stan/Ecoop2024_TIforWildFJ
1
0
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity

Change Prepare rule to simpler version

Browse Source
This commit is contained in:
Andreas Stadelmeier 2024-01-16 18:03:54 +01:00
parent 521dee7fa2
commit bc4fcaf43c
2 changed files with 45 additions and 20 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

53
soundness.tex
View File

@ -196,14 +196,16 @@ This rule is only applied for the outer wildcard environments for each type.
\begin{lemma} \begin{lemma}
The \unify{} algorithm only produces correct output for constraints not containing free variables. The \unify{} algorithm only produces correct output for constraints not containing free variables.
\begin{description} \begin{description}
\item[If] $(\sigma, \Delta) = \unify{}( \overline{ \type{S} \lessdot \type{T} } \cup \overline{ \type{S'} \lessdotCC \type{T'} } )$ %\cup \overline{ \type{S} \doteq \type{S'} })$ \item[If] $(\sigma, \Delta) = \unify{}( \Delta', \, \overline{ \type{S} \lessdot \type{T} } \cup \overline{ \type{S'} \lessdotCC \type{T'} } )$ %\cup \overline{ \type{S} \doteq \type{S'} })$
%\item[and] $fv(\overline{ \type{S} }) = \emptyset$, $fv(\overline{ \type{T} }) = \emptyset$ %\item[and] $fv(\overline{ \type{S} }) = \emptyset$, $fv(\overline{ \type{T} }) = \emptyset$
\item[Then] there exists a $\Delta'$ with: \item[Then] there exists a $\sigma'$ with:
$\Delta \vdash \overline{\sigma(\type{S}) <: \sigma(\type{T})}$ $\sigma \subseteq \sigma'$ and
and $\Delta, \Delta' \vdash \overline{\CC{}(\sigma(\type{S'})) <: \sigma(\type{T'})}$ $\Delta, \Delta' \vdash \overline{\sigma'(\type{S}) <: \sigma'(\type{T})}$
and $\Delta, \Delta', \overline{\Delta} \vdash \overline{\type{N} <: \sigma'(\type{T'})}$ where $\overline{\sigma(\type{S'}) = \wcNtype{\Delta}{N}}$,
otherwise $\Delta, \Delta' \vdash \overline{\sigma'(\type{S'}) <: \sigma'(\type{T'})}$
% and $\sigma(\type{T'}) = \sigma(\type{T'})$. % and $\sigma(\type{T'}) = \sigma(\type{T'})$.
The function $\CC{}$ is given as $\CC{}(\wcNtype{\Delta}{N}) = \type{N}$ % The function $\CC{}$ is given as $\CC{}(\wcNtype{\Delta}{N}) = \type{N}$
% Unify cannot guarantee that only wildcards declared on the left side are used. Example input: % Unify cannot guarantee that only wildcards declared on the left side are used. Example input:
@ -271,7 +273,25 @@ $\Delta \vdash \sigma(C') \implies \Delta \vdash \sigma(C)$
%\item[Capture, Reduce] are always applied together. We have to destinct between two cases: %\item[Capture, Reduce] are always applied together. We have to destinct between two cases:
\item[Prepare] \item[Prepare]
To show To show
$\Delta \vdash \wctype{\overline{\wildcard{B}{\type{U'}}{\type{L'}}}}{C}{\ol{S}} <: \wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}}$ by S-Exists we have to proof: $\Delta \vdash \sigma(\wctype{\overline{\wildcard{B}{\type{U}}{\type{L}}}}{C}{\ol{S}}) <: \sigma(\wcNtype{\Delta''}{N})$ by S-Exists we have to proof:
We set $\Delta' = \sigma(\wildcardEnv)$.
\begin{gather}
\Delta', \Delta \vdash [\ol{T}/\ol{\type{A}}]\ol{L} <: \ol{T} \\
\Delta', \Delta \vdash \ol{T} <: [\ol{T}/\ol{\type{X}}]\ol{U} \\
\label{rp:3}
\text{fv}(\ol{T}) \subseteq \text{dom}(\Delta, \overline{\wildcard{B}{U}{L}}) \\
\label{rp:4}
\text{dom}(\overline{\wildcard{B}{U}{L}}) \cap \text{fv}(\wctype{\ol{\wildcard{A}{U}{L}}}{C}{\ol{T}}) = \emptyset\\
[\ol{A}/\ol{T}] = \ol{T} %TODO: rename T
\end{gather}
\item[Prepare]
To show
$\Delta \vdash \sigma(\wctype{\overline{\wildcard{B}{\type{U'}}{\type{L'}}}}{C}{\ol{S}}) <: \sigma(\wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}})$ by S-Exists we have to proof:
\begin{gather} \begin{gather}
