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Add Solution rules

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This commit is contained in:
Andreas Stadelmeier 2024-06-12 14:37:35 +02:00
parent 0678f96ef4
commit 05260c286d
1 changed files with 91 additions and 45 deletions
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136
aspUnify.tex
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@ -126,6 +126,8 @@ Those can be directly translated to ASP.
\type{T}_2 \doteq \type{T}_1
}
\and
\inferrule[S-Object]{}{\tv{a} \lessdot \type{Object}}
\and
\inferrule[Match]{
\tv{a} \lessdot \type{N}_1 \\
\tv{a} \lessdot \type{N}_2 \\
@ -141,13 +143,13 @@ Those can be directly translated to ASP.
\tv{a} \lessdot \type{T}
}
\and
\inferrule[Subst-Param]{
\tv{a} \doteq \type{N} \\
\tv{a} = \type{T}_i \\
\exptype{C}{\type{T}_1 \ldots \type{T}_n} <: \type{T} \\
}{
\type{T}_i \doteq \type{N} \\
}
% \inferrule[Subst-Param]{
% \tv{a} \doteq \type{N} \\
% \tv{a} = \type{T}_i \\
% \exptype{C}{\type{T}_1 \ldots \type{T}_n} \lessdot \type{T} \\
% }{
% \type{T}_i \doteq \type{N} \\
% }
\and
\inferrule[Adapt]{
\type{N}_1 \lessdot \exptype{C}{\type{T}_1 \ldots \type{T}_n} \\
@ -189,6 +191,45 @@ Those can be directly translated to ASP.
\type{N} \lessdot \tv{b}
}
\end{mathpar}
Result:
\begin{mathpar}
\inferrule[Solution]{
\tv{a} \doteq \type{N} \\
\tv{a} \notin \type{N}
}{
\sigma(\tv{a}) = \type{N}
}
\and
\inferrule[Solution-Sub]{
\tv{a} \lessdot \exptype{C_1}{\ol{T_1}}, \ldots, \tv{a} \lessdot \exptype{C_n}{\ol{T_n}} \\
\forall i: \type{C_m} << \type{C_i} \\
\text{not}\ \tv{a} \doteq \type{N}
}{
\sigma(\tv{a}) = \exptype{C_m}{\ol{T_m}}
}
\and
\inferrule[Solution]{
\tv{a} \doteq \type{G}
}{
\sigma(\tv{a}) = \type{N}
}
\and
\inferrule[Unfold]{
\tv{b} \doteq \exptype{C}{\type{T}_1 \ldots \type{T}_n}
}{
\type{T}_i \doteq \type{T}_i
}
\and
\inferrule[Subst-Param]{
\tv{a} \doteq \type{G} \\
\type{T} \doteq \exptype{C}{\type{T}_1 \ldots, \tv{a}, \ldots \type{T}_n} \\
}{
\type{T} \doteq \exptype{C}{\type{T}_1, \ldots \type{G}, \ldots \type{T}_n}
}
\end{mathpar}
Fail:
\begin{mathpar}
\inferrule[Fail]{
@ -222,31 +263,18 @@ Those can be directly translated to ASP.
}
\end{mathpar}
Result:
\begin{mathpar}
\inferrule[Solution]{
\tv{a} \doteq \type{N} \\
\tv{a} \notin \type{N}
}{
\sigma(\tv{a}) = \type{N}
}
\and
\inferrule[Solution-Sub]{
\tv{a} \lessdot \type{N}_1, \ldots, \tv{a} \lessdot \type{N}_n \\
\forall i: \type{N} <: \type{N}_i \\
\text{not}\ \tv{a} \doteq \type{N}
}{
\sigma(\tv{a}) = \type{N}
}
\end{mathpar}
% Subst
% a =. N, a <. T, N <: T
% --------------
% N <. T
% a <. List<b>, b <. List<a>
% how to proof completeness and termination?
% TODO: how to proof termination?
