agda-spa/Lattice/FiniteValueMap.agda

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-- Because iterated products currently require both A and B to be of the same
-- universe, and the FiniteMap is written in a universe-polymorphic way,
-- specialize the FiniteMap module with Set-typed types only.
open import Lattice
open import Equivalence
open import Relation.Binary.PropositionalEquality as Eq using (_≡_; refl; sym; trans; cong; subst)
open import Relation.Binary.Definitions using (Decidable)
open import Agda.Primitive using (Level) renaming (_⊔_ to _⊔_)
open import Function.Definitions using (Inverseˡ; Inverseʳ)
module Lattice.FiniteValueMap (A : Set) (B : Set)
(_≈₂_ : B B Set)
(_⊔₂_ : B B B) (_⊓₂_ : B B B)
(≈-dec-A : Decidable (_≡_ {_} {A}))
(lB : IsLattice B _≈₂_ _⊔₂_ _⊓₂_) where
open import Data.List using (List; length; []; _∷_)
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
open import Data.Unit using (; tt)
open import Lattice.Unit using () renaming (_≈_ to _≈ᵘ_; _⊔_ to _⊔ᵘ_; _⊓_ to _⊓ᵘ_; ≈-dec to ≈ᵘ-dec; isLattice to isLatticeᵘ)
open import Lattice.IterProd _≈₂_ _≈ᵘ_ _⊔₂_ _⊔ᵘ_ _⊓₂_ _⊓ᵘ_ lB isLatticeᵘ as IP using (IterProd)
from : {ks : List A} FiniteMap ks IterProd (length ks)
from {[]} (([] , _) , _) = tt
from {k ks'} (((k' , v) kvs' , push _ uks') , refl) = (v , from ((kvs' , uks'), refl))
to : {ks : List A} Unique ks IterProd (length ks) FiniteMap ks
to {[]} _ = (([] , empty) , refl)
to {k ks'} (push k≢ks' uks') (v , rest)
with to uks' rest
... | ((kvs' , ukvs') , kvs'≡ks') =
let
-- This would be easier if we pattern matched on the equiality proof
-- to get refl, but that makes it harder to reason about 'to' when
-- the arguments are not known to be refl.
k≢kvs' = subst (λ ks All (λ k' ¬ k k') ks) (sym kvs'≡ks') k≢ks'
kvs≡ks = cong (k ∷_) kvs'≡ks'
in
(((k , v) kvs' , push k≢kvs' ukvs') , kvs≡ks)
private
_≈ᵐ_ : {ks : List A} FiniteMap ks FiniteMap ks Set
_≈ᵐ_ {ks} = _≈_ ks
_≈ⁱᵖ_ : {ks : List A} IterProd (length ks) IterProd (length ks) Set
_≈ⁱᵖ_ {ks} = IP._≈_ (length ks)
from-to-inverseˡ : {ks : List A} (uks : Unique ks)
Inverseˡ (_≈ᵐ_ {ks}) (_≈ⁱᵖ_ {ks}) (from {ks}) (to {ks} uks) -- from (to x) = x
from-to-inverseˡ {[]} _ _ = IsEquivalence.≈-refl (IP.≈-equiv 0)
from-to-inverseˡ {k ks'} (push k≢ks' uks') (v , rest)
with ((kvs' , ukvs') , refl) to uks' rest in p rewrite sym p =
(IsLattice.≈-refl lB , from-to-inverseˡ {ks'} uks' rest)
-- the rewrite here is needed because the IH is in terms of `to uks' rest`,
-- 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))