Big Cleanup (delete comments in soundness proof

soundness.tex

@@ -9,97 +9,6 @@ The first is lemma \ref{lemma:freeVariablesOnlyTravelOneHop} which ensures that
 travel one hop at the time through a constraint set.
 And the second one is the fact that normal type placeholders never contain free variables.
 
-% \begin{lemma}
-% A sound TypelessFJ program is also sound under LetFJ type rules.
-% \begin{description}
-%     \item[if:]
-% $\Gamma | \Delta \vdash \texttt{m}(\ol{x}) = \texttt{e} \ \ok \ \text{in}\ C \text{with} \ \generics{\ol{Y \triangleleft P}}$
-% \end{description}
-% \end{lemma}
-
-% TODO: Beforehand we have to show that $\Delta \cup \overline{\Delta} | \Theta \vdash \texttt{e} : \type{T} \mid \overline{\Delta}$
-% Here $\Delta$ does not contain every $\overline{\Delta}$ ever created.
-% %what prevents a free variable to emerge in \Delta.N example Y^Object |- C<String> <: X^Y.C<X>
-% % if the Y is later needed for an equals: same(id(x), x2)
-% Free wildcards do not move inwards. We can show that every new type is either well-formed and therefore does not contain any free variables.
-% Or it is a generic method call: is it possible to use any free wildcards here?
-% let empty 
-
-% <X> Box<X> empty()
-% same(Box<?>, empty())
-% let p1 : X.Box<X> = Box<?> in let 
-% X.Box<X> <. Box<x> 
-% Box<e> <. Box<x>
-
-% boxin(empty()), Box2<?>
-
-% Where can a problem arise? When we use free wildcards before they are freed.
-% But we can always CC them first. Exception two types: X.Pair<X, y> and Y.Pair<x, Y>
-% Here y = Y and x = X but 
-
-% <X,Y> void same(Pair<X,Y> a, Pair<X,Y> b){}
-% <X> Pair<?, X> left() { return null; }
-% <X> Pair<X, ?> right() { return null; }
-
-% <X> Box<X> id(Box<? extends Box<X>> x)
-% here it could be beneficial to use a free wildcard as the parameter X to have it later
-% Box<?> x = ...
-% same(id(x), id(x)) <- this will be accepted by TI
-
-% let left : X,Y.Pair<X,Y> = left() in
-%  let right : Pair<X,Y> = right() in
-
-% Compromise:
-% - Generate constraints so that they comply with LetFJ
-% - Propose a version which is close to Java
-
-% Version for LetFJ:
-% Is it still possible to do the capture conversion in form of constraints?
-% X.C<X> <. C<x>
-% T <. X.C<X>
-% how to proof: X.C<X> ok
-
-
-% If $\Delta \cup \overline{\Delta} | \Theta \vdash \texttt{e} : \type{T} \mid \overline{\Delta}$
-% then there exists a $|\texttt{e}|$ with $\Delta | \Theta \vdash |\texttt{e}| : \wcNtype{\Delta'}{N}$ in LetFJ.
-% This is possible by starting with the parameter types as the base case $\overline{\Delta} = \emptyset$.
-
-
-% Each type $\wcNtype{\Delta'}{N}$ can only use wildcards already freed.
-
-% \textit{Proof} by structural induction.
-% \begin{description}
-%     \item[$\texttt{e} = \texttt{x}$] $\Delta | \Theta \vdash \texttt{e} : \type{T} \mid \emptyset$
-%     $\Delta \vdash \type{T} \ \ok$ by \rulename{T-Method}
-%     and therefore $\Delta | \Theta \vdash \texttt{let}\ \texttt{e} : \type{T} = \texttt{x in } \texttt{e}$.
-
-% $|\texttt{x}, \texttt{e}| = \texttt{let}\ \texttt{e} : \type{T} = \texttt{x in } \texttt{e}$
-
-% \item[$\texttt{e} = \texttt{e}.\texttt{m}(\ol{e})$] there must be atleast one value in $\texttt{e}$ or $\ol{e}$
-% \item[$\texttt{e}.f$] given let x : T = e' in x 
-% let x : T = e' in let xf = x.f in xf
-
-% Required:
-% $ \Delta | \Theta \vdash e' : \type{T}_1$
-% $\Delta \vdash \type{T}_1 <: \wcNtype{\Delta'}{N}$
-% $\Delta, \Delta' | \Theta, x : \type{N} \vdash let xf = x.f in xf : \type{T}_2$
-
-% \end{description}
-
-
-% \textbf{Proof:} Every program complying with our type rules can be converted to a correct LetFJ program.
-% First we convert the program so that every wildcards used in an expression are in the $\Delta$ environment:
-% m(p) = e => let xp = p in [xp/p]e
-% x1.m(x2) => let xm = x1.m(x2=) in xm
-% x.f => let xf = x.f in xf
-% Then we have to proof there is never a wildcard used before it is declared.
-% Wildcards are introduced by the capture conversions and nothing else.
-
-
-% \begin{lemma}{Well-formedness:}
-% TODO: 
-% \end{lemma}
-
 \unify{} calculates solutions for all normal type placeholders.
 Those are used for all untyped method's argument and return type.
 A correct typing for method calls can be deducted from those type informations.
@@ -108,28 +17,9 @@ A correct typing for method calls can be deducted from those type informations.
     \unify{}'s type solutions for a constraint set generated by $\typeExpr{}$ are correct.
     \begin{description}
     \item[if] $\typeExpr{}(\mtypeEnvironment{}, \texttt{e}, \tv{a}) = (\Delta', C)$
-    and $(\Delta_u, \sigma) = \unify{}(\Delta', C)$ % and let $\Delta= \Delta_u \cup \Delta'$
-%    $\Delta, \Delta' \vdash $
-%    , with $C = \set{ \overline{ \type{S} \lessdot \type{T} } \cup \overline{ \type{S'} \lessdotCC \type{T'} } }$
-%    and $\vdash \ol{L} : \mtypeEnvironment{}$
-%    and $\Gamma \subseteq \mtypeEnvironment{}$
-%    \item[given] a $(\Delta, \sigma)$ with $\Delta \vdash \overline{\sigma(\type{S}) <: \sigma(\type{T})}$
-%    and there exists a $\Delta'$ with $\Delta, \Delta' \vdash \overline{\CC{}(\sigma(\type{S'})) <: \sigma(\type{T'})}$
-    %\item[then] there is a completion $|\texttt{e}|$ with $\Delta|\Gamma  \vdash |\texttt{e}| : \sigma(\tv{a})$
     \item[then] $\Delta|\Gamma  \vdash \texttt{e} : \sigma(\tv{a})$ where $\Delta = \Delta_u \cup \Delta'$
     \end{description}
 \end{lemma}
-% Regular type placeholders represent type annotations.
-% These are the only types a \wildFJ{} program needs to be correctly typed.
-% The type placeholders flagged as wildcard placeholders are intermediate types
-% used in let statements and as type parameters for generic method calls.
-
-%Unify needs to return S aswell and guarantee that the \Delta' environment are the wildcards
-% only used inside the constraint the wildcard variable occurs
-% should Unify also return the \Delta' environment? Otherwise the bounds of free wildcard variables are lost
-
-% Or is it possible to deduct the right \ol{S} directly from the types in the normal TPHs?
-
 \textit{Proof:}
 By structural induction over the expression $\texttt{e}$. 
 \begin{description}
@@ -175,14 +65,6 @@ we proof $\Delta, \Delta' | \Gamma, x : \type{N} \vdash \expr{x}.f_1 : \type{T}_
 \item $\Delta, \Delta' \vdash \type{T}_2 <: \type{T}$ by constraint %TODO: Rename constraints
 \end{itemize}
 
-% method call: a1 <c C<a>, a2 <c C<b>, a3 <c b?
