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