Proof doubleton_uniqueness_Out.
Section second_has_geq_2_elements_proof.
Variable A : Ens.
Hypothesis k : has_leq_1_elements A.
Let sA := Singleton A.
Let E := Second A.
(* recall zero_has_leq_0_elements : (has_leq_0_elements Zero) *)
Let step1 : has_geq_1_elements sA := Singletons_have_geq_1_elements A.
Section second_geq_2_a.
Hypothesis m : EQ Zero sA.
Let stepa_1 : has_leq_0_elements sA :=
Big_Extensionality has_leq_0_elements Zero sA zero_has_leq_0_elements m.
Definition second_geq_2_a_Out : False := step1 stepa_1.
End second_geq_2_a.
Let step2 : NEQ Zero sA := second_geq_2_a_Out.
Definition second_has_geq_2_elements_Out : has_geq_2_elements E :=
distinct_Doubletons_have_geq_2_elements Zero sA step2.
End second_has_geq_2_elements_proof.

Theorem Second_has_geq_2_elements :
forall A : Ens, has_geq_2_elements (Second A).