Everything is in English, proofs are verified until p9

acc.tex

@@ -18,7 +18,7 @@
 \begin{defi}[Accessible position]
  For $f\in\CObj$, where $\tau(f)=\Pi\overline{(x:T)}.C\,t_1\dots t_{\arity(C)}$,
  we define
- \[\Acc(f)=\enscond{i\<\arity(f)}{T_i\in\Froz_{\< C}}.\]
+ \[\Acc(f)=\enscond{i\<\arity(f)}{T_i\in\Froz_{\preced C}}.\]
 \end{defi}
 
 \begin{defi}[Order associated to accessible subterms]

cand.tex

 
-\section{Candidats}
+\section{Candidates}
 
-\begin{defi}[Candidats de réductibilité]
-  $S\in\Part{\Lambda}$ est un \emph{candidat de réductibilité} si
+\begin{defi}[Reducibility candidates]
+  $S\in\Part{\Lambda}$ is a \emph{reducibility candidate} if
   \begin{itemize}
    \item $S\subseteq\SN(\rew)$,
-   \item $\enscond{u}{\text{il y a }t\in S,t\rew u}\subseteq S$,
-   \item si $t$ est neutre et $\enscond{u}{t\rew u}\subseteq S$ alors $t\in S$.
+   \item $\enscond{u}{\text{there is a }t\in S,t\rew u}\subseteq S$,
+   \item if $t$ is neutral and $\enscond{u}{t\rew u}\subseteq S$ then $t\in S$.
   \end{itemize}
-  On note $\Candidates$ l'ensemble des candidats de réductibilité.
+  We denote by $\Candidates$ the set of reducibility candidates.
 \end{defi}
 
 \begin{lem}
- $\SN(\rew\cup\surterm_{acc})$ est candidat.
+ $\SN(\rew\cup\surterm_{acc})$ is a candidate.
 \end{lem}
 \begin{proof}
  \begin{itemize}
     \item
-     Par définition, on a $\SN(\rew\cup\surterm_{acc})\subseteq\SN(\rew)$.
+     By definition we have $\SN(\rew\cup\surterm_{acc})\subseteq\SN(\rew)$.
     \item
-     Soit $t\in\SN(\rew\cup\surterm_{acc})$ et $t'$ tel que $t\rew t'$.
-     On a $t'\in\SN(\rew\cup\surterm_{acc})$.
+     Let $t\in\SN(\rew\cup\surterm_{acc})$ and $t'$ such that $t\rew t'$.
+     We have $t'\in\SN(\rew\cup\surterm_{acc})$.
     \item
-     Soit $t$ un terme neutre tel que $\enscond{u}{t\rew u}\subseteq\SN(\rew\cup\surterm_{acc})$.
-     Comme $t$ est neutre, il n'a pas en tête un symbole de $\CObj$,
-     donc $\enscond{u}{t(\rew\cup\surterm_{acc}) u}=\enscond{u}{t\rew u}$.
-     Par conséquent, $t\in\SN(\rew\cup\surterm_{acc})$.
+     Let $t$ be a neutral term such that $\enscond{u}{t\rew u}\subseteq\SN(\rew\cup\surterm_{acc})$.
+     Since $t$ is neutral, it is not headed by an element of $\CObj$,
+     so $\enscond{u}{t(\rew\cup\surterm_{acc}) u}=\enscond{u}{t\rew u}$.
+     Hence, $t\in\SN(\rew\cup\surterm_{acc})$.\qedhere
    \end{itemize}
 \end{proof}
 
-\begin{lem}[Reductibilité de $\mcal{J}$]
- Pour tout $C\,\bar u$ avec $C\in\FTyp$ et $\valabs{u}=\arity(C)$, alors $\mcal{I}_{C\bar u}\in\Candidates$.
+
+\begin{lem}[Reducibility of $\mcal{I}_C$]
+ If $C\in\CTyp$, then $\mcal{I}_C\in\Candidates$.
 \end{lem}
 
 \begin{proof}
-\begin{itemize}
- \item
-   Si $C\in \CTyp$.
-   On a $\mcal{I}_{C\bar u}=K_{C\bar u}(\mcal{I}_{C\bar u})$.
-   \begin{itemize}
-   \item
-     Par définition de $K_{C\bar u}$, on a $K_{C\bar u}(\mcal{I}_{C\bar u})\subseteq\SN(\rew)$.
-   \item
-     Soit $t\in\mcal{I}_{C\bar u}$ et $u$ tel que $t\rew u$.
-     Comme $t\in\SN(\rew\cup\surterm_{acc})$, $u\in\SN(\rew\cup\surterm_{acc})$ aussi.
-     Si $u\reww c\,\bar v$, alors $t\reww c\,\bar v$ donc la contrainte de réductibilité des arguments accessibles est remplie.
-   \item
-     Let $t$ be a neutral term such that $\enscond{u}{t\rew u}\subseteq\mcal{I}_0(T)$.
-     Since $t$ is neutral, $\enscond{u}{t(\rew\cup\surterm_{acc}) u}=\enscond{u}{t\rew u}$, so $t\in\SN(\rew\cup\surterm_{acc})$.
-     If $t\reww c\,\bar v$, then there is a $u\in\enscond{u}{t\rew u}$ such that $u\reww c\,\bar v$ so the constraint on the reducibility of the accessible arguments are fulfilled.
-   \end{itemize}
- \item
-   
+ First note that $\mcal{I}_C=K_{\bar C}(\mcal{I}_C)$.
+ \begin{itemize}
+  \item
+   By definition of $K_{\bar C}$, we have $K_{\bar C}(\mcal{I}_C)\subseteq\SN(\rew)$.
+  \item
+   Let $t\in\mcal{I}_C$ and $u$ be such that $t\rew u$.
+   Since $t\in\SN(\rew\cup\surterm_{acc})$, $u\in\SN(\rew\cup\surterm_{acc})$ too.
+   If $u\reww c\,\bar v$, then $t\reww c\,\bar v$ so the constraint on the reducibility of the accessible arguments are fulfilled.
+  \item
+   Let $t$ be a neutral term such that $\enscond{u}{t\rew u}\subseteq\mcal{I}_C$.
+   Since $t$ is neutral, $\enscond{u}{t(\rew\cup\surterm_{acc}) u}=\enscond{u}{t\rew u}$, so $t\in\SN(\rew\cup\surterm_{acc})$.
+   If $t\reww c\,\bar v$, then there is a $u\in\enscond{u}{t\rew u}$ such that $u\reww c\,\bar v$ so the constraint on the reducibility of the accessible arguments are fulfilled.\qedhere
  \end{itemize}
 \end{proof}
 
+
 \begin{lem}[Product of candidates]\label{lem-product-candidate}
   If $P\in\Candidates$ and, for all $a\in P$, $Q(a)\in\Candidates$, then $\enscond{t}{\text{for all }a\in P. t\,a\in Q(a)}\in\Candidates$.
 \end{lem}

ordonnabilite.tex

 \section{Ordering}
 
-\begin{defi}[Type constructors]
+\begin{defi}[Type constants]
   \[\CTyp=\enscond{C\in\FTyp}{C\text{ is not the head of any rule}}\]
   We suppose given a well-founded pre-order $\preced$ on $\CTyp$ .
 \end{defi}
@@ -8,10 +8,10 @@
 \begin{defi}[Frozen type]
  For $C\in\CTyp$, we define the following grammars :
  \begin{align*}
-  T_{\<C}::=& \Pi(x:U_{<C}). T_{\<C}\,\mid\,C'\,u_1\dots u_{\arity(C')}&\text{ where }C'\< C\\
-  U_{<C}::=& \Pi(x:T_{\<C}). U_{<C}\,\mid\,C'\,u_1\dots u_{\arity(C')}&\text{ where }C'< C\\
+  T_{\preced C}::=& \Pi(x:U_{\prec C}). T_{\preced C}\,\mid\,C'\,u_1\dots u_{\arity(C')}&\text{ where }C'\preced C\\
+  U_{\prec C}::=& \Pi(x:T_{\preced C}). U_{\prec C}\,\mid\,C'\,u_1\dots u_{\arity(C')}&\text{ where }C'\prec C\\
  \end{align*}
- We denote those sets respectively by $\Froz_{\<C}$ and $\Froz_{<C}$.
+ We denote those sets respectively by $\Froz_{\preced C}$ and $\Froz_{\prec C}$.
 \end{defi}
 
 \todo{

typing.tex

@@ -138,7 +138,7 @@ A \emph{term} is $\Kind$, a \emph{kind}, a \emph{family} or an \emph{object}.
 \end{rmq}
 
 
-\begin{propo}[Stratification lemma]
+\begin{propo}[Stratification lemma]\label{lem-Stratification}
  If $\D\vdash t:A$ then we are in one of the following cases:
  \begin{itemize}
   \item $t$ is an object, $A$ a family and $\D\vdash A:\Type$,