题
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).
.
现在我必须证明一系列电感类型的属性,但我继续陷入困境。
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.
.
有些人可以帮助我进步。
解决方案
事实上,直接对子集判断进行诱导更容易。 但是,您需要尽可能一般,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.
.
“反演”是检查归纳项的策略,并为您提供构建此类术语的所有可能方法!!没有任何归因! 它只为您提供建设性的法案。
您可以通过L1上的诱导直接完成它,然后您必须通过手工构造正确的反转实例,因为您的归纳假设会真的很弱。
希望它有所帮助, v。
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