Fixed some spelling mistakes, thanks Henrik. again

thesis/sections/mu_security_of_eddsa/mu-gamez_implies_mu-uf-nma.tex

@@ -2,7 +2,7 @@
 
 This section shows that MU-\igame implies MU-UF-NMA security of the EdDSA signature scheme using the Random Oracle Model. The section starts by first providing an intuition of the proof followed by the detailed security proof.
 
-\paragraph{\underline{Introducing MU-\igame}} This game followed closely the definition of the \igame game. It again replaces the random oracle with the \ioracle oracle. The only difference to the \igame game is that the adversary gets access to $N$ public keys. The adversary again has to output a valid result for any commitment challenge pair generated by the \ioracle oracle for any of the public keys. The MU-\igame game is depicted in figure \ref{game:mu-igame}.
+\paragraph{\underline{Introducing MU-\igame}} This game follows closely the definition of the \igame game. It again replaces the random oracle with the \ioracle oracle. The only difference to the \igame game is that the adversary gets access to $N$ public keys. The adversary again has to output a valid result for any commitment challenge pair generated by the \ioracle oracle for any of the public keys. The MU-\igame game is depicted in figure \ref{game:mu-igame}.
 
 \begin{definition}[MU-\igame]
 	Let $n$ and $N$ be positive integers. For an adversary $\adversary{A}$, receiving $N$ public keys as input, we define its advantage in the MU-\igame as following:

thesis/sections/mu_security_of_eddsa/mu-uf-nma_implies_mu-suf-cma.tex

@@ -93,7 +93,7 @@ This section shows that the MU-UF-NMA security of the EdDSA signature scheme imp
 	
 	\[ \Pr[G_0^{\adversary{A}} \Rightarrow 1] = \Pr[G_1^{\adversary{A}} \Rightarrow 1]. \]
 	
-	\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 queries 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
+	\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 queried 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
 	
 	\[ |\Pr[G_1^{\adversary{A}} \Rightarrow 1] - \Pr[G_2^{\adversary{A}} \Rightarrow 1]| \leq \Pr[bad] \leq \frac{\oraclequeries \hashqueries}{2^{-\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}}. \]
 	
@@ -142,7 +142,7 @@ This section shows that the MU-UF-NMA security of the EdDSA signature scheme imp
 	
 	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,
+	Finally, consider $\adversary{A}$ output $(\m^*, \signature^* \assign (\encoded{R^*}, S^*))$. Every valid signature outputted by adversary $\adversary{A}$ has to fulfill the 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 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} \\
@@ -158,11 +158,11 @@ This section shows that the MU-UF-NMA security of the EdDSA signature scheme imp
 
 \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. Similar to the proof in the single-user setting, the SUF-CMA security of EdDSA with lax parsing cannot be shown, as there are multiple valid encodings of $S$ for one signature. This way the adversary would be able to generate a new valid signature from an obtained one by simply choosing a different encoding of $S$. This would result in the output of $H'(\encoded{R^*}|\encoded{A}|m^*)$ being programmed by the reduction itself and therefore the signature not being valid for the UF-NMA challenger.
+This section shows that MU-UF-NMA security of EdDSA implies the MU-EUF-CMA security of EdDSA with lax parsing used 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. Similar to the proof in the single-user setting, the SUF-CMA security of EdDSA with lax parsing cannot be shown, as there are multiple valid encodings of $S$ for one signature. This way the adversary would be able to generate a new valid signature from an obtained one by simply choosing a different encoding of $S$. This would result in the output of $H'(\encoded{R^*}|\encoded{A}|m^*)$ being programmed by the reduction itself and therefore the signature not being valid for the UF-NMA challenger.
 
 \begin{theorem}
 	\label{theorem:adv2_mu-uf-nma}
-	Let $n$ and $N$ be positive integer and $\adversary{A}$ an adversary against MU-EUF-CMA, receiving $N$ public keys and making at most $\hashqueries$ hash queries and $\oraclequeries$ oracle queries. Then,
+	Let $n$ and $N$ be positive integers and $\adversary{A}$ an adversary against MU-EUF-CMA, receiving $N$ public keys and 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}
@@ -210,7 +210,7 @@ This section shows that MU-UF-NMA security of EdDSA implies the MU-EUF-CMA secur
 
 	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,
+	Finally, consider $\adversary{A}$ output $(\m^*, \signature^* \assign (\encoded{R^*}, S^*))$. Every valid signature outputted by adversary $\adversary{A}$ has to fulfill the 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}\\
@@ -221,6 +221,6 @@ This section shows that MU-UF-NMA security of EdDSA implies the MU-EUF-CMA secur
 	
 	\item Since the adversary $\adversary{B}$ is the same as in the proof above, its runtime is roughly the same as the runtime of adversary $\adversary{A}$, for the same reason.
 	
-	\item This proofs theorem \ref{theorem:adv2_mu-uf-nma}.
+	\item This proves theorem \ref{theorem:adv2_mu-uf-nma}.
 
