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-\chapter[Detailed examples of tactics]{Detailed examples of tactics\label{Tactics-examples}}
-
-This chapter presents detailed examples of certain tactics, to
-illustrate their behavior.
-
-\section[\tt refine]{\tt refine\tacindex{refine}
-\label{refine-example}}
-
-This tactic applies to any goal. It behaves like {\tt exact} with a
-big difference : the user can leave some holes (denoted by \texttt{\_} or
-{\tt (\_:}{\it type}{\tt )}) in the term.
-{\tt refine} will generate as many
-subgoals as they are holes in the term. The type of holes must be
-either synthesized by the system or declared by an
-explicit cast like \verb|(\_:nat->Prop)|. This low-level
-tactic can be useful to advanced users.
-
-%\firstexample
-\Example
-
-\begin{coq_example*}
-Inductive Option : Set :=
- | Fail : Option
- | Ok : bool -> Option.
-\end{coq_example}
-\begin{coq_example}
-Definition get : forall x:Option, x <> Fail -> bool.
-refine
- (fun x:Option =>
- match x return x <> Fail -> bool with
- | Fail => _
- | Ok b => fun _ => b
- end).
-intros; absurd (Fail = Fail); trivial.
-\end{coq_example}
-\begin{coq_example*}
-Defined.
-\end{coq_example*}
-
-% \example{Using Refine to build a poor-man's ``Cases'' tactic}
-
-% \texttt{Refine} is actually the only way for the user to do
-% a proof with the same structure as a {\tt Cases} definition. Actually,
-% the tactics \texttt{case} (see \ref{case}) and \texttt{Elim} (see
-% \ref{elim}) only allow one step of elementary induction.
-
-% \begin{coq_example*}
-% Require Bool.
-% Require Arith.
-% \end{coq_example*}
-% %\begin{coq_eval}
-% %Abort.
-% %\end{coq_eval}
-% \begin{coq_example}
-% Definition one_two_or_five := [x:nat]
-% Cases x of
-% (1) => true
-% | (2) => true
-% | (5) => true
-% | _ => false
-% end.
-% Goal (x:nat)(Is_true (one_two_or_five x)) -> x=(1)\/x=(2)\/x=(5).
-% \end{coq_example}
-
-% A traditional script would be the following:
-
-% \begin{coq_example*}
-% Destruct x.
-% Tauto.
-% Destruct n.
-% Auto.
-% Destruct n0.
-% Auto.
-% Destruct n1.
-% Tauto.
-% Destruct n2.
-% Tauto.
-% Destruct n3.
-% Auto.
-% Intros; Inversion H.
-% \end{coq_example*}
-
-% With the tactic \texttt{Refine}, it becomes quite shorter:
-
-% \begin{coq_example*}
-% Restart.
-% \end{coq_example*}
-% \begin{coq_example}
-% Refine [x:nat]
-% <[y:nat](Is_true (one_two_or_five y))->(y=(1)\/y=(2)\/y=(5))>
-% Cases x of
-% (1) => [H]?
-% | (2) => [H]?
-% | (5) => [H]?
-% | n => [H](False_ind ? H)
-% end; Auto.
-% \end{coq_example}
-% \begin{coq_eval}
-% Abort.
-% \end{coq_eval}
-
-\section[\tt eapply]{\tt eapply\tacindex{eapply}
-\label{eapply-example}}
-\Example
-Assume we have a relation on {\tt nat} which is transitive:
-
-\begin{coq_example*}
-Variable R : nat -> nat -> Prop.
-Hypothesis Rtrans : forall x y z:nat, R x y -> R y z -> R x z.
-Variables n m p : nat.
-Hypothesis Rnm : R n m.
-Hypothesis Rmp : R m p.
-\end{coq_example*}
-
-Consider the goal {\tt (R n p)} provable using the transitivity of
-{\tt R}:
-
-\begin{coq_example*}
-Goal R n p.
-\end{coq_example*}
-
-The direct application of {\tt Rtrans} with {\tt apply} fails because
-no value for {\tt y} in {\tt Rtrans} is found by {\tt apply}:
-
-\begin{coq_eval}
-Set Printing Depth 50.
-(********** The following is not correct and should produce **********)
-(**** Error: generated subgoal (R n ?17) has metavariables in it *****)
-\end{coq_eval}
-\begin{coq_example}
-apply Rtrans.
-\end{coq_example}
-
-A solution is to rather apply {\tt (Rtrans n m p)}.
-
-\begin{coq_example}
-apply (Rtrans n m p).
-\end{coq_example}
-
-\begin{coq_eval}
-Undo.
-\end{coq_eval}
-
-More elegantly, {\tt apply Rtrans with (y:=m)} allows to only mention
-the unknown {\tt m}:
-
-\begin{coq_example}
-
- apply Rtrans with (y := m).
-\end{coq_example}
-
-\begin{coq_eval}
-Undo.
-\end{coq_eval}
-
-Another solution is to mention the proof of {\tt (R x y)} in {\tt
-Rtrans}...
-
-\begin{coq_example}
-
- apply Rtrans with (1 := Rnm).
-\end{coq_example}
-
-\begin{coq_eval}
-Undo.
-\end{coq_eval}
-
-... or the proof of {\tt (R y z)}:
-
-\begin{coq_example}
-
- apply Rtrans with (2 := Rmp).
-\end{coq_example}
-
-\begin{coq_eval}
-Undo.
-\end{coq_eval}
-
-On the opposite, one can use {\tt eapply} which postpone the problem
-of finding {\tt m}. Then one can apply the hypotheses {\tt Rnm} and {\tt
-Rmp}. This instantiates the existential variable and completes the proof.
-
-\begin{coq_example}
-eapply Rtrans.
-apply Rnm.
-apply Rmp.
-\end{coq_example}
-
-\begin{coq_eval}
-Reset R.
-\end{coq_eval}
-
-\section[{\tt Scheme}]{{\tt Scheme}\comindex{Scheme}
-\label{Scheme-examples}}
-
-\firstexample
-\example{Induction scheme for \texttt{tree} and \texttt{forest}}
-
-The definition of principle of mutual induction for {\tt tree} and
-{\tt forest} over the sort {\tt Set} is defined by the command:
-
-\begin{coq_eval}
-Reset Initial.
-Variables A B :
- Set.
-\end{coq_eval}
-
-\begin{coq_example*}
-Inductive tree : Set :=
- node : A -> forest -> tree
-with forest : Set :=
- | leaf : B -> forest
- | cons : tree -> forest -> forest.
-
-Scheme tree_forest_rec := Induction for tree Sort Set
- with forest_tree_rec := Induction for forest Sort Set.
