Add Prepare explanation. Restructure

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JanUlrich 2024-05-17 19:18:53 +02:00
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unify.tex
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@ -25,35 +25,10 @@ Each calculation step of the algorithm is expressed as a transformation rule con
The input is shown above the line, the output below, and additional premises are displayed on the right side.
If the wildcard environment $\wildcardEnv$ and the constraint set $C$ match the pattern declared as the input
the transformation will reduce them into the specified output.
The \rulename{Subst} rule for example takes a constraint set that has atleas one constraint of the form
The \rulename{Subst} rule for example takes a constraint set that has atleast one constraint of the form
$\ntv{a} \doteq \type{T}$ and replaces every occurence of $\ntv{a}$ by $\type{T}$
in the wildcard environment $\wildcardEnv{}$ aswell as the remaining constraint set $C$.
The \rulename{Upper} and \rulename{Lower} conversions (figure \ref{fig:wildcard-rules}) replace wildcards with their respective bounds when appearing in a subtype constraint.
\rulename{Lower} has to check if the wildcard is part of the input wildcards $\Delta_{in}$.
If that is the case the wildcard can be part of the type solution and stays in the constraint set.
\textit{Note:} The subtype constraints in these rules are annotated with numbers $\lessdot_1$.
All rule inputs containing subtype constraints $(\lessdot)$ are always meant for both,
subtype constraints and capture constraints ($\lessdotCC$) aswell.
If multiple constraints are stated in the input format they will be annotated with numbers which map them to the constraints used in the output of the rule.
Constraints with the same number stay the same kind.
So if the input to \rulename{Upper} is $\rwildcard{A} \lessdotCC \type{G}$ the output will be something like $\type{U} \lessdotCC \type{G}$.
If the input is a normal subtype constraint the output has to be a normal subtype constraint too.
%TODO: Rephrase
%The reason being that capture constraints are treated like regular subtype constraints in these rules.
%All transformations for subtype constraints work for capture constraints aswell.
%For clarification Subtype constraints are marked with a number.
But the \rulename{Adopt} rule for example takes multiple subtype constraints and also adds a new one.
Here the numbered annotations are necessary.
\textit{Example:} Having the constraints
$\ntv{a} \lessdotCC \wtv{b}, \ntv{a} \lessdot \type{String}, \wtv{b} \lessdot \type{Object}$
would lead to
$\wtv{b} \lessdot \type{String}, \ntv{a} \lessdotCC \wtv{b}, \ntv{a} \lessdot \type{String}, \wtv{b} \lessdotCC \type{Object}$
after applying \rulename{Adopt}.
Note that the new generated constraint $\wtv{b} \lessdot \type{String}$ is a normal subtype constraint.
%The type placeholders which are annotated as wildcard placeholders also stay wildcard placeholders.
%The only rule that replaces wildcard type placeholders with regular placeholders is the \rulename{Normalize} rule.
\begin{figure}
\begin{center}
\leavevmode
@ -114,6 +89,167 @@ Note that the new generated constraint $\wtv{b} \lessdot \type{String}$ is a nor
\caption{Wildcard reduce rules}\label{fig:wildcard-rules}
\end{figure}
The \rulename{Upper} and \rulename{Lower} conversions (figure \ref{fig:wildcard-rules}) replace wildcards with their respective bounds when appearing in a subtype constraint.
\rulename{Lower} has to check if the wildcard is part of the input wildcards $\Delta_{in}$.
If that is the case the wildcard can be part of the type solution and stays in the constraint set.
\textit{Note:} The subtype constraints in these rules are annotated with numbers $\lessdot_1$.
All rule inputs containing subtype constraints $(\lessdot)$ are always meant for both,
subtype constraints and capture constraints ($\lessdotCC$).
If multiple constraints are stated in the input format they will be annotated with numbers which map them to the constraints used in the output of the rule.
Constraints with the same number stay the same kind.
So if the input to \rulename{Upper} is $\rwildcard{A} \lessdotCC \type{G}$ the output will be something like $\type{U} \lessdotCC \type{G}$.
If the input is a normal subtype constraint the output has to be a normal subtype constraint too.
%TODO: Rephrase
%The reason being that capture constraints are treated like regular subtype constraints in these rules.
%All transformations for subtype constraints work for capture constraints aswell.
%For clarification Subtype constraints are marked with a number.
But the \rulename{Adopt} rule for example takes multiple subtype constraints and also adds a new one.
Here the numbered annotations are necessary.
\textit{Example:} Having the constraints
$\ntv{a} \lessdotCC \wtv{b}, \ntv{a} \lessdot \type{String}, \wtv{b} \lessdot \type{Object}$
would lead to
$\wtv{b} \lessdot \type{String}, \ntv{a} \lessdotCC \wtv{b}, \ntv{a} \lessdot \type{String}, \wtv{b} \lessdotCC \type{Object}$
after applying \rulename{Adopt}.
Note that the new generated constraint $\wtv{b} \lessdot \type{String}$ is a normal subtype constraint.
%The type placeholders which are annotated as wildcard placeholders also stay wildcard placeholders.
%The only rule that replaces wildcard type placeholders with regular placeholders is the \rulename{Normalize} rule.
\begin{figure}
\begin{center}
\leavevmode
\fbox{
\begin{tabular}[t]{l@{~}l}
\rulename{Prepare} %The lessdotCC constraint only ensures that the left side looses its wildcardEnvironment.
%It does not ensure that the left side doesn't contain free variables. If you want to ensure that you have to give the left side a normal placeholder
&
$
\begin{array}[c]{@{}ll}
\begin{array}[c]{l}
\wildcardEnv \vdash
C \cup \, \set{ \wctype{\Delta}{C}{\ol{S}} \lessdot \wctype{\Delta'}{C}{\ol{T}} } \\
\hline
\vspace*{-0.4cm}\\
\wildcardEnv \vdash
C \cup \, \set{ \wctype{\Delta}{C}{\ol{S}} \lessdotCC \wctype{\Delta'}{C}{\ol{T}} } \\
\end{array}
%\quad \ol{Y} = \textit{fresh}(\ol{X})
\quad \begin{array}[c]{l}
\text{fv}(\type{N'}) \subseteq \Delta_{in} \\
\text{wtv}(\type{N'}) = \emptyset
\end{array}
\end{array}
$
\\\\
\rulename{Trim}
&
$
\begin{array}[c]{@{}ll}
\begin{array}[c]{l}
\wildcardEnv \vdash
C \cup \, \set{ \wctype{\Delta,\Delta'}{C}{\ol{S}} \lessdot \type{T} } \\
\hline
\vspace*{-0.4cm}\\
\wildcardEnv \vdash
C \cup \, \set{ \wctype{\Delta}{C}{\ol{S}} \lessdot \type{T} } \\
\end{array}
%\quad \ol{Y} = \textit{fresh}(\ol{X})
\quad \begin{array}[c]{l}
\text{fv}(\ol{S}) \cap \Delta' = \emptyset
\end{array}
\end{array}
$
\\\\
\rulename{Clear}
&
$
\begin{array}[c]{@{}ll}
\begin{array}[c]{l}
\wildcardEnv \cup \set{\wildcard{A}{\type{U}}{\type{L}}} \vdash
C \cup \, \set{ \wctype{\Delta}{C}{\ol{S}} \lessdot \type{T} } \\
\hline
\vspace*{-0.4cm}\\
\subst{\type{U}}{\rwildcard{A}}\wildcardEnv \vdash
[\type{U}/\rwildcard{A}]C \cup \, [\type{U}/\rwildcard{A}]\set{ \wctype{\Delta}{C}{\ol{S}} \lessdot \type{T}, \type{U} \doteq \type{L} } \\
\end{array}
%\quad \ol{Y} = \textit{fresh}(\ol{X})
\quad \begin{array}[c]{l}
\Delta \neq \emptyset\\
\rwildcard{A} \in \text{fv}(\type{T})
\end{array}
\end{array}
$
\\\\
\rulename{Exclude}
&
$
\begin{array}[c]{@{}ll}
\begin{array}[c]{l}
\wildcardEnv \vdash
C \cup \, \set{ \wctype{\Delta}{C}{\ol{S}} \lessdot \type{T} } \\
\hline
\vspace*{-0.4cm}\\
\subst{\tv{a}}{\wtv{a}}\wildcardEnv \vdash
[\tv{a}/\wtv{a}]C \cup \, [\tv{a}/\wtv{a}]\set{ \wctype{\Delta}{C}{\ol{S}} \lessdot \type{T} } \\
\end{array}
%\quad \ol{Y} = \textit{fresh}(\ol{X})
\quad \begin{array}[c]{l}
\Delta \neq \emptyset\\
\wtv{a} \in \text{fv}(\type{T}), \tv{a} \ \text{fresh}
\end{array}
\end{array}
$
\end{tabular}}
\end{center}
\caption{Dealing with wildcard types on the left side of a subtype constraint}
\end{figure}
An existential type can never be a subtype of a regular type:
$\wctype{\rwildcard{X}}{C}{\rwildcard{X}} \nless: \exptype{C}{\rwildcard{X}}$.
