}}} = \prone{G_1^{\adversary{A}}}. \]
	
	\item \paragraph{\underline{$G_2:$}} $G_2$ replaces the blue boxes with the red ones. This change affects the discrete logarithm of the group elements in the prime order subgroup. The discrete logarithm is now represented as a multivariate polynomial. Each indeterminate of the polynomial represents the discrete logarithm of one of the group elements in the challenge to the adversary. The discrete logarithm of the group element in the challenge to the adversary is then instantiated with the indeterminate representing the discrete logarithm of that challenge, instead of the discrete logarithm itself. This change is only conceptual, since the polynomials are evaluated with the discrete logarithm vector of the group elements in the challenge before being compared in the Enc procedure. Hence,
	
	\[ \prone{G_1^{\adversary{A}}} = \prone{G_2^{\adversary{A}}}. \]
	
	\item \paragraph{\underline{$G_3:$}} $G_3$ introduces the $bad_1$ flag in the \textit{Reveal} query. Without loss of generality the following explanation assumes that the adversary queries the \textit{Reveal} oracle with input $j = N$. Each polynomial, generated by the challenger, is a linear multivariate polynomial of degree one. This is due to the fact that the challenger starts with linear multivariate polynomials of degree one in $\field{L}[Z_1, ..., Z_N]$ and only adds them to generate new polynomials. This means that each polynomial $P_i \in \field{L}[Z_1,...,Z_N]$, generated by the challenger, can be split into two polynomials $R_i \in \field{L}[Z_1,...,Z_{N-1}], S_i \in \field{L}[Z_N]$ so that $P_i = R_i + S_i$, simply by distributing the monials between the polynomials $R_i$ and $S_i$. The polynom $S_i$ only contains the monial $Z_n$, while the polynom $R_i$ contains the remaining monials and the constant. Now the polynomial $P_i$ can be partially evaluated by setting $P_i = R_i(\overset{\rightharpoonup}{a}) + S_i$. For the simulation to be correct, when replacing the polynomial $P_i$ with $R_i(\overset{\rightharpoonup}{a}) + S_i$, it has to be ensured that distinct polynomials stay distinct after being partially evaluated. To ensure this, it is necessary to check that no two distinct polynomials $R_i, R_j$ result in the same value when evaluated with $\overset{\rightharpoonup}{a}$. In the case of this happening the $bad_1$ flag is set to true. Afterward, each generated polynomial is partially evaluated as described and the table $\sum$, which stores the association between group elements and labels, is updated to reflect this partial evaluation as well. From now on, each polynomial used by the challenger is in $\field{L}[Z_N]$. This change is purely conceptual, since the polynomials still get fully evaluated before being compared in the Enc procedure. Therefore,
	
	\[ \prone{G_2^{\adversary{A}}} = \prone{G_3^{\adversary{A}}}. \]
	
	\item \paragraph{\underline{$G_4:$}} In $G_4$ the abort instruction in the orange box is introduced, which is executed after the $bad_1$ flag is set. The $bad_1$ flag is set if distinct polynomials result in the same polynomial, after being partially evaluated. To calculate the probability of this happening the Schwart-Zippel lemma can be utilized. For every $R_i, R_j \in \pset{R} \wedge R_i \neq R_j$ a polynomial $R^* \assign R_i - R_j$ can be constructed. If and only if $R_i(\overset{\rightharpoonup}{a}) = R_j(\overset{\rightharpoonup}{a})$ then $R^*(\overset{\rightharpoonup}{a}) = 0$. Since $R^* \neq 0$, the degree of $R^*$ being $1$ and $\overset{\rightharpoonup}{a}$ being chosen uniformly at random from $\{2^{n-1}, 2^{n-1} + 2^c, ..., 2^{n} - 2^c\}$ the Schwartz-Zippel lemma can be used to calculate the probability of $R^*(\overset{\rightharpoonup}{a}) = 0$, which is $\Pr[R^*(\overset{\rightharpoonup}{a}) = 0] \leq \frac{1}{2^{n - 1 - c}}$. The challenger can generate at most $\groupqueries + N + 2$ many polynomials, one per group operation query GOp and $N + 2$ for encoding the input to the adversary. By the Union bound over all $(\groupqueries + N + 2)^2$ possible pairs of polynomials an upper bound on the $bad_1$ flag being set can be calculated as $\Pr[bad_1] \leq \frac{(\groupqueries + N + 2)^2}{2^{n - 1 - c}}$. Since $G_3$ and $G_4$ are identical-until-bad games,
	
	\[ |\prone{G_3^{\adversary{A}}} - \prone{G_4^{\adversary{A}}}| \leq \frac{(\groupqueries + N + 2)^2}{2^{n - 1 - c}}. \]
	
	To improve the readability, $G_4$ is also depicted in figure \ref{fig:sdlog_games_ggm_2} by only including the black boxes. The following game-hops are illustrated in the same figure.
	
