Cleanup soundness. Add Delta environment to Unify input

soundness.tex

@@ -13,6 +13,10 @@ Type inference produces a correctly typed program.
 \end{description}
 \end{theorem}
 
+\begin{lemma}{Well-formedness:}
+TODO: 
+\end{lemma}
+
 \begin{lemma}{Soundness:}
     \unify{}'s type solutions for a constraint set generated by $\typeExpr{}$ are correct.
     \begin{description}
@@ -44,7 +48,7 @@ By structural induction over the expression $\texttt{e}$.
     then $\Delta|\Gamma  \vdash \texttt{e} : \wcNtype{\Delta_c}{N}$ by assumption.
     $\Delta', \Delta, \Delta_c \vdash \type{N} <: \sigma(\exptype{C}{\overline{\wtv{a}}})$ by premise.
     %Let $\sigma(\tv{r}) = \wcNtype{\Delta'}{N}$.
-    Let $\sigma([\ol{\wtv{a}}/\ol{X}]\type{T}) = \wcNtype{\Delta_t}{N_t}$.
+    %Let $\sigma([\ol{\wtv{a}}/\ol{X}]\type{T}) = \wcNtype{\Delta_t}{N_t}$.
     
     The completion of $|\texttt{e}.\texttt{f}|$ is $\texttt{let}\ \texttt{x} = \texttt{e} : \wcNtype{\Delta_c}{N}\ \texttt{in} \ \texttt{x}.\texttt{f}$
     
@@ -63,6 +67,8 @@ By structural induction over the expression $\texttt{e}$.
     %Easy, because unify only generates substitutions for normal type placeholders which are OK
 
     \item[$\texttt{e}.\texttt{m}(\ol{e})$]
+Lets have a look at the case where the receiver and parameter types are all named types.
+So $\sigma(\ol{\tv{r}}) = \ol{T} = \ol{\wcNtype{\Delta}{N}}$ and $\sigma(\tv{r}) = \type{T}_r = \wcNtype{\Delta'}{N}$:
     \begin{gather}
         \label{sp:1}
         \Delta, \Delta', \overline{\Delta} \vdash \ol{N} <: [\ol{S}/\ol{X}]\ol{U} \\
@@ -70,7 +76,7 @@ By structural induction over the expression $\texttt{e}$.
         \Delta, \Delta', \overline{\Delta} \vdash \ol{S} <: [\ol{S}/\ol{X}]\ol{U'} \\
         \Delta, \Delta', \overline{\Delta} \vdash \ol{S} \ \ok \\
         \label{sp:4}
-        \Delta \vdash \ol{T} <: \ol{\wcNtype{\Delta}{N}} \\
+        \Delta, \Delta' \vdash \ol{T} <: \ol{\wcNtype{\Delta}{N}} \\
     \end{gather}
     \ref{sp:1} is guaranteed by the constraints $\ol{r} \lessdot \ol{T}$.
     $\ol{\tv{r}} \lessdot \ol{T}$ says that there is a $\Delta'$ with 
@@ -78,6 +84,12 @@ By structural induction over the expression $\texttt{e}$.
 
     If $\overline{\sigma(\tv{r}) = \wcNtype{\Delta}{N}}$
     then \ref{sp:1} due to lemma \ref{lemma:unifyCC}.
+    \ref{sp:4} is not using $\ol{\Delta}$.
+    $\sigma(\ol{\tv{r}}) = \ol{T} = \ol{\wcNtype{\Delta}{N}}$
+
+Method calls generate multiple constraints that share the same wildcard placeholders ($\ol{\wtv{a}}$, $\ol{\wtv{b}}$).
+%TODO: show that only those wildcards in the parameters and receiver type are used ($\ol{\Delta}$)
+
     % X.List<X> <. List<a?>
     % set \sigma(r) = X.N
     % If there exists a subtype r <: T, then r <: X.N, N <: T with X.N == r
@@ -100,73 +112,32 @@ By structural induction over the expression $\texttt{e}$.
     % If $\ol{\tv{r}} \lessdot \ol{T}$ how can we say that \Delta' only uses
 \end{description}
 
