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