5

我有定义的归纳类型：

Inductive InL (A:Type) (y:A) : list A -> Prop := 
  | InHead : forall xs:list A, InL y (cons y xs) 
  | InTail : forall (x:A) (xs:list A), InL y xs -> InL y (cons x xs).

Inductive SubSeq (A:Type) : list A -> list A -> Prop :=
 | SubNil : forall l:list A, SubSeq nil l
 | SubCons1 : forall (x:A) (l1 l2:list A), SubSeq l1 l2 -> SubSeq l1 (x::l2)
 | SubCons2 : forall (x:A) (l1 l2:list A), SubSeq l1 l2 -> SubSeq (x::l1) (x::l2).

现在我必须证明该归纳类型的一系列属性，但我一直卡住。

Lemma proof1: forall (A:Type) (x:A) (l1 l2:list A), SubSeq l1 l2 -> InL x l1 -> InL x l2.
Proof.
 intros.
 induction l1.
 induction l2.
 exact H0.

Qed.

有人可以帮助我前进。

4

1 回答 1

8

其实直接对SubSet判断做归纳比较容易。但是，您需要尽可能笼统，所以这是我的建议：

Lemma proof1: forall (A:Type) (x:A) (l1 l2:list A), 
  SubSeq l1 l2 -> InL x l1 -> InL x l2.
(* first introduce your hypothesis, but put back x and In foo
   inside the goal, so that your induction hypothesis are correct*)
intros. 
revert x H0. induction H; intros.
(* x In [] is not possible, so inversion will kill the subgoal *)
inversion H0.

(* here it is straitforward: just combine the correct hypothesis *)
apply InTail; apply IHSubSeq; trivial.

(* x0 in x::l1 has to possible sources: x0 == x or x0 in l1 *)
inversion H0; subst; clear H0.
apply InHead.
apply InTail; apply IHSubSeq; trivial.
Qed.

“倒置”是一种检查归纳项的策略，并为您提供构建此类项的所有可能方法！！无需任何归纳假设！它只会给你建设性的前提。

您可以直接通过对 l1 然后 l2 进行归纳来完成它，但是您必须手动构建正确的反演实例，因为您的归纳假设会非常弱。

希望对你有帮助，V。

于 2010-07-11T15:18:49.283 回答