 ### Fix The Rel Rule

parent a195aa7d
 ... ... @@ -57,7 +57,7 @@ Note: $$\mathcal{T}_{\mathcal{\emptyset}}$$ is the set of closed terms. \begin{rules}{typing}{Typing rules} \inferrule*{ } {\wf{\emptyset}} {\wf{[]}} \and \inferrule*{ \wf{\Gamma} ... ... @@ -76,7 +76,7 @@ Note: $$\mathcal{T}_{\mathcal{\emptyset}}$$ is the set of closed terms. {\wf{\Gamma, \mathcal{R}}} \and \inferrule*[Right=Rel]{ \Gamma \vdash A : \ttype \Gamma \vdash A : s \\ \Gamma \vdash t : A \\ ... ... @@ -118,7 +118,7 @@ Note: $$\mathcal{T}_{\mathcal{\emptyset}}$$ is the set of closed terms. } {\Gamma \vdash t~u : \subst{B}{x}{u}} \and \inferrule*{ \inferrule*[Right=conv]{ \Gamma \vdash t : A \\ \Gamma \vdash B : s ... ... @@ -129,12 +129,28 @@ Note: $$\mathcal{T}_{\mathcal{\emptyset}}$$ is the set of closed terms. \end{rules} \begin{lemma} There is no context $$\Gamma$$ and term $$A$$ such that $$\Gamma \vdash \tkind : A$$ is derivable. \end{lemma} \begin{proof} By induction on the derivation of $$\Gamma \vdash \tkind : A$$. The only rule that can be applied is \textsc{conv}. We conclude this case by applying the induction hypothesis on the first premise. \end{proof} \begin{lemma} For any context $$\Gamma$$, if $$\tkind \convbg t$$ then $$t = \tkind$$. \end{lemma} \begin{proof} By induction on the length of $$\Gamma$$, we can prove that there is no $$u$$ such that $$\Gamma \vdash \wf{(\tkind,u)}$$ or the symmetrical case are derivable. From this, we can conclude the statement. \todo{Peut-on le prouver sans induction ?} By induction on the length of $$\Gamma$$. \begin{itemize} \item base case: $$\Gamma$$ is $$\empty$$ and the $$\beta$$ rule cannot be applied \item inductive case: \begin{itemize} \item $$\Gamma = \Gamma, x : A$$, then the congruence relation is the same \item $$\Gamma = \Gamma, (t,u)$$, TODO \end{itemize} \end{itemize} we can prove that there is no $$u$$ such that $$\Gamma \vdash \wf{(\tkind,u)}$$ or the symmetrical case are derivable. \end{proof} \begin{lemma} ... ...
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