329 lines
9.7 KiB
Coq
329 lines
9.7 KiB
Coq
Require Import Omega.
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Require Import Autosubst.Autosubst.
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Require Import AutosubstExtra. (* just for [upn_ren] *)
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Require Import MyTactics. (* TEMPORARY *)
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(* This file defines the construction [eos x t], which can be understood as
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an end-of-scope mark for [x] in the term [t]. *)
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(* It also defines the single-variable substitution t.[u // x], which is the
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substitution of [u] for [x] in [t]. *)
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(* -------------------------------------------------------------------------- *)
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Section EOS.
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Context {A} `{Ids A, Rename A, Subst A, SubstLemmas A}.
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(* The substitution [Var 0 .: Var 1 .: ... .: Var (x-1) .: Var (x+1) .: ...]
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does not have [Var x] in its codomain. Thus, applying this substitution
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to a term [t] can be understood as an end-of-scope construct: it means
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``end the scope of [x] in [t]''. We write [eos x t] for this construct.
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It is also known as [adbmal]: see Hendriks and van Oostrom,
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https://doi.org/10.1007/978-3-540-45085-6_11 *)
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(* There are at least two ways of defining the above substitution. One way
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is to define it in terms of AutoSubst combinators: *)
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Definition eos_var x : var -> var :=
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(iterate upren x (+1)).
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Definition eos x t :=
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t.[ren (eos_var x)].
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Lemma eos_eq:
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forall x t,
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t.[upn x (ren (+1))] = eos x t.
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Proof.
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intros. unfold eos, eos_var. erewrite upn_ren by tc. reflexivity.
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Qed.
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(* Another way is to define directly as a function of type [var -> var]. *)
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Definition lift_var x : var -> var :=
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fun y => if le_gt_dec x y then 1 + y else y.
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(* The two definitions coincide: *)
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Lemma upren_lift_var:
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forall x,
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upren (lift_var x) = lift_var (S x).
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Proof.
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intros. f_ext; intros [|y].
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{ reflexivity. }
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{ simpl. unfold lift_var, var. dblib_by_cases; omega. }
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Qed.
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Lemma eos_var_eq_lift_var:
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eos_var = lift_var.
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Proof.
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(* An uninteresting proof. *)
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f_ext; intros x.
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unfold eos_var.
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induction x.
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{ reflexivity. }
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{ rewrite iterate_S.
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rewrite IHx.
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rewrite upren_lift_var.
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reflexivity. }
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Qed.
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(* -------------------------------------------------------------------------- *)
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(* [eos] enjoys certain commutation laws. *)
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(* Ending the scope of variable [k], then the scope of variable [s], is the
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same as first ending the scope of variable [1 + s], then ending the scope
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of variable [k]. This holds provided [k <= s] is true, i.e., [k] is the
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most recently-introduced variable.*)
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Lemma lift_var_lift_var:
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forall k s,
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k <= s ->
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lift_var s >>> lift_var k = lift_var k >>> lift_var (S s).
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Proof.
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(* By case analysis. *)
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intros. f_ext; intros x. asimpl.
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unfold lift_var, var. dblib_by_cases; omega.
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Qed.
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Lemma eos_eos:
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forall k s t,
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k <= s ->
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eos k (eos s t) = eos (1 + s) (eos k t).
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Proof.
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intros. unfold eos. asimpl.
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rewrite eos_var_eq_lift_var.
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rewrite lift_var_lift_var by eauto.
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reflexivity.
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Qed.
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(* What about the case where [k] is the least recently-introduced variable?
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It is obtained by symmetry, of course. *)
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Lemma eos_eos_reversed:
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forall k s t,
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k >= s + 1 ->
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eos k (eos s t) = eos s (eos (k - 1) t).
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Proof.
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intros.
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replace k with (1 + (k - 1)) by omega.
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rewrite <- eos_eos by omega.
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replace (1 + (k - 1) - 1) with (k - 1) by omega.
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reflexivity.
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Qed.
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(* -------------------------------------------------------------------------- *)
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(* Single-variable substitutions. *)
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(* [subst_var u x] is the substitution of [u] for [x]. *)
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(* We give a direct definition of it as a function of type [var -> term],
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defined by cases. I don't know if it could also be nicely defined in
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terms of the basic combinators of de Bruijn algebra. Note that the
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candidate definition [upn x (t .: ids)] is WRONG when [x > 0]. *)
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Definition subst_var (u : A) (x y : var) : A :=
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match lt_eq_lt_dec y x with
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| inleft (left _) => ids y
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| inleft (right _) => u
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| inright _ => ids (y - 1)
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end.
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(* A nice notation: [t.[u // x]] is the substitution of [u] for [x] in [t]. *)
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Notation "t .[ u // x ]" := (subst (subst_var u x) t)
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(at level 2, u at level 200, left associativity,
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format "t .[ u // x ]") : subst_scope.
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(* The following laws serve as sanity checks: we got the definition right. *)
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Lemma subst_var_miss_1:
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forall x y u,
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y < x ->
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(ids y).[u // x] = ids y.
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Proof.
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intros. asimpl. unfold subst_var. dblib_by_cases. reflexivity.
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Qed.
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Lemma subst_var_match:
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forall x u,
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(ids x).[ u // x ] = u.
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Proof.
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intros. asimpl. unfold subst_var. dblib_by_cases. reflexivity.
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Qed.
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Lemma subst_var_miss_2:
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forall x y u,
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x < y ->
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(ids y).[u // x] = ids (y - 1).
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Proof.
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intros. asimpl. unfold subst_var. dblib_by_cases. reflexivity.
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Qed.
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(* In the special case where [x] is 0, the substitution [t // 0] can also
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be written [t/], which is an AutoSubst notation for [t .: ids]. *)
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Lemma subst_var_0:
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forall t u,
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t.[u // 0] = t.[u/].
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Proof.
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intros. f_equal. clear t.
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f_ext. intros [|x].
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{ reflexivity. }
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{ unfold subst_var. simpl. f_equal. omega. }
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Qed.
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(* -------------------------------------------------------------------------- *)
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(* A cancellation law: substituting for a variable [x] that does not occur in
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[t] yields just [t]. In other words, a substitution for [x] vanishes when
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it reaches [eos x _]. *)
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(* In informal syntax, this lemma would be written:
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t[u/x] = t
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under the hypothesis that x does not occur free in t.
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In de Bruijn style, the statement is just as short, and does not have a
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side condition. Instead, it requires an explicit [eos x _] to appear at the
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root of the term to which the substitution is applied; this may require
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rewriting before this lemma can be applied. *)
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Lemma subst_eos:
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forall x t u,
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(eos x t).[u // x] = t.
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Proof.
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intros.
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(* Again, let's simplify this first. *)
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unfold eos. asimpl.
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(* Aha! We can forget about [t], and focus on proving that two
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substitutions are equal. To do so, it is sufficient that
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their actions on a variable [y] are the same. *)
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rewrite <- subst_id.
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f_equal. clear t.
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f_ext. intro y.
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(* The proof is easy if we replace [eos_var] with [lift_var]. *)
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rewrite eos_var_eq_lift_var. simpl.
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unfold subst_var, lift_var. dblib_by_cases; f_equal; omega.
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Qed.
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(* The above property allows us to prove that [eos x _] is injective.
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Indeed, it has an inverse, namely [u // x], where [u] is arbitrary. *)
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Lemma eos_injective:
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forall x t1 t2,
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eos x t1 = eos x t2 ->
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t1 = t2.
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Proof.
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intros.
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pose (u := t1). (* dummy *)
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erewrite <- (subst_eos x t1 u).
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erewrite <- (subst_eos x t2 u).
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congruence.
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Qed.
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(* -------------------------------------------------------------------------- *)
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(* More commutation laws. *)
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Lemma eos_subst_1:
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forall k s t u,
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k <= s ->
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eos k (t.[u // s]) = (eos k t).[eos k u // s + 1].