-\end{coq_example*}
-
-You may now look at the type of {\tt tree\_forest\_rec}:
-
-\begin{coq_example}
-Check tree_forest_rec.
-\end{coq_example}
-
-This principle involves two different predicates for {\tt trees} and
-{\tt forests}; it also has three premises each one corresponding to a
-constructor of one of the inductive definitions.
-
-The principle {\tt forest\_tree\_rec} shares exactly the same
-premises, only the conclusion now refers to the property of forests.
-
-\begin{coq_example}
-Check forest_tree_rec.
-\end{coq_example}
-
-\example{Predicates {\tt odd} and {\tt even} on naturals}
-
-Let {\tt odd} and {\tt even} be inductively defined as:
-
-% Reset Initial.
-\begin{coq_eval}
-Open Scope nat_scope.
-\end{coq_eval}
-
-\begin{coq_example*}
-Inductive odd : nat -> Prop :=
- oddS : forall n:nat, even n -> odd (S n)
-with even : nat -> Prop :=
- | evenO : even 0
- | evenS : forall n:nat, odd n -> even (S n).
-\end{coq_example*}
-
-The following command generates a powerful elimination
-principle:
-
-\begin{coq_example}
-Scheme odd_even := Minimality for odd Sort Prop
- with even_odd := Minimality for even Sort Prop.
-\end{coq_example}
-
-The type of {\tt odd\_even} for instance will be:
-
-\begin{coq_example}
-Check odd_even.
-\end{coq_example}
-
-The type of {\tt even\_odd} shares the same premises but the
-conclusion is {\tt (n:nat)(even n)->(Q n)}.
-
-\subsection[{\tt Combined Scheme}]{{\tt Combined Scheme}\comindex{Combined Scheme}
-\label{CombinedScheme-examples}}
-
-We can define the induction principles for trees and forests using:
-\begin{coq_example}
-Scheme tree_forest_ind := Induction for tree Sort Prop
- with forest_tree_ind := Induction for forest Sort Prop.
-\end{coq_example}
-
-Then we can build the combined induction principle which gives the
-conjunction of the conclusions of each individual principle:
-\begin{coq_example}
-Combined Scheme tree_forest_mutind from tree_forest_ind, forest_tree_ind.
-\end{coq_example}
-
-The type of {\tt tree\_forest\_mutrec} will be:
-\begin{coq_example}
-Check tree_forest_mutind.
-\end{coq_example}
-
-\section[{\tt Functional Scheme} and {\tt functional induction}]{{\tt Functional Scheme} and {\tt functional induction}\comindex{Functional Scheme}\tacindex{functional induction}
-\label{FunScheme-examples}}
-
-\firstexample
-\example{Induction scheme for \texttt{div2}}
-
-We define the function \texttt{div2} as follows:
-
-\begin{coq_eval}
-Reset Initial.
-\end{coq_eval}
-
-\begin{coq_example*}
-Require Import Arith.
-Fixpoint div2 (n:nat) : nat :=
- match n with
- | O => 0
- | S O => 0
- | S (S n') => S (div2 n')
- end.
-\end{coq_example*}
-
-The definition of a principle of induction corresponding to the
-recursive structure of \texttt{div2} is defined by the command:
-
-\begin{coq_example}
-Functional Scheme div2_ind := Induction for div2 Sort Prop.
-\end{coq_example}
-
-You may now look at the type of {\tt div2\_ind}:
-
-\begin{coq_example}
-Check div2_ind.
-\end{coq_example}
-
-We can now prove the following lemma using this principle:
-
-
-\begin{coq_example*}
-Lemma div2_le' : forall n:nat, div2 n <= n.
-intro n.
- pattern n , (div2 n).
-\end{coq_example*}
-
-
-\begin{coq_example}
-apply div2_ind; intros.
-\end{coq_example}
-
-\begin{coq_example*}
-auto with arith.
-auto with arith.
-simpl; auto with arith.
-Qed.
-\end{coq_example*}
-
-We can use directly the \texttt{functional induction}
-(\ref{FunInduction}) tactic instead of the pattern/apply trick:
-
-\begin{coq_example*}
-Reset div2_le'.
-Lemma div2_le : forall n:nat, div2 n <= n.
-intro n.
-\end{coq_example*}
-
-\begin{coq_example}
-functional induction (div2 n).
-\end{coq_example}
-
-\begin{coq_example*}
-auto with arith.
-auto with arith.
-auto with arith.
-Qed.
-\end{coq_example*}
-
-\Rem There is a difference between obtaining an induction scheme for a
-function by using \texttt{Function} (see Section~\ref{Function}) and by
-using \texttt{Functional Scheme} after a normal definition using
-\texttt{Fixpoint} or \texttt{Definition}. See \ref{Function} for
-details.
-
-
-\example{Induction scheme for \texttt{tree\_size}}
-
-\begin{coq_eval}
-Reset Initial.
-\end{coq_eval}
-
-We define trees by the following mutual inductive type:
-
-\begin{coq_example*}
-Variable A : Set.
-Inductive tree : Set :=
- node : A -> forest -> tree
-with forest : Set :=
- | empty : forest
- | cons : tree -> forest -> forest.
-\end{coq_example*}
-
-We define the function \texttt{tree\_size} that computes the size
-of a tree or a forest. Note that we use \texttt{Function} which
-generally produces better principles.
-
-\begin{coq_example*}
-Function tree_size (t:tree) : nat :=
- match t with
- | node A f => S (forest_size f)
- end
- with forest_size (f:forest) : nat :=
- match f with
- | empty => 0
- | cons t f' => (tree_size t + forest_size f')
- end.
-\end{coq_example*}
-
-Remark: \texttt{Function} generates itself non mutual induction
-principles {\tt tree\_size\_ind} and {\tt forest\_size\_ind}:
-
-\begin{coq_example}
-Check tree_size_ind.
-\end{coq_example}
-
-The definition of mutual induction principles following the recursive
-structure of \texttt{tree\_size} and \texttt{forest\_size} is defined
-by the command:
-
-\begin{coq_example*}
-Functional Scheme tree_size_ind2 := Induction for tree_size Sort Prop
-with forest_size_ind2 := Induction for forest_size Sort Prop.
-\end{coq_example*}
-
-You may now look at the type of {\tt tree\_size\_ind2}:
-
-\begin{coq_example}
-Check tree_size_ind2.
-\end{coq_example}
-
-
-
-
-\section[{\tt inversion}]{{\tt inversion}\tacindex{inversion}
-\label{inversion-examples}}
-
-\subsection*{Generalities about inversion}
-
-When working with (co)inductive predicates, we are very often faced to
-some of these situations:
-\begin{itemize}
-\item we have an inconsistent instance of an inductive predicate in the
- local context of hypotheses. Thus, the current goal can be trivially
- proved by absurdity.
