From 79cc6c5806aa1edc4314fb70082d3f564bf7b819 Mon Sep 17 00:00:00 2001
From: Fabrice Mouhartem <fabrice.mouhartem@ens-lyon.org>
Date: Wed, 28 Feb 2018 18:02:06 +0100
Subject: [PATCH] + Restatable definition

---
 chap-proofs.tex  |  5 +++--
 sec-pairings.tex | 13 ++-----------
 2 files changed, 5 insertions(+), 13 deletions(-)

diff --git a/chap-proofs.tex b/chap-proofs.tex
index aa203af..ce3a9b6 100644
--- a/chap-proofs.tex
+++ b/chap-proofs.tex
@@ -128,12 +128,13 @@ To illustrate this, let us consider the two following assumptions:
   The \textit{discrete logarithm assumption} is the intractability of this problem.
 \end{definition}
 
-\begin{definition}[Decisional Diffie-Hellman] \label{de:DDH} \index{Discrete Logarithm!Decisional Diffie-Hellman}
+\begin{restatable}[Decisional Diffie-Hellman]{definition}{defDDH}
+    \index{Discrete Logarithm!Decisional Diffie-Hellman} \label{de:DDH}
   Let $\GG$ be a cyclic group of order $p$. The \emph{decisional Diffie-Hellman} ($\DDH$) problem is the following.
   Given the tuple $(g, g_1^{}, g_2^{}, g_3^{}) = (g, g^a_{}, g^b{}, g^c_{}) \in \GG^4_{}$, the goal is to decide whether $c = ab$ or $c$ is sampled uniformly in $\GG$.
 
   The \textit{\DDH assumption} is the intractability of the problem for any $\ppt$ algorithm.
-\end{definition}
+\end{restatable}
 
 The discrete logarithm assumption is implied by the decisional Diffie-Hellman assumption for instance.
 Indeed, if one is able to solve the discrete logarithm problem, then it suffices to compute the discrete logarithm of $g_1$, let us say $\alpha$, and then check whether $g_2^\alpha = g_3^{}$ or not.
diff --git a/sec-pairings.tex b/sec-pairings.tex
index 2f00c6e..ba70acd 100644
--- a/sec-pairings.tex
+++ b/sec-pairings.tex
@@ -23,18 +23,9 @@ In the following, we rely on the black-box definition of cryptographic pairings
 For cryptographic purpose, pairings are usually defined over elliptic curves, hence $\GT$ is a multiplicative subgroup of the multiplicative group of a finite field.
 
 The most standard assumptions over pairings are derived from the equivalent of the Diffie-Hellman assumptions from cyclic groups,
-described in Definition~\ref{de:DDH} and recalled here.
+described in \cref{de:DDH} and recalled here.
 
-\begin{definition}[$\DDH$ (recall)]  \index{Discrete Logarithm!Decisional Diffie-Hellman}
-  Let $\GG$ be a cyclic group of order $p$. The \emph{decisional Diffie-Hellman} ($\DDH$) problem is the following.
-  Given $(g, g^a, g^b, g^c) \in \GG^4$, the goal is to decide if $c = ab$ or if $c$ is sampled uniformly in $\GG$.
-
-  The DDH assumption is the intractability of the problem for any $\ppt$ algorithm.
-
-%  Let us now define the $\DDH$ language as
-%  $L_\DDH = \bigl\{ (g, g^a, g^b, g^{c}) \in \GG^4 \mid c = a \cdot b \bigr\}.$
-%  Thus the $\DDH$ problem is equivalently the question of whether $L_\DDH \in \mathsf{PP}$ or not.
-\end{definition}
+\defDDH*
 
 This hypothesis, from which the Diffie-Hellman key exchange relies its security on, is then used to defined the $\SXDH$ assumption.