-% here lemma:freeVariablesOnlyTravelOneHop can be used too
-
-    %TODO: use a lemma that says if Unify succeeds, then it also succeeds if the capture converted types are used.
-    % but it also works with a subset of the initial constraints.
-    % the generated constraints do not share wildcard placehodlers with other constraints.
-    % can they contain free variables from other places? They could, but isolation prevents that.
-    % TODO: but how to proof?
 %generated constraints: t1 <. x, x <. N, T <. t2
         We are allowed to use capture conversion for $\expr{v}$ here.
         $\Delta \vdash \expr{v} : \sigma(\tv{a})$ by assumption.
@@ -190,70 +72,22 @@ we proof $\Delta, \Delta' | \Gamma, x : \type{N} \vdash \expr{x}.f_1 : \type{T}_
         $\Delta \vdash \type{U}_i <: \sigma(\tv{a})$,
     because of the constraints $[\overline{\wtv{a}}/\ol{X}]\type{T} \lessdot \tv{a}$, $\tv{r} \lessdotCC \exptype{C}{\ol{\wtv{a}}}$ and lemma \ref{lemma:unifySoundness}.
     $\textit{fields}(\sigma(\exptype{C}{\overline{\wtv{a}}})) = \sigma([\overline{\wtv{a}}/\ol{X}]\type{T})$.
-    % \item[$\texttt{let}\ \texttt{x} = \texttt{e} \ \texttt{in} \ \texttt{x}.\texttt{f}$] 
-    % $\Delta|\Gamma \vdash \expr{e}: \type{T}$ by assumption.
-    % $\text{dom}(\Delta') \subseteq \text{fv}(\type{N})$ by lemma \ref{lemma:wildcardWellFormedness}.
-    % $\Delta, \Delta' | \Gamma, \expr{x} : \type{T} \vdash \texttt{x}.\texttt{f}$
-    % \item[$\texttt{e}.\texttt{f}$] Let $\sigma(\tv{r}) = \wcNtype{\Delta_c}{N}$,
-    % 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 $\sigma([\ol{\wtv{a}}/\ol{X}]\type{T}) = \wcNtype{\Delta_t}{N_t}$.
-    
-    % The completion of $|\texttt{e}.\texttt{f}|$ is $\texttt{let}\ \texttt{x} = \texttt{e} : \wcNtype{\Delta_c}{N}\ \texttt{in} \ \texttt{x}.\texttt{f}$
-    
-    % We now show
-    % $\Delta|\Gamma \vdash \texttt{let}\ \texttt{x} = \texttt{e} : \wcNtype{\Delta_c}{N}\ \texttt{in} \ \texttt{x}.\texttt{f} : \sigma(a)$
-    % by the T-Field rule.
-    % $\Delta \vdash \wcNtype{\Delta_c}{N} <: \wcNtype{\Delta_c}{N}$ by S-Refl.
-    % $\Delta, \Delta_c \vdash \type{U}_i <: \sigma(\tv{a})$,
-    % because of the constraint $[\overline{\wtv{a}}/\ol{X}]\type{T} \lessdot \tv{a}$ and lemma \ref{lemma:unifySoundness}.
-    % $\textit{fields}(\sigma(\exptype{C}{\overline{\wtv{a}}})) = \sigma([\overline{\wtv{a}}/\ol{X}]\type{T})$
-    % and $\text{fv}(\type{U}_i) \subseteq \text{fv}(\type{N})$ by definition of $\textit{fields}$.
-
-    % $\text{dom}(\Delta_c) \subseteq \text{fv}{\type{N}}$ by lemma \ref{lemma:tvsNoFV}.
