\subsection{MU-UF-NMA $\overset{\text{ROM}}{\Rightarrow}$ $\text{MU-SUF-CMA}_{\text{EdDSA sp}}$}

This section shows that the MU-UF-NMA security of the EdDSA signature scheme implies the MU-SUF-CMA security of the EdDSA signature scheme using the Random Oracle Model. The section starts with providing an intuition of the proof followed by the detailed security proof.

\begin{theorem}
	\label{theorem:adv_mu-uf-nma}
	Let $n$ and $N$ be positive integer and $\adversary{A}$ an adversary against MU-SUF-CMA, making at most $\hashqueries$ hash queries and $\oraclequeries$ oracle queries. Then,
	
	\[ \advantage{\adversary{A}}{\text{MU-\cma}}(\secparamter) \leq \advantage{\adversary{B}}{\text{MU-UF-NMA}}(\secparamter) + \frac{\oraclequeries \hashqueries}{2^{-\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}}. \]
\end{theorem}

\paragraph{\underline{Proof Overview}} This proof follows closely the proof in section \ref{proof:uf-nma_implies_suf-cma}. The only difference of both security notions is the missing \Osign oracle in MU-UF-NMA. For this reason the reduction has to simulate the \Osign oracle without the knowledge of the private keys.

Again the programmability of the random oracle together with the \simalg algorithm is used to generate valid signatures. The different games are depicted in figure \ref{fig:mu-uf-nma_implies_mu-suf-cma_games}.

\paragraph{\underline{Formal Proof}}

\begin{figure}
	\hrule
	\begin{multicols}{2}
		\large
		\begin{algorithmic}[1] 
			\Statex \underline{\game $G_0$ / \textcolor{blue}{$G_1$} / \textcolor{red}{$G_2$} / \textcolor{green}{$G_3$}}
			\State \textbf{for} $j \in \{1,2,...,N\}$
			\State \quad $(h_{j_0}, h_{j_1}, ..., h_{j_{2b-1}}) \randomsample \{0,1\}^{2b}$
			\State \quad $s_j \leftarrow 2^n + \sum_{i=c}^{n-1} 2^i h_{j_i}$
			\State \quad $\groupelement{A_j} \assign s_j \groupelement{B}$
			\State $(\m^*, \signature^*) \randomassign \adversary{A}^{H(\inp), \sign(\inp, \inp)}(\groupelement{A_1}, \groupelement{A_2},...,\groupelement{A_N})$
			\State \Return $\exists j \in \{1,2,...,N\}: \verify(\groupelement{A_j}, \m^*,\signature^*) \wedge (\groupelement{A_j}, \m^*, \signature^*) \notin \pset{Q}$
		\end{algorithmic}
		\columnbreak
		\begin{algorithmic}[1] 
			\Statex \underline{\oracle \sign($j \in \{1,2,...,N\}$, $\m \in \messagespace$)}
			\Comment{$G_0 - G_2$}
			\State $(r'_0, r'_1, ..., r'_{2b-1}) = RF(h_{j_b} | ... | h_{j_{2b-1}} | \m)$
			\State $r \assign \sum_{i=0}^{2b-1} 2^i r'_i$
			\State $R \assign rB$
			\BeginBox[draw=black]
			\State $S \assign (r + sH(\encoded{R} | \encoded{A_j} | \m)) \pmod L$
			\Comment{$G_0$}
			\EndBox
			\BeginBox[draw=blue]
			\State $\textbf{if } \sum[\encoded{R} | \encoded{A_j} | \m] \neq \bot \textbf{ then}$
			\Comment{$G_1 - G_2$}
			\State \quad $bad \assign true$
			\BeginBox[draw=red,dashed]
			\State \quad $abort$
			\Comment{$G_2$}
			\EndBox
			\State $\textbf{if } \sum[\encoded{R} | \encoded{A_j} | \m] = \bot \textbf{ then}$
			\State \quad $\sum[\encoded{R} | \encoded{A_j} | \m] \randomsample \{0,1\}^{2b}$
			\State $S \assign (r + s\sum[\encoded{R} | \encoded{A_j} | \m]) \pmod L$
			\EndBox
			\State $\signature \assign (\encoded{R}, S)$
			\State $\pset{Q} \assign \pset{Q} \cup \{(\groupelement{A_j}, \m, \signature)\}$
			\State \Return $\signature$
		\end{algorithmic}
	\end{multicols}
	\begin{multicols}{2}
		\begin{algorithmic}[1] 
			\Statex \underline{\oracle $H(\m \in \{0,1\}^*)$}
			\State $\textbf{if } \sum[\m] = \bot \textbf{ then}$
			\State \quad $\sum[\m] \randomsample \{0,1\}^{2b}$
			\State \Return $\sum[\m]$
		\end{algorithmic}
		\columnbreak
		\begin{algorithmic}[1]
			%TODO: Nummer vor Oracle
			\BeginBox[draw=green]
			\State \underline{\oracle \sign($j \in \{1,2,...,N\}$, $\m \in \messagespace$)}
			\Comment{$G_3$}
			\State $(R,\textbf{ch},S) \randomassign \simalg(\groupelement{A_j})$
			\State $\textbf{if } \sum[\encoded{R} | \encoded{A_j} | \m] \neq \bot \textbf{ then}$
			\State \quad $bad \assign true$
			\State \quad $abort$
			\State $\sum[\encoded{R} | \encoded{A_j} | \m] = \textbf{ch}$
			\State $\signature \assign (\encoded{R}, S)$
			\State $\pset{Q} \assign \pset{Q} \cup \{(\groupelement{A_j}, \m, \signature)\}$
			\State \Return $\signature$
			\EndBox
		\end{algorithmic}
	\end{multicols}
	\hrule
	\caption{Games $G_0 - G_3$}
	\label{fig:mu-uf-nma_implies_mu-suf-cma_games}
\end{figure}

