180 lines
3.9 KiB
Coq
180 lines
3.9 KiB
Coq
Require Import MyTactics.
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Require Import LambdaCalculusSyntax.
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(* The syntactic subcategory of values is decidable. *)
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Definition if_value {A} (t : term) (a1 a2 : A) : A :=
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match t with
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| Var _ | Lam _ => a1
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| _ => a2
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end.
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Definition is_value (t : term) :=
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if_value t True False.
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Hint Extern 1 (is_value _) => (simpl; tauto) : is_value obvious.
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(* A term either is or is not a value. *)
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Lemma value_or_nonvalue:
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forall t,
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is_value t \/ ~ is_value t.
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Proof.
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destruct t; simpl; eauto.
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Qed.
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(* Simplification rules for [if_value]. *)
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Lemma if_value_value:
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forall A v (a1 a2 : A),
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is_value v ->
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if_value v a1 a2 = a1.
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Proof.
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destruct v; simpl; tauto.
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Qed.
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Lemma if_value_nonvalue:
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forall A t (a1 a2 : A),
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~ is_value t ->
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if_value t a1 a2 = a2.
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Proof.
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destruct t; simpl; tauto.
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Qed.
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(* The syntactic subcategory of values is preserved by renamings. *)
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Lemma is_value_renaming:
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forall v xi,
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is_value v ->
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is_value v.[ren xi].
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Proof.
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destruct v; simpl; eauto.
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Qed.
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Lemma is_nonvalue_renaming:
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forall v xi,
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~ is_value v ->
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~ is_value v.[ren xi].
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Proof.
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destruct v; simpl; eauto.
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Qed.
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Hint Resolve is_value_renaming is_nonvalue_renaming : is_value obvious.
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Ltac is_value :=
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eauto with is_value.
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(* The tactic [not_a_value] can be used to prove that the current case
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is impossible, because we have a hypothesis of the form [~ is_value v],
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where [v] clearly is a value. *)
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Ltac not_a_value :=
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solve [ false; is_value ].
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Ltac if_value :=
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repeat first [ rewrite if_value_value by is_value |
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rewrite if_value_nonvalue by is_value ].
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(* The tactic [value_or_nonvalue t] creates two cases: either [t] is a value,
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or it isn't. *)
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Ltac value_or_nonvalue t :=
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destruct (value_or_nonvalue t);
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if_value.
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(* The tactic [value_or_app_or_let] creates three cases: either [t] is a value,
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or it is an application, or it is a [let] construct. *)
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Ltac value_or_app_or_let t :=
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value_or_nonvalue t; [|
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destruct t as [ ? | ? | t1 t2 | t1 t2 ]; [ not_a_value | not_a_value | |]
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(* In principle, we should not fix the names [t1] and [t2] here,
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as it might cause name clashes. *)
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].
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(* The predicate [is_value_subst sigma] holds if every term in the
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codomain of the substitution [sigma] is a value. *)
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Definition is_value_subst (sigma : var -> term) :=
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forall x, is_value (sigma x).
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Lemma is_value_subst_ids:
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is_value_subst ids.
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Proof.
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intros x. is_value.
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Qed.
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Lemma is_value_subst_cons:
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forall v sigma,
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is_value v ->
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is_value_subst sigma ->
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is_value_subst (v .: sigma).
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Proof.
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intros. intros [|x]; is_value.
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Qed.
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Definition is_value_subst_up:
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forall sigma,
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is_value_subst sigma ->
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is_value_subst (up sigma).
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Proof.
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intros sigma h. intros [|x]; asimpl.
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{ simpl. eauto. }
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{ is_value. }
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Qed.
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Definition is_value_subst_upn:
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forall sigma i,
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is_value_subst sigma ->
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is_value_subst (upn i sigma).
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Proof.
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induction i; intros; asimpl.
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{ eauto. }
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{ rewrite <- fold_up_upn. eauto using is_value_subst_up. }
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Qed.
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Lemma is_value_subst_renaming:
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forall sigma i,
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is_value_subst sigma ->
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is_value_subst (sigma >> ren (+i)).
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Proof.
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intros. intro x. asimpl. is_value.
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Qed.
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Hint Resolve is_value_subst_up is_value_subst_upn is_value_subst_renaming
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: is_value obvious.
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Lemma values_are_preserved_by_value_substitutions:
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forall v sigma,
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is_value v ->
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is_value_subst sigma ->
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is_value v.[sigma].
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Proof.
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destruct v; simpl; intros; eauto with is_value.
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Qed.
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Lemma nonvalues_are_preserved_by_substitutions:
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forall t sigma,
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~ is_value t ->
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~ is_value t.[sigma].
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Proof.
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destruct t; simpl; tauto.
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Qed.
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Hint Resolve
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is_value_subst_ids
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is_value_subst_cons
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values_are_preserved_by_value_substitutions
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nonvalues_are_preserved_by_substitutions
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: is_value obvious.
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Lemma is_ren_is_value_subst:
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forall sigma,
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is_ren sigma ->
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is_value_subst sigma.
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Proof.
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intros ? [ xi ? ]. subst. eauto with is_value.
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Qed.
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Hint Resolve is_ren_is_value_subst : is_value obvious.
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