1

假设我有

  • T型
  • 有根据的关系 R:T->T->Prop
  • 函数 F1:T->T 使参数“更小”
  • 条件 C:T->Prop 描述 R 的“起始值”
  • 函数 F2:T->T 使参数“更大”

如何使 Fixpoint 看起来与此类似:

Fixpoint Example (n:T):X :=
  match {C n} + {~C n} with
    left _ => ... |
    right _ => Example (F1 n)
  end.

以及如何使战术“归纳”(或类似)的以下用法成为可能:

Theorem ...
Proof.
 ...
 induction n F.
(* And now I have two goals:
   the first with assumption C n and goal P n,
   the second with assumption P n and goal P (F2 n) *)
 ...
Qed.

我尝试使用 nz 类型来做到这一点:{n:nat | n<>O}(查看 Certiified Programming with Dependent Types 的第 7.1 章),但仅此而已:

Require Import Omega.

Definition nz: Set := {n:nat | n<>O}.
Theorem nz_t1 (n:nat): S n<>O. Proof. auto. Qed.

Definition nz_eq (n m:nz) := eq (projT1 n) (projT1 m).
Definition nz_one: nz := exist _ 1 (nz_t1 O).
Definition nz_lt (n m:nz) := lt (projT1 n) (projT1 m).

Definition nz_pred (n:nz): nz := exist _ (S (pred (pred (projT1 n)))) (nz_t1 _).

Theorem nz_Acc: forall (n:nz), Acc nz_lt n.
Proof.
 intro. destruct n as [n pn], n as [|n]. omega.
 induction n; split; intros; destruct y as [y py]; unfold nz_lt in *; simpl in *.
   omega.
   assert (y<S n\/y=S n). omega. destruct H0.
    assert (S n<>O); auto.
    assert (nz_lt (exist _ y py) (exist _ (S n) H1)). unfold nz_lt; simpl; assumption.
    fold nz_lt in *. apply Acc_inv with (exist (fun n0:nat=>n0<>O) (S n) H1). apply IHn.
    unfold nz_lt; simpl; assumption.
    rewrite <- H0 in IHn. apply IHn.
Defined.

Theorem nz_lt_wf: well_founded nz_lt. Proof. exact nz_Acc. Qed.

Lemma pred_wf: forall (n m:nz), nz_lt nz_one n -> m = nz_pred n -> nz_lt m n.
Proof.
 intros. unfold nz_lt, nz_pred in *. destruct n as [n pn], m as [m pm]. simpl in *.
 destruct n, m; try omega. simpl in *. inversion H0. omega. 
Defined.

我无法理解进一步发生了什么,因为这对我来说太复杂了。

PS 正如我所见——对于初学者来说,关于 Coq 中的一般递归和归纳的教程还不够好。至少我能找到。:(

4

1 回答 1

2

稍后我会尝试写一个更完整的答案,但是 Coq 有一个名为 Function 的命令,可以更轻松地编写参数根据某种良好排序而减少的函数。在参考手册(http://coq.inria.fr/distrib/current/refman/Reference-Manual004.html#hevea_command48)上查找命令,特别是“wf”变体。

于 2013-09-16T14:06:09.637 回答