为什么函数组合和应用程序在Agda中有依赖的实现?

为什么要进行函数合成( ∘ $ )使实现在中可用 https://github.com/agda/agda-stdlib/blob/master/src/Function.agda#L74-L76

_∘_ : ∀ {a b c}
        {A : Set a} {B : A → Set b} {C : {x : A} → B x → Set c} →
        (∀ {x} (y : B x) → C y) → (g : (x : A) → B x) →
        ((x : A) → C (g x))
f ∘ g = λ x → f (g x)

_∘'_ : ∀ {a b c} {A : Set a} {B : Set b} {C : Set c} →
         (B → C) → (A → B) → (A → C)
f ∘' g = λ x → f (g x)

_$_ : ∀ {a b} {A : Set a} {B : A → Set b} →
      ((x : A) → B x) → ((x : A) → B x)
f $ x = f x

_$'_ : ∀ {a b} {A : Set a} {B : Set b} →
       (A → B) → (A → B)
f $' x = f x

我最初认为这背后的理由是 $ 能够处理 $' :: ,其中B依赖于A。但经过大量测试,我无法给出一个示例来说明 $' $ $' 不能处理吗?(类似地,什么样的情况会发生 ∘ 处理好 ∘'

open import Agda.Builtin.Nat public
open import Agda.Primitive public

--data List {a} (A : Set a) : Set a where
--  []  : List A
--  _∷_ : (x : A) (xs : List A) → List A

data Vec {a} (A : Set a) : Nat → Set a where
  []  : Vec A zero
  _∷_ : ∀ {n} (x : A) (xs : Vec A n) → Vec A (suc n)

tail : ∀ {a n} {A : Set a} → Vec A (suc n) → Vec A n
tail (x ∷ s) = s

_$_ : ∀ {a b} {A : Set a} {B : A → Set b} →
      ((x : A) → B x) → ((x : A) → B x)
f $ x = f x

_$'_ : ∀ {a b} {A : Set a} {B : Set b} →
       (A → B) → (A → B)
f $' x = f x

_∘_ : ∀ {a b c}
        {A : Set a} {B : A → Set b} {C : {x : A} → B x → Set c} →
        (∀ {x} (y : B x) → C y) → (g : (x : A) → B x) →
        ((x : A) → C (g x))
f ∘ g = λ x → f (g x)

_∘'_ : ∀ {a b c} {A : Set a} {B : Set b} {C : Set c} →
         (B → C) → (A → B) → (A → C)
f ∘' g = λ x → f (g x)

Vecc : ∀ {a} → Nat → (A : Set a) → (Set a)
Vecc x y = Vec y x

data Pair {a b} (A : Set a) (B : A → Set b) : Set (a ⊔ b) where
  _,_ : (x : A) → (y : B x) → Pair A B

-- Dependent Pair attempt
--fst : ∀ {a b} {A : Set a} {B : A → Set b} → Pair A B → A
--fst (a , b) = a
--
--f : Pair Nat $' Vec Nat
--f = _,_ zero $' []
--
--g : Pair (Pair Nat $' Vec Nat) $' λ x → Nat
--g = _,_ (_,_ zero $' []) $' zero

-- Some other attempt
--f : ∀ {a n} {A : Set a} → Vec A ((suc ∘' suc) n) → Vec A n
--f {a} = tail {a} ∘' tail {a}

-- Vec attempt
--f : ∀ {a} (A : Set a) → (Set a)
--f {a} = Vecc {a} (suc zero) ∘' Vecc {a} (suc zero)
--
--h = f Nat
--
--x : h
--x = (zero ∷ []) ∷ []

-- List attempt
--f : ∀ {a} (A : Set a) → (Set a)
--f {a} = List {a} ∘' List {a}
--
--g : ∀ {a} (A : Set a) → (Set a)
--g {a} = List {a} ∘ List {a}
--
--h = f Nat
--i = g Nat
--
--x : h
--x = (zero ∷ []) ∷ []