-\begin{lemma}
+% \begin{lemma} Unify does add free variables to types not containing free variables (or wildcard placeholders)
-\begin{description}
-    \item[If] $(\sigma, \Delta) = \unify{}( C )$
-    \item[Then] $\forall x \in \set{\type{S} \mid \type{S} \in C, \text{fv}(\type{S}) = \emptyset }: \text{fv}(\sigma(\type{S})) \subseteq \text{dom}(\Delta)$
-\end{description}
-\end{lemma}
-\textit{Proof:} by induction over the \unify{} algorithm.
-A unifier $\sigma(\tv{a}) = \type{T}$, where the type variable $\tv{a}$ is not flagged as a wildcard will always hold
-$\text{fv}(\type{T}) \subseteq \text{dom}(\Delta)$.
-%UNIFY fails when there are free variables on the right side of a a =. T constraint
-
-\begin{lemma}
-    \unify{} only produces well-formed type substitutions.
-    % Could this be done via the GenSigma rule? just check if substitutions are well-formed?
-    % L <. U
-    % not possible because well-formedness depends on the input, which is not checked (should we check it? is that possible?)
-    %only type variables have to be OK
-
-    % TODO: Problem: how to proof \ol{S} ok for T-Call.
-    % a <. List<x?> --> here the type substituted for x? must be ok. Proof that by changing Unify to not return type insertions for this type and assume there is a correct substitution for x
-\end{lemma}
-
-% %This lemma can be used to proof Normalize rule!
-% \begin{lemma}\label{lemma:wildcardReplacement}
-%     Wildcards with the same upper and lower bound can be replaced by their bounds without breaking subtype relations.
-%     \begin{description}
-%         \item[If] $\Delta \cup \set{\wildcard{X}{\type{U}}{\type{L}}} \vdash \type{T} <: \type{S}$
-%         \item[and] $\type{U} = \type{L}$ and $\text{fv}(\type{U}) = \emptyset$
-%         \item[Then] $\Delta \vdash [\type{U}/\type{X}]\type{T} <: [\type{U}/\type{X}]\type{S}$
-%         \end{description}
-% \end{lemma}
-% \textit{Proof:} %TODO
-% %By structural induction over the subtype relation
-% %S-Refl: by assumption L <: L implies [\type{U}/\type{X}]L
-
-% \begin{lemma}\label{lemma:noAdditionalFV}
-%     Type solution $\sigma$ does not add additional free variables.
-%     \begin{description}
-%         \item[If] $(\sigma, \Delta) = \unify{}( \overline{ \type{S} \lessdot \type{T} } \cup C)$
-%         \item[and] $fv(\overline{ \type{S} }) = \emptyset$, $fv(\overline{ \type{T} }) = \emptyset$
-%         \item[Then] $fv(\sigma(\overline{ \type{S} })) = \emptyset, fv(\sigma(\overline{ \type{T} })) = \emptyset$
-%         \end{description}
-% \end{lemma}
-
-% \begin{lemma}
-% % https://math.libretexts.org/Bookshelves/Mathematical_Logic_and_Proof/Book%3A_Mathematical_Reasoning__Writing_and_Proof_(Sundstrom)/07%3A_Equivalence_Relations/7.03%3A_Equivalence_Classes
-% $\doteq$ is an equivalence relation on a constraint set $C$.
-
 % \begin{description}
-% \item[If] $\Delta \vdash \overline{\type{T} <: \type{S}}$ and $\type{T} \doteq \type{S}$
+%     \item[If] $(\sigma, \Delta) = \unify{}( \Delta', C )$
-% \item[Then] $\Delta \vdash [\type{T}/\type{S}](\overline{\type{T} <: \type{S}})$
+%     \item[Then] $\forall x \in \set{\type{S} \mid \type{S} \in C, \text{fv}(\type{S}) = \emptyset }: \text{fv}(\sigma(\type{S})) \subseteq \text{dom}(\Delta)$
 % \end{description}
 % \end{lemma}
+% \textit{Proof:} by induction over the \unify{} algorithm.
+% A unifier $\sigma(\tv{a}) = \type{T}$, where the type variable $\tv{a}$ is not flagged as a wildcard will always hold
+% $\text{fv}(\type{T}) \subseteq \text{dom}(\Delta)$.
+% %UNIFY fails when there are free variables on the right side of a a =. T constraint
 
 \begin{lemma} \label{lemma:tvsNoFV}
     \unify{} does not add free variables to types not containing free variables.
     \begin{description}
-        \item[If] $(\sigma, \Delta) = \unify{} (C)$
+        \item[If] $(\sigma, \Delta) = \unify{} (\Delta', C)$
         \item[and] $\tv{a}$ being a type placeholders used in $C$
-        \item[then] $\text{fv}(\sigma(\tv{a})) = \emptyset$
+        \item[then] $\text{fv}(\sigma(\tv{a})) \subseteq \Delta, \Delta'$
     \end{description}
 \end{lemma}
 \textit{Proof:}
-Trivial
+Trivial. \unify{} fails when a constraint $\tv{a} \doteq \rwildcard{X}$ arises.
 