 \end{proof}

thesis/sections/mu_security_of_eddsa/omdl'_implies_mu-gamez.tex

@@ -6,7 +6,7 @@ This section shows that \somdl implies MU-\igame using the algebraic group model
 
 \begin{definition}[\somdl]
 	\label{def:somdl}
-	Let $n$ and $N$ be positive integer. For an adversary $\adversary{A}$, receiving $N$ challenge group elements, we define its advantage in the \somdl game as following:
+	Let $n$ and $N$ be positive integers. For an adversary $\adversary{A}$, receiving $N$ challenge group elements, we define its advantage in the \somdl game as following:
 	
 	\[ \advantage{\adversary{A}}{\text{\somdl}}(\secparamter) \assign | \Pr[\text{\somdl}^{\adversary{A}} \Rightarrow 1] |. \]
 \end{definition}
@@ -43,7 +43,7 @@ This section shows that \somdl implies MU-\igame using the algebraic group model
 	\[ \advantage{\group{G},\adversary{A}}{\text{MU-\igame}}(\secparamter) \leq \advantage{\group{G},\adversary{B}}{\somdl}(\secparamter) + \frac{\oraclequeries N}{2^{-\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}}. \]
 \end{theorem}
 
-\paragraph{\underline{Proof Overview}} In the multi-user setting the adversary gets access to not only the generator $\groupelement{B}$ and one public key $\groupelement{A}$ but rather a set of public keys $\groupelement{A_1}$ to $\groupelement{A_N}$. For this reason the representation of a group element, the adversary has to provide, looks the following: $\groupelement{R} = r_1 \groupelement{B} + r_2 \groupelement{A_1} + ... + r_{N+1} \groupelement{A_N}$. Since there are multiple group elements with unknown discrete logarithms it is not possible to directly calculate the discrete logarithm of one of the public keys given a valid forgery of a signature. Upon receiving a valid solution the \textit{DL} oracle can be used to get the discrete logarithm of all the public keys except the one for which the solution is valid. This way it again possible to construct a representation looking like $\groupelement{R} = r_1 \groupelement{B} + r_2 \groupelement{A_i}$. Then it is again possible to calculate the discrete logarithm of $\groupelement{A_i}$ and win the \somdl game.
+\paragraph{\underline{Proof Overview}} In the multi-user setting the adversary gets access to not only the generator $\groupelement{B}$ and one public key $\groupelement{A}$ but rather a set of public keys $\groupelement{A_1}$ to $\groupelement{A_N}$. For this reason, the representation of a group element, the adversary has to provide, looks the following: $\groupelement{R} = r_1 \groupelement{B} + r_2 \groupelement{A_1} + ... + r_{N+1} \groupelement{A_N}$. Since there are multiple group elements with unknown discrete logarithms it is not possible to directly calculate the discrete logarithm of one of the public keys given a valid forgery of a signature. Upon receiving a valid solution the \textit{DL} oracle can be used to get the discrete logarithm of all the public keys except the one for which the solution is valid. This way it is again possible to construct a representation looking like $\groupelement{R} = r_1 \groupelement{B} + r_2 \groupelement{A_i}$. Then it is again possible to calculate the discrete logarithm of $\groupelement{A_i}$ and win the \somdl game.
 
 \paragraph{\underline{Formal Proof}}
 
@@ -88,7 +88,7 @@ This section shows that \somdl implies MU-\igame using the algebraic group model
 	
 	\[ \advantage{\group{G},\adversary{A}}{\text{MU-\igame}}(\secparamter) = \Pr[\text{MU-\igame}^{\adversary{A}} \Rightarrow 1] = \Pr[G_0^{\adversary{A}} \Rightarrow 1]. \]
 	
-	\item \paragraph{\underline{$G_1$:}} $G_1$ is defined by including the if condition in the blue box, which sets a bad flag if the randomly chosen value $\ch$ fulfills $2^c \ch \equiv - r_i \pmod L$ for any $i \in \{2,3,...,N+1\}$. This represents challenges $\ch$ to which the solution might not be usable to break the discrete logarithm of one of the public keys, due to $(r_i + 2^c \ch)$ not being invertible in $\field{L}$. Since only the bad flag being introduced this change does not influence the behavior of the game and is therefore only conceptual.
+	\item \paragraph{\underline{$G_1$:}} $G_1$ is defined by including the if condition in the blue box, which sets a bad flag if the randomly chosen value $\ch$ fulfills $2^c \ch \equiv - r_i \pmod L$ for any $i \in \{2,3,...,N+1\}$. This represents challenges $\ch$ to which the solution might not be usable to break the discrete logarithm of one of the public keys, due to $(r_i + 2^c \ch)$ not being invertible in $\field{L}$. Since only the bad flag is introduced this change does not influence the behavior of the game and is therefore only conceptual.
 	
 	\[ \Pr[G_0^{\adversary{A}} \Rightarrow 1] = \Pr[G_1^{\adversary{A}} \Rightarrow 1]. \]