Proof Singleton_th2.
Definition has_leq_1_elements (A : Ens) :=
forall (X Y : Ens) (k : IN X A) (l : IN Y A), EQ X Y.
Section has_leq_0_implies_leq_1.
Variable A : Ens.
Hypothesis q : has_leq_0_elements A.
Variable X : Ens.
Variable Y : Ens.
Hypothesis k : IN X A.
Let step1 : False.
Proof q X k.
Let step2 : EQ X Y := false_implies_everything (EQ X Y) step1.
Definition has_leq_0_implies_leq_1_Out (l : IN Y A) : EQ X Y := step2.
(**** we had to do that above because otherwise the sectioning mechanism would have ignored l which we don't actually use ***)
End has_leq_0_implies_leq_1.

Theorem has_leq_0_elements_implies_leq_1 :
forall (A : Ens) (q : has_leq_0_elements A), has_leq_1_elements A.