Proving decidability of subset in Agda

Question

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)

This subset definition doesn't allow that. You can just have `const t : Subset Q` for some `t` that is not decidable. — András Kovács, Dec 09 '15 at 17:39

score 1 · Accepted Answer · answered Dec 10 '15 at 10:02

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.

Proving decidability of subset in Agda

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