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Agda-2.3.2.2: notes/papers/implicit/proof.tex

The correctness of the algorithm relies on the fact that we only compute with
well-typed terms.  This guarantees the existence of normal forms, and hence,
ensures the termination of the type checking algorithm.

The proof will be done in two stages: first we prove soundness in the absence
of constraint solving, and then we prove that constraint solving is sound.

\subsection{Soundness without constraint solving}

There are a number of things we need to prove: that type checking preserves
well-formed signatures, that it produces well-typed terms, that conversion
checking is sound, and that new signatures respect the old signatures.
Unfortunately these properties are all interdependent, so we cannot prove them
separately.

\begin{definition}[Signature extension] \label{defSigExt}
    We say that $\Sigma'$ {\em extends} $\Sigma$ if for any {\Core}
    judgement $J$, $\vdash_\Sigma J$ implies $\vdash_{\Sigma'} J$.
\end{definition}

Note that this definition admits both simple extensions--adding a new
constant--and refinement, where we give a definition to a constant.
\if \DetailedProofs 1
This is expressed by the following two lemmas.

\begin{lemma}[Signature weakening] \label{lemWeakenSig}
    If $\Sigma' = \Ext \Sigma {c:A}$ and $\IsSigCS {\Sigma'}$ then $\Extends
    {\Sigma'} \Sigma$.
\end{lemma}

\begin{lemma}[Signature refinement] \label{lemRefineSig}
    Giving a definition to a constant in a signature is an extension of the
    signature.
\if \DetailedProofs 1
    More precisely, if
    \begin{itemize}
	\item $\HasTypeCS {\Sigma_1} {} M A$
	\item $\Sigma ~ = ~ \Ext {\Sigma_1} \Ext {\MetaDecl c A} \Sigma_2$
	\item $\Sigma' ~ = ~ \Ext {\Sigma_1} \Ext {\IMetaDecl c A M} \Sigma_2$
    \end{itemize}
    then $\Extends {\Sigma'} \Sigma$.
\fi
\end{lemma}
\fi

\if \DetailedProofs 1
\begin{proof}[Proofs]
    In both lemmas any derivation using $\Sigma$ is immediately valid also with $\Sigma'$.
\end{proof}
\fi

\if \DetailedProofs 1
To express the soundness of conversion checking we need to define when a
constraint is well-formed. Note that this is not the same as being true. For
instance, $\TypeConstr {\mathit{Nat}} {\mathit{Bool}}$ is a well-formed
constraint given $\mathit{Nat} : \SET$ and $\mathit{Bool} : \SET$ in the
signature.

\begin{definition}[Well-formed constraint]
    A constraint if well-formed if the terms (or types) under consideration are
    well-typed. 
    \if \DetailedProofs 1
    \[
	\infer{ \ValidConstr \Sigma {\TypeConstr A B}}
	{\begin{array}{l}
	    \IsTypeCS \Sigma \Gamma A \\
	    \IsTypeCS \Sigma \Gamma B 
	\end{array}}
    \qquad
	\infer{ \ValidConstr \Sigma {\TermConstr M N A}}
	{\begin{array}{l}
	    \HasTypeCS \Sigma \Gamma M A \\
	    \HasTypeCS \Sigma \Gamma N A \\
	\end{array}}
    \]
    \else
    For instace, $\TermConstr M N A$ is well-formed if $\HasTypeC
    \Gamma M A$ and $\HasTypeC \Gamma N A$.
    \fi
\end{definition}
\fi

Now we are ready to state the soundness of the type checking algorithm in the
absence of constraint solving.

\begin{theorem}[Soundness of type checking] \label{thmSoundNoCs}
    Type checking produces well-typed terms, conversion checking produces
    well-formed constraints and if no constraints are produced, the conversion
    is valid in {\Core}. Also, all rules produce well-formed extensions of the
    signature.
\if \DetailedProofs 1
    \begin{itemize}

	% is type
	\item
	    $\begin{array}[t]{l}
		\ExplicitJudgement \Sigma {\IsType e A} {\Sigma'}
		~ \wedge ~ \IsCtxCS \Sigma \Gamma \\
		{} \implies \Extends {\Sigma'} {\Sigma}
		~ \wedge ~ \IsTypeCS {\Sigma'} \Gamma A
	    \end{array}$

	% check type
	\item
	    $\begin{array}[t]{l}
		\ExplicitJudgement \Sigma {\CheckType e A M} {\Sigma'}
		~ \wedge ~ \IsTypeCS \Sigma \Gamma A \\
		{} \implies \Extends {\Sigma'} {\Sigma}
		~ \wedge ~ \HasTypeCS {\Sigma'} \Gamma M A
	    \end{array}$

