Prove the other direction for inverses.

Signed-off-by: Danila Fedorin <danila.fedorin@gmail.com>

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))