-\item we have a hypothesis that is an instance of an inductive
- predicate, and the instance has some variables whose constraints we
- would like to derive.
-\end{itemize}
-
-The inversion tactics are very useful to simplify the work in these
-cases. Inversion tools can be classified in three groups:
-
-\begin{enumerate}
-\item tactics for inverting an instance without stocking the inversion
- lemma in the context; this includes the tactics
- (\texttt{dependent}) \texttt{inversion} and
- (\texttt{dependent}) \texttt{inversion\_clear}.
-\item commands for generating and stocking in the context the inversion
- lemma corresponding to an instance; this includes \texttt{Derive}
- (\texttt{Dependent}) \texttt{Inversion} and \texttt{Derive}
- (\texttt{Dependent}) \texttt{Inversion\_clear}.
-\item tactics for inverting an instance using an already defined
- inversion lemma; this includes the tactic \texttt{inversion \ldots using}.
-\end{enumerate}
-
-As inversion proofs may be large in size, we recommend the user to
-stock the lemmas whenever the same instance needs to be inverted
-several times.
-
-\firstexample
-\example{Non-dependent inversion}
-
-Let's consider the relation \texttt{Le} over natural numbers and the
-following variables:
-
-\begin{coq_eval}
-Reset Initial.
-\end{coq_eval}
-
-\begin{coq_example*}
-Inductive Le : nat -> nat -> Set :=
- | LeO : forall n:nat, Le 0 n
- | LeS : forall n m:nat, Le n m -> Le (S n) (S m).
-Variable P : nat -> nat -> Prop.
-Variable Q : forall n m:nat, Le n m -> Prop.
-\end{coq_example*}
-
-For example, consider the goal:
-
-\begin{coq_eval}
-Lemma ex : forall n m:nat, Le (S n) m -> P n m.
-intros.
-\end{coq_eval}
-
-\begin{coq_example}
-Show.
-\end{coq_example}
-
-To prove the goal we may need to reason by cases on \texttt{H} and to
- derive that \texttt{m} is necessarily of
-the form $(S~m_0)$ for certain $m_0$ and that $(Le~n~m_0)$.
-Deriving these conditions corresponds to prove that the
-only possible constructor of \texttt{(Le (S n) m)} is
-\texttt{LeS} and that we can invert the
-\texttt{->} in the type of \texttt{LeS}.
-This inversion is possible because \texttt{Le} is the smallest set closed by
-the constructors \texttt{LeO} and \texttt{LeS}.
-
-\begin{coq_example}
-inversion_clear H.
-\end{coq_example}
-
-Note that \texttt{m} has been substituted in the goal for \texttt{(S m0)}
-and that the hypothesis \texttt{(Le n m0)} has been added to the
-context.
-
-Sometimes it is
-interesting to have the equality \texttt{m=(S m0)} in the
-context to use it after. In that case we can use \texttt{inversion} that
-does not clear the equalities:
-
-\begin{coq_example*}
-Undo.
-\end{coq_example*}
-
-\begin{coq_example}
-inversion H.
-\end{coq_example}
-
-\begin{coq_eval}
-Undo.
-\end{coq_eval}
-
-\example{Dependent Inversion}
-
-Let us consider the following goal:
-
-\begin{coq_eval}
-Lemma ex_dep : forall (n m:nat) (H:Le (S n) m), Q (S n) m H.
-intros.
-\end{coq_eval}
-
-\begin{coq_example}
-Show.
-\end{coq_example}
-
-As \texttt{H} occurs in the goal, we may want to reason by cases on its
-structure and so, we would like inversion tactics to
-substitute \texttt{H} by the corresponding term in constructor form.
-Neither \texttt{Inversion} nor {\tt Inversion\_clear} make such a
-substitution.
-To have such a behavior we use the dependent inversion tactics:
-
-\begin{coq_example}
-dependent inversion_clear H.
-\end{coq_example}
-
-Note that \texttt{H} has been substituted by \texttt{(LeS n m0 l)} and
-\texttt{m} by \texttt{(S m0)}.
-
-\example{using already defined inversion lemmas}
-
-\begin{coq_eval}
-Abort.
-\end{coq_eval}
-
-For example, to generate the inversion lemma for the instance
-\texttt{(Le (S n) m)} and the sort \texttt{Prop} we do:
-
-\begin{coq_example*}
-Derive Inversion_clear leminv with (forall n m:nat, Le (S n) m) Sort
- Prop.
-\end{coq_example*}
-
-\begin{coq_example}
-Check leminv.
-\end{coq_example}
-
-Then we can use the proven inversion lemma:
-
-\begin{coq_example}
-Show.
-\end{coq_example}
-
-\begin{coq_example}
-inversion H using leminv.
-\end{coq_example}
-
-\begin{coq_eval}
-Reset Initial.
-\end{coq_eval}
-
-\section[\tt dependent induction]{\tt dependent induction\label{dependent-induction-example}}
-\def\depind{{\tt dependent induction}~}
-\def\depdestr{{\tt dependent destruction}~}
-
-The tactics \depind and \depdestr are another solution for inverting
-inductive predicate instances and potentially doing induction at the
-same time. It is based on the \texttt{BasicElim} tactic of Conor McBride which
-works by abstracting each argument of an inductive instance by a variable
-and constraining it by equalities afterwards. This way, the usual
-{\tt induction} and {\tt destruct} tactics can be applied to the
-abstracted instance and after simplification of the equalities we get
-the expected goals.
-
-The abstracting tactic is called {\tt generalize\_eqs} and it takes as
-argument an hypothesis to generalize. It uses the {\tt JMeq} datatype
-defined in {\tt Coq.Logic.JMeq}, hence we need to require it before.
-For example, revisiting the first example of the inversion documentation above:
-
-\begin{coq_example*}
-Require Import Coq.Logic.JMeq.
-\end{coq_example*}
-\begin{coq_eval}
-Require Import Coq.Program.Equality.
-\end{coq_eval}
-
-\begin{coq_eval}
-Inductive Le : nat -> nat -> Set :=
- | LeO : forall n:nat, Le 0 n
- | LeS : forall n m:nat, Le n m -> Le (S n) (S m).
-Variable P : nat -> nat -> Prop.
-Variable Q : forall n m:nat, Le n m -> Prop.
-\end{coq_eval}
-
-\begin{coq_example*}
-Goal forall n m:nat, Le (S n) m -> P n m.
-intros n m H.
-\end{coq_example*}
-\begin{coq_example}
-generalize_eqs H.