\begin{proof}
	\item \paragraph{\underline{$G_0:$}} Let $G_0$ be defined in figure \ref{fig:mu-uf-nma_implies_mu-suf-cma_games} by excluding all boxes except the black one and $G_0$ be MU-SUF-CMA. By definition,
	
	\[ \advantage{\text{EdDSA},\adversary{A}}{\text{MU-}\cma}(\secparamter) = \Pr[\text{\text{MU-}\cma}^{\adversary{A}} \Rightarrow 1] = \Pr[G_0^{\adversary{A}} \Rightarrow 1]. \]
	
	\item \paragraph{\underline{$G_1:$}} $G_1$ now is defined by replacing the black box with the blue one. This change inlines the call to the hash function and introduces a bad flag, which is set if the hash value is already set. This change is only conceptual, since it does not alter the behavior of the oracle. Therefore,
	
	\[ \Pr[G_0^{\adversary{A}} \Rightarrow 1] = \Pr[G_1^{\adversary{A}} \Rightarrow 1]. \]
	
	\item \paragraph{\underline{$G_2:$}} $G_2$ is defined by also introducing the abort instruction in the red box. Again without loss of generality it is assumed that the adversary only quries each public key message pair only once since the signatures are deterministic and the attacker would not gain any additional information by querying the \Osign oracle multiple times with the same input. Since the commitment $\groupelement{R}$ is the only unknown input to the hash function the probability of the bad flag being set for each individual \Osign query is at most $\frac{\hashqueries}{2^{-\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}}$. By the Union bound over all oracle queries $\oraclequeries$ we obtain $\Pr[bad] \leq \frac{\oraclequeries \hashqueries}{2^{-\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}}$. Since $G_1$ and $G_2$ are identical-until-bad games, we have
	
	\[ |\Pr[G_1^{\adversary{A}} \Rightarrow 1] - \Pr[G_2^{\adversary{A}} \Rightarrow 1]| \leq \Pr[bad] \leq \frac{\oraclequeries \hashqueries}{2^{-\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}}. \]
	
	\item \paragraph{\underline{$G_3:$}} In $G_3$ the \Osign oracle is replaced by the \Osign oracle in the green box. Instead of calculating the response using the secret key the \simalg algorithm is used to generate a tuple of commitment, challenge and response. Then the random oracle is programmed to output the specific challenge given $\encoded{R} | \encoded{A_j} | \m$ as an input. This change is only conceptual, since \simalg outputs a correctly distributed set and it was ruled out in earlier games that the random oracle was previously queries with this input. Hence,
	
	\[ \Pr[G_2^{\adversary{A}} \Rightarrow 1] = \Pr[G_3^{\adversary{A}} \Rightarrow 1]. \]
	
	\item Finally, Game $G_3$ is well prepared to show that there exists an adversary $\adversary{B}$ satisfying
	
	\begin{align}
		\Pr[G_3^{\adversary{A}} \Rightarrow 1] = \advantage{\adversary{B}}{\text{MU-UF-NMA}}(\secparamter). \label{eq:adv_mu-uf-nma}
	\end{align}

