419 lines
23 KiB
TeX
419 lines
23 KiB
TeX
\section{Soundness}
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\newcommand{\CC}{\text{CC}}
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\begin{theorem}\label{testenv-theorem}
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Type inference produces a correctly typed program.
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\begin{description}
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\item[If] $\fjtypeinference(\mv{\Pi}, \texttt{class}\ \exptype{C}{\ol{X}
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\triangleleft \ol{N}} \triangleleft \type{N}\ \{ \overline{\type{T} \ f};\ \ol{M} \}) = \mtypeEnvironment{}'$
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\item[Then] $\texttt{class}\ \exptype{C}{\ol{X}
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\triangleleft \ol{N}} \triangleleft \type{N}\ \{ \overline{\type{T} \ f};\ \ol{M} \} \text{ok}$,
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with $\ol{M} = $
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\end{description}
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\end{theorem}
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\begin{lemma}{Well-formedness:}
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TODO:
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\end{lemma}
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\begin{lemma}{Soundness:}
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\unify{}'s type solutions for a constraint set generated by $\typeExpr{}$ are correct.
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\begin{description}
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\item[if] $\typeExpr{}(\mtypeEnvironment{}, \texttt{e}, \tv{a}) = C$
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and $(\Delta_u, \sigma) = \unify{}(\Delta', C)$ and let $\Delta = \Delta_u \cup \Delta'$
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% , with $C = \set{ \overline{ \type{S} \lessdot \type{T} } \cup \overline{ \type{S'} \lessdotCC \type{T'} } }$
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% and $\vdash \ol{L} : \mtypeEnvironment{}$
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% and $\Gamma \subseteq \mtypeEnvironment{}$
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% \item[given] a $(\Delta, \sigma)$ with $\Delta \vdash \overline{\sigma(\type{S}) <: \sigma(\type{T})}$
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% and there exists a $\Delta'$ with $\Delta, \Delta' \vdash \overline{\CC{}(\sigma(\type{S'})) <: \sigma(\type{T'})}$
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\item[then] there is a completion $|\texttt{e}|$ with $\Delta|\Gamma \vdash |\texttt{e}| : \sigma(\tv{a})$
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\end{description}
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\end{lemma}
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Regular type placeholders represent type annotations.
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These are the only types a \wildFJ{} program needs to be correctly typed.
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The type placeholders flagged as wildcard placeholders are intermediate types
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used in let statements and as type parameters for generic method calls.
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%Unify needs to return S aswell and guarantee that the \Delta' environment are the wildcards
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% only used inside the constraint the wildcard variable occurs
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% should Unify also return the \Delta' environment? Otherwise the bounds of free wildcard variables are lost
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% Or is it possible to deduct the right \ol{S} directly from the types in the normal TPHs?
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\textit{Proof:}
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By structural induction over the expression $\texttt{e}$.
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\begin{description}
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\item[$\texttt{e}.\texttt{f}$] Let $\sigma(\tv{r}) = \wcNtype{\Delta_c}{N}$,
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then $\Delta|\Gamma \vdash \texttt{e} : \wcNtype{\Delta_c}{N}$ by assumption.
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$\Delta', \Delta, \Delta_c \vdash \type{N} <: \sigma(\exptype{C}{\overline{\wtv{a}}})$ by premise.
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%Let $\sigma(\tv{r}) = \wcNtype{\Delta'}{N}$.
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%Let $\sigma([\ol{\wtv{a}}/\ol{X}]\type{T}) = \wcNtype{\Delta_t}{N_t}$.
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The completion of $|\texttt{e}.\texttt{f}|$ is $\texttt{let}\ \texttt{x} = \texttt{e} : \wcNtype{\Delta_c}{N}\ \texttt{in} \ \texttt{x}.\texttt{f}$
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We now show
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$\Delta|\Gamma \vdash \texttt{let}\ \texttt{x} = \texttt{e} : \wcNtype{\Delta_c}{N}\ \texttt{in} \ \texttt{x}.\texttt{f} : \sigma(a)$
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by the T-Field rule.
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$\Delta \vdash \wcNtype{\Delta_c}{N} <: \wcNtype{\Delta_c}{N}$ by S-Refl.