-\end{coq_example}
-
-The index {\tt S n} gets abstracted by a variable here, but a
-corresponding equality is added under the abstract instance so that no
-information is actually lost. The goal is now almost amenable to do induction
-or case analysis. One should indeed first move {\tt n} into the goal to
-strengthen it before doing induction, or {\tt n} will be fixed in
-the inductive hypotheses (this does not matter for case analysis).
-As a rule of thumb, all the variables that appear inside constructors in
-the indices of the hypothesis should be generalized. This is exactly
-what the \texttt{generalize\_eqs\_vars} variant does:
-
-\begin{coq_eval}
-Undo 1.
-\end{coq_eval}
-\begin{coq_example}
-generalize_eqs_vars H.
-induction H.
-\end{coq_example}
-
-As the hypothesis itself did not appear in the goal, we did not need to
-use an heterogeneous equality to relate the new hypothesis to the old
-one (which just disappeared here). However, the tactic works just a well
-in this case, e.g.:
-
-\begin{coq_eval}
-Admitted.
-\end{coq_eval}
-
-\begin{coq_example}
-Goal forall n m (p : Le (S n) m), Q (S n) m p.
-intros n m p ; generalize_eqs_vars p.
-\end{coq_example}
-
-One drawback of this approach is that in the branches one will have to
-substitute the equalities back into the instance to get the right
-assumptions. Sometimes injection of constructors will also be needed to
-recover the needed equalities. Also, some subgoals should be directly
-solved because of inconsistent contexts arising from the constraints on
-indexes. The nice thing is that we can make a tactic based on
-discriminate, injection and variants of substitution to automatically
-do such simplifications (which may involve the K axiom).
-This is what the {\tt simplify\_dep\_elim} tactic from
-{\tt Coq.Program.Equality} does. For example, we might simplify the
-previous goals considerably:
-% \begin{coq_eval}
-% Abort.
-% Goal forall n m:nat, Le (S n) m -> P n m.
-% intros n m H ; generalize_eqs_vars H.
-% \end{coq_eval}
-
-\begin{coq_example}
-induction p ; simplify_dep_elim.
-\end{coq_example}
-
-The higher-order tactic {\tt do\_depind} defined in {\tt
- Coq.Program.Equality} takes a tactic and combines the
-building blocks we have seen with it: generalizing by equalities
-calling the given tactic with the
-generalized induction hypothesis as argument and cleaning the subgoals
-with respect to equalities. Its most important instantiations are
-\depind and \depdestr that do induction or simply case analysis on the
-generalized hypothesis. For example we can redo what we've done manually
-with \depdestr:
-
-\begin{coq_eval}
-Abort.
-\end{coq_eval}
-\begin{coq_example*}
-Require Import Coq.Program.Equality.
-Lemma ex : forall n m:nat, Le (S n) m -> P n m.
-intros n m H.
-\end{coq_example*}
-\begin{coq_example}
-dependent destruction H.
-\end{coq_example}
-\begin{coq_eval}
-Abort.
-\end{coq_eval}
-
-This gives essentially the same result as inversion. Now if the
-destructed hypothesis actually appeared in the goal, the tactic would
-still be able to invert it, contrary to {\tt dependent
- inversion}. Consider the following example on vectors:
-
-\begin{coq_example*}
-Require Import Coq.Program.Equality.
-Set Implicit Arguments.
-Variable A : Set.
-Inductive vector : nat -> Type :=
-| vnil : vector 0
-| vcons : A -> forall n, vector n -> vector (S n).
-Goal forall n, forall v : vector (S n),
- exists v' : vector n, exists a : A, v = vcons a v'.
- intros n v.
-\end{coq_example*}
-\begin{coq_example}
- dependent destruction v.
-\end{coq_example}
-\begin{coq_eval}
-Abort.
-\end{coq_eval}
-
-In this case, the {\tt v} variable can be replaced in the goal by the
-generalized hypothesis only when it has a type of the form {\tt vector
- (S n)}, that is only in the second case of the {\tt destruct}. The
-first one is dismissed because {\tt S n <> 0}.
-
-\subsection{A larger example}
-
-Let's see how the technique works with {\tt induction} on inductive
-predicates on a real example. We will develop an example application to the
-theory of simply-typed lambda-calculus formalized in a dependently-typed style:
-
-\begin{coq_example*}
-Inductive type : Type :=
-| base : type
-| arrow : type -> type -> type.
-Notation " t --> t' " := (arrow t t') (at level 20, t' at next level).
-Inductive ctx : Type :=
-| empty : ctx
-| snoc : ctx -> type -> ctx.
-Notation " G , tau " := (snoc G tau) (at level 20, t at next level).
-Fixpoint conc (G D : ctx) : ctx :=
- match D with
- | empty => G
- | snoc D' x => snoc (conc G D') x
- end.
-Notation " G ; D " := (conc G D) (at level 20).
-Inductive term : ctx -> type -> Type :=
-| ax : forall G tau, term (G, tau) tau
-| weak : forall G tau,
- term G tau -> forall tau', term (G, tau') tau
-| abs : forall G tau tau',
- term (G , tau) tau' -> term G (tau --> tau')
-| app : forall G tau tau',
- term G (tau --> tau') -> term G tau -> term G tau'.
-\end{coq_example*}
-
-We have defined types and contexts which are snoc-lists of types. We
-also have a {\tt conc} operation that concatenates two contexts.
-The {\tt term} datatype represents in fact the possible typing
-derivations of the calculus, which are isomorphic to the well-typed
-terms, hence the name. A term is either an application of:
-\begin{itemize}
-\item the axiom rule to type a reference to the first variable in a context,
-\item the weakening rule to type an object in a larger context
-\item the abstraction or lambda rule to type a function
-\item the application to type an application of a function to an argument
-\end{itemize}
-
-Once we have this datatype we want to do proofs on it, like weakening:
-
-\begin{coq_example*}
-Lemma weakening : forall G D tau, term (G ; D) tau ->
- forall tau', term (G , tau' ; D) tau.
-\end{coq_example*}
-\begin{coq_eval}
- Abort.
-\end{coq_eval}
-
-The problem here is that we can't just use {\tt induction} on the typing
-derivation because it will forget about the {\tt G ; D} constraint
-appearing in the instance. A solution would be to rewrite the goal as:
-\begin{coq_example*}
-Lemma weakening' : forall G' tau, term G' tau ->
- forall G D, (G ; D) = G' ->
- forall tau', term (G, tau' ; D) tau.
-\end{coq_example*}
-\begin{coq_eval}
- Abort.