 \begin{lemma} \label{lemma:unifyCC}
 Free variables do not leave their scope. TODO
+Wildcard variables are only substituted by wildcards inside the same constraint.
+
 \begin{description}
     \item[If] $(\sigma, \Delta) = \unify{}( C \cup \set{ \type{T} \lessdot \type{S} } \cup \set{ \overline{\type{T} \lessdotCC \type{S} } } )$
     \item[and] $\text{fv}(\type{T}, \type{S}) \subseteq \text{fv}(\ol{T}, \ol{S})$, 
@@ -180,11 +151,6 @@ Free variables do not leave their scope. TODO
 
 Their scope is the set of constraints, which contain the same free variable placeholders.
 
-    % Wildcard place holders only use free variables occuring in the same constraint as them.
-    % \begin{description}
-    %     \item[If] $(\sigma, \Delta) = \unify{}( C )$
-    %     \item[Then] $\text{fv}(\sigma(\tv{a})) \subseteq \set{ \text{fv}(\type{T}) \cup \text{fv}(\type{S}) \mid \tv{a} \in \text{tph}(\type{T},\type{S}), (\type{T} \lessdot \type{S}) \in C }$
-    % \end{description}
 \end{lemma}
 \textit{Proof:}
 %TODO: is it true even?
@@ -192,20 +158,6 @@ The input set does not contain free variables, only free variable placeholders.
 The only time \unify{} add a wildcard to the $\wildcardEnv$ is the \rulename{Capture} rule.
 This rule is only applied for the outer wildcard environments for each type.
 
-%where to get a Box<Box<a?>> ?
-
-% Box<Box<a?>> <. Box<? super Box<?>>
-
-% %What is with wildcards in the lower bounds of type
-% Box<Box<a?>> <. X_Y.Box<Y>.Box<X>
-% Y.Box<Y> <. Box<a?>
-% Y =. a?
-
-% Box<Y.Box<Y>> <. Box<? extends Box<a>>
-% Y.Box<Y> <. Box<a>
-
-% Problem: für tvs dürfen keine Wildcards eingesetzt werden, außer für die 
-
 \begin{lemma}\label{lemma:unifySoundness}
 The \unify{} algorithm only produces correct output.
 \begin{description}
@@ -213,14 +165,13 @@ The \unify{} algorithm only produces correct output.
 %\item[and] $fv(\overline{ \type{S} }) = \emptyset$, $fv(\overline{ \type{T} }) = \emptyset$
 \item[Then] there exists a $\sigma'$ with:
 $\sigma \subseteq \sigma'$ and
-$\Delta, \Delta' \vdash \overline{\sigma'(\type{S}) <: \sigma'(\type{T})}$
+$\Delta, \Delta', \overline{\Delta} \vdash \overline{\sigma'(\type{S}) <: \sigma'(\type{T})}$
 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 $\sigma(\type{T'}) = \sigma(\type{T'})$.
 
 % 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:
 % List<b?> <. List<b?>
 % b? =. X

unify.tex

@@ -5,7 +5,8 @@
 The \unify{} algorithm computes the type solution.
 
 \begin{description}
-  \item[Input:] List of constraints $C = \set{ \type{T} \lessdot \type{T}, \type{T} \doteq \type{T} \ldots}$
+  \item[Input:] An environment $\Delta'$
+and a List of constraints $C = \set{ \type{T} \lessdot \type{T}, \type{T} \doteq \type{T} \ldots}$
 
 The input constraints must be of the following format:
 
@@ -18,6 +19,7 @@ The input constraints must be of the following format:
 \noindent
 Additional requirements:
 \begin{itemize}
+\item Every wildcard $\rwildcard{X} \in \Delta'$ has to have an unique name which is not defined anywhere in the constraint set $C$.
   \item The input only consists of $\lessdot$ constraints
 %   \item No free variables in type parameters.
 % A constraint like $\tv{a} \lessdot \exptype{List}{\rwildcard{X}}$ is prohibited.
@@ -91,7 +93,7 @@ C \cup \set{\tv{a} \doteq \type{T}}
                      \end{array}
                    \quad \begin{array}{c}
                     \tv{a} \notin \type{T} \\
-                    \text{fv}(\type{T}) = \emptyset
+                    \text{fv}(\type{T}) \subseteq \Delta'
                    \end{array}$\\
 \\
 \rulename{Subst-WC} &$