Fixed some spelling mistakes, thanks Henrik. again

thesis/sections/security_of_eddsa/uf-nma_implies_suf-cma.tex

@@ -107,7 +107,7 @@ This method of simulating the \Osign oracle and the resulting loss of advantage
 \end{figure}
 
 \begin{proof}
-	\item The proof begins by providing an algorithm that generates a correctly distributed tuple of commitment, challenge, and response. This algorithm is called \simalg and is shown in the figure \ref{fig:sim}. This procedure is taken from \cite{SP:BCJZ21}. A proof can be found in the same paper. The formula for the min-entropy of the commitment $\groupelement{R}$ is also taken from that paper.
+	\item The proof begins by providing an algorithm that generates a correctly distributed tuple of commitment, challenge, and response. This algorithm is called \simalg and is shown in figure \ref{fig:sim}. This procedure is taken from \cite{SP:BCJZ21}. A proof can be found in the same paper. The formula for the min-entropy of the commitment $\groupelement{R}$ is also taken from that paper.
 	
 	\item \paragraph{\underline{$G_0:$}} Let $G_0$ be defined in figure \ref{fig:uf-nma_implies_suf-cma_games} by excluding all boxes except the black one. Clearly $G_0$ is the game $\text{\cma}$ for EdDSA. By definition,
 	
@@ -164,9 +164,9 @@ This method of simulating the \Osign oracle and the resulting loss of advantage
 		\label{fig:adversarybuf-nma}
 	\end{figure}
 
-	To prove (\ref{eq:adv_uf-nma}), an adversary $\adversary{B}$ is defined that attacks $\text{UF-NMA}$ simulating the view of $\adversary{A}$ in $G_3$. The adversary $\adversary{B}$ formally defined in figure \ref{fig:adversarybuf-nma} is run in the $\text{UF-NMA}$ game, and the adversary $\adversary{B}$ simulates \Osign for the adversary $\adversary{A}$. \Osign is simulated perfectly. The hash queries of $\adversary{A}$ are forwarded to the UF-NMA challenger.
+	To prove (\ref{eq:adv_uf-nma}), an adversary $\adversary{B}$ is defined that attacks $\text{UF-NMA}$ simulating the view of $\adversary{A}$ in $G_3$. The adversary $\adversary{B}$, formally defined in figure \ref{fig:adversarybuf-nma}, is run in the $\text{UF-NMA}$ game, and the adversary $\adversary{B}$ simulates \Osign for the adversary $\adversary{A}$. \Osign is simulated perfectly. The hash queries of $\adversary{A}$ are forwarded to the UF-NMA challenger.
 	
-	Finally, consider $\adversary{A}$'s output $(\m^*, \signature^* \assign (\encoded{R^*}, S^*))$. It is known that $2^c S^* \groupelement{B} = 2^c \groupelement{R^*} + 2^c H'(\encoded{R^*}|\encoded{A}|m^*) \groupelement{A}$. If strict parsing is used to decode $S$ from the signature, it is known that $0 \leq S < L$. Therefore, for every $R$, $\m$ pair, there is only one valid encoded $S$ that satisfies the equation. This means that a new and valid signature cannot be generated by simply changing the $S$ value of an already valid signature. Therefore, $R$ or $m$ must also be changed in order to create a new valid signature from another. Since $R$ and $m$ are inputs to the hash query used to generate the challenge, the result of this hash query is passed from the $H$ hash oracle provided to the adversary $\adversary{B}$ instead of being set by the $\adversary{B}$ itself. Also, the existence of multiple encodings of the commitment $\groupelement{R}$ does not pose a problem, since if another representation of the same $\groupelement{R}$ is chosen, its hash value has not been set by $\adversary{B}$ and therefore must have passed from the UF-NMA challenger. For this reason, $H'(\encoded{R^*}|\encoded{A}|m^*) = H(\encoded{R^*}|\encoded{A}|m^*)$. Therefore,
+	Finally, consider $\adversary{A}$'s output $(\m^*, \signature^* \assign (\encoded{R^*}, S^*))$. It is known that $2^c S^* \groupelement{B} = 2^c \groupelement{R^*} + 2^c H'(\encoded{R^*}|\encoded{A}|m^*) \groupelement{A}$. If strict parsing is used to decode $S$ from the signature, it is known that $0 \leq S < L$. Therefore, for every $R$, $\m$ pair, there is only one valid encoded $S$ that satisfies the equation. This means that a new and valid signature cannot be generated by simply changing the $S$ value of an already valid signature. Therefore, $R$ or $m$ must also be changed in order to create a new valid signature from another. Since $R$ and $m$ are inputs to the hash query used to generate the challenge, the result of this hash query is passed from the $H$ hash oracle provided to the adversary $\adversary{B}$ instead of being set by $\adversary{B}$ itself. Also, the existence of multiple encodings of the commitment $\groupelement{R}$ does not pose a problem, since if another representation of the same $\groupelement{R}$ is chosen, its hash value has not been set by $\adversary{B}$ and therefore must have been passed from the UF-NMA challenger. For this reason, $H'(\encoded{R^*}|\encoded{A}|m^*) = H(\encoded{R^*}|\encoded{A}|m^*)$. Therefore,
 	
 	\begin{align*}
 		2^c S^* \groupelement{B} &= 2^c \groupelement{R^*} + 2^c H'(\encoded{R^*}|\encoded{A}|m^*) \groupelement{A}\\