This is a problem for constraints of the form $\wctype{\rwildcard{X}}{C}{\rwildcard{X}} \lessdot \type{N}$.
Take the constraint $\wctype{\rwildcard{X}}{C}{\rwildcard{X}} \lessdot \exptype{C}{\wtv{a}}$ for example.
After applying a reduction we get $\rwildcard{X} \doteq \wtv{a}$ which is not a valid solution.
The problem is that we loose the information of the left side being an existential type with the reduction step.
Therefore we have to assure beforehand that errors (like $\rwildcard{X} \doteq \wtv{a}$ in this example) cannot occur.
But we alsod don't want to exclude reductions for these kind of constraints in general.
The solution is to check if there are any free variables or wildcard placeholders on the right side of the constraint.
If that is the case one of the rules \rulename{Trim}, \rulename{Clear}, or \rulename{Exclude} have to be applied.
In our example this would be the \rulename{Exclude} rule replacing the wildcard placeholder with $\wtv{x}$ with a normal placeholder.
Afterwards \rulename{Prepare} can be used eventually leading to the erasure of the wildcard $\rwildcard{X}$ by equalizing its upper and lower bounds:
\begin{displaymath}
\prftree[r]{\rulename{Contract}}{
\prftree[r]{\rulename{Reduce}}{
\prftree[r]{\rulename{Equals}}{
\prftree[r]{\rulename{Reduce}}{
\prftree[r]{\rulename{Capture}}{
\prftree[r]{\rulename{Prepare}}{
\wctype{\wildcard{X}{\tv{u}}{\tv{l}}}{List}{\exptype{List}{\rwildcard{X}}} \lessdot \exptype{List}{\exptype{List}{\ntv{x}}}
}
{
\wctype{\wildcard{X}{\tv{u}}{\tv{l}}}{List}{\exptype{List}{\rwildcard{X}}} \lessdotCC \exptype{List}{\exptype{List}{\ntv{x}}}
}
}
{
\wildcard{X}{\tv{u}}{\tv{l}} \vdash \exptype{List}{\exptype{List}{\rwildcard{X}}} \lessdot \exptype{List}{\exptype{List}{\ntv{x}}}
}
}
{
\wildcard{X}{\tv{u}}{\tv{l}} \vdash \exptype{List}{\rwildcard{X}} \doteq \exptype{List}{\ntv{x}}
}}
{
\wildcard{X}{\tv{u}}{\tv{l}} \vdash \exptype{List}{\rwildcard{X}} \lessdot \exptype{List}{\ntv{x}}, \exptype{List}{\ntv{x}} \lessdot \exptype{List}{\rwildcard{X}}
}
}
{
\wildcard{X}{\tv{u}}{\tv{l}} \vdash \ntv{x} \doteq \rwildcard{X}
}}{
\text{equalize upper and lower bound of }\rwildcard{X}: \quad
\ntv{x} \doteq \tv{u}, \tv{u} \doteq \tv{l}
}
\end{displaymath}
Note that the \rulename{Prepare} rule is always applied together with the \rulename{Capture} and the \rulename{Reduce} rule:
\rulename{Trim} removes unused wildcard declarations
whereas \rulename{Clear} and \rulename{Exclude} remove wildcard placeholders or wildcards to
allow the constraint to be processed by \rulename{Prepare}.
