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我想在 Coq 中使用类型类来定义 Maybe monad。 Monad继承Functor.

我想证明Some (f x') = fmap f (Some x')，这是单子定律之一。我使用compute,reflexivity和destruct option_functor，但我无法证明这一点。我不能fmap适当地简化。

Class Functor (F: Type -> Type) := {
   fmap : forall {A B}, (A -> B) -> (F A -> F B)
 ; homo_id : forall {A} (x : F A), x = fmap (fun x' => x') x
 ; homo_comp : forall {A B C} (f : A -> B) (g : B -> C) (x : F A),
     fmap (fun x' => g (f x')) x = fmap g (fmap f x)
}.

Class Monad (M: Type -> Type) := {
   functor :> Functor M
 ; unit : forall {A}, A -> M A
 ; join : forall {A}, M (M A) -> M A
 ; unit_nat : forall {A B} (f : A -> B) (x : A), unit (f x) = fmap f (unit x)
 ; join_nat : forall {A B} (f : A -> B) (x : M (M A)), join (fmap (fmap f) x) = fmap f (join x)
 ; identity : forall {A} (x : M A), join (unit x) = x /\ x = join (fmap unit x)
 ; associativity : forall {A} (x : M (M (M A))), join (join x) = join (fmap join x)
}.

Instance option_functor : Functor option := {
   fmap A B f x :=
     match x with
     | None => None
     | Some x' => Some (f x')
     end
}.
Proof.
  intros. destruct x; reflexivity.
  intros. destruct x; reflexivity.
Qed.

Instance option_monad : Monad option := {
   unit A x := Some x
 ; join A x :=
     match x with
     | Some (Some x') => Some x'
     | _ => None
     end
}.
Proof.
  Admitted.
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1 回答 1

5

您的问题源于您结束了option_functionwithQed而不是 的定义Defined。

使用Qed，您以某种方式“隐藏” fmap. 然后你不能再展开它的定义（例如使用unfoldandsimpl策略）。使用Defined而不是Qed让你告诉 Coq 你打算使用fmap后者的定义，所以它应该是可展开的。

于 2014-03-10T08:25:42.843 回答