Commit ec4d67e4 by Guillaume GENESTIER

### To \mbb{U}

parent 57a438e4
 ... ... @@ -2,7 +2,7 @@ \section{Dependency pairs} \begin{defi}[Dependency pairs] Let $f\bar l>g\bar m$ if there are a rule $f\bar l\rul r\in\mcal{R}$, Let $f\bar l>g\bar m$ if there is a rule $f\bar l\rul r\in\mcal{R}$, such that $g\,\bar m$ occurs in $r$, $\valabs{\bar m}\<\arity(g)$ and if $\valabs{\bar m}<\arity(g)$ then $\bar m$ are all the arguments to which $g$ is applied (ie. $\bar m$ are the arguments to which $g$ is applied truncated to the arity of $g$). \end{defi} ... ... @@ -13,23 +13,95 @@ and for all $k\g\,\bar u$, then $f\sqsupseteq g$. \end{itemize} \end{defi} \begin{defi}[Computability closure] Given $f\bar l$, let $\vdash_{f\bar l}$ be the relation defined as $\vdash$, except the rules (app) and (fun) replaced by \begin{cond}{cond-symb-order-wf} $\sqsubset$ is well-founded. \todo{Note that it is free if $\Func$ is finite.} \end{cond} \todo{ Should we add function symbol occuring at accessible position in the lhs? } \begin{defi}[Ordered typing] Let $f\in\Func$. Let $\vdash_{\sqsubset f}$ be the relation defined as $\vdash$, but where the rule (fun) is replaced by: \begin{prooftree} \AxiomC{$\Gamma\vdash_{f\bar l} t: \Pi(x:A). B$} \AxiomC{$\Gamma\vdash_{f\bar l} u:A$} \RightLabel{ \begin{tabular}{l} $t$ not of the form $g\,\bar l$\\ with $g$ a symbol of the signature \end{tabular}} \BinaryInfC{$\Gamma\vdash_{f\bar l} t\,u:\subst{B}{x}{u}$} \AxiomC{$\G\vdash_{\sqsubset f}\tau(g):s(g)$} \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} \AxiomC{$\phantom{\Gamma\vdash A}$} \UnaryInfC{$[]$ well-formed${}_{\sqsubset f}$} \DisplayProof&& \AxiomC{$\Gamma\vdash_{\sqsubset f} A:s$} \RightLabel{$\sys{x\notin\support(\Gamma)}{s\in\Sort}$} \UnaryInfC{$\Gamma,x:A$ well-formed${}_{\sqsubset f}$} \DisplayProof \end{tabular} \end{center} \end{defi} \begin{defi}[Computability closure] Let $f\,\bar l$ be the left-hand side of a rewrite rule. Let $\vdash_{f\bar l}$ be the relation defined by: \begin{center} \begin{tabular}{rcl} \AxiomC{$\Gamma$ well-formed${}_{\sqsubset f}$} \UnaryInfC{$\Gamma\vdash_{f\bar l} \Type:\Kind$} \DisplayProof&& \AxiomC{$\Gamma$ well-formed${}_{\sqsubset f}$} \RightLabel{$x:A\in\Gamma$} \UnaryInfC{$\Gamma\vdash_{f\bar l} x:A$} \DisplayProof \end{tabular} \end{center} \begin{prooftree} \AxiomC{$\Gamma\vdash_{f\bar l} A:\Type$} \AxiomC{$\Gamma,x:A\vdash_{f\bar l} B:s$} \RightLabel{$s\in\Sort$} \BinaryInfC{$\Gamma\vdash_{f\bar l} \Pi(x:A).B:s$} \end{prooftree} \begin{prooftree} \AxiomC{$\Gamma\vdash_{\sqsubset f}\Pi(x:A).B:s$} \AxiomC{$\Gamma,x:A\vdash_{f\bar l} t:B$} \RightLabel{$s\in\Sort$} \BinaryInfC{$\Gamma\vdash_{f\bar