Add well-formedness description

soundness.tex

@@ -46,6 +46,7 @@ By structural induction over the expression $\texttt{e}$.
 %    $\Delta, \Delta' \vdash \type{U}_i <: \type{S}$ by premise.
 
     $\Delta \vdash \type{S}, \wcNtype{\Delta'}{N} \ \ok$ %TODO
+    %Easy, because unify only generates substitutions for normal type placeholders which are OK
 
     \item[$\texttt{e}.\texttt{m}(\ol{e})$] Given 
 \end{description}
@@ -59,11 +60,17 @@ By structural induction over the expression $\texttt{e}$.
 \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
 $\text{fv}(\type{T}) \subseteq \text{dom}(\Delta)$.
-%TODO: Unify subst rule does not fail when there are free type variables on the right side
+%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!
@@ -146,7 +153,7 @@ $\Delta \vdash  \sigma(C') \implies \Delta \vdash \sigma(C)$
 \begin{description}
 \item[AddDelta] $C$ is not changed
 \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 $C$)
-\item[GenSigma] by definition.
+\item[GenSigma] by definition and S-Refl.
 % holds for $\set{\tv{a} \doteq \type{G}}$ by definition.
 % Holds for $C$ by assumption and because $\tv{a} \notin C$ by solved form definition ($[\type{G}/\tv{a}]C = C$).
 \item[Ground] Assumption and S-Bot

tRules.tex

@@ -139,6 +139,28 @@ $\begin{array}{l}
 We change the type rules to require the well-formedness instead of the witnessed property.
 See figure \ref{fig:well-formedness}.
 
+Our well-formedness criteria is more restrictive than the ones used for \wildFJ{}.
+\cite{WildcardsNeedWitnessProtection} works with the two different judgements $\ok$ and \texttt{witnessed}.
+With \texttt{witnessed} being the stronger one.
+We rephrased the $\ok$ judgement to include \texttt{witnessed} aswell.
+$\type{T} \ \ok$ in this paper means $\type{T} \ \ok$ and $\type{T} \ \texttt{witnessed}$ in the sense of the \WildFJ{} type rules.
+Java's type system is complex enough as it is. Simplification, when possible, is always appreciated.
+Our $\ok$ rule may not be able to express every corner of the Java type system, but is sufficient to show soundness regarding type inference for a Java core calculus.
+
+The \rulename{WF-Class} rule requires upper and lower bounds to be direct subtypes of each other $\type{L} <: \type{U}$.
+The type rules from \cite{WildcardsNeedWitnessProtection} use a witness type instead.
+Stating that a type is well formed (\texttt{witnessed}) if there exists atleast one simple type $\type{N}$ as a possible subtype:
+$\Delta \vdash \type{N} <: \wctype{\Delta'}{C}{\ol{T}}$.
+
+%TODO: do we need \Delta' \Delta \vdash T, L ,U ok ? or is \Delta \vdashh ... sufficient?
+%<A> m(List<? extends A> l)
+%This is important because it means different things maybe for the proof of our OK being more restrictive than "witnessed"
+
+Everything else regarding subtyping stays the same as in \cite{WildcardsNeedWitnessProtection}.
+\begin{lemma}
+    If $\Delta \vdash $
+\end{lemma}
+
 Figure \ref{fig:tletexpr} shows type rules for fields and method calls.
 They have been merged with let statements and simplified.
 

unify.tex

@@ -949,19 +949,23 @@ Otherwise the generation rules \rulename{GenSigma} and \rulename{GenDelta} will
             \wildcardEnv \vdash [\type{B}/\tv{b}]C \implies \Delta \cup \set{\wildcard{B}{\type{N}}{\bot}}, \sigma \cup \set{\tv{b} \to \type{B}}
             } \quad
             \begin{array}{l}
-            \tph(\type{N}) = \emptyset, \text{fv}(\type{N}) = \emptyset \\
+            \tph(\type{N}) = \emptyset, \text{fv}(\type{N}) \subseteq \Delta \\
-            \rwildcard{B} \ \text{fresh}, \tv{b} \notin \text{dom}(\sigma)
+            \rwildcard{B} \ \text{fresh}, \tv{b} \notin \text{dom}(\sigma), \Delta \vdash \type{N} \ \ok
             \end{array}
           $
         \\\\ 
-          \rulename{AddSigma}
+          \rulename{AddSigma} %This rule adds the substitutions for a? variables
           & $
           \deduction{
-          \wildcardEnv \vdash C \cup \set{\tv{a} \doteq \rwildcard{X}} \implies \Delta, \sigma
+          \wildcardEnv \vdash C \cup \set{\tv{a} \doteq \type{T}} \implies \Delta, \sigma
           }{
-          \wildcardEnv \vdash C \cup \set{\tv{a} \doteq \rwildcard{X}} \implies \Delta, \sigma
+          \wildcardEnv \vdash C \cup \set{\tv{a} \doteq \type{T}} \implies \Delta, \sigma
           \cup \set{\tv{a} \to \rwildcard{X}}
-          }
+          } \quad 
+          \begin{array}{l}
+          \tph(\type{T}) = \emptyset \\
+          \tv{a} \notin \text{dom}(\sigma),\, \Delta \vdash \type{T} \ \ok
+          \end{array}
             $
         \\\\ 
       \rulename{GenSigma}
@@ -974,8 +978,8 @@ Otherwise the generation rules \rulename{GenSigma} and \rulename{GenDelta} will
       \set{\tv{a} \to \type{T} }
       } \quad 
       \begin{array}{l}
-      \tph(\type{T}) = \emptyset \\
+      \tph(\type{T}) = \emptyset \\ %,\, \text{fv}(\type{T}) \subseteq \Delta \\ % T ok implies that
-      \tv{a} \notin \text{dom}(\sigma)
+      \tv{a} \notin \text{dom}(\sigma),\, \Delta \vdash \type{T} \ \ok
       \end{array}
       $
     \\\\