Prove that restrict needs the key in both maps
Signed-off-by: Danila Fedorin <danila.fedorin@gmail.com>
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Map.agda
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Map.agda
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@ -329,6 +329,22 @@ private module ImplInsert (f : B → B → B) where
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Unique (keys l₂) → Unique (keys (intersect l₁ l₂))
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Unique (keys l₂) → Unique (keys (intersect l₁ l₂))
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intersect-preserves-Unique {l₁} u = restrict-preserves-Unique (updates-preserve-Unique {l₁} u)
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intersect-preserves-Unique {l₁} u = restrict-preserves-Unique (updates-preserve-Unique {l₁} u)
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restrict-needs-both : ∀ {k : A} {v : B} {l₁ l₂ : List (A × B)} →
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(k , v) ∈ restrict l₁ l₂ → (k ∈k l₁ × (k , v) ∈ l₂)
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restrict-needs-both {k} {v} {l₁} {[]} ()
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restrict-needs-both {k} {v} {l₁} {(k' , v') ∷ xs} k,v∈l₁l₂
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with ∈k-dec k' l₁ | k,v∈l₁l₂
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... | yes k'∈kl₁ | here k,v≡k',v'
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rewrite cong proj₁ k,v≡k',v' rewrite cong proj₂ k,v≡k',v' =
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(k'∈kl₁ , here refl)
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... | yes _ | there k,v∈l₁xs =
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let (k∈kl₁ , k,v∈xs) = restrict-needs-both k,v∈l₁xs
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in (k∈kl₁ , there k,v∈xs)
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... | no k'∉kl₁ | k,v∈l₁xs =
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let (k∈kl₁ , k,v∈xs) = restrict-needs-both k,v∈l₁xs
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in (k∈kl₁ , there k,v∈xs)
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Map : Set (a ⊔ b)
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Map : Set (a ⊔ b)
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Map = Σ (List (A × B)) (λ l → Unique (keys l))
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Map = Σ (List (A × B)) (λ l → Unique (keys l))
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