Question

我想在Coq中使用类型类定义Maybe monad。 Monad继承Functor。

我想证明Some (f x') = fmap f (Some x')，这是monad法律之一。我使用了compute，reflexivity和destruct option_functor，但我无法证明这一点。我无法恰当地简化fmap。

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.

Answer 1

您的问题源于您使用option_function而不是Qed结束Defined的定义。

使用Qed，您以某种方式'隐藏'fmap的内部。然后，您无法再展开其定义（例如，使用unfold和simpl策略）。使用Defined而不是Qed让您告诉Coq您打算使用fmap后者的定义，因此它应该是可展开的。

在Coq中定义Maybe monad

1 个答案: