问题
Suppose I have this definition of Subset in Agda
Subset : ∀ {α} → Set α → {ℓ : Level} → Set (α ⊔ suc ℓ)
Subset A {ℓ} = A → Set ℓ
and I have a set
data Q : Set where
a : Q
b : Q
Is it possible to prove that all subset of q is decidable and why?
Qs? : (qs : Subset Q {zero}) → Decidable qs
Decidable is defined here:
-- Membership
infix 10 _∈_
_∈_ : ∀ {α ℓ}{A : Set α} → A → Subset A → Set ℓ
a ∈ p = p a
-- Decidable
Decidable : ∀ {α ℓ}{A : Set α} → Subset A {ℓ} → Set (α ⊔ ℓ)
Decidable as = ∀ a → Dec (a ∈ as)
回答1:
Not for that definition of Subset, since decidability would require to check whether "p a" is inhabited or not, i.e. excluded middle.
Decidable subsets would exactly be maps into Bool:
Subset : ∀ {α} (A : Set α) -> Set
Subset A = A → Bool
_∈_ : ∀ {α}{A : Set α} → A → Subset A → Set
a ∈ p = T (p a)
But if you want more flexibility on the shape of the membership proofs you could use your definition of Subset and carry around a proof that it is Decidable.
来源：https://stackoverflow.com/questions/34183349/proving-decidability-of-subset-in-agda