-\end{coq_eval}
-
-With this proper separation of the index from the instance and the right
-induction loading (putting {\tt G} and {\tt D} after the inducted-on
-hypothesis), the proof will go through, but it is a very tedious
-process. One is also forced to make a wrapper lemma to get back the
-more natural statement. The \depind tactic alleviates this trouble by
-doing all of this plumbing of generalizing and substituting back automatically.
-Indeed we can simply write:
-
-\begin{coq_example*}
-Require Import Coq.Program.Tactics.
-Lemma weakening : forall G D tau, term (G ; D) tau ->
- forall tau', term (G , tau' ; D) tau.
-Proof with simpl in * ; simpl_depind ; auto.
- intros G D tau H. dependent induction H generalizing G D ; intros.
-\end{coq_example*}
-
-This call to \depind has an additional arguments which is a list of
-variables appearing in the instance that should be generalized in the
-goal, so that they can vary in the induction hypotheses. By default, all
-variables appearing inside constructors (except in a parameter position)
-of the instantiated hypothesis will be generalized automatically but
-one can always give the list explicitly.
-
-\begin{coq_example}
- Show.
-\end{coq_example}
-
-The {\tt simpl\_depind} tactic includes an automatic tactic that tries
-to simplify equalities appearing at the beginning of induction
-hypotheses, generally using trivial applications of
-reflexivity. In cases where the equality is not between constructor
-forms though, one must help the automation by giving
-some arguments, using the {\tt specialize} tactic.
-
-\begin{coq_example*}
-destruct D... apply weak ; apply ax. apply ax.
-destruct D...
-\end{coq_example*}
-\begin{coq_example}
-Show.
-\end{coq_example}
-\begin{coq_example}
- specialize (IHterm G empty).
-\end{coq_example}
-
-Then the automation can find the needed equality {\tt G = G} to narrow
-the induction hypothesis further. This concludes our example.
-
-\begin{coq_example}
- simpl_depind.
-\end{coq_example}
-
-\SeeAlso The induction \ref{elim}, case \ref{case} and inversion \ref{inversion} tactics.
-
-\section[\tt autorewrite]{\tt autorewrite\label{autorewrite-example}}
-
-Here are two examples of {\tt autorewrite} use. The first one ({\em Ackermann
-function}) shows actually a quite basic use where there is no conditional
-rewriting. The second one ({\em Mac Carthy function}) involves conditional
-rewritings and shows how to deal with them using the optional tactic of the
-{\tt Hint~Rewrite} command.
-
-\firstexample
-\example{Ackermann function}
-%Here is a basic use of {\tt AutoRewrite} with the Ackermann function:
-
-\begin{coq_example*}
-Reset Initial.
-Require Import Arith.
-Variable Ack :
- nat -> nat -> nat.
-Axiom Ack0 :
- forall m:nat, Ack 0 m = S m.
-Axiom Ack1 : forall n:nat, Ack (S n) 0 = Ack n 1.
-Axiom Ack2 : forall n m:nat, Ack (S n) (S m) = Ack n (Ack (S n) m).
-\end{coq_example*}
-
-\begin{coq_example}
-Hint Rewrite Ack0 Ack1 Ack2 : base0.
-Lemma ResAck0 :
- Ack 3 2 = 29.
-autorewrite with base0 using try reflexivity.
-\end{coq_example}
-
-\begin{coq_eval}
-Reset Initial.
-\end{coq_eval}
-
-\example{Mac Carthy function}
-%The Mac Carthy function shows a more complex case:
-
-\begin{coq_example*}
-Require Import Omega.
-Variable g :
- nat -> nat -> nat.
-Axiom g0 :
- forall m:nat, g 0 m = m.
-Axiom
- g1 :
- forall n m:nat,
- (n > 0) -> (m > 100) -> g n m = g (pred n) (m - 10).
-Axiom
- g2 :
- forall n m:nat,
- (n > 0) -> (m <= 100) -> g n m = g (S n) (m + 11).
-\end{coq_example*}
-
-\begin{coq_example}
-Hint Rewrite g0 g1 g2 using omega : base1.
-Lemma Resg0 :
- g 1 110 = 100.
-autorewrite with base1 using reflexivity || simpl.
-\end{coq_example}
-
-\begin{coq_eval}
-Abort.
-\end{coq_eval}
-
-\begin{coq_example}
-Lemma Resg1 : g 1 95 = 91.
-autorewrite with base1 using reflexivity || simpl.
-\end{coq_example}
-
-\begin{coq_eval}
-Reset Initial.
-\end{coq_eval}
-
-\section[\tt quote]{\tt quote\tacindex{quote}
-\label{quote-examples}}
-
-The tactic \texttt{quote} allows to use Barendregt's so-called
-2-level approach without writing any ML code. Suppose you have a
-language \texttt{L} of
-'abstract terms' and a type \texttt{A} of 'concrete terms'
-and a function \texttt{f : L -> A}. If \texttt{L} is a simple
-inductive datatype and \texttt{f} a simple fixpoint, \texttt{quote f}
-will replace the head of current goal by a convertible term of the form
-\texttt{(f t)}. \texttt{L} must have a constructor of type: \texttt{A
- -> L}.
-
-Here is an example:
-
-\begin{coq_example}
-Require Import Quote.
-Parameters A B C : Prop.
-Inductive formula : Type :=
- | f_and : formula -> formula -> formula (* binary constructor *)
- | f_or : formula -> formula -> formula
- | f_not : formula -> formula (* unary constructor *)
- | f_true : formula (* 0-ary constructor *)
- | f_const : Prop -> formula (* constructor for constants *).
-Fixpoint interp_f (f:
- formula) : Prop :=
- match f with
- | f_and f1 f2 => interp_f f1 /\ interp_f f2
- | f_or f1 f2 => interp_f f1 \/ interp_f f2
- | f_not f1 => ~ interp_f f1
- | f_true => True
- | f_const c => c
- end.
-Goal A /\ (A \/ True) /\ ~ B /\ (A <-> A).
-quote interp_f.
-\end{coq_example}
-
-The algorithm to perform this inversion is: try to match the
-term with right-hand sides expression of \texttt{f}. If there is a
-match, apply the corresponding left-hand side and call yourself
-recursively on sub-terms. If there is no match, we are at a leaf:
-return the corresponding constructor (here \texttt{f\_const}) applied
-to the term.
-
-\begin{ErrMsgs}
-\item \errindex{quote: not a simple fixpoint} \\
- Happens when \texttt{quote} is not able to perform inversion properly.
-\end{ErrMsgs}
-
-\subsection{Introducing variables map}
-
-The normal use of \texttt{quote} is to make proofs by reflection: one
-defines a function \texttt{simplify : formula -> formula} and proves a
-theorem \texttt{simplify\_ok: (f:formula)(interp\_f (simplify f)) ->
- (interp\_f f)}. Then, one can simplify formulas by doing:
-\begin{verbatim}
- quote interp_f.
