问题
Consider the following simple expression language:
Inductive Exp : Set :=
| EConst : nat -> Exp
| EVar : nat -> Exp
| EFun : nat -> list Exp -> Exp.
and its wellformedness predicate:
Definition Env := list nat.
Inductive WF (env : Env) : Exp -> Prop :=
| WFConst : forall n, WF env (EConst n)
| WFVar : forall n, In n env -> WF env (EVar n)
| WFFun : forall n es, In n env ->
Forall (WF env) es ->
WF env (EFun n es).
which basically states that every variable and function symbols must be defined in the environment. Now, I want to define a function that states the decidability of WF predicate:
Definition WFDec (env : Env) : forall e, {WF env e} + {~ WF env e}.
refine (fix wfdec e : {WF env e} + {~ WF env e} :=
match e as e' return e = e' -> {WF env e'} + {~ WF env e'} with
| EConst n => fun _ => left _ _
| EVar n => fun _ =>
match in_dec eq_nat_dec n env with
| left _ _ => left _ _
| right _ _ => right _ _
end
| EFun n es => fun _ =>
match in_dec eq_nat_dec n env with
| left _ _ => _
| right _ _ => right _ _
end
end (eq_refl e)) ; clear wfdec ; subst ; eauto.
The trouble is how to state that WF predicate holds or not for a list of expressions in the EFun case. My obvious guess was:
...
match Forall_dec (WF env) wfdec es with
...
But Coq refuses it, arguing that the recursive call wfdec is ill-formed. My question is: Is it possible to define decidability of such wellformedness predicate without changing the expression representation?
The complete working code is at the following gist.
回答1:
The problem is that Forall_dec is defined as opaque in the standard library (that is, with Qed instead of Defined). Because of that, Coq does not know that the use of wfdec is valid.
The immediate solution to your problem is to redefine Forall_dec so that it is transparent. You can do this by printing the proof term that Coq generates and pasting it in your source file. I've added a gist here with a complete solution.
Needless to say, this approach lends itself to bloated, hard to read, and hard to maintain code. As ejgallego was pointing out in his answer, your best bet in this case is probably to define a Boolean function that decides WF, and use that instead of WFDec. The only problem with his approach, as he said, is that you will need to write your own induction principle to Exp in order to prove that the Boolean version indeed decides the inductive definition. Adam Chlipala's CPDT has a chapter on inductive types that gives an example of such an induction principle; just look for "nested inductive types".
回答2:
As a temporal workaround you can define wf as:
Definition wf (env : Env) := fix wf (e : Exp) : bool :=
match e with
| EConst _ => true
| EVar v => v \in env
| EFun v l => [&& v \in env & all wf l]
end.
which is usually way more convenient to use. However, this definition will be pretty useless due to Coq generating the wrong induction principle for exp, as it doesn't detect the list. What I usually do is to fix the induction principle manually, but this is costly. Example:
From Coq Require Import List.
From mathcomp Require Import all_ssreflect.
Set Implicit Arguments.
Unset Printing Implicit Defensive.
Import Prenex Implicits.
Section ReflectMorph.
Lemma and_MR P Q b c : reflect P b -> reflect Q c -> reflect (P /\ Q) (b && c).
Proof. by move=> h1 h2; apply: (iffP andP) => -[/h1 ? /h2 ?]. Qed.
Lemma or_MR P Q b c : reflect P b -> reflect Q c -> reflect (P \/ Q) (b || c).
Proof. by move=> h1 h2; apply: (iffP orP) => -[/h1|/h2]; auto. Qed.
End ReflectMorph.
Section IN.
Variables (X : eqType).
Lemma InP (x : X) l : reflect (In x l) (x \in l).
Proof.
elim: l => [|y l ihl]; first by constructor 2.
by apply: or_MR; rewrite // eq_sym; exact: eqP.
Qed.
End IN.
Section FORALL.
Variables (X : Type) (P : X -> Prop).
Variables (p : X -> bool).
Lemma Forall_inv x l : Forall P (x :: l) -> P x /\ Forall P l.
Proof. by move=> U; inversion U. Qed.
Lemma ForallP l : (forall x, In x l -> reflect (P x) (p x)) -> reflect (Forall P l) (all p l).
Proof.
elim: l => [|x l hp ihl /= ]; first by constructor.
have/hp {hp}hp : forall x : X, In x l -> reflect (P x) (p x).
by move=> y y_in; apply: ihl; right.
have {ihl} ihl := ihl _ (or_introl erefl).
by apply: (iffP andP) => [|/Forall_inv] [] /ihl hx /hp hall; constructor.
Qed.
End FORALL.
Inductive Exp : Type :=
| EConst : nat -> Exp
| EVar : nat -> Exp
| EFun : nat -> list Exp -> Exp.
Lemma Exp_rect_list (P : Exp -> Type) :
(forall n : nat, P (EConst n)) ->
(forall n : nat, P (EVar n)) ->
(forall (n : nat) (l : seq Exp), (forall x, In x l -> P x) -> P (EFun n l)) ->
forall e : Exp, P e.
Admitted.
Definition Env := list nat.
Definition wf (env : Env) := fix wf (e : Exp) : bool :=
match e with
| EConst _ => true
| EVar v => v \in env
| EFun v l => [&& v \in env & all wf l]
end.
Inductive WF (env : Env) : Exp -> Prop :=
| WFConst : forall n, WF env (EConst n)
| WFVar : forall n, In n env -> WF env (EVar n)
| WFFun : forall n es, In n env ->
Forall (WF env) es ->
WF env (EFun n es).
Lemma WF_inv env e (wf : WF env e ) :
match e with
| EConst n => True
| EVar n => In n env
| EFun n es => In n env /\ Forall (WF env) es
end.
Proof. by case: e wf => // [n|n l] H; inversion H. Qed.
Lemma wfP env e : reflect (WF env e) (wf env e).
Proof.
elim/Exp_rect_list: e => [n|n|n l ihe] /=; try repeat constructor.
by apply: (iffP idP) => [/InP|/WF_inv/InP //]; constructor.
apply: (iffP andP) => [[/InP ? /ForallP H]|/WF_inv[/InP ? /ForallP]].
by constructor => //; exact: H.
by auto.
Qed.
来源：https://stackoverflow.com/questions/43453365/problems-with-dependent-types-in-coq-proof-assistant