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Coq 逻辑：均衡双控

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Coq 逻辑：均衡双控,coq,logical-foundations,Coq,Logical Foundations,我被困在这里： Theorem evenb_double_conv : forall n, exists k, n = if evenb n then double k else S (double k). Proof. (* Hint: Use the [evenb_S] lemma from [Induction.v]. *) intros n. induction n as [|n' IHn']. - simpl. exists O. sim

我被困在这里：

Theorem evenb_double_conv : forall n,
  exists k, n = if evenb n then double k
                else S (double k).
Proof.
  (* Hint: Use the [evenb_S] lemma from [Induction.v]. *)
  intros n. induction n as [|n' IHn'].
  - simpl. exists O. simpl. reflexivity.
  - rewrite -> evenb_S. destruct (evenb n') as [H1 | H2].
    + simpl.
我们可以使用归纳假设将（双k）重写为n’，或者在目标上使用注入，然后应用归纳假设

但是我不能做这些，因为
存在


rewrite我们需要打破
存在的
与destruct的假设：
destruct IHn'as[k HE]。


n' : nat
IHn' : exists k : nat, n' = double k
============================
exists k : nat, S n' = S (double k)



注射在这里不起作用，因为它只在假设条件下起作用。
这不是课程中练习的答案吗？公布答案有意义吗？@Yves在保留对困难部分的提示的同时，我删除了解决方案的最后一部分。

Theorem evenb_double_conv : forall n,
  exists k, n = if evenb n then double k
                else S (double k).
Proof.
  (* Hint: Use the [evenb_S] lemma from [Induction.v]. *)
  intros n. induction n as [|n' IHn'].
  - simpl. exists O. simpl. reflexivity.
  - rewrite -> evenb_S. destruct IHn' as [k HE].  destruct (evenb n').
    (* Now find out which k we need to insert into the goal for every branch *)