Commit 96cbc902 by Gaspard Ferey

### Fixed lemma 2.

parent ccb3ba8f
 ... @@ -137,25 +137,19 @@ Note: $$\mathcal{T}_{\mathcal{\emptyset}}$$ is the set of closed terms. ... @@ -137,25 +137,19 @@ Note: $$\mathcal{T}_{\mathcal{\emptyset}}$$ is the set of closed terms. \end{proof} \end{proof} \begin{lemma} \begin{lemma} For any well-formed context\, $\Gamma$, term $t$, term $A$ such that $\Gamma \vdash t : A$, there exists $t'$ such that $t \convbg t'$ and $\tkind$ is not a subterm of $t'$. Let $t$ be a term. If $t$ is typable in a well-formed context then $\tkind$ is not a subterm of $t$. The contrapositive states that if $\tkind$ is a subterm of $t$ then $t$ isn't typable in any well-formed context. \end{lemma} \end{lemma} \begin{proof} \begin{proof} By induction on the derivation of $\Gamma \vdash t : A$. By induction on the height of the derivation of $\Gamma \vdash t : A$. If $\tkind$ is already not a subterm of $t$ we are done. \\ If $t$ is a variable or $\ttype$ then we are done. Otherwise $t$ is neither a variable nor $\ttype$ so only three rules can be applied: Otherwise: \begin{itemize} \begin{itemize} \item \textsc{conv} : \item \textsc{conv} : The induction hypothesis on the first premise gives us a suitable $t'$ such that $t \convbg t'$. By induction hypothesis on the first premise, $\tkind$ is not a subterm of $t$. \item \textsc{app} : $t = u~v$ and $\tkind$ can be a subterm of $u$ or $u$. \item \textsc{app} : $t = u~v$ and by induction hypothesis on the first and second premises, $\tkind$ is neither a subterm of $u$ nor a subterm of $v$ so it can't be a subterm of $t$ either. The induction hypothesis on the first and second premises gives us $u'$ and $v'$ containing no subterm $\tkind$ and such that $u \convbg u'$ and $v \convbg v'$. It follows from Definition~\ref{def:convbg} that $t = u~v \convbg u'~v' = t'$ with $\tkind$ not a subterm of $t'$. \item \textsc{lambda} : $t = \tabs{x}{A}{u}$ and by induction hypothesis \item \textsc{lambda} : $t = \tabs{x}{A}{u}$ and $\tkind$ can be a subterm of $A$ and $u$. on the first and third premises, $\tkind$ is neither a subterm of $A$ nor a subterm of $u$ so it can't be a subterm of $t$ either. The induction hypothesis on the first premise gives us a $\tkind$-free $A'$ such that $A' \convbg A$. Two solutions to conclude for $u$ : \begin{itemize} \item Consider that the induction on the height of the derivation is "forall context $\Gamma$". \item Prove that adding a variable to the context doesn't enrich $\conv$.\\ i.e. $\conv_{\beta\Gamma} \ = \ \conv_{\beta\Gamma, x : A}$. \end{itemize} \end{itemize} \end{itemize} \end{proof} \end{proof} ... ...
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