proofs for the system without type schemes

proofs.tex

@@ -1407,7 +1407,7 @@ theorem for the algorithmic type system presented in \ref{sec:algorules}.
             A derivation of the declarative type system is said positive iff it does not contain any rule \Rule{Abs-}.
           \end{definition}
         
-          \begin{theorem}[Completeness for positive derivations]
+          \begin{theorem}[Completeness for positive derivations]\label{completenessAtsPositive}
             For every $\Gamma$, $e$, $t$ such that we have a positive derivation of $\Gamma \vdash e:t$,
             there exists a global parameter $n_o$ with which $\tsrep{\tyof e \Gamma} \leq t$.
         
@@ -1710,7 +1710,7 @@ theorem for the algorithmic type system presented in \ref{sec:algorules}.
     \end{theorem}
 
     \begin{proof}
-      Trivial by using the previous lemma and the theorem \ref{soundnessAts}.
+      Trivial by using the theorem \ref{soundnessAts} and the previous lemma.
     \end{proof}
 
     \subsubsection{Completeness}
@@ -1723,6 +1723,32 @@ theorem for the algorithmic type system presented in \ref{sec:algorules}.
         \textbf{Positive expression} & e_+ & ::= & c\alt x\alt e_+ e_+\alt\lambda^{t^\lambda_+} x.e_+\alt \pi_j e_+\alt(e_+,e_+)\alt\tcase{e_+}{t_{s}}{e_+}{e_+}
       \end{array}
     \]
+
+    \begin{lemma}
+      If we restrict the language to positive expressions $e_+$,
+      then we have the following property:
+
+      $\forall \Gamma, e_+, \ts.\ \Gamma \vdashAts e_+:\ts \Rightarrow \Gamma \vdashA e_+:\tsrep{\ts}$
+    \end{lemma}
+
+    \begin{proof}
+      We can prove it by induction over the structure of $e_+$.
+
+      The main idea of this proof is that, as $e_+$ is a positive expression, the rule \Rule{Abs-} is not needed anymore
+      because the negative part of functional types (i.e. the $N_i$ part of their DNF) become useless:
+  
+      \begin{itemize}
+        \item When typing an application $e_1 e_2$, the negative part of the type of $e_1$ 
+        is ignored by the operator $\apply {} {}$.
+        \item Moreover, as there is no negated arrows in the domain of lambda-abstractions,
+        the negative arrows of the type of $e_2$ can also be ignored.
+        \item Similarly, negative arrows can be ignored when refining an application ($\worra {} {}$ also ignore the negative part
+        of the type of $e_1$).
+        \item Finally, as the only functional type that we can test is $\arrow \Empty \Any$, a functional type
+        cannot be refined to $\Empty$ due to its negative part, and thus we can ignore its negative part
+        (it makes no difference relatively to the rule \Rule{Efq}).
+      \end{itemize}
+    \end{proof}
   
     \begin{theorem}[Completeness of the algorithmic type system for positive expressions]\label{completenessA}
       If we restrict the language to positive expressions $e_+$,
@@ -1732,21 +1758,6 @@ theorem for the algorithmic type system presented in \ref{sec:algorules}.
       $\forall \Gamma, e_+, t.\ \Gamma \vdash e_+:t \Rightarrow \exists t'.\ \Gamma \vdashA e_+ : t'$
     \end{theorem}
   
-    \begin{proof} TODO: ADAPT IT. INDUCTION HYPOTHESIS MUST BE STRONGER.
-
-      With such restrictions, the rule \Rule{Abs-} is not needed anymore
-      because the negative part of functional types (i.e. the $N_i$ part of their DNF) is useless.
-  
-      Indeed, when typing an application $e_1 e_2$, the negative part of the type of $e_1$ 
-      is ignored by the operator $\apply {} {}$.
-  
-      Moreover, as there is no negated arrows in the domain of lambda-abstractions,
-      the negative arrows of the type of $e_2$ can also be ignored.
-  
-      Similarly, negative arrows can be ignored when refining an application ($\worra {} {}$ also ignore the negative part
-      of the type of $e_1$).
-  
-      Finally, as the only functional type that we can test is $\arrow \Empty \Any$, a functional type
-      cannot be refined to $\Empty$ due to its negative part, and thus we can ignore its negative part
-      (it makes no difference relatively to the rule \Rule{Efq}).
+    \begin{proof}
+      Trivial by using the theorem \ref{completenessAtsPositive} and the previous lemma.
     \end{proof}