Proof second_has_geq_2_elements_Out.
Section second_uniqueness_proof.
Variable A : Ens.
Variable B : Ens.
Hypothesis k : EQ (Second A) (Second B).
(*** recall Second A = Doubleton Zero (Singleton A)  ***)
Let sA := Singleton A.
Let sB := Singleton B.
Let step1 : has_geq_1_elements sA := Singletons_have_geq_1_elements A.
Let step2 : has_geq_1_elements sB := Singletons_have_geq_1_elements B.
Let step3 : pairwise_EQ Zero sA Zero sB :=
Doubleton_uniqueness has_leq_0_elements Zero sA Zero sB
zero_has_leq_0_elements step1 zero_has_leq_0_elements step2 k.
Definition second_uniqueness_Out : EQ A B :=
Singleton_uniqueness A B (pairwise_EQ_bd Zero sA Zero sB step3).
End second_uniqueness_proof.

Theorem Second_uniqueness :
forall (A B : Ens) (k : EQ (Second A) (Second B)), EQ A B.