- apply simplify_ok.
- compute.
-\end{verbatim}
-But there is a problem with leafs: in the example above one cannot
-write a function that implements, for example, the logical simplifications
-$A \land A \ra A$ or $A \land \lnot A \ra \texttt{False}$. This is
-because the \Prop{} is impredicative.
-
-It is better to use that type of formulas:
-
-\begin{coq_eval}
-Reset formula.
-\end{coq_eval}
-\begin{coq_example}
-Inductive formula : Set :=
- | f_and : formula -> formula -> formula
- | f_or : formula -> formula -> formula
- | f_not : formula -> formula
- | f_true : formula
- | f_atom : index -> formula.
-\end{coq_example*}
-
-\texttt{index} is defined in module \texttt{quote}. Equality on that
-type is decidable so we are able to simplify $A \land A$ into $A$ at
-the abstract level.
-
-When there are variables, there are bindings, and \texttt{quote}
-provides also a type \texttt{(varmap A)} of bindings from
-\texttt{index} to any set \texttt{A}, and a function
-\texttt{varmap\_find} to search in such maps. The interpretation
-function has now another argument, a variables map:
-
-\begin{coq_example}
-Fixpoint interp_f (vm:
- varmap Prop) (f:formula) {struct f} : Prop :=
- match f with
- | f_and f1 f2 => interp_f vm f1 /\ interp_f vm f2
- | f_or f1 f2 => interp_f vm f1 \/ interp_f vm f2
- | f_not f1 => ~ interp_f vm f1
- | f_true => True
- | f_atom i => varmap_find True i vm
- end.
-\end{coq_example}
-
-\noindent\texttt{quote} handles this second case properly:
-
-\begin{coq_example}
-Goal A /\ (B \/ A) /\ (A \/ ~ B).
-quote interp_f.
-\end{coq_example}
-
-It builds \texttt{vm} and \texttt{t} such that \texttt{(f vm t)} is
-convertible with the conclusion of current goal.
-
-\subsection{Combining variables and constants}
-
-One can have both variables and constants in abstracts terms; that is
-the case, for example, for the \texttt{ring} tactic (chapter
-\ref{ring}). Then one must provide to \texttt{quote} a list of
-\emph{constructors of constants}. For example, if the list is
-\texttt{[O S]} then closed natural numbers will be considered as
-constants and other terms as variables.
-
-Example:
-
-\begin{coq_eval}
-Reset formula.
-\end{coq_eval}
-\begin{coq_example*}
-Inductive formula : Type :=
- | f_and : formula -> formula -> formula
- | f_or : formula -> formula -> formula
- | f_not : formula -> formula
- | f_true : formula
- | f_const : Prop -> formula (* constructor for constants *)
- | f_atom : index -> formula.
-Fixpoint interp_f
- (vm: (* constructor for variables *)
- varmap Prop) (f:formula) {struct f} : Prop :=
- match f with
- | f_and f1 f2 => interp_f vm f1 /\ interp_f vm f2
- | f_or f1 f2 => interp_f vm f1 \/ interp_f vm f2
- | f_not f1 => ~ interp_f vm f1
- | f_true => True
- | f_const c => c
- | f_atom i => varmap_find True i vm
- end.
-Goal
-A /\ (A \/ True) /\ ~ B /\ (C <-> C).
-\end{coq_example*}
-
-\begin{coq_example}
-quote interp_f [ A B ].
-Undo.
- quote interp_f [ B C iff ].
-\end{coq_example}
-
-\Warning Since function inversion
-is undecidable in general case, don't expect miracles from it!
-
-\begin{Variants}
-
-\item {\tt quote {\ident} in {\term} using {\tac}}
-
- \tac\ must be a functional tactic (starting with {\tt fun x =>})
- and will be called with the quoted version of \term\ according to
- \ident.
-
-\item {\tt quote {\ident} [ \ident$_1$ \dots\ \ident$_n$ ] in {\term} using {\tac}}
-
- Same as above, but will use \ident$_1$, \dots, \ident$_n$ to
- chose which subterms are constants (see above).
-
-\end{Variants}
-
-% \SeeAlso file \texttt{theories/DEMOS/DemoQuote.v}
-
-\SeeAlso comments of source file \texttt{plugins/quote/quote.ml}
-
-\SeeAlso the \texttt{ring} tactic (Chapter~\ref{ring})
-
-
-
-\section{Using the tactical language}
-
-\subsection{About the cardinality of the set of natural numbers}
-
-A first example which shows how to use the pattern matching over the proof
-contexts is the proof that natural numbers have more than two elements. The
-proof of such a lemma can be done as %shown on Figure~\ref{cnatltac}.
-follows:
-%\begin{figure}
-%\begin{centerframe}
-\begin{coq_eval}
-Reset Initial.
-Require Import Arith.
-Require Import List.
-\end{coq_eval}
-\begin{coq_example*}
-Lemma card_nat :
- ~ (exists x : nat, exists y : nat, forall z:nat, x = z \/ y = z).
-Proof.
-red; intros (x, (y, Hy)).
-elim (Hy 0); elim (Hy 1); elim (Hy 2); intros;
- match goal with
- | [_:(?a = ?b),_:(?a = ?c) |- _ ] =>
- cut (b = c); [ discriminate | apply trans_equal with a; auto ]
- end.
-Qed.
-\end{coq_example*}
-%\end{centerframe}
-%\caption{A proof on cardinality of natural numbers}
-%\label{cnatltac}
-%\end{figure}
-
-We can notice that all the (very similar) cases coming from the three
-eliminations (with three distinct natural numbers) are successfully solved by
-a {\tt match goal} structure and, in particular, with only one pattern (use
-of non-linear matching).
-
-\subsection{Permutation on closed lists}
-
-Another more complex example is the problem of permutation on closed lists. The
-aim is to show that a closed list is a permutation of another one.
-
-First, we define the permutation predicate as shown in table~\ref{permutpred}.
-
-\begin{figure}
-\begin{centerframe}
-\begin{coq_example*}
-Section Sort.
-Variable A : Set.
-Inductive permut : list A -> list A -> Prop :=
- | permut_refl : forall l, permut l l
- | permut_cons :
- forall a l0 l1, permut l0 l1 -> permut (a :: l0) (a :: l1)
- | permut_append : forall a l, permut (a :: l) (l ++ a :: nil)
- | permut_trans :
- forall l0 l1 l2, permut l0 l1 -> permut l1 l2 -> permut l0 l2.
