Question

我如何在coq中证明函数f接受bool true|false并返回bool true|false（如下所示），当应用两次bool {{ 1}}总会返回相同的值true|false：

true|false

例如，函数(f:bool -> bool)只能做4件事，让我们调用函数f的输入：

始终返回b
始终返回true
返回false（如果b为真则返回true，反之亦然）
返回b（即如果b为真，则返回false）

因此，如果函数始终返回true：

not b

如果函数总是返回false，我们会得到：

f (f bool) = f true = true

对于其他情况，我们假设该函数返回f (f bool) = f false = false

not b

在两种可能的输入情况下，我们总是最终得到原始输入。如果我们假设函数返回f (f true) = f false = true f (f false) = f true = false，则同样成立。

那么您如何在coq中证明这一点？

Answer 1

Goal forall (f:bool -> bool) (b:bool), f (f (f b)) = f b.
Proof.
intros.
remember (f true) as ft.
remember (f false) as ff.
destruct ff ; destruct ft ; destruct b ; 
    try rewrite <- Heqft ; try rewrite <- Heqff ; 
    try rewrite <- Heqft ; try rewrite <- Heqff ; auto.
Qed.

Answer 2

一点点证明：

Require Import Sumbool.

Goal forall (f : bool -> bool) (b:bool), f (f (f b)) = f b.
Proof.
  destruct b;                             (* case analysis on [b] *)
    destruct (sumbool_of_bool (f true));  (* case analysis on [f true] *)
    destruct (sumbool_of_bool (f false)); (* case analysis on [f false] *)
    congruence.                           (* equational reasoning *)
Qed.

Answer 3

在SSReflect中：

Require Import ssreflect.

Goal forall (f:bool -> bool) (b:bool), f (f (f b)) = f b.
Proof.
move=> f.
by case et:(f true); case ef:(f false); case; rewrite ?et ?ef // !et ?ef.
Qed.

Answer 4

感谢您的精彩任务！这么可爱的定理！

这是使用Coq的C-zar声明性证明样式的证明。它比命令式的要长得多（虽然因为我的技能太低而可能是这样）。

Theorem bool_cases : forall a, a = true \/ a = false.
proof.
    let a:bool.
    per cases on a.
    suppose it is false.
        thus thesis.
    suppose it is true.
        thus thesis.
    end cases.
end proof. Qed.

Goal forall (b:bool), f (f (f b)) = f b.
proof.
    let b:bool.
    per cases on b.

    suppose it is false.
        per cases of (f false = false \/ f false = true) by bool_cases.
        suppose (f false = false).
            hence (f (f (f false)) = f false).
        suppose H:(f false = true).
            per cases of (f true = false \/ f true = true) by bool_cases.
            suppose (f true = false).
                hence (f (f (f false)) = f false) by H.
            suppose (f true = true).
                hence (f (f (f false)) = f false) by H.
            end cases.
        end cases.

    suppose it is true.
        per cases of (f true = false \/ f true = true) by bool_cases.
        suppose H:(f true = false).
            per cases of (f false = false \/ f false = true) by bool_cases.
            suppose (f false = false).
                hence (f (f (f true)) = f true) by H.
            suppose (f false = true).
                hence (f (f (f true)) = f true) by H.
            end cases.
        suppose (f true = true).
            hence (f (f (f true)) = f true).
        end cases.

end cases.
end proof. Qed.

证明f（f bool）= bool

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