Can one create such function in Agda ?

ℕ→ℕ-undecidable : ¬ ( (f g :  ℕ → ℕ ) → Dec (f ≡ g)) 
ℕ→ℕ-undecidable = ?

I am particularly interested in proof using cubical Agda.


1 Answer 1


ℕ→ℕ-undecidable is not provable in Agda. If we postulate the law of excluded middle (LEM), it follows that equality on every set is decidable, contradicting ℕ→ℕ-undecidable. Since Agda is consistent with LEM, it follows that ℕ→ℕ-undecidable is not provable in base Agda. This holds the same for cubical and vanilla Agda.

  • $\begingroup$ What about coq? Is it possible to prove it there? $\endgroup$
    – srghma
    Sep 5, 2019 at 9:11
  • 1
    $\begingroup$ @srghma no, the same thing holds for Coq. Consistency with LEM (classical logic) is a common expected feature in proof assistants. $\endgroup$ Sep 5, 2019 at 17:47

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