diff --git a/Lattice/FiniteValueMap.agda b/Lattice/FiniteValueMap.agda index 7eed51b..5f43e3c 100644 --- a/Lattice/FiniteValueMap.agda +++ b/Lattice/FiniteValueMap.agda @@ -20,8 +20,10 @@ open import Utils using (Unique; push; empty) open import Data.Product using (_,_) open import Data.List.Properties using (∷-injectiveʳ) open import Data.List.Relation.Unary.All using (All) +open import Data.List.Relation.Unary.Any using (Any; here; there) open import Relation.Nullary using (¬_) +open import Lattice.Map A B _≈₂_ _⊔₂_ _⊓₂_ ≈-dec-A lB using (subset-impl) open import Lattice.FiniteMap A B _≈₂_ _⊔₂_ _⊓₂_ ≈-dec-A lB public module IterProdIsomorphism where @@ -65,3 +67,27 @@ module IterProdIsomorphism where -- but we end up with the 'unpacked' form (kvs', ...). So, put it back -- in the 'packed' form after we've performed enough inspection -- to know we take the cons branch of `to`. + + -- The map has its own uniqueness proof, but the call to 'to' needs a standalone + -- uniqueness proof too. Work with both proofs as needed to thread things through. + from-to-inverseʳ : ∀ {ks : List A} (uks : Unique ks) → + Inverseʳ (_≈ᵐ_ {ks}) (_≈ⁱᵖ_ {ks}) (from {ks}) (to {ks} uks) -- to (from x) = x + from-to-inverseʳ {[]} _ (([] , empty) , kvs≡ks) rewrite kvs≡ks = ((λ k v ()) , (λ k v ())) + from-to-inverseʳ {k ∷ ks'} uks@(push k≢ks'₁ uks'₁) fm₁@(m₁@((k , v) ∷ kvs'₁ , push k≢ks'₂ uks'₂) , refl) + with to uks'₁ (from ((kvs'₁ , uks'₂) , refl)) | from-to-inverseʳ {ks'} uks'₁ ((kvs'₁ , uks'₂) , refl) + ... | ((kvs'₂ , ukvs'₂) , _) | (kvs'₂⊆kvs'₁ , kvs'₁⊆kvs'₂) = (m₂⊆m₁ , m₁⊆m₂) + where + kvs₁ = (k , v) ∷ kvs'₁ + kvs₂ = (k , v) ∷ kvs'₂ + + m₁⊆m₂ : subset-impl kvs₁ kvs₂ + m₁⊆m₂ k' v' (here refl) = (v' , (IsLattice.≈-refl lB , here refl)) + m₁⊆m₂ k' v' (there k',v'∈kvs'₁) = + let (v'' , (v'≈v'' , k',v''∈kvs'₂)) = kvs'₁⊆kvs'₂ k' v' k',v'∈kvs'₁ + in (v'' , (v'≈v'' , there k',v''∈kvs'₂)) + + m₂⊆m₁ : subset-impl kvs₂ kvs₁ + m₂⊆m₁ k' v' (here refl) = (v' , (IsLattice.≈-refl lB , here refl)) + m₂⊆m₁ k' v' (there k',v'∈kvs'₂) = + let (v'' , (v'≈v'' , k',v''∈kvs'₁)) = kvs'₂⊆kvs'₁ k' v' k',v'∈kvs'₂ + in (v'' , (v'≈v'' , there k',v''∈kvs'₁))