当我试图证明一个关于递归函数定理(见下文),我在还原表现结束了如何使勒柯克评估特定归约(或 - ?它为什么在这种情况下拒绝)
(fix picksome L H := match A with .... end) L1 H1 = RHS
我想展开match
表达式,但Coq拒绝。做simpl
只是将右侧扩展成难以理解的混乱。 Coq为什么不能用simpl; reflexivity
完成证明,应该如何指导Coq精确地扩展redex,并完成证明?
功能是一个递归函数pick
,需要一个list nat
并采取第一nat
称为a
,下降从列表中选择以下a
项目,以及递归其余的名单上。即
pick [2;3;4;0;1;3])=[2; 0; 1]
我试图证明的这个定理是函数在只包含零的列表上“什么都不做”。下面是通向这一问题的发展:
Require Import Arith.
Require Import List.
Import ListNotations.
Fixpoint drop {T} n (l:list T) :=
match n,l with
| S n', cons _ l' => drop n' l'
| O, _ => l
| _, _ => nil
end.
第一引理:
Lemma drop_lemma_le : forall {T} n (l:list T), length (drop n l) <= (length l).
Proof.
intros; generalize n; induction l; intros; destruct n0; try reflexivity;
apply le_S; apply IHl.
Defined.
二引理:
Lemma picksome_term: forall l l' (a :nat),
l = a::l' -> Acc lt (length l) -> Acc lt (length (drop a l')).
Proof.
intros; apply H0; rewrite H; simpl; apply le_lt_n_Sm; apply drop_lemma_le.
Defined.
一些更多的定义:
Fixpoint picksome (l:list nat) (H : Acc lt (length l)) {struct H}: list nat :=
match l as m return l=m -> _ with
| nil => fun _ => nil
| cons a l' => fun Hl =>
cons a (picksome (drop a l')
(picksome_term _ _ _ Hl H))
end
(eq_refl _).
Definition pick (l:list nat) : list nat := picksome l (lt_wf (length l)).
Inductive zerolist : list nat -> Prop :=
| znil : zerolist nil
| hzlist : forall l, zerolist l -> zerolist (O::l).
现在,我们可以证明我们的定理˚F我们有引理H
:
Theorem pickzero': (forall k, pick (0::k) = 0::pick k) ->
forall l, zerolist l -> pick l = l.
Proof.
intros H l H0; induction H0; [ | rewrite H; rewrite IHzerolist]; reflexivity.
Qed.
(* but trying to prove the lemma *)
Lemma pickzero_lemma : forall k, pick (0::k) = 0::pick k.
induction k; try reflexivity.
unfold pick at 1.
unfold picksome.
这是我们的目标和背景:
a : nat
k : list nat
IHk : pick (0 :: k) = 0 :: pick k
============================
(fix picksome (l : list nat) (H : Acc lt (length l)) {struct H} :
list nat :=
match l as m return (l = m -> list nat) with
| [] => fun _ : l = [] => []
| a0 :: l' =>
fun Hl : l = a0 :: l' =>
a0 :: picksome (drop a0 l') (picksome_term l l' a0 Hl H)
end eq_refl) (0 :: a :: k) (lt_wf (length (0 :: a :: k))) =
0 :: pick (a :: k)
是否有可能只是减少匹配而不是运行整个'simp'? – larsr 2014-12-08 14:35:20
我不知道任何“简单”的方法来做到这一点。在这种情况下,我将单步缩减写为一个引理,大多数情况下我都会使用“简单”和“反身性”,然后用它重写。 – Vinz 2014-12-08 15:26:52