(** To be equivalent is an equivalence relation *)
Section Equiv.
Context {T : Type} {K : AbsRing} {V : NormedModule K}.

Lemma equiv_carac_0 :
forall {F : (T -> Prop) -> Prop} {FF : Filter F} (f g : T -> V),
is_equiv F f g ->
{o : T -> V | (forall x : T, f x = plus (g x) (o x)) /\ is_domin F g o }.