On 2011-08-09 16:34, Andrés Sicard-Ramírez wrote: > abstract > one : ℕ > one = suc 0 > > thm₁ : one ≡ suc 0 > thm₁ = refl > > thm₂ : one ≡ suc 0 > thm₂ = {!refl!} -- Doesn't type check > > I am stuck in prove that one ≡ suc 0 outside the abstract block. Is it > possible? Yes, using thm₁. -- /NAD