Var Rule Subtyping. Add Termination Theorem
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Andreas Stadelmeier
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05260c286d
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71ffb2c766
88
aspUnify.tex
88
aspUnify.tex
@ -87,6 +87,8 @@ We want to bring type inference for Java to the next level.
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\type{T}_1 <: \type{T}_3
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}
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\and
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\inferrule[Var]{}{\type{A} <: \Delta(\type{A})}
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\and
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\inferrule[Class]{\texttt{class}\ \exptype{C}{\ol{X}} \triangleleft \type{N}}{
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\exptype{C}{\ol{T}} <: [\ol{T}/\ol{X}]\type{N}
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}
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@ -126,6 +128,19 @@ Those can be directly translated to ASP.
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\type{T}_2 \doteq \type{T}_1
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}
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\and
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\inferrule[Unfold]{
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\tv{b} \doteq \exptype{C}{\type{T}_1 \ldots \type{T}_n}
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}{
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\type{T}_i \doteq \type{T}_i
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}
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\and
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\inferrule[Subst-Param]{
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\tv{a} \doteq \type{G} \\
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\type{T} \doteq \exptype{C}{\type{T}_1 \ldots, \tv{a}, \ldots \type{T}_n} \\
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}{
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\type{T} \doteq \exptype{C}{\type{T}_1, \ldots \type{G}, \ldots \type{T}_n}
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}
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\and
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\inferrule[S-Object]{}{\tv{a} \lessdot \type{Object}}
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\and
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\inferrule[Match]{
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@ -166,6 +181,7 @@ Those can be directly translated to ASP.
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\end{mathpar}
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\begin{mathpar}
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\text{Apply only once per constraint:}\quad
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\inferrule[Super]{
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\type{T} \lessdot \tv{a}\\
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\type{T} <: \type{N}
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@ -174,6 +190,9 @@ Those can be directly translated to ASP.
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}
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\end{mathpar}
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\begin{center}
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Apply one or the other:
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\end{center}
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\begin{mathpar}
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\inferrule[Split-L]{
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\tv{a} \lessdot \tv{b}\\
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@ -181,9 +200,9 @@ Those can be directly translated to ASP.
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}{
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\tv{b} \lessdot \type{N}
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}
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\and
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\quad \quad
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\vline
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\and
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\quad \quad
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\inferrule[Split-R]{
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\tv{a} \lessdot \tv{b}\\
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\tv{a} \lessdot \type{N}\\
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@ -194,9 +213,14 @@ Those can be directly translated to ASP.
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Result:
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\begin{mathpar}
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% \inferrule[Solution]{
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% \tv{a} \doteq \type{N} \\
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% \tv{a} \notin \type{N}
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% }{
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% \sigma(\tv{a}) = \type{N}
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% }
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\inferrule[Solution]{
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\tv{a} \doteq \type{N} \\
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\tv{a} \notin \type{N}
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\tv{a} \doteq \type{G}
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}{
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\sigma(\tv{a}) = \type{N}
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}
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@ -206,39 +230,20 @@ Result:
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\forall i: \type{C_m} << \type{C_i} \\
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\text{not}\ \tv{a} \doteq \type{N}
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}{
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\sigma(\tv{a}) = \exptype{C_m}{\ol{T_m}}
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}
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\and
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\inferrule[Solution]{
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\tv{a} \doteq \type{G}
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}{
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\sigma(\tv{a}) = \type{N}
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}
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\and
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\inferrule[Unfold]{
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\tv{b} \doteq \exptype{C}{\type{T}_1 \ldots \type{T}_n}
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}{
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\type{T}_i \doteq \type{T}_i
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}
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\and
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\inferrule[Subst-Param]{
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\tv{a} \doteq \type{G} \\
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\type{T} \doteq \exptype{C}{\type{T}_1 \ldots, \tv{a}, \ldots \type{T}_n} \\
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}{
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\type{T} \doteq \exptype{C}{\type{T}_1, \ldots \type{G}, \ldots \type{T}_n}
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\Delta(\type{A}) = \exptype{C_m}{\ol{T_m}} \\ \sigma(\tv{a}) = \type{A}
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}
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\end{mathpar}
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Fail:
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\begin{mathpar}
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\inferrule[Fail]{
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\type{T} \lessdot \type{N}\\
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\type{T} \nless : \type{N}
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}{
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\emptyset
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}
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\and
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% \inferrule[Fail]{
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% \type{T} \lessdot \type{N}\\
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% \type{T} \nless : \type{N}
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% }{
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% \emptyset
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% }
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% \and
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\inferrule[Fail]{
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\exptype{C}{\ldots} \doteq \exptype{D}{\ldots}\\
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\type{C} \neq \type{D}
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@ -338,6 +343,15 @@ Due to Match there must be $\type{N}_1 \lessdot \type{N}_2 \ldots \lessdot \type
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And more text.
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\end{proof}
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\begin{lemma}{Substitution}
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\begin{description}
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\item[If] $\tv{a} \doteq \type{N}$ with $\tv{a} \neq \type{N}$
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\item[Then] for every $\type{T} \doteq \type{T}$ there exists a $[\type{N}/\tv{a}]\type{T} \doteq [\type{N}/\tv{a}]\type{T}$
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\item[Then] for every $\type{T} \lessdot \type{T}$ there exists a $[\type{N}/\tv{a}]\type{T} \lessdot [\type{N}/\tv{a}]\type{T}$
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\end{description}
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\end{lemma}
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\textit{Proof:}
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TODO
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\begin{lemma} \label{lemma:subtypeOnly}
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If $\sigma(\tv{a}) = \emptyset$ then $\tv{a}$ appears only on the left side of $\tv{a} \lessdot \type{T}$ constraints.
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@ -350,6 +364,18 @@ The Solution-Sub rule is always correct.
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Proof:
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\begin{theorem}{Termination}
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%jede nichtendliche Menge von Constraints bleibt endlich. Die Regeln können nicht unendlich oft angewendet werden
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%Trivial. The only possibility would be if we allow a =. C<a> constraints!
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\end{theorem}
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TODO: For completeness we have to proof that not $\tv{a} \doteq \type{N}$ only is the case if $\tv{a}$ only appears on the left side of $\tv{a} \lessdot \type{T}$ constraints.
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Problem: a <. List<a>
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a <. List<b>
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then a =. b
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Solution: Keep the a <. N constraints and apply the Step 6 from the GTFGJ paper.
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Then we have to proof that only a <. N constraints remain with sigma(a) = empty. Or Fail
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\begin{theorem}{Completeness}
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$\forall \tv{a} \in C_{input}: \sigma(\tv{a}) = \type{N}$, if there is a solution for $C_{input}$
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and every type placeholder has an upper bound $\tv{a} \lessdot \type{N}$.
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