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Prove the other direction for inverses.

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Signed-off-by: Danila Fedorin <danila.fedorin@gmail.com>
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Danila Fedorin 2024-02-25 13:57:45 -08:00
parent 99fc21cef2
commit b96bac5518
1 changed files with 26 additions and 0 deletions
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Lattice/FiniteValueMap.agda
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@ -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))