	\begin{figure}
		\hrule
		\begin{multicols}{2}
			\large
			\begin{algorithmic}[1]
				\Statex \underline{\textbf{Adversary} $\adversary{B}^{H(\inp)}(\groupelement{A_1}, \groupelement{A_2}, ..., \groupelement{A_N})$}
				\State $(\m^*, \signature^*) \randomassign \adversary{A}^{H'(\inp), \sign(\inp, \inp)}(\groupelement{A_1}, \groupelement{A_2}, ..., \groupelement{A_N})$
				\State \Return $(\m^*, \signature^*)$
			\end{algorithmic}
			\columnbreak
			\begin{algorithmic}[1] 
				\Statex \underline{\oracle \sign($j \in \{1,2,...,N\}$, $\m \in \messagespace$)}
				\State $(R,\textbf{ch},S) \randomassign \simalg(\groupelement{A_j})$
				\State $\textbf{if } \sum[\encoded{R} | \encoded{A_j} | m] \neq \bot \textbf{ then}$
				\State \quad $bad \assign true$
				\State \quad $abort$
				\State $\sum[\encoded{R} | \encoded{A_j} | m] = \textbf{ch}$
				\State $\signature \assign (\encoded{R}, S)$
				\State $\pset{Q} \assign \pset{Q} \cup \{(\groupelement{A_j}, \m, \signature)\}$
				\State \Return $\signature$
			\end{algorithmic}
		\end{multicols}
		\begin{algorithmic}[1] 
			\Statex \underline{\oracle $H'(m \in \{0,1\}^*)$}
			\State $\textbf{if } \sum[m] = \bot \textbf{ then}$
			\State \quad $\sum[m] \assign H(m)$
			\State \Return $\sum[m]$
		\end{algorithmic}
		\hrule
		\caption{Adversary $\adversary{B}$ breaking $\text{MU-UF-NMA}$}
		\label{fig:adversaryb_mu-uf-nma}
	\end{figure}
	
	To prove (\ref{eq:adv_mu-uf-nma}), we define an adversary $\adversary{B}$ attacking $\text{MU-UF-NMA}$ that simulates $\adversary{A}$'s view in $G_2$. Adversary $\adversary{B}$ formally defined in figure \ref{fig:adversaryb_mu-uf-nma} is run in the $\text{MU-UF-NMA}$ game and adversary $\adversary{B}$ simulates \Osign for adversary $\adversary{A}$. \Osign is simulated perfectly.
	
	Finally, consider $\adversary{A}$ output $(\m^*, \signature^* \assign (\encoded{R^*}, S^*))$. Every valid signature outputted by adversary $\adversary{A}$ has to fulfill following equation for one public key $\groupelement{A_i}$: $2^c S \groupelement{B} = 2^c R + 2^c H'(\encoded{R}|\encoded{A_i}|m) \groupelement{A_i}$. Again there is only one valid encoded $S$ for each $\groupelement{R}$, $m$, $\groupelement{A_i}$ tuple that satisfies the verification equation. For the signature to be a valid forgery it must not be outputted by the \Osign oracle for this specific $m^*$ and $\groupelement{A_i}$. No new valid signature can be generated from a valid one by just changing the $S$ value. This means that at either $\groupelement{R}$, $m$ or $\groupelement{A_i}$ have to be changed to generate a new valid signature from an already valid signature. Since all these parameters are part of the hash query to generate the challenge the resulting hash value has to be forwarded from the $H$ hash oracle provided to the adversary $\adversary{B}$. For this reason $H'(\encoded{R^*}|\encoded{A_i}|m^*) = H(\encoded{R^*}|\encoded{A_i}|m^*)$. Hence,
	
	\begin{align*}
		2^c S^* \groupelement{B} &= 2^c \groupelement{R^*} + 2^c H'(\encoded{R^*}|\encoded{A_i}|m^*) \groupelement{A_i} \\
		\Leftrightarrow 2^c S^* \groupelement{B} &= 2^c \groupelement{R^*} + 2^c H(\encoded{R^*}|\encoded{A_i}|m^*) \groupelement{A_i}
	\end{align*}

	Since the public keys and the results of the hash queries are forwarded from the MU-UF-NMA challenger the forged signature from $\adversary{A}$ in the MU-\cma game is also a valid forgery for the MU-UF-NMA challenger.
	
	\item This proves theorem \ref{theorem:adv_mu-uf-nma}.
\end{proof}

\subsection{MU-UF-NMA $\overset{\text{ROM}}{\Rightarrow}$ $\text{MU-EUF-CMA}_{\text{EdDSA lp}}$}

This section shows that MU-UF-NMA security of EdDSA implies the MU-EUF-CMA security of EdDSA with lax parsing using in the random oracle model. This proof is very similar to the proof MU-SUF-CMA proof of EdDSA with strict parsing. The modification to the games are the same as in the proof above with the only modifications being in the win condition, which is $\exists j \in \{1,2,...,N\}: \verify(\groupelement{A_j}, \m^*) \wedge (\groupelement{A_j}, \m^*) \notin \pset{Q}$. For this reason this proof starts at showing the existence of an adversary $\adversary{B}$ breaking MU-UF-NMA security.

