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Rework Implementation rules. Add comments to Completeness

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JanUlrich 2024-06-10 17:59:40 +02:00
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130
aspUnify.tex
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@ -76,23 +76,86 @@ Currently Java only has local type inference.
We want to bring type inference for Java to the next level. We want to bring type inference for Java to the next level.
\section{Unify} \section{Unify}
Input: Every type placeholder must have an upper bound.
Output: Every $\tv{a} \lessdot \type{T}$ constraint gets a
We have to formulate the unification algorithm with implication rules. We have to formulate the unification algorithm with implication rules.
Those can be directly translated to ASP. Those can be directly translated to ASP.
\begin{mathpar} \begin{mathpar}
\inferrule[Subst-L]{ \inferrule[Subst-L]{
\tv{a} \doteq \type{N} \\ \tv{a} \doteq \type{T}_1 \\
\tv{a} \lessdot \type{T} \tv{a} \lessdot \type{T}_2
}{ }{
\type{N} \lessdot \type{T} \type{T}_1 \lessdot \type{T}_2
} }
\and \and
\inferrule[Subst-R]{ \inferrule[Subst-R]{
\tv{a} \doteq \type{N} \\ \tv{a} \doteq \type{T}_1 \\
\type{T} \lessdot \tv{a} \type{T}_2 \lessdot \tv{a}
}{ }{
\type{T} \lessdot \type{N} \type{T}_2 \lessdot \type{T}_1
} }
\and \and
\inferrule[Subst-Equal]{
\tv{a} \doteq \type{T}_1 \\
\tv{a} \doteq \type{T}_2
}{
\type{T}_1 \doteq \type{T}_2
}
\and
\inferrule[Swap]{
\type{T}_1 \doteq \type{T}_2
}{
\type{T}_2 \doteq \type{T}_1
}
\and
\inferrule[Match]{
\tv{a} \lessdot \type{N}_1 \\
\tv{a} \lessdot \type{N}_2 \\
\type{N}_1 << \type{N}_2
}{
\type{T}_1 \lessdot \type{T}_2
}
\and
\inferrule[Adopt]{
\tv{a} \lessdot \tv{b} \\
\tv{b} \lessdot \type{T}
}{
\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} \\
}
\and
\inferrule[Adapt]{
\type{N}_1 \lessdot \exptype{C}{\type{T}_1 \ldots \type{T}_n} \\
\type{N}_1 <: \exptype{C}{\type{S}_1 \ldots \type{S}_n} \\
}{
\exptype{C}{\type{S}_1 \ldots \type{S}_n} \doteq \exptype{C}{\type{T}_1 \ldots \type{T}_n} \\
}
\and
\inferrule[Reduce]{
\exptype{C}{\type{S}_1 \ldots \type{S}_n} \doteq \exptype{C}{\type{T}_1 \ldots \type{T}_n} \\
}{
\type{S}_i \doteq \type{T}_i \\
}
\end{mathpar}
\begin{mathpar}
\inferrule[Super]{
\type{T} \lessdot \tv{a}\\
\type{T} <: \type{N}
}{
\tv{a} \doteq \type{N}
}
\end{mathpar}
\begin{mathpar}
\inferrule[Fail]{ \inferrule[Fail]{
\type{T} \lessdot \type{N}\\ \type{T} \lessdot \type{N}\\
\type{T} \nless : \type{N} \type{T} \nless : \type{N}
@ -108,35 +171,13 @@ Those can be directly translated to ASP.
} }
\and \and
\inferrule[Fail]{ \inferrule[Fail]{
\tv{a} \lessdot \type{T}_1 \\ \tv{a} \lessdot \type{N}_1 \\
\tv{a} \lessdot \type{T}_2 \\ \tv{a} \lessdot \type{N}_2 \\
\type{T}_1 \nless : \type{T}_2 \\ \text{not}\ \type{N}_1 << \type{N}_2 \\
\type{T}_2 \nless : \type{T}_1 \text{not}\ \type{N}_2 << \type{N}_1
}{ }{
\emptyset \emptyset
} }
\and
\inferrule[Swap]{
\type{T}_1 \doteq \type{T}_2
}{
\type{T}_2 \doteq \type{T}_1
}
\and
\inferrule[Adopt]{
\tv{a} \lessdot \type{T}_1 \\
\tv{a} \lessdot \type{T}_2 \\
\type{T}_1 <: \type{T}_2
}{
\type{T}_1 \lessdot \type{T}_2
}
\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} \\
}
\end{mathpar} \end{mathpar}
% Subst % Subst
% a =. N, a <. T, N <: T % a =. N, a <. T, N <: T
@ -149,4 +190,29 @@ Those can be directly translated to ASP.
\section{Completeness} \section{Completeness}
To proof completeness we have to show that every type can be replaced by a placeholder in a correct constraint set. To proof completeness we have to show that every type can be replaced by a placeholder in a correct constraint set.
Completeness -> we never exclude a solution
Following constraints stay: $\tv{a} \lessdot \type{T}$ if $\tv{a}$ is never on a right side of another constraint.
Every other type placeholder will be reduced to $\tv{a} \doteq \type{T}$, if there is a solution.
Proof:
%Induction over every possible constraint variation:
a =. T -> induction start
a <. T -> if no other constraint then it can stay otherwise there is either a =. T or a <. T
in latter case: a <. T, a <. T'
Proof that every type can be replaced by a Type Placeholder.
% Whats with a =. T, can T be replaced by a Type Placeholder?
% What is our finish condition? a <. T constraints stay, a =. b constraints stay.
% Algorithm does not fail -> \emptyset if a solution exists
% Otherwise there exists a substitution. If the algorithm succeeds we have to pick one of the possible solutions
% by: a <. T -> a =.T
% a =. b, b =. T -> use the solution generation from other paper
% TODO: try to include solution generation in the algorithm and proof that this solution is valid and will always occur as long as there is a solution
Soundness -> we never make a wrong implication
%$\tv{a} \doteq \type{T}$ means that $\[type{T}/\tv{a}]C$ is correct
If it succeeds then we can substitute all $\tv{a} \doteq \type{T}$
constraints in the original constraint set and
there exists a typing for the remaining type placeholders
so that the constraint set is satisfied.
\end{document} \end{document}