在逻辑章节中,介绍了反向列表函数的尾递归版本。我们需要证明它可以正常工作:
Fixpoint rev_append {X} (l1 l2 : list X) : list X :=
match l1 with
| [] => l2
| x :: l1' => rev_append l1' (x :: l2)
end.
(* Tail recursion rev *)
Definition tr_rev {X} (l : list X) : list X :=
rev_append l [].
但在证明之前,我想证明一个引理:
Lemma rev_append_app: forall (X: Type) (x: X) (l : list X),
rev_append l [x] = rev_append l [] ++ [x].
Proof.
intros X x l. induction l as [| h t IH].
- simpl. reflexivity.
- simpl.
我在这里卡住了:
X : Type
x, h : X
t : list X
IH : rev_append t [x] = rev_append t [ ] ++ [x]
============================
rev_append t [h; x] = rev_append t [h] ++ [x]
接下来做什么?