The algorithm terminates if every type placeholder in the input constraint set has an assigned type.
\section{Completeness}
To proof completeness we have to show that every type can be replaced by a placeholder in a correct constraint set.
@ -280,10 +308,36 @@ if $\type{T} \lessdot \type{T'}$ and $\sigma(\tv{a}) = \type{N}$
then $[\type{N}/\tv{a}]\type{T} <: [\type{N}/\tv{a}]\type{T'}$.
\end{theorem}
\begin{theorem}{Completeness}
$\forall \tv{a} \in C_{input}: \sigma(\tv{a}) = \type{N}$, if there is a solution for $C_{input}$.
\end{theorem}
%Problem: We do not support multiple inheritance
\SetEnumitemKey{ncases}{itemindent=!,before=\let\makelabel\ncasesmakelabel}
\newcommand*\ncasesmakelabel[1]{Case #1}
\newenvironment{subproof}
{\def\proofname{Subproof}%
\def\qedsymbol{$\triangleleft$}%
\proof}
{\endproof}
Due to Match there must be $\type{N}_1 \lessdot \type{N}_2 \ldots \lessdot \type{N}_n$
\begin{proof}
\begin{enumerate}[ncases]
\item $\tv{a} \lessdot \exptype{C}{\ol{T}}$.
Solution-Sub
Let $\sigma(\tv{a}) = \type{N}$. Then $\type{N} <: \exptype{C}{[\type{N}/\tv{a}]}$
\item $\tv{a} \doteq \type{N}$.
Solution
\item $\tv{a} \lessdot \tv{b}$.
There must be a $\tv{a} \lessdot \type{N}$
\begin{subproof}
$\sigma(\tv{a}) = \type{Object}$,
$\sigma(\tv{b}) = \type{Object}$.
\end{subproof}
\item $\type{N} \lessdot \tv{a}$.
\begin{subproof}
$2$
\end{subproof}
\end{enumerate}
And more text.
\end{proof}
\begin{lemma} \label{lemma:subtypeOnly}
If $\sigma(\tv{a}) = \emptyset$ then $\tv{a}$ appears only on the left side of $\tv{a} \lessdot \type{T}$ constraints.
@ -295,28 +349,20 @@ Then either the Solution-Sub generates a $\sigma$ or the Solution rule can be us
The Solution-Sub rule is always correct.
Proof:
\SetEnumitemKey{ncases}{itemindent=!,before=\let\makelabel\ncasesmakelabel}
\newcommand*\ncasesmakelabel[1]{Case #1}
\newenvironment{subproof}
{\def\proofname{Subproof}%
\def\qedsymbol{$\triangleleft$}%
\proof}
{\endproof}
\begin{theorem}{Completeness}
$\forall \tv{a} \in C_{input}: \sigma(\tv{a}) = \type{N}$, if there is a solution for $C_{input}$
and every type placeholder has an upper bound $\tv{a} \lessdot \type{N}$.
\end{theorem}
%Problem: We do not support multiple inheritance
\begin{proof}
\begin{enumerate}[ncases]
\item $\tv{a} \lessdot \type{N}$.
\begin{subproof}
If $\tv{a} \notin \type{N}$ then $\sigma(\tv{a}) = \type{N}$ otherwise there is no type solution.
$\sigma(\tv{a}) = \type{B}$ with $type{B} \triangleleft [\type{B}/\tv{a}]\type{N}$
\end{subproof}
Solution-Sub
\item $\tv{a} \doteq \type{N}$.
\begin{subproof}
$\sigma(\tv{a}) = \type{N}$
\end{subproof}
\item $C \cup \tv{a} \lessdot \tv{b}$.
If
Solution
\item $\tv{a} \lessdot \tv{b}$.
There must be a $\tv{a} \lessdot \type{N}$
\begin{subproof}
$\sigma(\tv{a}) = \type{Object}$,
$\sigma(\tv{b}) = \type{Object}$.