	% infer type
	\item
	    $\begin{array}[t]{l}
		\ExplicitJudgement \Sigma {\InferType e A M} {\Sigma'}
		~ \wedge ~ \IsCtxCS \Sigma \Gamma \\
		{} \implies \Extends {\Sigma'} {\Sigma}
		~ \wedge ~ \HasTypeCS {\Sigma'} \Gamma M A
	    \end{array}$

	% equal types
	\item
	    $\begin{array}[t]{l}
		\ExplicitJudgement \Sigma {\EqualType A B \Cs} {\Sigma'}
		~ \wedge ~ \IsTypeCS \Sigma \Gamma A
		~ \wedge ~ \IsTypeCS \Sigma \Gamma B
		\\
		{} \implies \Extends {\Sigma'} {\Sigma}
		~ \wedge ~ \ValidConstr {\Sigma'} \Cs
		~ \wedge ~ (\Cs = \emptyset \implies \EqualTypeCS {\Sigma'} \Gamma A B)
	    \end{array}$

	% equal terms
	\item
	    $\begin{array}[t]{l}
		\ExplicitJudgement \Sigma {\Equal M N A \Cs} {\Sigma'}
		~ \wedge ~ \HasTypeCS \Sigma \Gamma M A
		~ \wedge ~ \HasTypeCS \Sigma \Gamma N A
		\\
		{} \implies \Extends {\Sigma'} {\Sigma}
		~ \wedge ~ \ValidConstr {\Sigma'} \Cs
		~ \wedge ~ (\Cs = \emptyset \implies \EqualCS {\Sigma'} \Gamma M N A)
	    \end{array}$

    \end{itemize}

    The statements for weak head normal form conversion ($\EqualWhnf M N A
    \Cs$) and term sequence conversion ($\Equal {\bar M} {\bar N} \Delta \Cs$)
    are equivalent to that of term conversion.
\fi
\end{theorem}

\begin{proof}
    By induction on the derivation.
\if \DetailedProofs 1
    Some interesting cases:
    \begin{itemize}

	\item {\em (TODO: Not so interesting)} In the type conversion case for
	function spaces where the domains produce constraints, we have to use
	Lemma~\ref{lemCoreSubstType}.

	\item In the term conversion case where the terms are weak head
	normalised we need subject reduction for weak head normalisation
	(Lemma~\ref{lemCoreSubjectReduction}).

	\item When checking conversion of terms with the same head we need an
	inversion principle for application (Lemma~\ref{lemCoreAppInv}).

	\item The most interesting case is the meta-variable instantiation
	case, so let us spell that out in more detail.

	The instantiation rule does not produce any constraints, so the only
	thing we have to prove is that it constructs a valid extension of the
	signature. This follows from the signature refinement lemma
	(Lemma~\ref{lemRefineSig}) which can be applied if we prove that if
	$\Sigma = \Ext {\Sigma_1} \Ext {\MetaDecl \alpha {B'}} \Sigma_2$ then
	$\LAM {\bar x} M : {B'}$.

	We have $\HasTypeCS \Sigma \Gamma {\alpha \, \bar x} A$ so $B'$ must
	have the form $\PI {\bar x} \Delta B$. By Lemma~\ref{lemCoreAppInv} we
	conclude that $\HasTypeCS \Sigma \Gamma {\bar x} \Delta$ and thus
	$\HasTypeCS \Sigma \Gamma {\alpha \, \bar x} B$. Then by
	Lemma~\ref{lemCoreEqType} $\EqualTypeCS \Sigma \Gamma A B$.

	From $\HasTypeCS \Sigma \Gamma M A$ we get $\HasTypeCS \Sigma \Gamma M
	B$ and using Lemma~\ref{lemCoreShadow}  $\HasTypeCS \Sigma \Gamma {\LAM
	{\bar x} M} {\Delta \to B}$. We know that $\Delta \to B$ is a closed
	type, and since $\FV{M} \subseteq \bar x$, $\LAM {\bar x} M$ is also
	closed. Thus by strenghtening (Lemma~\ref{lemCoreStrengthen})
	$\HasTypeCS \Sigma {} {\LAM {\bar x} M} {\Delta \to B}$. We have
	$\IsTypeCS {\Sigma_1} {} {\Delta \to B}$ and $\InScope \alpha M$ so
	$\HasTypeCS {\Sigma_1} {} {\LAM {\bar x} M} {\Delta \to B}$ which is
	what we set out to prove.