\subsection{Adding Wildcards to the mix}
\unify{} is able to create wildcard solutions even when the input set of constraints do not contain any wildcard variables.
@ -899,95 +1035,6 @@ Prepare, Capture, Reduce, Trim, Clear, Exclude, Adapt
\caption{Constraint reduce rules}\label{fig:reduce-rules}
\end{figure}
\begin{figure}
\begin{center}
\leavevmode
\fbox{
\begin{tabular}[t]{l@{~}l}
\rulename{Prepare} %The lessdotCC constraint only ensures that the left side looses its wildcardEnvironment.
%It does not ensure that the left side doesn't contain free variables. If you want to ensure that you have to give the left side a normal placeholder
&
$
\begin{array}[c]{@{}ll}
\begin{array}[c]{l}
\wildcardEnv \vdash
C \cup \, \set{ \wctype{\Delta}{C}{\ol{S}} \lessdot \wctype{\Delta'}{C}{\ol{T}} } \\
\hline
\vspace*{-0.4cm}\\
\wildcardEnv \vdash
C \cup \, \set{ \wctype{\Delta}{C}{\ol{S}} \lessdotCC \wctype{\Delta'}{C}{\ol{T}} } \\
\end{array}
%\quad \ol{Y} = \textit{fresh}(\ol{X})
\quad \begin{array}[c]{l}
\text{fv}(\wctype{\Delta'}{C}{\ol{T}}) \subseteq \Delta_{in}
\end{array}
\end{array}
$
\\\\
\rulename{Trim}
&
$
\begin{array}[c]{@{}ll}
\begin{array}[c]{l}
\wildcardEnv \vdash
C \cup \, \set{ \wctype{\Delta,\Delta'}{C}{\ol{S}} \lessdot \type{T} } \\
\hline
\vspace*{-0.4cm}\\
\wildcardEnv \vdash
C \cup \, \set{ \wctype{\Delta}{C}{\ol{S}} \lessdot \type{T} } \\
\end{array}
%\quad \ol{Y} = \textit{fresh}(\ol{X})
\quad \begin{array}[c]{l}
\text{fv}(\ol{S}) \cap \Delta' = \emptyset
\end{array}
\end{array}
$
\\\\
\rulename{Clear}
&
$
\begin{array}[c]{@{}ll}
\begin{array}[c]{l}
\wildcardEnv \cup \set{\wildcard{A}{\type{U}}{\type{L}}} \vdash
C \cup \, \set{ \wctype{\Delta}{C}{\ol{S}} \lessdot \type{T} } \\
\hline
\vspace*{-0.4cm}\\
\subst{\type{U}}{\rwildcard{A}}\wildcardEnv \vdash
[\type{U}/\rwildcard{A}]C \cup \, [\type{U}/\rwildcard{A}]\set{ \wctype{\Delta}{C}{\ol{S}} \lessdot \type{T}, \type{U} \doteq \type{L} } \\
\end{array}
%\quad \ol{Y} = \textit{fresh}(\ol{X})
\quad \begin{array}[c]{l}
\Delta \neq \emptyset\\
\rwildcard{A} \in \text{fv}(\type{T})
\end{array}
\end{array}
$
\\\\
\rulename{Exclude}
&
$
\begin{array}[c]{@{}ll}
\begin{array}[c]{l}
\wildcardEnv \vdash
C \cup \, \set{ \wctype{\Delta}{C}{\ol{S}} \lessdot \type{T} } \\
\hline
\vspace*{-0.4cm}\\
\subst{\tv{a}}{\wtv{a}}\wildcardEnv \vdash
[\tv{a}/\wtv{a}]C \cup \, [\tv{a}/\wtv{a}]\set{ \wctype{\Delta}{C}{\ol{S}} \lessdot \type{T} } \\
\end{array}
%\quad \ol{Y} = \textit{fresh}(\ol{X})
\quad \begin{array}[c]{l}
\Delta \neq \emptyset\\
\wtv{a} \in \text{fv}(\type{T}), \tv{a} \ \text{fresh}
\end{array}
\end{array}
$
\end{tabular}}
\end{center}
\caption{Dealing with wildcard types on the left side of a subtype constraint}
\end{figure}
\begin{figure}
If we find an illicit constraint assigning a type containing free variables to a type placeholder not flagged as a wildcard placeholder the algorithm fails.
@ -1275,152 +1322,10 @@ $
\label{fig:generation-rules}
\end{figure}
\begin{figure}
\begin{center}
\fbox{
\begin{tabular}[t]{l@{~}l}
\rulename{Equals} %TODO
& $
\begin{array}[c]{l}
\wildcardEnv \vdash C \cup \, \set{ \wcNtype{\Delta}{N} \doteq \wcNtype{\Delta'}{N'} } \\
\hline
\vspace*{-0.4cm}\\
\wildcardEnv \vdash C \cup \,
\set{
\wcNtype{\Delta}{N} \lessdot \wcNtype{\Delta'}{N'}, \wcNtype{\Delta'}{N'} \lessdot \wcNtype{\Delta}{N}
}
\end{array} %\quad |\Delta| = |\Delta'|
% \quad \text{fv}(\type{N}) = \text{fv}(\type{N'}) = \emptyset
$
\\\\
\ruleReduceWC{}
&
$
\begin{array}[c]{@{}ll}
\begin{array}[c]{l}
\wildcardEnv \vdash
C \cup \, \set{ \exptype{C}{\ol{S}} \lessdot
\wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}} } \\
\hline
\vspace*{-0.4cm}\\
\wildcardEnv
\vdash C \cup \, \set{
\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}
%\quad \ol{Y} = \textit{fresh}(\ol{X})
\quad \begin{array}[c]{l}
\ol{\wtv{a}} \ \text{fresh}\\
%\text{fv}(\exptype{C}{\ol{S}}) \subseteq \text{dom}(\overline{\wildcard{B}{\type{U'}}{\type{L'}}})
%\text{dom}(\overline{\wildcard{A}{\type{U}}{\type{L}}}) \subseteq \text{fv}(\exptype{C}{\ol{T}}) \\
%\text{fv}(\wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}}) = \emptyset
\end{array}
\end{array}
$
\\\\
\rulename{Capture}
&
$
\begin{array}[c]{@{}ll}
\begin{array}[c]{l}
\wildcardEnv \vdash
C \cup \, \set{ \wctype{\overline{\wildcard{B}{\type{U}}{\type{L}}}}{C}{\ol{S}} \lessdotCC \type{T} } \\
\hline
\vspace*{-0.4cm}\\
\wildcardEnv \cup \overline{\wildcard{C}{[\ol{\rwildcard{C}}/\ol{\rwildcard{B}}]\type{U}}{[\ol{\rwildcard{C}}/\ol{\rwildcard{B}}]\type{L}}}
\vdash C \cup \, \set{
[\ol{\rwildcard{C}}/\ol{\rwildcard{B}}] \exptype{C}{\ol{S}} \lessdot \type{T} }
\end{array}
%\quad \ol{Y} = \textit{fresh}(\ol{X})
\quad \begin{array}[c]{l}
\ol{\rwildcard{C}} \ \text{fresh}\\
%\text{fv}(\type{T}) \neq \emptyset
\end{array}
\end{array}
$
\\\\
\rulename{Prepare} %The lessdotCC constraint only ensures that the left side looses its wildcardEnvironment.
%It does not ensure that the left side doesn't contain free variables. If you want to ensure that you have to give the left side a normal placeholder
&
$
\begin{array}[c]{@{}ll}
\begin{array}[c]{l}
\wildcardEnv \vdash
C \cup \, \set{ \type{N} \lessdot \type{N'} } \\
\hline
\vspace*{-0.4cm}\\
\wildcardEnv \vdash
C \cup \, \set{ \type{N} \lessdotCC \type{N'} } \\
\end{array}
%\quad \ol{Y} = \textit{fresh}(\ol{X})
\quad \begin{array}[c]{l}
\text{fv}(\type{N'}) \subseteq \Delta_{in} \\
\text{wtv}(\type{N'}) = \emptyset
\end{array}
\end{array}
$
\end{tabular}}
\end{center}
\caption{Type Reduction rules}\label{fig:reductionRules}
\end{figure}
\subsection{Explanations}
%The type reduction is done by the rules in figure \ref{fig:reductionRules}
\begin{description}
\item[Prepare]
The \rulename{Prepare} transformation is always applied together with the \rulename{Reduce} transformation.