-End Sort.
-\end{coq_example*}
-\end{centerframe}
-\caption{Definition of the permutation predicate}
-\label{permutpred}
-\end{figure}
-
-A more complex example is the problem of permutation on closed lists.
-The aim is to show that a closed list is a permutation of another one.
-First, we define the permutation predicate as shown on
-Figure~\ref{permutpred}.
-
-\begin{figure}
-\begin{centerframe}
-\begin{coq_example}
-Ltac Permut n :=
- match goal with
- | |- (permut _ ?l ?l) => apply permut_refl
- | |- (permut _ (?a :: ?l1) (?a :: ?l2)) =>
- let newn := eval compute in (length l1) in
- (apply permut_cons; Permut newn)
- | |- (permut ?A (?a :: ?l1) ?l2) =>
- match eval compute in n with
- | 1 => fail
- | _ =>
- let l1' := constr:(l1 ++ a :: nil) in
- (apply (permut_trans A (a :: l1) l1' l2);
- [ apply permut_append | compute; Permut (pred n) ])
- end
- end.
-Ltac PermutProve :=
- match goal with
- | |- (permut _ ?l1 ?l2) =>
- match eval compute in (length l1 = length l2) with
- | (?n = ?n) => Permut n
- end
- end.
-\end{coq_example}
-\end{centerframe}
-\caption{Permutation tactic}
-\label{permutltac}
-\end{figure}
-
-Next, we can write naturally the tactic and the result can be seen on
-Figure~\ref{permutltac}. We can notice that we use two toplevel
-definitions {\tt PermutProve} and {\tt Permut}. The function to be
-called is {\tt PermutProve} which computes the lengths of the two
-lists and calls {\tt Permut} with the length if the two lists have the
-same length. {\tt Permut} works as expected. If the two lists are
-equal, it concludes. Otherwise, if the lists have identical first
-elements, it applies {\tt Permut} on the tail of the lists. Finally,
-if the lists have different first elements, it puts the first element
-of one of the lists (here the second one which appears in the {\tt
- permut} predicate) at the end if that is possible, i.e., if the new
-first element has been at this place previously. To verify that all
-rotations have been done for a list, we use the length of the list as
-an argument for {\tt Permut} and this length is decremented for each
-rotation down to, but not including, 1 because for a list of length
-$n$, we can make exactly $n-1$ rotations to generate at most $n$
-distinct lists. Here, it must be noticed that we use the natural
-numbers of {\Coq} for the rotation counter. On Figure~\ref{ltac}, we
-can see that it is possible to use usual natural numbers but they are
-only used as arguments for primitive tactics and they cannot be
-handled, in particular, we cannot make computations with them. So, a
-natural choice is to use {\Coq} data structures so that {\Coq} makes
-the computations (reductions) by {\tt eval compute in} and we can get
-the terms back by {\tt match}.
-
-With {\tt PermutProve}, we can now prove lemmas as
-% shown on Figure~\ref{permutlem}.
-follows:
-%\begin{figure}
-%\begin{centerframe}
-
-\begin{coq_example*}
-Lemma permut_ex1 :
- permut nat (1 :: 2 :: 3 :: nil) (3 :: 2 :: 1 :: nil).
-Proof. PermutProve. Qed.
-Lemma permut_ex2 :
- permut nat
- (0 :: 1 :: 2 :: 3 :: 4 :: 5 :: 6 :: 7 :: 8 :: 9 :: nil)
- (0 :: 2 :: 4 :: 6 :: 8 :: 9 :: 7 :: 5 :: 3 :: 1 :: nil).
-Proof. PermutProve. Qed.
-\end{coq_example*}
-%\end{centerframe}
-%\caption{Examples of {\tt PermutProve} use}
-%\label{permutlem}
-%\end{figure}
-
-
-\subsection{Deciding intuitionistic propositional logic}
-
-\begin{figure}[b]
-\begin{centerframe}
-\begin{coq_example}
-Ltac Axioms :=
- match goal with
- | |- True => trivial
- | _:False |- _ => elimtype False; assumption
- | _:?A |- ?A => auto
- end.
-\end{coq_example}
-\end{centerframe}
-\caption{Deciding intuitionistic propositions (1)}
-\label{tautoltaca}
-\end{figure}
-
-
-\begin{figure}
-\begin{centerframe}
-\begin{coq_example}
-Ltac DSimplif :=
- repeat
- (intros;
- match goal with
- | id:(~ _) |- _ => red in id
- | id:(_ /\ _) |- _ =>
- elim id; do 2 intro; clear id
- | id:(_ \/ _) |- _ =>
- elim id; intro; clear id
- | id:(?A /\ ?B -> ?C) |- _ =>
- cut (A -> B -> C);
- [ intro | intros; apply id; split; assumption ]
- | id:(?A \/ ?B -> ?C) |- _ =>
- cut (B -> C);
- [ cut (A -> C);
- [ intros; clear id
- | intro; apply id; left; assumption ]
- | intro; apply id; right; assumption ]
- | id0:(?A -> ?B),id1:?A |- _ =>
- cut B; [ intro; clear id0 | apply id0; assumption ]
- | |- (_ /\ _) => split
- | |- (~ _) => red
- end).
-Ltac TautoProp :=
- DSimplif;
- Axioms ||
- match goal with
- | id:((?A -> ?B) -> ?C) |- _ =>
- cut (B -> C);
- [ intro; cut (A -> B);
- [ intro; cut C;
- [ intro; clear id | apply id; assumption ]
- | clear id ]
- | intro; apply id; intro; assumption ]; TautoProp
- | id:(~ ?A -> ?B) |- _ =>
- cut (False -> B);
- [ intro; cut (A -> False);
- [ intro; cut B;
- [ intro; clear id | apply id; assumption ]
- | clear id ]
- | intro; apply id; red; intro; assumption ]; TautoProp
- | |- (_ \/ _) => (left; TautoProp) || (right; TautoProp)
- end.
-\end{coq_example}
-\end{centerframe}
-\caption{Deciding intuitionistic propositions (2)}
-\label{tautoltacb}
-\end{figure}
-
-The pattern matching on goals allows a complete and so a powerful
-backtracking when returning tactic values. An interesting application
-is the problem of deciding intuitionistic propositional logic.
-Considering the contraction-free sequent calculi {\tt LJT*} of
-Roy~Dyckhoff (\cite{Dyc92}), it is quite natural to code such a tactic
-using the tactic language as shown on Figures~\ref{tautoltaca}
-and~\ref{tautoltacb}. The tactic {\tt Axioms} tries to conclude using
-usual axioms. The tactic {\tt DSimplif} applies all the reversible
-rules of Dyckhoff's system. Finally, the tactic {\tt TautoProp} (the
-main tactic to be called) simplifies with {\tt DSimplif}, tries to
-conclude with {\tt Axioms} and tries several paths using the
-backtracking rules (one of the four Dyckhoff's rules for the left
-implication to get rid of the contraction and the right or).