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$\Delta, \Delta_c \vdash \type{U}_i <: \sigma(\tv{a})$,
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because of the constraint $[\overline{\wtv{a}}/\ol{X}]\type{T} \lessdot \tv{a}$ and lemma \ref{lemma:unifySoundness}.
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$\textit{fields}(\sigma(\exptype{C}{\overline{\wtv{a}}})) = \sigma([\overline{\wtv{a}}/\ol{X}]\type{T})$
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and $\text{fv}(\type{U}_i) \subseteq \text{fv}(\type{N})$ by definition of $\textit{fields}$.
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$\text{dom}(\Delta_c) \subseteq \text{fv}{\type{N}}$ by lemma \ref{lemma:tvsNoFV}.
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% X.List<X> <. List<a?>
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% $\sigma(\ol{\tv{r}}) = \overline{\wcNtype{\Delta}{N}}$,
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% $\ol{N} <: [\ol{S}/\ol{X}]\ol{U}$,
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% TODO: S ok? We could proof $\Delta, \Delta' \overline{\Delta} \vdash \ol{S} \ \ok$
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% by proofing every substitution in Unify is ok aslong as every type in the inputs is ok
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% S ok when all variables are in the environment and every L <: U and U <: Class-bound
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% This can be given by the constraints generated. We can proof if T ok and S <: T and T <: S' then S ok and S' ok
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% If S ok and T <. S , then Unify generates a T ok
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S typeinference:
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T <: [S/Y]U
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We apply the following lemma
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Lemma
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if T ok and T <: S then S ok
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until
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T = [S/Y]U
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and then we can say by
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Lemma:
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If [S/Y]U ok then S ok (TODO: proof!)
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So we do not have to proof S ok (but T)
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% T_r <: C<T> (S is in T)
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% Is C<T> ok?
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% if every type environment \Delta supplied to Unify is ok (L <: U), then \sigma(a) = \Delta'.N implies \Delta' conforms to (L <: U)
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% this together with the X <. N constraints proofs T_r ok
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$\Delta \vdash \sigma(\tv{a}), \wcNtype{\Delta_c}{N} \ \ok$ %TODO
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%Easy, because unify only generates substitutions for normal type placeholders which are OK
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\item[$\texttt{e}.\texttt{m}(\ol{e})$]
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Lets have a look at the case where the receiver and parameter types are all named types.
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So $\sigma(\ol{\tv{r}}) = \ol{T} = \ol{\wcNtype{\Delta_u}{N}}$ and $\sigma(\tv{r}) = \type{T}_r = \wcNtype{\Delta_u}{N}$
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and a $\Delta'$, $\overline{\Delta}$ where $\Delta' \subseteq \Delta_u$, $\overline{\Delta \subseteq \Delta_u}$
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and $\text{dom}(\Delta') = (\text{fv}(\type{N})/\Delta)$, $\overline{\text{dom}(\Delta) = (\text{fv}(\type{N})/\Delta)}$ in
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$\texttt{let}\ \texttt{x} : \wcNtype{\Delta'}{N} = \texttt{e} \ \texttt{in}\
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\texttt{let}\ \ol{x} : \overline{\wcNtype{\Delta'}{N}} = \ol{e} \ \texttt{in}\ \texttt{x}.\texttt{m}(\ol{x})$
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%TODO: show that this type exists \wcNtype{\Delta'}{N} (a \Delta' which only contains free variables of the type N)
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% and which is a supertype of T_r (so no free variables in T_r)
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% Solution: Make this a lemma and guarantee that Unify does only create types Delta.T with \Delta subset fv(T)
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% This is possible because free variables (except those in \Delta_in) are never used in wildcard environments
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By lemma \ref{lemma:unifyWeakening} we know that
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$\unify{}(\Delta, [\ol{\tv{a}}/\ol{X}][\ol{\tv{b}}/\ol{Y}]\set{ \overline{\wcNtype{\Delta}{N}} \lessdotCC \ol{T}, \wcNtype{\Delta'}{N} \lessdotCC \exptype{C}{\ol{X}}})$
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has a solution.
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Also $\overline{\wcNtype{\Delta}{N}} \lessdot \ol{T}$ and $\wcNtype{\Delta'}{N} \lessdotCC \exptype{C}{\ol{X}}$ will be converted to
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$\ol{N} \lessdot \ol{T}$ and $\type{N} \lessdot \exptype{C}{\ol{X}}$ by the \rulename{Capture} rule.
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Which then results in the same as calling $\unify{}((\Delta, \Delta', \overline{\Delta}), [\ol{\tv{a}}/\ol{X}][\ol{\tv{b}}/\ol{Y}]\set{\ol{N} \lessdot \ol{T}, \type{N} \lessdot \exptype{C}{\ol{X}}})$,
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which shows $\Delta, \Delta', \overline{\Delta} \vdash \overline{N} <: [\ol{S}/\ol{X}]\overline{U}$.
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This implies $\text{dom}(\Delta') \subseteq \text{fv}(\type{N})$ and $\text{dom}(\Delta') \subseteq \text{fv}(\type{N})$
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$\Delta, \Delta', \overline{\Delta} \vdash [\ol{S}/\ol{X}]\type{U} \type{T}$,
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$\Delta, \Delta', \overline{\Delta} \vdash \ol{N} <: [\ol{S}/\ol{X}]\ol{U}$,
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and $\Delta, \Delta', \overline{\Delta} \vdash \ol{S} <: [\ol{S}/\ol{X}]\ol{U'}$
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are guaranteed by the generated constraints and lemma \ref{lemma:unifySoundness} and \ref{lemma:unifyCC}.
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$\Delta, \Delta' \vdash \ol{T} <: \ol{\wcNtype{\Delta}{N}}$ due to $\ol{T} = \ol{\wcNtype{\Delta}{N}}$ and S-Refl. $\Delta, \Delta' \vdash \type{T}_r <: \wcNtype{\Delta'}{N}$ accordingly.
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$\Delta|\Gamma \vdash \texttt{t}_r : \type{T}_r$ and $\Delta|\Gamma \vdash \ol{t} : \ol{T}_r$ by assumption.
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$\Delta, \Delta' \vdash \ol{\wcNtype{\Delta}{N}} \ \ok$ by lemma \ref{lemma:unifyWellFormedness},
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therefore $\Delta, \Delta', \overline{\Delta} \vdash \ol{N} \ \ok$, which implies $\Delta, \Delta', \overline{\Delta} \vdash \ol{U} \ \ok$
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by lemma \ref{lemma:wfHereditary} and $\Delta, \Delta', \overline{\Delta} \vdash \ol{S} \ \ok $ by premise of WF-Class.
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The same goes for $\wcNtype{\Delta'}{N}$.
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% \begin{gather}
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% \label{sp:0}
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% \Delta, \Delta', \overline{\Delta} \vdash [\ol{S}/\ol{X}]\type{U} \type{T} \\
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% \label{sp:1}
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% \Delta, \Delta', \overline{\Delta} \vdash \ol{N} <: [\ol{S}/\ol{X}]\ol{U} \\
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% \label{sp:2}
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% \Delta, \Delta', \overline{\Delta} \vdash \ol{S} <: [\ol{S}/\ol{X}]\ol{U'} \\
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% \label{sp:3}
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% \Delta, \Delta', \overline{\Delta} \vdash \ol{S} \ \ok \\
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% \label{sp:4}
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% \Delta, \Delta' \vdash \ol{T} <: \ol{\wcNtype{\Delta}{N}}
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% \end{gather}
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Method calls generate multiple constraints that share the same wildcard placeholders ($\ol{\wtv{a}}$, $\ol{\wtv{b}}$).
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\end{description}
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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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% \item[Then] $\forall x \in \set{\type{S} \mid \type{S} \in C, \text{fv}(\type{S}) = \emptyset }: \text{fv}(\sigma(\type{S})) \subseteq \text{dom}(\Delta)$
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% \end{description}
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% \end{lemma}
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% \textit{Proof:} by induction over the \unify{} algorithm.
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% A unifier $\sigma(\tv{a}) = \type{T}$, where the type variable $\tv{a}$ is not flagged as a wildcard will always hold
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% $\text{fv}(\type{T}) \subseteq \text{dom}(\Delta)$.
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% %UNIFY fails when there are free variables on the right side of a a =. T constraint
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\begin{lemma}\label{lemma:unifyNoExcessWildcards}
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\begin{description}
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\item[If] $(\Delta, \sigma) = \unify{}(\Delta', C)$
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\item[Then] $\sigma(\tv{a}) = \wcNtype{\Delta''}{N} \implies \text{dom}(\Delta'') \subseteq \text{fv}(\type{N})$
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\end{description}
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\end{lemma}
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\textit{Proof:} Easy -
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All types given in the input to the \unify{} algorithm already comply with this requirement
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and none of the rules change.
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Note: the \rulename{Super} rule removes unnecessary wildcards.
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\begin{lemma}\label{lemma:unifyWellFormedness}
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\unify{} generates well-formed types as long as well-formed types are supplied.
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\begin{description}
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\item[If] $(\Delta, \sigma) = \unify{}(\Delta', C)$ %and $\sigma(\tv{a}) = \wcNtype{\ol{\wildcard{X}{\type{U}}{\type{L}}}}{N}$
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\item[and] $\wcNtype{\Delta}{N} \in C \implies \Delta' \vdash \wcNtype{\Delta}{N} \ \ok$
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%\item[then] $\Delta \vdash \ol{L <: U}$ and $\Delta \vdash \ol{U <: U'}$ % (with U' being upper limit by class definition)
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\item[then] $\Delta \vdash \sigma(\tv{a}) \ \ok$
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\end{description}
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\end{lemma}
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%Only the \rulename{General} rule generates fresh wildcards.
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%By lemma \ref{lemma:unifySoundness} we get $\Delta \vdash \sigma(\type{T}) <: \sigma(\tv{a})$ with $\sigma(\tv{a}) = \wctype{\ol{\wildcard{X}{\sigma(U)}{\sigma(L)}}}{C}{\ol{X}}$
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%By S-Exists and S-Trans we can say $\Delta \vdash \sigma(\type{L}) <: \sigma(\type{U})$
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\textit{Proof:}
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by induction over every step of the \unify{} algorithm.
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Hint: a type placeholder $\tv{a}$ will never be replaced by a free variable or a type containing free variables.
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This fact together with the presumption that every supplied type is well-formed we can easily show that this lemma is true.
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\textit{Proof: (Variant 2)}
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The GenSigma and GenDelta rules check for well-formedness.
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% All types supplied to the Unify algorithm by TYPEs only contain ? extends or ? super wildcards. (X : [bot..T] or X : [T .. Object]).
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% Both are well formed!
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\begin{lemma}\label{lemma:wfHereditary}
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Well-formedness is hereditary
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\begin{description}
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\item[If] $\triangle \vdash \exptype{C}{\ol{X}} \ \ok$ and $\triangle \vdash \exptype{C}{\ol{X}} <: \type{S}$
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\item[Then] $\triangle \vdash \type{S} \ \ok$
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\end{description}
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\end{lemma}
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\textit{Proof:}
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\begin{lemma}
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\begin{description}
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\item[If] $\Delta \vdash \wctype{\Delta'}{C}{\ol{T}} \ \ok$
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\item[Then] $\Delta, \Delta' \vdash \ok{T} \ \ok$
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\end{description}
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\end{lemma}
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\textit{Proof:} by definition of WF-Class
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\begin{lemma} \label{lemma:tvsNoFV}
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\unify{} does not add free variables to types not containing free variables.
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\begin{description}
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\item[If] $(\sigma, \Delta) = \unify{} (\Delta', C)$
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\item[and] $\tv{a}$ being a type placeholders used in $C$
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\item[then] $\text{fv}(\sigma(\tv{a})) \subseteq \Delta, \Delta'$
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\end{description}
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\end{lemma}
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\textit{Proof:}
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Trivial. \unify{} fails when a constraint $\tv{a} \doteq \rwildcard{X}$ arises.
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% do we even have to proof that?
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% \begin{lemma}
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% C = Delta.N <c T
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% is the same for Unify as:
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% Delta |- N <. T
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% \end{lemma}
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\begin{lemma} \label{lemma:unifyWeakening}
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Removing constraints does not render \unify{} impossible.
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If there is a solution for a constraint set $C$, then there is also a solution for a subset of that constraint set.
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\begin{description}
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\item[If] $(\sigma, \Delta) = \unify{}( C \cup C')$
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\item[then] $ (\sigma', \Delta') = \unify{}( C')$
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\end{description}
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\textit{Proof:}
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%TODO
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\end{lemma}
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\textit{Proof:}
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%TODO: is it true even?
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The input set does not contain free variables, only free variable placeholders.
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The only time \unify{} add a wildcard to the $\wildcardEnv$ is the \rulename{Capture} rule.
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This rule is only applied for the outer wildcard environments for each type.
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\begin{lemma}\label{lemma:unifySoundness}
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The \unify{} algorithm only produces correct output.
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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[and] $fv(\overline{ \type{S} }) = \emptyset$, $fv(\overline{ \type{T} }) = \emptyset$
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\item[Then] there exists a $\sigma'$ with:
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$\sigma \subseteq \sigma'$ and
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$\Delta, \Delta' \vdash \overline{\sigma'(\type{S}) <: \sigma'(\type{T})}$
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and $\Delta, \Delta', \overline{\Delta} \vdash \overline{\type{N} <: \sigma'(\type{T'})}$ where $\overline{\sigma(\type{S'}) = \wcNtype{\Delta}{N}}$,
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otherwise $\Delta, \Delta' \vdash \overline{\sigma'(\type{S'}) <: \sigma'(\type{T'})}$
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% and $\sigma(\type{T'}) = \sigma(\type{T'})$.
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\end{description}
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\end{lemma}
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% \begin{lemma} % a lemma where we distinguis between free variable on the left or the right side of a constraint (not needed anymore)
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% The \unify{} algorithm only produces correct output for constraints not containing free variables.
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% \begin{description}
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% \item[If] $(\sigma, \Delta) = \unify{}( \overline{ \type{S} \lessdot \type{T} } \cup \overline{ \wcNtype{\Delta'}{N} \lessdot \type{T'} } \cup \overline{ \type{S'} \lessdot \wcNtype{\Delta'}{N'} })$
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% \item[and] $fv(\overline{ \type{S} }) = \emptyset, fv(\overline{ \type{T} }) = \emptyset, fv(\overline{ \type{T'} }) = \emptyset, fv(\overline{ \type{S'} }) = \emptyset$
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% \item[Then] $\Delta \vdash \overline{\sigma(\type{S}) <: \sigma(\type{T})}$
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% and $\Delta, \Delta' \vdash \overline{\sigma(\type{N}) <: \sigma(\type{T'})}, \overline{\sigma(\type{S'}) <: \sigma(\type{N'})}$
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% %TODO: Rephrase (\Delta' is used three times!)
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% %The function $\CC{}$ is given as $\CC{}(\wcNtype{\Delta}{N}) = \type{N}$
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% \end{description}
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% \end{lemma}
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\textit{Proof:}
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%(we are going backwards over the algorithm)
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%first we have to determine the \Delta'' -> it's only the wildcards which are free in N
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% during this proof we can use Delta'' as we like
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For every step in the \unify{} algorithm:
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Assuming the unifier $\sigma$ is correct for a constraint set $C'$, the unifier is also correct for the
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constraint set $C$ before the transformation.
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% \begin{description}
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% \item[Assumption:] $\unify{}(C) = (\Delta, \sigma)$, with $\Delta \vdash \sigma(C)$
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% \item[Induction step:] For every case $C'$ which can be transformed to $C$ we have to show $\Delta \vdash \sigma(C')$
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% \end{description}
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\unify{} terminates with $C = \emptyset$ for which the preposition holds:
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$\Delta \vdash \sigma(\emptyset)$
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We now show that for every transformation of a constraint set $C$ to a constraint set $C'$
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the preposition holds for $C$ using the assumption that it holds for $C'$ :
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$\Delta \vdash \sigma(C') \implies \Delta \vdash \sigma(C)$
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\begin{description}
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\item[AddDelta] $C$ is not changed
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\item[GenDelta] by definition, S-Var-Left, and S-Trans %The generated type variable is unique due to the solved form property. %and the solved form property (no $\tv{a}$ in $C$)
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\item[GenSigma] by definition and S-Refl.
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% holds for $\set{\tv{a} \doteq \type{G}}$ by definition.
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% Holds for $C$ by assumption and because $\tv{a} \notin C$ by solved form definition ($[\type{G}/\tv{a}]C = C$).
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\item[Ground] Assumption and S-Bot
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\item[Sub-Elim] Assumption and S-Refl
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\item[Force] by assumption and $\rwildcard{X} = \type{U}$ %TODO: step 5 should remove all X^T_T with T (make wildcards with same upper and lower bounds to normal types)
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\item[Raise] Assumption, S-Trans
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\item[Settle] Assumption, S-Trans
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\item[Super] S-Extends ($\vdash \wctype{\Delta}{C}{\ol{T}} <: \wctype{\Delta}{D}{[\ol{T}/\ol{X}]\ol{N}}$), S-Trans
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\item[\generalizeRule{}] by Assumption, because $C \subset C'$
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\item[Adapt] Assumption, S-Extends, S-Trans
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\item[Adopt] Assumption, because $C \subset C'$
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%\item[Capture, Reduce] are always applied together. We have to destinct between two cases:
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\item[Prepare]
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Given a $\wctype{\Delta_c}{C}{\ol{S}} \lessdot \wcNtype{\Delta''}{N}$
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we get $\Delta, \Delta', \overline{\Delta} \vdash \exptype{C}{\ol{S}} <: \wcNtype{\Delta''}{N}$ with $\Delta_c \subseteq \overline{\Delta}$.
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$\text{fv}(\sigma'(\wctype{\Delta_c}{C}{\ol{S}})) = \emptyset$ implies $\text{fv}(\ol{S})\subseteq \text{dom}(\Delta, \Delta_c)$
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and $\text{fv}(\sigma'(\wcNtype{\Delta''}{N})) = \emptyset$ implies $\text{dom}(\Delta_c) \cap \text{fv}(\sigma'(\wcNtype{\Delta''}{N})) = \emptyset$.
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No free variables on both sides also mean we do not need $\overline{\Delta}$.
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Therefore we can say $\Delta, \Delta' \vdash \wctype{\Delta_c}{C}{\ol{S}} <: \wcNtype{\Delta''}{N}$.
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\item[Capture]
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Everytime the \rulename{Capture} rule is invoked we add the freshly generated free variables to the global environment $\wildcardEnv$.
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We get a $\sigma'$ %and a $\Delta'$
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with $\Delta, \Delta' \vdash \sigma'([\ol{\rwildcard{C}}/\ol{\rwildcard{B}}] \exptype{C}{\ol{S}}) <: \sigma'(\wctype{\Delta}{C}{\ol{T}})$
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where $\sigma'(\ol{\wildcard{C}{U}{L}}) \subseteq \Delta'$ by assumption.
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\unify{} performs a capture conversion only on $\lessdotCC$ constraints.
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Therefore we can say that $\Delta, \Delta', \overline{\Delta} \vdash \sigma'(\exptype{C}{\ol{S}})
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<: \sigma'(\wctype{\Delta}{C}{\ol{T}})$ with $\overline{\Delta}$ being all the fresh wildcards generated by \rulename{Capture}.
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\item[Reduce]
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%Assumption and S-Exists.
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% Three different cases of the constraint $\exptype{C}{\ol{S}} \lessdot \wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}}$:
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% \begin{description}
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% \item[$\text{fv}(\exptype{C}{\ol{S}}) = \emptyset, \text{fv}(\wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}})$:]
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% the preposition holds by Assumption and S-Exists.
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% \item[$\text{fv}(\wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}}) = \emptyset$:]
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% then $\text{fv}(\exptype{C}{\ol{S}}) \subseteq \Delta'$ with $\Delta' = \overline{\wildcard{A}{\type{U}}{\type{L}}}$
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% \item[$\text{fv}(\exptype{C}{\ol{S}}) = \emptyset$] $\Delta' = \emptyset$
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% \end{description}
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% List<X> <. Y.List<Y>, free variables are either in
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If $\text{fv}(\exptype{C}{\ol{S}}) = \text{fv}(\wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}}) = \emptyset$ the preposition holds by Assumption and S-Exists.
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Otherwise
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$\Delta' \vdash \CC{}(\sigma(\exptype{C}{\ol{S}})) <: \sigma(\wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}})$
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holds with any $\Delta'$ so that $(\text{fv}(\exptype{C}{\ol{S}}) \cup \text{fv}(\wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}}) ) \subseteq \text{dom}(\Delta') $.
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\item[Match] Assumption, S-Trans
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\item[Trim] Assumption and S-Exists
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\item[Remove] $C$ is not changed
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\item[Circle] S-Refl
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\item[Swap] by definition
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\item[Erase] S-Refl
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\item[Equals] by definition
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% \item[Reduce]
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% The renaming from $\rwildcard{C}$ to $\rwildcard{B}$ is not a problem. It's allowed to rename wildcards inside a type.