	\item \paragraph{\underline{$G_5:$}} $G_5$ introduces the check in the blue box. This check ensures that after the adversary provided its solution no two distinct polynomials where generated by the challenger that evaluate to the same value, when evaluated with the vector of discrete logarithms. If this happens the $bad_2$ flag is set. This change is only conceptual, as it only changes internal variables, which have no effect on the behavior of the challenger. Hence,
	
	\[ \prone{G_4^{\adversary{A}}} = \prone{G_5^{\adversary{A}}}. \]
	
	\item \paragraph{\underline{$G_6:$}} $G_6$ aborts if the $bad_2$ flag is set. The $bad_2$ flag is set if any two distinct polynomials evaluate to the same value, when evaluated with the vector of discrete logarithms. There are two cases. The first case is that the adversary has queried the \textit{Reveal} oracle. The second case is that the adversary did not queried the \textit{Reveal} oracle.
	
	In the first case the adversary got the discrete logarithms of all but one challenge. Without loss of generality it is assumed that the adversary queried the discrete logarithm of all but the $N$th group element. In this case all polynomials in $\pset{P}$ are in $\field{L}[Z_N]$, since at the time of the \textit{Reveal} query all polynomials, generated up to this point, are partially evaluated and are in $\field{Z}[Z_N]$. All polynomials that are generated after this point are generated by the addition of the existing polynomials and are therefore also in $\field{L}[Z_N]$. In this case the Schwartz-Zippel lemma can be applied since the adversary has no information on the remaining discrete logarithm. This is the same scenario as in the \sdlog proof.
	
	In the case where the adversary did not queried the \textit{Reveal} oracle the adversary has no information on any of the discrete logarithms. All polynomials in $\pset{P}$ are in $\field{Z}[N_1, ..., Z_N]$. In this case the Schwartz-Zippel lemma can be applied, since the all discrete logarithms are chosen uniformly at random and the adversary has no information on them, prior to them being chosen.
	
	The probability of $bad_2$ being true can be calculated using the Schwartz-Zippel lemma, as described in the game-hop to $G_4$. With the Union bound over all polynomial pairs in $\pset{P}$ the probability of $bad_2$ being true is $\Pr[bad_2] \leq \frac{(\groupqueries + N + 2)^2}{2^{n - 1 - c}}$. $G_5$ and $G_6$ are identical-until-bad games, therefore:

	\[ |\prone{G_5^{\adversary{A}}} - \prone{G_6^{\adversary{A}}}| \leq \frac{(\groupqueries + N + 2)^2}{2^{n - 1 - c}}. \]
	
	\item \paragraph{\underline{$G_7:$}} $G_7$ removes the evaluation of polynomials in the Enc procedure. It is argued that this change is only conceptual. When the evaluation of polynomials is removed, the polynomials are compared directly. Group elements represented by different polynomials are assigned different labels by the challenger. This is equivalent to the original definition as long as different polynomials do not evaluate to the same value, when evaluated with the discrete logarithms. This inconsistency in the simulation can be detected by the adversary when it gets some information on the discrete logarithms. This can either be during the query to the \textit{Reveal} oracle or after the adversary provided its solution. In both cases there is an if condition checking for this inconsistency. If such an inconsistency is detected the game aborts. This change is only conceptual, since the different polynomials correspond to different group elements, in the cases where the game does not abort, and since the adversary only sees the labels it cannot detect whether the challenger works with polynomials or concrete discrete logarithms. Hence,
	
	\[ \prone{G_6^{\adversary{A}}} = \prone{G_7^{\adversary{A}}}. \]
	
	\item \paragraph{\underline{$G_8:$}} In $G_8$ the discrete logarithms of the challenge are only generated right before they are used. Since the discrete logarithms are not used during the Enc function anymore they the challenger can generate them not at the start of the game but only right before they are used. The discrete logarithms are only used during the inconsistency checks in the \textit{Reveal} oracle or after the adversary has provided its solution. $N - 1$ discrete logarithms are used in the \textit{Reveal} oracle to check for inconsistencies and to partially evaluate the polynomials. After the adversary provided its solution the remaining discrete logarithms can chosen to fully evaluate all polynomials. This can be either all discrete logarithm, in the case that the adversary did not queried the \textit{Reveal} oracle, or the remaining one, in the case that the adversary did queried the \textit{Reveal} oracle. This change is only conceptual, since the initialization of variables is only moved right before the variable is used. Therefore,
	
	\[ \prone{G_7^{\adversary{A}}} = \prone{G_8^{\adversary{A}}}. \]
	
	\item Since at least one discrete logarithm is chosen after the adversary provided its solution, its only chance is to  guess it. Therefore, the probability of the adversary of winning $G_7$ is upper bounded by the probability of it guessing that discrete logarithm. Hence,
	
	\[ \prone{G_8^{\adversary{A}}} \leq \frac{1}{2^{n - 1 - c}}. \]
	
	\item This proves theorem \ref{theorem:somdl_ggm}.
\end{proof}