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Proof.
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intros. unfold eos. asimpl. f_equal. clear t.
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rewrite eos_var_eq_lift_var.
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f_ext. intros y.
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asimpl.
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unfold subst_var, lift_var.
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dblib_by_cases; asimpl; dblib_by_cases; solve [ eauto | f_equal; omega ].
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Qed.
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Lemma eos_subst_2:
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forall k s t u,
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s <= k ->
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eos k (t.[u // s]) = (eos (k + 1) t).[eos k u // s].
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Proof.
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intros. unfold eos. asimpl. f_equal. clear t.
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rewrite eos_var_eq_lift_var.
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f_ext. intros y.
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asimpl.
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unfold subst_var, lift_var.
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dblib_by_cases; asimpl; dblib_by_cases; solve [ eauto | f_equal; omega ].
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Qed.
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Lemma subst_subst:
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forall t k v s w,
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k <= s ->
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t.[w // k].[v // s] =
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t.[eos k v // 1 + s].[w.[v // s] // k].
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Proof.
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(* First, get rid of [t]. It is sufficient to consider the action of
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these substitutions at a variable [y]. *)
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intros. asimpl. f_equal. clear t. f_ext. intros y.
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(* Replace [eos_var] with [lift_var], whose definition is more direct. *)
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unfold eos. rewrite eos_var_eq_lift_var.
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(* Then, use brute force (case analysis) to prove that the goal holds. *)
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unfold subst_var. simpl.
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dblib_by_cases; asimpl; dblib_by_cases;
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(* This case analysis yields 5 cases, of which 4 are trivial... *)
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eauto.
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(* ... thus, one case remains. *)
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(* Now get rid of [v]. It is again sufficient to consider the action
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of these substitutions at a variable [z]. *)
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replace v with v.[ids] at 1 by autosubst.
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f_equal. f_ext. intros z. simpl.
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(* Again, use brute force. *)
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unfold lift_var. dblib_by_cases; f_equal. unfold var. omega.
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(* Not really proud of this proof. *)
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Qed.
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Lemma pun_1:
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forall t x,
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(eos x t).[ ids x // x + 1 ] = t.
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Proof.
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(* First, get rid of [t]. It is sufficient to consider the action of
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these substitutions at a variable [y]. *)
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intros. unfold eos. asimpl.
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replace t with t.[ids] at 2 by autosubst.
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f_equal. clear t. f_ext. intros y.
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(* Replace [eos_var] with [lift_var], whose definition is more direct. *)
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rewrite eos_var_eq_lift_var.
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(* Then, use brute force (case analysis) to prove that the goal holds. *)
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simpl. unfold subst_var, lift_var. dblib_by_cases; f_equal; unfold var; omega.
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Qed.
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Lemma pun_2:
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forall t x,
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(eos (x + 1) t).[ ids x // x ] = t.
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Proof.
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(* First, get rid of [t]. It is sufficient to consider the action of
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these substitutions at a variable [y]. *)
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intros. unfold eos. asimpl.
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replace t with t.[ids] at 2 by autosubst.
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f_equal. clear t. f_ext. intros y.
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(* Replace [eos_var] with [lift_var], whose definition is more direct. *)
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rewrite eos_var_eq_lift_var.
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(* Then, use brute force (case analysis) to prove that the goal holds. *)
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simpl. unfold subst_var, lift_var. dblib_by_cases; f_equal; unfold var; omega.
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Qed.
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End EOS.
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(* Any notations defined in the above section must now be repeated. *)
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Notation "t .[ u // x ]" := (subst (subst_var u x) t)
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(at level 2, u at level 200, left associativity,
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format "t .[ u // x ]") : subst_scope.
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(* The tactic [subst_var] attempts to simplify applications of [subst_var]. *)
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Ltac subst_var :=
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first [
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rewrite subst_var_miss_1 by omega
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| rewrite subst_var_match by omega
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| rewrite subst_var_miss_2 by omega
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].
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