Agda中使用標准庫的對/列表的字典排序

Question

Agda標准庫包含一些模塊Relation.Binary.*.(Non)StrictLex （目前僅用於Product和List ）。 我們可以使用這些模塊輕松構造一個實例，例如IsStrictTotalOrder用於成對的自然數（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

這將使用逐點相等_≡_ ×-Rel _≡_創建一個實例。 在命題相等的情況下，這應該是等同於只使用命題平等。

是否有一種簡單的方法可以使用正常的命題相等將上面的實例轉換為IsStrictTotalOrder _≡_ (×-Lex _≡_ _<_ _<_)類型的實例？

Answer 1

所需的套件不太難組裝：

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

然后我們可以用它來解決你原來的問題：

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