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% Removing $\rwildcard{C}$ from the environment does not change anything because it was freshly generated and is used nowhere.
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% The rest follows directly from S-Exists.
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% We can say: $\text{fv}(\wctype{\overline{\wildcard{A}{\type{U}}{\type{L}}}}{C}{\ol{T}}) = \emptyset$,
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% because the input to the \unify{} algorithm has no free type variables and we never substitute a type with free type variables
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% and none of the other steps of the algorithm generates a $\lessdot$ constraint containing free type variables on the right side. %TODO: proof
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% $\text{fv}(\ol{T}) \subseteq \text{dom}(\wildcardEnv \cup \overline{\wildcard{B}{\type{U'}}{\type{L'}}})$
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% TODO: The capture conversion has to be when substituting a $\wtv{a}$ variable. Then we have to rename!
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% %Lets first try it without the capture conversion. And involve the wtvs in the second step
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% % The algorithm works by never substituting wildcards
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% \unify{} cannot guarantee the premise $\text{dom}(\Delta, \Delta') \cap \text{fv}(\wcNtype{\overline{\wildcard{X}{\type{U}}{\type{L}}}}{N}) = \emptyset$.
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% We loosen the soundness requirements and allow a arbitrary environment $\Delta''$ to be added to the right side of the subtype relation.
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% This is still sufficient to proof soundness for the whole algorithm.
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% We show that the need for the additional environment $\Delta''$ can be satisfied by a let statement.
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% \begin{itemize}
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% \item $\Delta, \Delta' \vdash [\ol{T}/\ol{X}]\ol{L} <: \ol{T}$: S-Exists
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% \item $\Delta, \Delta' \vdash \ol{T} <: [\ol{T}/\ol{X}]\ol{U}$: S-Exists
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% \item $\textit{fv}(\ol{T}) \subseteq \text{dom}{\Delta, \Delta'}$: $\wildcardEnv$ holds all variables %TODO: Proof
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% \end{itemize}
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\item[Normalize] Assumption and lemma 5 \emph{substitution preserves subtyping}.%\ref{lemma:wildcardReplacement}. (Or Lemma 5 from the wildcard paper. \emph{substitution preserves subtyping})
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% The GenSigma step replaces both sides of $\rwildcard{A} \doteq \rwildcard{B}$ with the upper bound $\type{U}$.
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% This works for every constraint, whether it contains free variables or not.
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% It does not add to free variables of constraints because the upper bound does not contain any.
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The GenSigma and Gen Delta steps remove Wildcards which have the same upper and lower bound.
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$\rwildcard{A},\rwildcard{B} \notin \sigma(C)$
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% sigma(T) = sigma(U) we have to show that T = U means \Delta \vdash [T/U]C \implies \Delta \vdash [U/T]C
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% the constraints L <. U, U <. L lead to L =. U
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%If L is List<X> with X being free wildcard
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%then U <. L will fail if U is type variable
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% this is because bounds never contain free variables (is that true?)
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%This type contains free variables when A is replaced by an CC wildcard
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%This must fail:
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\begin{verbatim}
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|
<A> A m(List<? extends List<A>> l, A)
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m(List<List<? super String>> l, "hi")
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\end{verbatim}
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%This fails because of Equals rule (TODO: proof)
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\item[Tame] same reasoning as Normalize
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\item[Bot] S-Bot
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\item[Pit] S-Bot
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\item[Upper] S-Trans and S-VarLeft
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\item[Lower] S-Trans and S-VerRight
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\item[Subst-WC] by S-Refl
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\item[Subst]
|
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$\sigma(C \cup \set{\tv{a} \doteq \type{T}}) = \sigma([\type{T}/\tv{a}]C \cup \set{\tv{a} \doteq \type{T}})$
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and
|
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$\sigma(\wildcardEnv) = \sigma([\type{T}/\tv{a}]\wildcardEnv)$
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\item[Subst-WC]
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%Proof by Lemma 5 \emph{Type substitution preserves subtyping} from \cite{WildcardsNeedWitnessProtection}.
|
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Same as Subst
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\end{description}
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