\Delta', \Delta \vdash [\ol{T}/\ol{\type{A}}]\ol{L} <: \ol{T} \\ \Delta', \Delta \vdash [\ol{T}/\ol{\type{A}}]\ol{L} <: \ol{T} \\
\Delta', \Delta \vdash \ol{T} <: [\ol{T}/\ol{\type{X}}]\ol{U} \\ \Delta', \Delta \vdash \ol{T} <: [\ol{T}/\ol{\type{X}}]\ol{U} \\
@ -284,11 +304,12 @@ $\Delta \vdash \wctype{\overline{\wildcard{B}{\type{U'}}{\type{L'}}}}{C}{\ol{S}}
We know We know
\begin{gather} \begin{gather}
\ol{S} = [\ol{\wtv{a}}/\ol{A}]\ol{T}\\ \sigma(\ol{S}) = [\ol{\wtv{a}}/\ol{A}]\ol{T}\\
\ol{\wtv{a}} \lessdot [\ol{\wtv{a}}/\ol{A}]\ol{U}\\
[\ol{\wtv{a}}/\ol{A}]\ol{L} \lessdot \ol{\wtv{a}}\\
\text{fv}(\wctype{\overline{\wildcard{B}{\type{U'}}{\type{L'}}}}{C}{\ol{S}}) = \emptyset\\ \text{fv}(\wctype{\overline{\wildcard{B}{\type{U'}}{\type{L'}}}}{C}{\ol{S}}) = \emptyset\\
\label{rp:fv2} \label{rp:fv2}
\text{fv}(\wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}}) = \emptyset \text{fv}(\wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}}) = \emptyset
%TODO
\end{gather} \end{gather}
\ref{rp:fv2} implies \ref{rp:4}. \ref{rp:fv2} implies \ref{rp:4}.
@ -297,10 +318,18 @@ $\text{fv}(\type{T}) \subseteq \text{dom}(\overline{\wildcard{B}{\type{U}}{\type
and therefore \ref{rp:3}. and therefore \ref{rp:3}.
\item[Capture] \item[Capture]
If $\text{fv}(\wctype{\Delta}{C}{\ol{T}}) = \emptyset$ the preposition holds by Assumption and S-Exists. Everytime the \rulename{Capture} rule is invoked we add the freshly generated free variables to the global environment $\wildcardEnv$.
If $\text{fv}(\wctype{\Delta}{C}{\ol{T}}) \neq \emptyset$ the preposition holds because They are from this point on treated like global variables and therefore we can assume $\sigma(\wildcardEnv \cup \ol{\wildcard{C}{U}{L}}) \subseteq \Delta'$ in
$\Delta, \Delta' \vdash [\ol{\rwildcard{C}}/\ol{\rwildcard{B}}] \exptype{C}{\ol{S}} <: \wctype{\Delta}{C}{\ol{T}}$ $\implies$ $\Delta, \Delta' \vdash \sigma([\ol{\rwildcard{C}}/\ol{\rwildcard{B}}] \exptype{C}{\ol{S}})
$\Delta' \Delta \vdash \text{CC}(\sigma(\wctype{\overline{\wildcard{B}{\type{U'}}{\type{L'}}}}{C}{\ol{S}})) <: \sigma(\wctype{\Delta}{C}{\ol{T}})$ <: \sigma(\wctype{\Delta}{C}{\ol{T}})$,
which implies $\Delta, \Delta'/\set{\ol{\wildcard{C}{U}{L}}} \vdash \type{N} <: \sigma(\wctype{\Delta}{C}{\ol{T}})$,
with $\wcNtype{\overline{\wildcard{C}{\type{U}}{\type{L}}}}{N} = \sigma(\wctype{\overline{\wildcard{C}{\type{U}}{\type{L}}}}{C}{\ol{S}})$
We can rename a fresh wildcard $\rwildcard{C}$ freely.
% If $\text{fv}(\wctype{\Delta}{C}{\ol{T}}) = \emptyset$ the preposition holds by Assumption and S-Exists.
% If $\text{fv}(\wctype{\Delta}{C}{\ol{T}}) \neq \emptyset$ the preposition holds because
% $\Delta, \Delta' \vdash [\ol{\rwildcard{C}}/\ol{\rwildcard{B}}] \exptype{C}{\ol{S}} <: \wctype{\Delta}{C}{\ol{T}}$ $\implies$
% $\Delta' \Delta \vdash \text{CC}(\sigma(\wctype{\overline{\wildcard{B}{\type{U'}}{\type{L'}}}}{C}{\ol{S}})) <: \sigma(\wctype{\Delta}{C}{\ol{T}})$
%we have to show $\Delta' \vdash \text{CC}(\sigma(\wctype{\overline{\wildcard{B}{\type{U'}}{\type{L'}}}}{C}{\ol{S}})) <: \sigma(\wctype{\Delta}{C}{\ol{T}})$, %we have to show $\Delta' \vdash \text{CC}(\sigma(\wctype{\overline{\wildcard{B}{\type{U'}}{\type{L'}}}}{C}{\ol{S}})) <: \sigma(\wctype{\Delta}{C}{\ol{T}})$,
%which holds by assumption with $\Delta'$ chosen in a way that $\text{fv}(\exptype{C}{\ol{S}}) \subseteq \Delta'$. The variables $\ol{C}$ in $\ol{S}$ can be renamed to $\ol{B}$, because $\ol{C}$ are fresh. %which holds by assumption with $\Delta'$ chosen in a way that $\text{fv}(\exptype{C}{\ol{S}}) \subseteq \Delta'$. The variables $\ol{C}$ in $\ol{S}$ can be renamed to $\ol{B}$, because $\ol{C}$ are fresh.