-
-For example, with {\tt TautoProp}, we can prove tautologies like
- those:
-% on Figure~\ref{tautolem}.
-%\begin{figure}[tbp]
-%\begin{centerframe}
-\begin{coq_example*}
-Lemma tauto_ex1 : forall A B:Prop, A /\ B -> A \/ B.
-Proof. TautoProp. Qed.
-Lemma tauto_ex2 :
- forall A B:Prop, (~ ~ B -> B) -> (A -> B) -> ~ ~ A -> B.
-Proof. TautoProp. Qed.
-\end{coq_example*}
-%\end{centerframe}
-%\caption{Proofs of tautologies with {\tt TautoProp}}
-%\label{tautolem}
-%\end{figure}
-
-\subsection{Deciding type isomorphisms}
-
-A more tricky problem is to decide equalities between types and modulo
-isomorphisms. Here, we choose to use the isomorphisms of the simply typed
-$\lb{}$-calculus with Cartesian product and $unit$ type (see, for example,
-\cite{RC95}). The axioms of this $\lb{}$-calculus are given by
-table~\ref{isosax}.
-
-\begin{figure}
-\begin{centerframe}
-\begin{coq_eval}
-Reset Initial.
-\end{coq_eval}
-\begin{coq_example*}
-Open Scope type_scope.
-Section Iso_axioms.
-Variables A B C : Set.
-Axiom Com : A * B = B * A.
-Axiom Ass : A * (B * C) = A * B * C.
-Axiom Cur : (A * B -> C) = (A -> B -> C).
-Axiom Dis : (A -> B * C) = (A -> B) * (A -> C).
-Axiom P_unit : A * unit = A.
-Axiom AR_unit : (A -> unit) = unit.
-Axiom AL_unit : (unit -> A) = A.
-Lemma Cons : B = C -> A * B = A * C.
-Proof.
-intro Heq; rewrite Heq; apply refl_equal.
-Qed.
-End Iso_axioms.
-\end{coq_example*}
-\end{centerframe}
-\caption{Type isomorphism axioms}
-\label{isosax}
-\end{figure}
-
-A more tricky problem is to decide equalities between types and modulo
-isomorphisms. Here, we choose to use the isomorphisms of the simply typed
-$\lb{}$-calculus with Cartesian product and $unit$ type (see, for example,
-\cite{RC95}). The axioms of this $\lb{}$-calculus are given on
-Figure~\ref{isosax}.
-
-\begin{figure}[ht]
-\begin{centerframe}
-\begin{coq_example}
-Ltac DSimplif trm :=
- match trm with
- | (?A * ?B * ?C) =>
- rewrite <- (Ass A B C); try MainSimplif
- | (?A * ?B -> ?C) =>
- rewrite (Cur A B C); try MainSimplif
- | (?A -> ?B * ?C) =>
- rewrite (Dis A B C); try MainSimplif
- | (?A * unit) =>
- rewrite (P_unit A); try MainSimplif
- | (unit * ?B) =>
- rewrite (Com unit B); try MainSimplif
- | (?A -> unit) =>
- rewrite (AR_unit A); try MainSimplif
- | (unit -> ?B) =>
- rewrite (AL_unit B); try MainSimplif
- | (?A * ?B) =>
- (DSimplif A; try MainSimplif) || (DSimplif B; try MainSimplif)
- | (?A -> ?B) =>
- (DSimplif A; try MainSimplif) || (DSimplif B; try MainSimplif)
- end
- with MainSimplif :=
- match goal with
- | |- (?A = ?B) => try DSimplif A; try DSimplif B
- end.
-Ltac Length trm :=
- match trm with
- | (_ * ?B) => let succ := Length B in constr:(S succ)
- | _ => constr:1
- end.
-Ltac assoc := repeat rewrite <- Ass.
-\end{coq_example}
-\end{centerframe}
-\caption{Type isomorphism tactic (1)}
-\label{isosltac1}
-\end{figure}
-
-\begin{figure}[ht]
-\begin{centerframe}
-\begin{coq_example}
-Ltac DoCompare n :=
- match goal with
- | [ |- (?A = ?A) ] => apply refl_equal
- | [ |- (?A * ?B = ?A * ?C) ] =>
- apply Cons; let newn := Length B in
- DoCompare newn
- | [ |- (?A * ?B = ?C) ] =>
- match eval compute in n with
- | 1 => fail
- | _ =>
- pattern (A * B) at 1; rewrite Com; assoc; DoCompare (pred n)
- end
- end.
-Ltac CompareStruct :=
- match goal with
- | [ |- (?A = ?B) ] =>
- let l1 := Length A
- with l2 := Length B in
- match eval compute in (l1 = l2) with
- | (?n = ?n) => DoCompare n
- end
- end.
-Ltac IsoProve := MainSimplif; CompareStruct.
-\end{coq_example}
-\end{centerframe}
-\caption{Type isomorphism tactic (2)}
-\label{isosltac2}
-\end{figure}
-
-The tactic to judge equalities modulo this axiomatization can be written as
-shown on Figures~\ref{isosltac1} and~\ref{isosltac2}. The algorithm is quite
-simple. Types are reduced using axioms that can be oriented (this done by {\tt
-MainSimplif}). The normal forms are sequences of Cartesian
-products without Cartesian product in the left component. These normal forms
-are then compared modulo permutation of the components (this is done by {\tt
-CompareStruct}). The main tactic to be called and realizing this algorithm is
-{\tt IsoProve}.
-
-% Figure~\ref{isoslem} gives
-Here are examples of what can be solved by {\tt IsoProve}.
-%\begin{figure}[ht]
-%\begin{centerframe}
-\begin{coq_example*}
-Lemma isos_ex1 :
- forall A B:Set, A * unit * B = B * (unit * A).
-Proof.
-intros; IsoProve.
-Qed.
-
-Lemma isos_ex2 :
- forall A B C:Set,
- (A * unit -> B * (C * unit)) =
- (A * unit -> (C -> unit) * C) * (unit -> A -> B).
-Proof.
-intros; IsoProve.
-Qed.
-\end{coq_example*}
-%\end{centerframe}
-%\caption{Type equalities solved by {\tt IsoProve}}
-%\label{isoslem}
-%\end{figure}
-
-%%% Local Variables:
-%%% mode: latex
-%%% TeX-master: "Reference-Manual"
-%%% End:
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