    \end{itemize}
\else
    The most interesting case is the meta-variable instantiation case where we
    have to prove that the instantiation constructs a valid extension of the
    signature. This is proven by showing that the instantiation is well-typed.
%     This follows from Lemma~\ref{lemRefineSig} once we show that
%     the instantiation is well-typed.
\fi
\end{proof}

Since well-typed terms in {\Core} have normal forms we get the existence of
normal forms for type checked terms and hence the type checking algorithm is
terminating.

\begin{corollary}
    The type checking algorithm is terminating.
\end{corollary}

Note that type checking terminates with one of three answers: {\em yes it is
type correct}, {\em no it is not correct}, or {\em it might be correct if the
meta-variables are instantiated properly}. The algorithm is not complete, since
finding correct instantiations to the meta-variables is undecidable in the
general case.

\subsection{Soundness of constraint solving}

In the previous section we proved type checking sound and decidable in the
absence of constraint solving. We also mostly ignored the constraints, only
requiring them to be well-formed. In this section we prove that the terms
produced by the type checker stay well-typed under constraint solving. This is
done by showing that constraint solving is a signature extension operator in
the sense of Definition~\ref{defSigExt}.

Previously we only ensured that the {\Core} restriction of the signature was
well-formed. Now, since we are going to update and remove the constraints of
guarded constants we have to strengthen the requirements and demand {\em
consistent} signatures. A signature is consistent if the solution of a guard is
a sufficient condition for the well-typedness of the definition it is guarding.

\if \DetailedProofs 1

\begin{definition}[Ensures] \label{defEnsures}
    A set of constraints $\Cs$ {\em ensures} a {\Core} judgement $J$ in a signature
    $\Sigma$ if, for any $\Sigma_1$ such that
    $\Extends {\Sigma_1} \Sigma$ and
    $\ExplicitJudgement {\Sigma_1} {\CheckConstr \Cs \emptyset} {\Sigma_2}$
    it is the case that $\vdash_{\CoreSig{\Sigma_2}} J$.
\end{definition}

\begin{remark} \label{lemExtendEnsures}
    If $\Cs$ ensures $J$ in $\Sigma$ and $\Extends {\Sigma'} \Sigma$ then $\Cs$
    ensures $J$ in $\Sigma'$.
\end{remark}

\begin{definition}[Consistent signature] \label{defConsistentSig}
    A signature $\Sigma$ is said to be {\em consistent} if $\Sigma = \Ext
    {\Sigma_1} \Ext {\ConstDecl p A M \Cs} \Sigma_2$ implies that $\Cs$ ensures
    $\HasTypeC {} M A$ in $\Sigma_1$.
\end{definition}

\fi

In order to prove that type checking preserves consistency, we first need to
know that the constraints we produce are sound.

\begin{lemma}[Soundness of generated constraints] \label{lemSoundConstraints}
    The constraints generated during conversion checking ensures that the
    checked terms are convertible. For instance, if $\EqualType A B \Cs$, then
    solving $\Cs$ guarantees that $\EqualTypeC \Gamma A B$ in {\Core}.
\if \DetailedProofs 1
    More precisely,
    \begin{itemize}
	\item \( \begin{array}[t]{l}
		    \IsTypeCS \Sigma \Gamma A
		    ~ \wedge ~ \IsTypeCS \Sigma \Gamma B
		    ~ \wedge ~ \ExplicitJudgement \Sigma {\EqualType A B \Cs} {\Sigma'}
		    \\ {} \implies \mbox{$\Cs$ ensures $\EqualTypeC \Gamma A B$ in $\Sigma'$}
	      \end{array} \)
	\item \( \begin{array}[t]{l}
		    \HasTypeCS \Sigma \Gamma M A
		    ~ \wedge ~ \HasTypeCS \Sigma \Gamma N A
		    ~ \wedge ~ \ExplicitJudgement \Sigma {\Equal M N A \Cs} {\Sigma'}
		    \\ {} \implies \mbox{$\Cs$ ensures $\EqualC \Gamma M N A$ in $\Sigma'$}
	      \end{array} \)
    \end{itemize}
\fi
\end{lemma}

\if \DetailedProofs 1
\begin{proof}
    Again we highlight some interesting cases.
    \begin{itemize}
	\item 
	
	The only difficult case is the case of conversion for function types
	where a new constant $p$ is introduced. There we need to prove
	$\EqualTypeCS {\Sigma_2} {\Ext \Gamma x : A_1} {B_1} {B_2}$ from
	$\EqualTypeCS {\Sigma_2} {\Ext \Gamma x : A_1} {B_1} {\SubstD {B_2} {p
	\, \Gamma \, x}}$

	If $\Sigma_2$ has an empty guard for $p$ then $p \, \Gamma \, x$
	reduces to $x$ and we are done. If the guard is non-empty we can apply
	Lemma~\ref{lemCoreEqualDummySubst}, since $p \, \Gamma \, x$ is on weak
	head normal form, and $p$ is a fresh constant which does not appear in
	$B_1$.