\unify{} cannot reduce constraints without checking a few prerequisites.
%Take the constraint $\wctype{\rwildcard{X}}{C}{\rwildcard{X}} \lessdot \exptype{C}{\wtv{a}}$ for example.
%If we apply a reduction here we get $\rwildcard{X} \doteq \wtv{a}$.
%The resulting $\sigma(\wtv{a}) = \rwildcard{X}$ seems like a correct substitution,
%but by S-Exists $\wctype{\rwildcard{X}}{C}{\rwildcard{X}} \nless: \exptype{C}{\rwildcard{X}}$.
%Reason: Free variables on the right side of a subtype relations are not allowed to show up as bound variables on the left side.
%$\rwildcard{X}$ in this case.
%Therefore the \rulename{Reduce} rule only reduces constraints where the left side does not declare any wildcards.
%But if the right side neither contains wildcard type placeholders nor free variables the constraint can be reduced anyways.
%The \rulename{Prepare} rule then converts this constraint to a capture constraint.
%Afterwards the \rulename{Capture} rule removes the wildcard declarations on the left side an the constraint can be reduced.
%We loose information during the unification process.
When reducing the constraint
$\wctype{\rwildcard{X}}{List}{\exptype{List}{\rwildcard{X}}} \lessdot \exptype{List}{\exptype{List}{\wtv{x}}}$
it turns into $\exptype{List}{\rwildcard{X}} \doteq \exptype{List}{\wtv{x}}$
and now it seems that $\wtv{x} \doteq \rwildcard{X}$ is a correct solution.
This is indeed wrong because $\wctype{\rwildcard{X}}{List}{\exptype{List}{\rwildcard{X}}} \nless: \exptype{List}{\exptype{List}{\rwildcard{X}}}$.
A correct solution is to remove the wildcard $\rwildcard{X}$ if possible.
% X.List<X> <. List<x> // incorrect
% X.List<List<X>> <. List<List<x>> // incorrect
Therefore the \rulename{Prepare} rule checks if there are any free variables or wildcard placeholders on the right side of the constraint.
If that is the case one of the rules \rulename{Trim}, \rulename{Clear}, or \rulename{Exclude} have to be applied.
In our example this would be the \rulename{Exclude} rule replacing the wildcard placeholder with $\wtv{x}$ with a normal placeholder.
Afterwards \rulename{Prepare} can be used eventually leading to the erasure of the wildcard $\rwildcard{X}$ by equalizing its upper and lower bounds:
\begin{displaymath}
\prftree[r]{\rulename{Contract}}{
\prftree[r]{\rulename{Reduce}}{
\prftree[r]{\rulename{Equals}}{
\prftree[r]{\rulename{Prepare}}{
\wctype{\wildcard{X}{\tv{u}}{\tv{l}}}{List}{\exptype{List}{\rwildcard{X}}} \lessdot \exptype{List}{\exptype{List}{\ntv{x}}}
}
{
\wildcard{X}{\tv{u}}{\tv{l}} \vdash \exptype{List}{\rwildcard{X}} \doteq \exptype{List}{\ntv{x}}
}}
{
\exptype{List}{\rwildcard{X}} \lessdot \exptype{List}{\ntv{x}}, \exptype{List}{\ntv{x}} \lessdot \exptype{List}{\rwildcard{X}}
}
}
{
\wildcard{X}{\tv{u}}{\tv{l}} \vdash \ntv{x} \doteq \rwildcard{X}
}}{
\text{equalize upper and lower bound of }\rwildcard{X}: \quad
\ntv{x} \doteq \tv{u}, \tv{u} \doteq \tv{l}
}
\end{displaymath}
\end{description}
\subsection{Examples}
\textit{Example} of the type reduction rules in figure \ref{fig:reductionRules} with the input
$\wctype{\rwildcard{X}}{Pair}{\wctype{\rwildcard{Y}}{List}{\rwildcard{Y}}, \rwildcard{X}} \lessdot \exptype{Pair}{\wctype{\rwildcard{Y}}{List}{\rwildcard{Y}}, \wtv{a}}$