Before revision: of F IdIndsc → Refl Proof: For any F and any x, Fx ⇔ Fx; hence x = x by IdIndsc IndscId ⇸ Refl Proof: The empty relation satisfies IndscId, but not Refl.
After revision: of F IdIndsc → Refl Proof: For any F and any x, Fx ⇔ Fx; hence x = x by IdIndsc