(** - Once we know that some subset of $V$ has an element, we must be able to get it
 if we want to prove theorem 1.11 - hence this axiom. It is a kind of choice principle,
 hence the name I gave it. I know AC+EM+impredicativity = inconsistency, but type isn't
 impredicative so we should be OK *)
Axiom
  AC' :
    forall S : part_set V,
    (exists v : V, in_part v S) -> sigT (fun v : V => in_part v S).

Lemma ACcomplement : forall S : part_set V, (exists v : V, ~ in_part v S) -> sigT (fun v : V => ~ in_part v S).