Adding to Soundness proof: Adding the old Unify-Soundness. Start writing lemmas to proof free variables cannot travel out of the scope of a let statement. They can only travel one hop at a time -> cant go through normal type placeholders

soundness.tex

@@ -86,11 +86,17 @@ We have to show T-Let and T-Call which leaves us with:
 \begin{itemize}
     \item $\Delta | \Gamma \vdash \expr{e} : \sigma(\tv{e})$ and $\Delta | \Gamma \vdash \overline{\expr{e} : \sigma(\tv{e})}$ by assumption
     \item $\Delta \vdash \type{T}_1 <: \wcNtype{\Delta'}{N}$ and $\Delta \vdash \overline{\type{T}_1 <: \wcNtype{\Delta'}{N}}$ by constraints $\tv{e} \lessdot \tv{x}$, $\overline{\tv{e} \lessdot \tv{x}}$ and lemma \ref{lemma:unifySoundness}
-    \item $\Delta, \Delta', \overline{\Delta'} | \Gamma, \expr{x} : \type{N}, \overline{\expr{x} : \type{N}} \vdash \expr{x}.\texttt{m}(\ol{x}) : \sigma(\tv{r})$ by T-Call, because the following promises are satisfied
+    \item $\Delta, \Delta', \overline{\Delta'} | \Gamma, \expr{x} : \type{N}, \overline{\expr{x} : \type{N}} \vdash \expr{x}.\texttt{m}(\ol{x}) : \sigma(\tv{r})$ by T-Call and lemma \ref{lemma:freeVariableScope}, because the following promises are satisfied.
+    Note: The lemma \ref{lemma:freeVariableScope} and the fact that the wildcard placeholders $\overline{\wtv{b}}$ are
+    solely used here guarantees that no other than the variables declared in $\Delta, \Delta', \overline{\Delta}$ appear in $\sigma'(\overline{\wtv{b}})$.
     \begin{itemize}
         \item $\generics{\ol{X} \triangleleft \ol{U'}} \ol{U} \to \type{U} \in \Pi(\texttt{m})$ is given, otherwise the program fails at the constraint generation step.
         \item $\Delta, \Delta', \overline{\Delta} \vdash \ol{S} <: [\ol{S}/\ol{X}]\ol{U'}$ by constraints $\ol{\wtv{b}} \lessdot [\ol{\wtv{b}}/\ol{X}]\ol{N}$ and lemma \ref{lemma:unifySoundness}
-        TODO: does not work because left side is a wildcard variable (maybe just change lemma soundness a bit)
+        there must be some $\ol{S}$ satisfying this premise.
+        
+        % what about we apply Unify again. With the known types. TODO: proof that unify can be applied again, with the capture converted versions of the constraints
+        %TODO: does not work because left side is a wildcard variable (maybe just change lemma soundness a bit)
+        \item 
     \end{itemize}
 \end{itemize}
     \item[$\texttt{v}.\texttt{m}(\ol{v})$]
@@ -156,12 +162,56 @@ We have to show T-Let and T-Call which leaves us with:
 
 \end{description}
 
+
+During the unification process free variables can only move one hop at a time.
+Free variables originate from capture constraints.
+By applying the capture rule to a constraint of the form $\wcNtype{\Delta}{N} \lessdotCC \type{T}$ the variables declared in $\Delta$
+are now free in the constraint $\type{N} \lessdot \type{T}$.
+The other way of free variables traveling into a constraint is by substitution.
+This is only possible if a constraint contains wildcard placeholders.
+And only free variables which reside in the same constraint as a wildcard placeholders have the possibility of being set in.
+
+This effect is transitive. Free variables travel from constraint to constraint.
+If there is no connection of constraints containing wildcard type placeholders between a free variable and a constraint,
+then free variables cannot travel there.
+
 \begin{lemma}
-    Free variables never leave their scope.
+    Free variables travel only 
-    If $\tv{e} \lessdot \tv{x}, \overline{\tv{e} \lessdot \tv{x}},
-    \tv{r} \lessdot \tv{a}$ % TODO
 \end{lemma}
 
+\begin{lemma}{Free variables}\label{lemma:freeVariableScope}
+    Free variables never leave their scope.
+    If a set of constraints does not share a single wildcard type placeholder with the rest of the constraint set,
+    then it will never contain free variables used outside of it.
+$(\Delta, \sigma) = \unify{}(\Delta_{in}, C \cup C')$
+
+and $\text{wtv}(C') \cap \text{wtv}(C) = \emptyset$
+and $\overline{\sigma(\tv{x}) = \wcNtype{\Delta}{N}}$
+
+then $\Delta, \Delta', \overline{\Delta}$ are all the variables used in $C'$.
+\end{lemma}
+\textit{Proof:} This is due to free variables only travel inbetween constraints and themselves via substitution.
+
+If $\type{S} \lessdotCC \type{T}$ spawns free variables, they will be contained within the reach of the wildcard type placeholders
+used for the method call.
+No other free variables can move into that scope.
+With this lemma we can further proof that during a method call only the generated free variables are used.
+
+% \begin{lemma}
+%     Free variables never leave their scope.
+%     $C' = \set{
+%         \overline{\tv{e} \lessdot \tv{x}}, \overline{\tv{x} \lessdotCC \tv{x}},
+%         \overline{\tv{r} \lessdot \tv{a}}
+%     }$
+% $(\Delta, \sigma) = \unify{}(\Delta_{in}, C \cup C')$
+
+% and $\text{wtv}(C') \cap \text{wtv}(C) = \emptyset$
+% and $\overline{\sigma(\tv{x}) = \wcNtype{\Delta}{N}}$
+
+% then $\Delta, \Delta', \overline{\Delta}$ are all the variables used in $C'$.
+% %if a wildcard type placeholder does not appear anywhere else, then only the free variables of its scope can be used for him
+% \end{lemma}
+
 % \begin{lemma} Unify does add free variables to types not containing free variables (or wildcard placeholders)
 % \begin{description}
 %     \item[If] $(\sigma, \Delta) = \unify{}( \Delta', C )$
@@ -288,11 +338,24 @@ Trivial. \unify{} fails when a constraint $\tv{a} \doteq \rwildcard{X}$ arises.
 %TODO
 % does this really work. We have to show that the algorithm never gets stuck as long as there is a solution
 % maybe there are substitutions with types containing free variables that make additional solutions possible. Check!
-
 \begin{lemma}{\unify{} Soundness:}\label{lemma:unifySoundness}
 \unify{}'s type solutions are correct respective to the subtyping rules defined in figure \ref{fig:subtyping}.
 \begin{description}
 \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[Then] there exists a substitution $\sigma'$ with:
+\begin{itemize}
+\item $\sigma \subseteq \sigma'$
+\item $\Delta, \Delta' \vdash \overline{\sigma'(\type{S}) <: \sigma'(\type{T})}$
+\item $\Delta, \Delta' \vdash \overline{\sigma'(\type{S'}) <: \wcNtype{\Delta}{N}}$
+\item $\Delta, \Delta', \overline{\Delta} \vdash \overline{\type{N} <: \sigma'(\type{T'})}$
+\end{itemize}
+\end{description}
+\end{lemma}
+
+\begin{lemma}{\unify{} Soundness:}%\label{lemma:unifySoundness}
+\unify{}'s type solutions are correct respective to the subtyping rules defined in figure \ref{fig:subtyping}.
+\begin{description}
+\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'} })$
 with $\text{wtv}(\ol{S}) = \emptyset$ and $\text{wtv}(\ol{T}) = \emptyset$
 \item[Then] $\Delta, \Delta' \vdash \overline{\sigma(\type{S}) <: \sigma(\type{T})}$
 \item[and] either $\Delta, \Delta' \vdash \overline{\sigma(\type{S'}) <: \sigma(\type{T'})}$