Commit 9d1d3259 by Guillaume GENESTIER

### k

parent ec4d67e4
 ... ... @@ -23,7 +23,7 @@ \begin{cond}{cond-symb-order-wf} $\sqsubset$ is well-founded. \todo{Note that it is free if $\Func$ is finite.} \end{cond} ... ... @@ -40,7 +40,7 @@ \AxiomC{$g\sqsubset f$} \BinaryInfC{$\G\vdash_{\sqsubset f}g:\tau(g)$} \end{prooftree} Note that it introduces predicates well-formed${}_{\sqsubset f}$. \begin{center} \begin{tabular}{rcl} ... ... @@ -168,14 +168,18 @@ we have that $\rew_{arg}\cup{\call}$ terminates on $\mbb{U}$. We now prove that, $f\,\bar t\in\interp{T'\pi}$ by induction on ${\rew_{arg}}\cup{\call}$. Because of the assumptions we have on rules, $f\,\bar t$ is neutral. Hence, it suffices to prove that, for all $u$ such that $f\,\bar t\rew u$, we have $u\in\interp{T'\pi}$. There are two cases: \begin{itemize} \item $u=f\bar u$ with $\bar t\rew\bar u$. \begin{itemize} \item If $f\in\FObj\setminus\CObj$, $f\,\bar t$ is neutral. Hence, it suffices to prove that, for all $u$ such that $f\,\bar t\rew u$, we have $u\in\interp{T'\pi}$. There are two cases: \begin{itemize} \item $u=f\,\bar u$ with $f\,\bar t\rew_{arg}f\,\bar u$. Then, we can conclude by induction hypothesis. \item There are $fl_1\dots l_k\rul r\in\mcal{R}$ and $\sg$ such that \item There are $fl_1\dots l_k\rul r\in\mcal{R}$ and $\sg$ such that $u=(r\sg)\,t_{k+1}\dots t_n$ and, for all $i\in\{1,\dots,k\}$, $t_i=l_i\sg$. Then, $\crochet{\overline{\sur{l}{x}}}\sg\vDash\G$. ... ... @@ -183,11 +187,17 @@ We now prove that, for all $\G$, $u$ and $U$, if $\G\vdash_{f\bar l} u:U$ and $\sg\vDash\G$, then $u\sg\in\interp{U\sg}$. The proof is the same as for \indthm{TypImpliInterp} except for (fun) replaced by (fun'). In this case, for all $i$, we have $\G\vdash_{f\bar l} u_i:U_i\g$. By induction hypothesis, $u_i\sg\in\interp{U_i\g\sg}$. So, $\g\sg\vDash\Sg$ and $g\,\bar u\sg\in\mbb{U}$. Now, $g\,\bar u\sg\tilde{<}f\,\bar l\sg$ since $g\,\bar u  \usepackage[T1]{fontenc} %Permet d'imprimer les caractères spéciaux \usepackage[utf8]{inputenc} %Permet de taper directement les caractères spéciaux dans le fichier Tex \usepackage[francais]{babel} %Écrit Démonstration et non Proof \frenchbsetup{StandardLists=true} %Met des puces rodes plutôt que des - pour les listes \usepackage[english]{babel} %Écrit Démonstration et non Proof %\frenchbsetup{StandardLists=true} %Met des puces rodes plutôt que des - pour les listes \usepackage{amsthm} %Permet de définir les différents types de théorèmes, remarques... \usepackage{amssymb} %Ajoute \mathbb ... ... @@ -23,7 +23,7 @@ \usepackage[toc,page]{appendix} % Pour gérer les annexes \bibliographystyle{apalike} % Type de bibliographie \usepackage{multirow} \usepackage{imakeidx} %\usepackage{imakeidx} \usepackage[refpage]{nomencl} \usepackage{listings} \usepackage{enumitem} ... ... @@ -34,24 +34,24 @@ \usepackage{diagbox} \theoremstyle{plain} \newtheorem{thm}{Th\'{e}or\`{e}me}[section] \newtheorem{thm}{Theorem}[section] \newtheorem{propo}[thm]{Proposition} \newtheorem{coro}[thm]{Corollaire} \newtheorem{lem}[thm]{Lemme} \newtheorem{propri}[thm]{Propri\'et\'e} \newtheorem{propris}[thm]{Propri\'et\'es} \newtheorem{defi}[thm]{D\'efinition} \newtheorem{algo}[thm]{Algorithme} \newtheorem{coro}[thm]{Corollary} \newtheorem{lem}[thm]{Lemma} \newtheorem{propri}[thm]{Property} \newtheorem{propris}[thm]{Properties} \newtheorem{defi}[thm]{Definition} \newtheorem{algo}[thm]{Algorithm} \theoremstyle{definition} \newtheorem{exo}[thm]{Exercice} \newtheorem{exo}[thm]{Exercise} \theoremstyle{remark} \newtheorem*{nots}{Notations} \newtheorem*{rmq}{Remarque} \newtheorem*{expl}{Exemple} \newtheorem*{expls}{Exemples} \newtheorem{sub}{But local} \newtheorem{rest}{But restant} \newtheorem*{locProof}{Démonstration} \newtheorem*{rmq}{Remark} \newtheorem*{expl}{Example} \newtheorem*{expls}{Examples} \newtheorem{sub}{Local goal} \newtheorem{rest}{Remaining goal} \newtheorem*{locProof}{Proof} \newcommand{\sur}[2]{\raisebox{0.4ex}{$#1$} / \raisebox{-0.7ex}{$#2\$}} ... ...
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