Question

我如何证明这两个陈述是平等的：

Val.shru（Val。和a（Vint b））（Vint c）= Vint？3434 / \？3434＆lt;＆gt; d
Val.shru（Val。和a（Vint b））（Vint c）＆lt;＆gt; d

这个概念非常简单，但坚持找到解决问题的正确策略。这实际上是我要证明的引理：

Require Import compcert.common.Values.
Require Import compcert.lib.Coqlib.
Require Import compcert.lib.Integers.

Lemma val_remains_int:
forall (a : val) (b c d: int),
(Val.shru (Val.and a (Vint b)) (Vint c)) <> (Vint d) ->
(exists (e : int), (Val.shru (Val.and a (Vint b)) (Vint c)) = (Vint e) /\ e <> d).

Proof.
  intros.
  eexists.
  ...
Admitted.

谢谢，

Answer 1

如果您可以在下面的示例中构造int（i0）类型的值，则此引理不成立：

Require Import compcert.lib.Coqlib.
Require Import compcert.lib.Integers.
Require Import compcert.common.Values.

Variable i0 : int.

Fact counter_example_to_val_remains_int:
  ~ forall (a : val) (b c d: int),
      (Val.shru (Val.and a (Vint b)) (Vint c)) <> (Vint d) ->
      (exists (e : int),
          (Val.shru (Val.and a (Vint b)) (Vint c)) = (Vint e)
        /\ e <> d).
Proof.
  intro H.
  assert (Vundef <> Vint i0) as H0 by easy.
  specialize (H Vundef i0 i0 i0 H0); clear H0.
  simpl in H.
  destruct H as (? & contra & _).
  discriminate contra.
Qed.

至少有两个原因：

Val.and和Val.shru返回Vundef所有不是Vint的参数（它是GIGO原则的实例）;
你也不能在C中过多地移位 - 结果是未定义的（这个约为Val.shru）。

至于您在评论中提到的修改后的引理，简单的reflexivity会：

Lemma val_remains_int: forall a b c d: int,
    Vint (Int.shru (Int.and a b) c) <> Vint d ->
    exists (e : int), Vint (Int.shru (Int.and a b) c) = Vint e /\ e <> d.
Proof.
  intros a b c d Hneq.
  eexists. split.
  - reflexivity.
  - intro Heq. subst. auto.
Qed.

解决目标，coq代码中的平等/不平等问题

1 个答案: