Is this relationship between forall and exists provable in Coq/intuitionistic logic?

Question

Is the following theorem provable in Coq? And if not, is there a way to prove it isn\'t provable?

Theorem not_for_all_is_exists:
    forall (X : Set) (P : X          


        

Answer 1

It's not provable, because it's equivalent to double negation elimination (and the other classical axioms).

My Coq skills are very rusty currently, but I can quickly illustrate why your theorem implies double negation elimination.

In your theorem, instantiate X to unit and P to fun _ => X for an arbitrary X : Prop. Now we have ~(unit -> ~ X) -> exists (u : unit), X. But because of the triviality of unit this is equivalent to ~ ~ X -> X.

The backwards implication can be proved with a straightforward application of double negation elimination on ~ ~ (exists x, P x).

My Agda is much better, so I can at least show the proofs there (don't know if that's helpful, but it might back up my claims a bit):

open import Relation.Nullary
open import Data.Product
open import Data.Unit
open import Data.Empty
open import Function

∀∃ : Set _
∀∃ = (A : Set)(P : A → Set) → ¬ (∀ x → ¬ P x) → ∃ P

Dneg : Set _
Dneg = (A : Set) → ¬ ¬ A → A

to : ∀∃ → Dneg
to ∀∃ A ¬¬A = proj₂ (∀∃ ⊤ (const A) (λ f → ¬¬A (f tt)))

fro : Dneg → ∀∃
fro dneg A P f = dneg (∃ P) (f ∘ curry)