使用标准库在 Agda 中对/列表进行字典排序

Question

Agda 标准库包含一些模块Relation.Binary.*.(Non)StrictLex（目前仅适用于Product 和List）。我们可以使用这些模块轻松地构造一个实例，例如，IsStrictTotalOrder 用于自然数对（即ℕ × ℕ）。

open import Data.Nat as ℕ using (ℕ; _<_)
open import Data.Nat.Properties as ℕ
open import Relation.Binary using (module StrictTotalOrder; IsStrictTotalOrder)
open import Relation.Binary.PropositionalEquality using (_≡_)
open import Relation.Binary.Product.StrictLex using (×-Lex; _×-isStrictTotalOrder_)
open import Relation.Binary.Product.Pointwise using (_×-Rel_)

ℕ-isSTO : IsStrictTotalOrder _≡_ _<_
ℕ-isSTO = StrictTotalOrder.isStrictTotalOrder ℕ.strictTotalOrder

ℕ×ℕ-isSTO : IsStrictTotalOrder (_≡_ ×-Rel _≡_) (×-Lex _≡_ _<_ _<_)
ℕ×ℕ-isSTO = ℕ-isSTO ×-isStrictTotalOrder ℕ-isSTO

这将使用逐点相等_≡_ ×-Rel _≡_ 创建一个实例。在命题相等的情况下，这应该等同于使用 just 命题相等。

有没有一种简单的方法可以将上面的实例转换为IsStrictTotalOrder _≡_ (×-Lex _≡_ _<_ _<_) 类型的实例，使用正常的命题相等？

Answer 1

所需的套件并不难组装：

open import Data.Product
open import Function using (_∘_; case_of_)
open import Relation.Binary

_⇔₂_ : ∀ {a ℓ₁ ℓ₂} {A : Set a} → Rel A ℓ₁ → Rel A ℓ₂ → Set _
_≈_ ⇔₂ _≈′_ = (∀ {x y} → x ≈ y → x ≈′ y) × (∀ {x y} → x ≈′ y → x ≈ y)

-- I was unable to write this nicely using Data.Product.map... 
-- hence it is moved here to a toplevel where it can pattern-match
-- on the product of proofs
transform-resp : ∀ {a ℓ₁ ℓ₂ ℓ} {A : Set a} {≈ : Rel A ℓ₁} {≈′ : Rel A ℓ₂} {< : Rel A ℓ} →
                 ≈ ⇔₂ ≈′ →
                 < Respects₂ ≈ → < Respects₂ ≈′
transform-resp (to ,  from) = λ { (resp₁ , resp₂) → (resp₁ ∘ from , resp₂ ∘ from) }

transform-isSTO : ∀ {a ℓ₁ ℓ₂ ℓ} {A : Set a} {≈ : Rel A ℓ₁} {≈′ : Rel A ℓ₂} {< : Rel A ℓ} →
                  ≈ ⇔₂ ≈′ →
                  IsStrictTotalOrder ≈ < → IsStrictTotalOrder ≈′ <
transform-isSTO {≈′ = ≈′} {< = <} (to , from) isSTO = record
  { isEquivalence = let open IsEquivalence (IsStrictTotalOrder.isEquivalence isSTO)
                    in record { refl = to refl
                              ; sym = to ∘ sym ∘ from
                              ; trans = λ x y → to (trans (from x) (from y))
                              }
  ; trans = IsStrictTotalOrder.trans isSTO
  ; compare = compare
  ; <-resp-≈ = transform-resp (to , from) (IsStrictTotalOrder.<-resp-≈ isSTO)
  }
  where
    compare : Trichotomous ≈′ <
    compare x y with IsStrictTotalOrder.compare isSTO x y
    compare x y | tri< a ¬b ¬c = tri< a (¬b ∘ from) ¬c
    compare x y | tri≈ ¬a b ¬c = tri≈ ¬a (to b) ¬c
    compare x y | tri> ¬a ¬b c = tri> ¬a (¬b ∘ from) c

那么我们可以用这个来解决你原来的问题：

ℕ×ℕ-isSTO′ : IsStrictTotalOrder _≡_ (×-Lex _≡_ _<_ _<_)
ℕ×ℕ-isSTO′ = transform-isSTO (to , from) ℕ×ℕ-isSTO
  where
    open import Function using (_⟨_⟩_)
    open import Relation.Binary.PropositionalEquality

    to : ∀ {a b} {A : Set a} {B : Set b}
         {x x′ : A} {y y′ : B} → (x , y) ⟨ _≡_ ×-Rel _≡_ ⟩ (x′ , y′) → (x , y) ≡ (x′ , y′)
    to (refl , refl) = refl

    from : ∀ {a b} {A : Set a} {B : Set b}
           {x x′ : A} {y y′ : B} → (x , y) ≡ (x′ , y′) → (x , y) ⟨ _≡_ ×-Rel _≡_ ⟩ (x′ , y′)
    from refl = refl , refl