Commit 33884413 by Guillaume GENESTIER

### Everything is in English, proofs are verified until p9

parent 1e6b820d
 ... ... @@ -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] ... ...
 \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} ... ...
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 \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_{\
 ... ... @@ -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 $\D\vdash A:\Type$, ... ...
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