Metamath Proof Explorer


Theorem elpssex

Description: A class has an element just in case it has a proper subset. (Contributed by Adrian Ducourtial, 26-Oct-2026)

Ref Expression
Assertion elpssex
|- ( E. x x e. A <-> E. y y C. A )

Proof

Step Hyp Ref Expression
1 0ss
 |-  (/) C_ A
2 1 a1i
 |-  ( x e. A -> (/) C_ A )
3 ne0i
 |-  ( x e. A -> A =/= (/) )
4 3 necomd
 |-  ( x e. A -> (/) =/= A )
5 df-pss
 |-  ( (/) C. A <-> ( (/) C_ A /\ (/) =/= A ) )
6 2 4 5 sylanbrc
 |-  ( x e. A -> (/) C. A )
7 0ex
 |-  (/) e. _V
8 psseq1
 |-  ( y = (/) -> ( y C. A <-> (/) C. A ) )
9 7 8 spcev
 |-  ( (/) C. A -> E. y y C. A )
10 6 9 syl
 |-  ( x e. A -> E. y y C. A )
11 10 exlimiv
 |-  ( E. x x e. A -> E. y y C. A )
12 pssnel
 |-  ( y C. A -> E. x ( x e. A /\ -. x e. y ) )
13 exsimpl
 |-  ( E. x ( x e. A /\ -. x e. y ) -> E. x x e. A )
14 12 13 syl
 |-  ( y C. A -> E. x x e. A )
15 14 exlimiv
 |-  ( E. y y C. A -> E. x x e. A )
16 11 15 impbii
 |-  ( E. x x e. A <-> E. y y C. A )