Question

我的证明脚本给了我愚蠢的类型，如nat = bool或 nat = list unit我需要用来解决相互矛盾的目标。

在正常的数学中，这将是微不足道的。给定集bool := { true, false }和 nat := { 0, 1, 2, ... }我知道true ∈ bool，但是true ∉ nat，因此bool ≠ nat。在Coq中，我甚至不知道如何陈述true :̸ nat。

问题

有没有办法证明这些平等是错误的？或许，这不可能吗？

（编辑：删除了一长串失败的attemts，仍然可以在历史中看到。）

Answer 1

扩展部分解决方案

作为参考，这是我对nat = bool -> False的证明。（这很长，但我希望很容易看到这个证明的一般结构。）

Goal nat = bool -> False.
  (* For any two types, if they are actually identical, the identity is an
     isomorphism. *)
  assert (forall (T U : Set), T = U ->
              exists (f : T -> U) (g : U -> T),
              (forall t, (g (f t)) = t) /\ (forall u, (f (g u)) = u))
          as Hiso
  by (intros T U H; rewrite H; exists (@id U); exists (@id U);
          split; intro; reflexivity).
  (* our nat = bool *)
  intro HC.
  (* combining the facts gives an iso between nat and bool *)
  pose proof (Hiso nat bool HC); clear HC Hiso.
  inversion H as [phi [phi_inv [Hl Hr]]]; clear H Hr.
  (* this breaks because ||bool|| = 2 while ||nat|| > 2 -- we get collisions *)
  assert (forall m n o,
              phi m = phi n \/ phi n = phi o \/ phi m = phi o)
   by (intros m n o;
        case (phi m); case (phi n); case (phi o); clear; tauto).
  (* building the collision for 0, 1 and 2 *)
  pose proof (H 0 1 2) as HC; clear H.
  (* (false) identity preservation for 0, 1, 2 *)
  pose proof (Hl 0) as H0; pose proof (Hl 1) as H1;
  pose proof (Hl 2) as H2; clear Hl.
  (* case analysis on phi calls yields equalities on non-equal numbers... *)
  destruct (phi 0); destruct (phi 1); destruct (phi 2);
  (* ...rewriting leads to an equality '0 = 2' or '0 = 1' or '1 = 2'... *)
  try (rewrite H2 in H0); try (rewrite H1 in H0); try (rewrite H2 in H1);
  (* ...which can be used to solve by constructor inequality *)
  try inversion H0; inversion H1.
Qed.

正如您所看到的，这对于大型有限类型（即使是自动化的）也不是真的可用 - 这些术语太大了。对此的任何改进都会很棒。

如何解决Coq中无效类型等于的目标？

问题

1 个答案:

扩展部分解决方案