12
unify.tex
View File

@ -380,19 +380,15 @@ Their upper and lower bounds are fresh type variables.
\begin{array}[c]{@{}ll} \begin{array}[c]{@{}ll}
\begin{array}[c]{l} \begin{array}[c]{l}
\wildcardEnv \vdash \wildcardEnv \vdash
C \cup \, \set{ \wctype{\overline{\wildcard{B}{\type{U'}}{\type{L'}}}}{C}{\ol{S}} \lessdot \wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}} } \\ C \cup \, \set{ \wcNtype{\Delta}{S} \lessdot \wcNtype{\Delta'}{N} } \\
\hline \hline
\vspace*{-0.4cm}\\ \vspace*{-0.4cm}\\
\wildcardEnv \cup \overline{\wildcard{B}{\type{U'}}{\type{L'}}} \wildcardEnv \vdash
\vdash C \cup \, \set{ C \cup \, \set{ \wcNtype{\Delta}{S} \lessdotCC \wcNtype{\Delta'}{N} } \\
\ol{\type{S}} \doteq [\ol{\wtv{a}}/\overline{\rwildcard{A}}]\ol{\type{T}},
\ol{\wtv{a}} \lessdot [\ol{\wtv{a}}/\overline{\rwildcard{A}}]\ol{U}, [\ol{\wtv{a}}/\overline{\rwildcard{A}}]\ol{L} \lessdot \ol{\wtv{a}} }
\end{array} \end{array}
%\quad \ol{Y} = \textit{fresh}(\ol{X}) %\quad \ol{Y} = \textit{fresh}(\ol{X})
\quad \begin{array}[c]{l} \quad \begin{array}[c]{l}
\ol{\rwildcard{C}} \ \text{fresh}\\ \text{fv}(\wcNtype{\Delta}{S}, \wcNtype{\Delta'}{N}) = \emptyset
\text{fv}(\wctype{\overline{\wildcard{B}{\type{U}}{\type{L}}}}{C}{\ol{S}}) = \emptyset\\
\text{fv}(\wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}}) = \emptyset
\end{array} \end{array}
\end{array} \end{array}
$ $