-
-    % X.List<X> <. List<a?>
-    % $\sigma(\ol{\tv{r}}) = \overline{\wcNtype{\Delta}{N}}$,
-    % $\ol{N} <: [\ol{S}/\ol{X}]\ol{U}$,
-    % TODO: S ok? We could proof $\Delta, \Delta' \overline{\Delta} \vdash \ol{S} \ \ok$
-    % by proving every substitution in Unify is ok aslong as every type in the inputs is ok
-    % S ok when all variables are in the environment and every L <: U and U <: Class-bound
-    % 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
-
-    % If S ok and T <. S , then Unify generates a T ok
-
-    % S typeinference:
-    % T <: [S/Y]U
-    % We apply the following lemma
-    % Lemma
-    % if T ok and T <: S then S ok
-
-    % until
-    % T = [S/Y]U
-
-    % and then we can say by 
-    % Lemma:
-    % If [S/Y]U ok then S ok (TODO: proof!)
-
-    % So we do not have to proof S ok (but T)
-
-    % % T_r <: C<T> (S is in T)
-    % % Is C<T> ok?
-    % % if every type environment \Delta supplied to Unify is ok (L <: U), then \sigma(a) = \Delta'.N implies \Delta' conforms to (L <: U)
-    % % this together with the X <. N constraints proofs T_r ok
-
-    % $\Delta \vdash \sigma(\tv{a}), \wcNtype{\Delta_c}{N} \ \ok$ %TODO
-    % %Easy, because unify only generates substitutions for normal type placeholders which are OK
 
 \item[$\text{let}\ \expr{x} = \expr{e} \ \text{in}\ 
 \text{let}\ \overline{\expr{x} = \expr{e}} \ \text{in}\ \texttt{x}.\texttt{m}(\ol{x})$]
 generates constraints $\tv{e} \lessdot \tv{x}, \overline{\tv{e} \lessdot \tv{x}},
 \tv{r} \lessdot \tv{a}, \ol{\tv{x}} \lessdotCC \ol{T}, \type{T} \lessdot \tv{r}, \ol{\wtv{b}} \lessdot \ol{N}$.
-We need to proof $\text{let}\ \expr{x} : \wcNtype{\Delta'}{N} = \expr{e}, \,
+%We need to proof $\text{let}\ \expr{x} : \wcNtype{\Delta'}{N} = \expr{e}, \,
-\overline{\expr{x} : \wcNtype{\Delta'}{N} = \expr{e}} \ \text{in}\ \texttt{x}.\texttt{m}(\ol{x}) : \type{T}_2$
+%\overline{\expr{x} : \wcNtype{\Delta'}{N} = \expr{e}} \ \text{in}\ \texttt{x}.\texttt{m}(\ol{x}) : \type{T}_2$
-where $\sigma(\tv{x}) = \wcNtype{\Delta'}{N}$, $\sigma(\ol{\tv{x}}) = \ol{\wcNtype{\Delta'}{N}}$,  $\sigma(\tv{a}) = \type{T}_2$.
+%where $\sigma(\tv{x}) = \wcNtype{\Delta'}{N}$, $\sigma(\ol{\tv{x}}) = \ol{\wcNtype{\Delta'}{N}}$,  $\sigma(\tv{a}) = \type{T}_2$.
-
+We omit the case where a capture conversion is not needed and
+assume $\sigma(\tv{x}) = \wcNtype{\Delta'}{N}$, $\sigma(\ol{\tv{x}}) = \ol{\wcNtype{\Delta'}{N}}$.
+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', \ol{\Delta'} | \Gamma, \expr{x} : \type{N}, \overline{\expr{x} : \type{N}} \vdash $
+\end{itemize}
     \item[$\texttt{v}.\texttt{m}(\ol{v})$]
     Proof is analog to field access, except the $\Delta \vdash \ol{S}\ \ok$ premise.
     We know that $\unify{}(\Delta, [\overline{\wtv{b}}/\ol{Y}]\set{