\begin{theorem}
	\label{theorem:adv2_mu-uf-nma}
	Let $n$ and $N$ be positive integer and $\adversary{A}$ an adversary against MU-EUF-CMA, making at most $\hashqueries$ hash queries and $\oraclequeries$ oracle queries. Then,
	
	\[ \advantage{\adversary{A}}{\text{MU-EUF-CMA}}(\secparamter) \leq \advantage{\adversary{B}}{\text{MU-UF-NMA}}(\secparamter) + \frac{\oraclequeries \hashqueries}{2^{-\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}}. \]
\end{theorem}

\paragraph{\underline{Formal Proof}}

\begin{proof}
	\item 
	\begin{align}
		\Pr[G_3^{\adversary{A}} \Rightarrow 1] = \advantage{\adversary{B}}{\text{MU-UF-NMA}}(\secparamter). \label{eq:adv2_mu-uf-nma}
	\end{align}
	
	\begin{figure}
		\hrule
		\begin{multicols}{2}
			\large
			\begin{algorithmic}[1]
				\Statex \underline{\textbf{Adversary} $\adversary{B}^{H(\inp)}(\groupelement{A_1}, \groupelement{A_2}, ..., \groupelement{A_N})$}
				\State $(\m^*, \signature^*) \randomassign \adversary{A}^{H'(\inp), \sign(\inp, \inp)}(\groupelement{A_1}, \groupelement{A_2}, ..., \groupelement{A_N})$
				\State \Return $(\m^*, \signature^*)$
			\end{algorithmic}
			\columnbreak
			\begin{algorithmic}[1] 
				\Statex \underline{\oracle \sign($j \in \{1,2,...,N\}$, $\m \in \messagespace$)}
				\State $(R,\textbf{ch},S) \randomassign \simalg(\groupelement{A_j})$
				\State $\textbf{if } \sum[\encoded{R} | \encoded{A_j} | m] \neq \bot \textbf{ then}$
				\State \quad $bad \assign true$
				\State \quad $abort$
				\State $\sum[\encoded{R} | \encoded{A_j} | m] = \textbf{ch}$
				\State $\signature \assign (\encoded{R}, S)$
				\State $\pset{Q} \assign \pset{Q} \cup \{(\groupelement{A_j}, \m)\}$
				\State \Return $\signature$
			\end{algorithmic}
		\end{multicols}
		\begin{algorithmic}[1] 
			\Statex \underline{\oracle $H'(m \in \{0,1\}^*)$}
			\State $\textbf{if } \sum[m] = \bot \textbf{ then}$
			\State \quad $\sum[m] \assign H(m)$
			\State \Return $\sum[m]$
		\end{algorithmic}
		\hrule
		\caption{Adversary $\adversary{B}$ breaking $\text{MU-UF-NMA}$}
		\label{fig:adversary_b_mu-uf-nma}
	\end{figure}

	To prove (\ref{eq:adv2_mu-uf-nma}), we define an adversary $\adversary{B}$ attacking MU-UF-NMA that simulates $\adversary{A}$'s view on $G_3$. Adversary $\adversary{B}$, formally defined in figure \ref{fig:adversary_b_mu-uf-nma}, is run in the MU-UF-NMA game and simulates \Osign for adversary $\adversary{A}$. \Osign is simulated perfectly.
	
	Finally, consider $\adversary{A}$ output $(\m^*, \signature^* \assign (\encoded{R^*}, S^*))$. Every valid signature outputted by adversary $\adversary{A}$ has to fulfill following equation for one public key $\groupelement{A_i}$: $2^c S \groupelement{B} = 2^c R + 2^c H'(\encoded{R}|\encoded{A_i}|m) \groupelement{A_i}$. Like in the single-user setting the adversary can create a new valid signature from an already valid one by choosing a different bitstring representation of the $S$ value that maps to the same $S \pmod L$. Since we are in the MU-EUF-CMA setting the adversary has to forge a signature for a message $m^*$ and public key $A_i$ to which it has not queried a signature before. For this reason the output of $H'(\encoded{R}|\encoded{A_i}|m)$ has not been set by the adversary $\adversary{B}$ but was forwarded from the $H$ hash oracle provided by the MU-UF-NMA challenger. For this reason $H'(\encoded{R^*}|\encoded{A_i}|m^*) = H(\encoded{R^*}|\encoded{A_i}|m^*)$. Therefore,
	
	\begin{align*}
		2^c S^* \groupelement{B} &= 2^c \groupelement{R^*} + 2^c H'(\encoded{R^*}|\encoded{A_i}|m^*) \groupelement{A_i}\\
		\Leftrightarrow 2^c S^* \groupelement{B} &= 2^c \groupelement{R^*} + 2^c H(\encoded{R^*}|\encoded{A_i}|m^*) \groupelement{A_i}.
	\end{align*}

	This shows that the forged signature from adversary $\adversary{A}$ is also a valid forged signature for the MU-UF-NMA challenger.
	
	\item This proofs theorem \ref{theorem:adv2_mu-uf-nma}.

\end{proof}