	\item In the case where $\Cs$ is known (for instance, in the rule for
	blocked terms), we can apply soundness of conversion checking
	(Theorem~\ref{thmSoundNoCs}) to get $\TrueConstr {\Sigma_2} \Cs$.

    \end{itemize}
\end{proof}
\fi

\if \DetailedProofs 1
\begin{lemma} \label{lemRefineConsistent}
    Refinement preserves consistent signatures.
    More precisely, if
    \begin{itemize}
	\item $\HasTypeCS {\Sigma_1} {} M A$
	\item $\Sigma ~ = ~ \Ext {\Sigma_1} \Ext {\MetaDecl c A} \Sigma_2$
	\item $\Sigma' ~ = ~ \Ext {\Sigma_1} \Ext {\IMetaDecl c A M} \Sigma_2$
	\item $\Sigma$ is consistent
    \end{itemize}
    then $\Sigma'$ is consistent.
\end{lemma}

\begin{proof}
    There are two cases to consider: refinement to the left and to the right of
    a guard. In the latter case the proof is trivial, and in the former case
    consistency follows from the fact that refinement extends a signature
    (Lemma~\ref{lemRefineSig}).
\end{proof}
\fi

\begin{lemma}[Type checking preserves consistency] \label{lemTypeCheckConsistent}
Type checking and conversion checking preserves consistent signatures.
\if \DetailedProofs 1
More precisely,
    \begin{itemize}

	% is type
	\item
	    $\begin{array}[t]{l}
		\ExplicitJudgement \Sigma {\IsType e A} {\Sigma'}
		~ \wedge ~ \IsCtxCS \Sigma \Gamma
		~ \wedge ~ \mbox{$\Sigma$ is consistent}
		\\ {} \implies \mbox{$\Sigma'$ is consistent}
	    \end{array}$

	% check type
	\item
	    $\begin{array}[t]{l}
		\ExplicitJudgement \Sigma {\CheckType e A M} {\Sigma'}
		~ \wedge ~ \IsTypeCS \Sigma \Gamma A
		~ \wedge ~ \mbox{$\Sigma$ is consistent}
		\\ {} \implies \mbox{$\Sigma'$ is consistent}
	    \end{array}$

	% infer type
	\item
	    $\begin{array}[t]{l}
		\ExplicitJudgement \Sigma {\InferType e A M} {\Sigma'}
		~ \wedge ~ \IsCtxCS \Sigma \Gamma
		~ \wedge ~ \mbox{$\Sigma$ is consistent}
		\\ {} \implies \mbox{$\Sigma'$ is consistent}
	    \end{array}$

	% equal types
	\item
	    $\begin{array}[t]{l}
		\ExplicitJudgement \Sigma {\EqualType A B \Cs} {\Sigma'}
		~ \wedge ~ \IsTypeCS \Sigma \Gamma A
		~ \wedge ~ \IsTypeCS \Sigma \Gamma B \\
		~ \wedge ~ \mbox{$\Sigma$ is consistent}
		\\ {} \implies \mbox{$\Sigma'$ is consistent}
	    \end{array}$

	% equal terms
	\item
	    $\begin{array}[t]{l}
		\ExplicitJudgement \Sigma {\Equal M N A \Cs} {\Sigma'}
		~ \wedge ~ \HasTypeCS \Sigma \Gamma M A
		~ \wedge ~ \HasTypeCS \Sigma \Gamma N A \\
		~ \wedge ~ \mbox{$\Sigma$ is consistent}
		\\ {} \implies \mbox{$\Sigma'$ is consistent}
	    \end{array}$