l}\lambda(x:A).t: \Pi(x:A).B$} \end{prooftree} \begin{prooftree} \AxiomC{$\Gamma\vdash_{f\bar l} t: \Pi(x:A).B$} \AxiomC{$\Gamma\vdash_{f\bar l} u:A$} \AxiomC{$\Gamma\vdash_{\sqsubset f}\Pi(x:A).B:s$} \TrinaryInfC{$\Gamma\vdash_{f\bar l} t\,u:\subst{B}{x}{u}$} \end{prooftree} \begin{prooftree} \AxiomC{$\Gamma\vdash_{f\bar l} t:A$} \AxiomC{$\Gamma\vdash_{\sqsubset f} A:s$} \AxiomC{$\Gamma\vdash_{\sqsubset f} B:s$} \RightLabel{$\sys{A\downarrow B}{s\in\Sort}$} \TrinaryInfC{$\Gamma\vdash_{f\bar l} t:B$} \end{prooftree} \begin{prooftree} \def\defaultHypSeparation{\hskip 0em} \AxiomC{$\vdash_{f\bar l} \tau(g):s(g)$} \AxiomC{$\vdash_{\sqsubset f} \tau(g):s(g)$} \AxiomC{$\overline{\G\vdash_{f\bar l} u:U\g}$} \RightLabel{ $\left\{\begin{matrix} ... ... @@ -41,6 +113,13 @@ \end{prooftree} \end{defi} \begin{prooftree} \AxiomC{$\G\vdash_{\sqsubset f}\tau(g):s(g)$} \RightLabel{$g$undefined} \UnaryInfC{$\G\vdash_{f\bar l}g:\tau(g)$} \end{prooftree} \begin{defi}[Valid rule] A rule$f\,\bar l\rul r$with$\tau(f)=\Pi\overline{(x:T)}. U$,$\G=\overline{(x:T)}$and$\pi=\crochet{\overline{\sur{l}{x}}}$is \emph{valid} if ... ... @@ -60,12 +139,19 @@ \tau(f)\in\interp{s(f)},\\ \tau(f)=\Pi\overline{(x:T)}. T',\\ T'\text{ is not an arrow}\\ \valabs{\bar t}=\arity(f)\\ \crochet{\overline{\sur{t}{x}}}\vDash\overline{(x:T)} \end{matrix} } \] \end{defi} \begin{defi}[$\rew_{arg}$] Let$f\,\bar t\in\mbb{U}$.$f\,\bar t\rew_{arg}u$if$u=f\,\bar t'$, there is a$i$such that$t_i\rew t'_i$and for all$j\neq i$,$t_j=t'_j$. \end{defi} \begin{thm}$\tau$is valid if$\call$terminates on$\mbb{U}$and all rules in$\mcal{R}$are valid. ... ... @@ -78,7 +164,7 @@ Let$\pi=\crochet{\overline{\sur{t}{x}}}$. We want to prove that$f\in\interp{\Pi\overline{(x:T)}. T'}$which by definition of the interpretation of a product is equivalent to show that$f\,\bar t\in\interp{T'\pi}$. Since$\rew_{arg}$and$\call$terminate on$\mbb{U}$and$\rew_{arg}\call \subseteq{\call}$, Since$\rew_{arg}$and$\call$terminate on$\mbb{U}$and$(\rew_{arg}\call) \subseteq{\call}$, 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${\rew_{arg}}\cup{\call}$. ... ...  ... ... @@ -20,7 +20,7 @@ \begin{proof} Let$\sg$be such that$\sg\vDash\G$. We prove this lemma by induction on$\G\vdash t:T$: \begin{itemize} \begin{description} \item[(axSort)]$\Type\in\interp{\Kind\sg}=\interp{\Kind}$by definition, \item[(axVar)] ... ... @@ -64,5 +64,5 @@ so$\interp{A\sg}=\interp{B\sg}$and since by induction hypothesis$t\sg\in\interp{A\sg}$, we have$t\sg\in\interp{B\sg}\$. \qedhere \end{itemize} \end{description} \end{proof}
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