[英]Quick Chick eqBoolArrowA_correct theorem
Passing Quick Chick course from Software foundations I am stuck at the following theorem:通过软件基础的 Quick Chick 课程我坚持以下定理:
Class Eq A :=
{
eqb: A -> A -> bool;
}.
Instance eqBoolArrowA {A : Type} `{Eq A} : Eq (bool -> A) :=
{
eqb f1 f2 :=
(andb (eqb (f1 false) (f2 false)) (eqb (f1 true) (f2 true)))
}.
Theorem eqBoolArrowA_correct : forall (f1 f2 : bool -> bool -> nat),
(eqb f1 f2) = true <-> true = true.
Proof.
intros. split.
- destruct (f1 =? f2).
+ intro. assumption.
+ intros H. discriminate H.
- intros _. destruct (f1 =? f2).
+ reflexivity.
+
Here we get:在这里我们得到:
1 subgoal (ID 164)
f1, f2 : bool -> bool -> nat
============================
false = true
Proving true = true -> (eqb f1 f2) = true
seems impossible, because (eqb f1 f2)
can be false, and in this case we get false = true
in empty context, which is unprovable.证明true = true -> (eqb f1 f2) = true
似乎是不可能的,因为(eqb f1 f2)
可能为 false,在这种情况下,我们在空上下文中得到false = true
,这是无法证明的。
Is this theorem incorrect or am I missing something?这个定理是不正确的还是我遗漏了什么?
That theorem is not valid.该定理无效。
Any proposition P <-> true = true
is equivalent to P
, and it is not valid that eqb f1 f2 = true
for all functions f1, f2 : bool -> A
(at least when A
has two or more inhabitants).任何命题P <-> true = true
等价于P
,并且对于所有函数f1, f2 : bool -> A
eqb f1 f2 = true
是无效f1, f2 : bool -> A
(至少当A
有两个或更多居民时)。
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