Metamath Proof Explorer


Theorem breng

Description: Equinumerosity relation. This variation of bren does not require the Axiom of Union. (Contributed by BTernaryTau, 23-Sep-2024)

Ref Expression
Assertion breng
|- ( ( A e. V /\ B e. W ) -> ( A ~~ B <-> E. f f : A -1-1-onto-> B ) )

Proof

Step Hyp Ref Expression
1 f1oeq2
 |-  ( x = A -> ( f : x -1-1-onto-> y <-> f : A -1-1-onto-> y ) )
2 1 exbidv
 |-  ( x = A -> ( E. f f : x -1-1-onto-> y <-> E. f f : A -1-1-onto-> y ) )
3 f1oeq3
 |-  ( y = B -> ( f : A -1-1-onto-> y <-> f : A -1-1-onto-> B ) )
4 3 exbidv
 |-  ( y = B -> ( E. f f : A -1-1-onto-> y <-> E. f f : A -1-1-onto-> B ) )
5 df-en
 |-  ~~ = { <. x , y >. | E. f f : x -1-1-onto-> y }
6 2 4 5 brabg
 |-  ( ( A e. V /\ B e. W ) -> ( A ~~ B <-> E. f f : A -1-1-onto-> B ) )