帮助进行子序列的Coq证明
我有定义的归纳类型：帮助进行子序列的Coq证明,coq,proof-system,Coq,Proof System,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 :=
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).
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.
有人能帮我进步吗。事实上，直接对子集判断进行归纳比较容易。 但是，您需要尽可能地通俗易懂，因此我的建议如下：
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.