I have the defined inductive types:
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).
Now I have to prove a series of properties of that inductive type, but I keep getting stuck.
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.
Can some one help me advance.
In fact, it is easier to do an induction on the SubSet judgment directly. However, you need to be as general as possible, so here is my advice:
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.
"inversion" is a tactic that checks an inductive term and gives you all the possible way to build such a term !!without any induction hypothesis!! It only gives you the constructive premices.
You could have done it directly by induction on l1 then l2, but you would have to construct by hand the correct instance of inversion because your induction hypothesis would have been really weak.
Hope it helps, V.