    \end{itemize}

    The statements for weak head normal form conversion ($\EqualWhnf M N A
    \Cs$) and term sequence conversion ($\Equal {\bar M} {\bar N} \Delta \Cs$)
    are equivalent to that of term conversion.
\fi
\end{lemma}

\begin{proof}
\if \DetailedProofs 1
    By induction on the derivation. We only need to consider the cases where
    the signature changes. Adding a (non-guarded) constant trivially preserves
    consistency, instantiating a meta-variable preserves consistency by
    Lemma~\ref{lemRefineConsistent}. What remains is to check that new guarded
    constants are consistent. There are two cases: the conversion rule and
    conversion checking of function types. In both cases consistency follows
    from soundness of conversion checking (Lemma~\ref{lemSoundConstraints}).
\else
    We have to prove that when we introduce a guarded constant the guard
    ensures the well-typedness of the value. This follows from
    Lemma~\ref{lemSoundConstraints}.
\fi
\end{proof}

\begin{lemma}[Constraint solving is sound] \label{lemSolveConsistent}
    If $\Sigma$ is consistent and the solving the constraints yields a
    signature $\Sigma'$, then $\Sigma'$ is consistent and $\Extends {\Sigma'}
    \Sigma$.
\end{lemma}

\begin{proof}
    Follows from Theorem~\ref{thmSoundNoCs}, Lemma~\ref{lemSoundConstraints},
    and Lemma~\ref{lemTypeCheckConsistent}.
\end{proof}

From this follows that we can mix type checking and constraint solving freely,
so we can add a constraint solving rule to the type checking algorithm. In
order to obtain optimal approximations we have to solve constraints eagerly,
i.e as soon as a meta-variable has been instantiated.

\subsection{Relating user expressions and checked terms}

An important property of the type checking algorithm is that the type correct terms
produced correspond to the expressions being type checked. The correspondance
is expressed by stating that the only operations the type checker is allowed
when constructing a term is replacing a $?$ by a term ({\em refinement}) and
replacing a term by a guarded constant ({\em approximation}).

\if \DetailedProofs 1

\begin{definition}[Approximation]
    A term $M$ {\em approximates} $M'$ if $M$ can be obtained by replacing
    subterms of $M'$ by guarded constants $p \, \bar x$.
\end{definition}

\begin{definition}[Refinement]
    A term $M$ is a {\em refinment} of a user expression $e$ if $M$ can be
    obtained by replacing the $?$ in $e$ by concrete terms.
\end{definition}

\fi

\begin{lemma} \label{lemApproxRefine}
    If $\CheckType e A M$ then $M$ approximates a refinement of $e$. This
    property is preserved when unfolding instantiated meta-variables and
    guarded constants in $M$.
\end{lemma}

\begin{lemma}
    If $\CheckType e A M$ then $M$ is an {\em optimal approximation} of a
    refinement $M'$ of $e$.
\end{lemma}

\begin{proof}
The proof relies on the fact that we only introduce guarded constants when
absolutely necessary and solve the constraints eagerly. This is proven by
showing that the constraints produced by conversion checking are not only
sufficient but also necessary for the validity of the judgement.
\end{proof}

\subsection{Main result}

We now prove the main soundness theorem stating that if all meta-variables are
instantiated and all guards solved, then the term produced by the type checker
(extended with constraint solving) is valid in the original signature after
unfolding the definitions of the meta-variables and guarded constants
introduced during type checking.

\begin{theorem}[Soundness of type checking] \label{thmMain}
    If $\Sigma$ is a well-formed {\Core} signature and $\ExplicitJudgement
    \Sigma {\CheckType e A M} {\Sigma'}$, then if all meta-variables have been
    instantiated and all guards are empty in $\Sigma'$, then $\HasTypeCS \Sigma
    \Gamma {M\sigma} A$ where $\sigma$ is the substitution inlining the meta
    variables and constants in $\Sigma'$. Moreover, $M\sigma$ is a refinement
    of $e$.
\end{theorem}

\if \DetailedProofs 1
\begin{proof}
    From Theorem~\ref{thmSoundNoCs} follows that $\HasTypeCS {\Sigma'} \Gamma M
    A$ and Lemma~\ref{lemTypeCheckConsistent} and
    Lemma~\ref{lemSolveConsistent} give that $\Sigma'$ is a consistent
    extension of $\Sigma$.  Thus $\sigma$ is well-typed and we have $\HasTypeCS
    {\Sigma'} \Gamma {M\sigma} A$.
    Since $M\sigma$ only uses constants from $\Sigma$ we can strengthen the
    signature to obtain $\HasTypeCS \Sigma \Gamma {M\sigma} A$.

    By Lemma~\ref{lemApproxRefine} we have that $M\sigma$ approximates a
    refinement of $e$. Since $M\sigma$ does not contain any guarded constants
    it is a refinement of $e$.
\end{proof}
\fi