Metamath Proof Explorer


Theorem brdom2g

Description: Dominance relation. This variation of brdomg does not require the Axiom of Union. (Contributed by BTernaryTau, 29-Nov-2024)

Ref Expression
Assertion brdom2g
|- ( ( A e. V /\ B e. W ) -> ( A ~<_ B <-> E. f f : A -1-1-> B ) )

Proof

Step Hyp Ref Expression
1 f1eq2
 |-  ( x = A -> ( f : x -1-1-> y <-> f : A -1-1-> y ) )
2 1 exbidv
 |-  ( x = A -> ( E. f f : x -1-1-> y <-> E. f f : A -1-1-> y ) )
3 f1eq3
 |-  ( y = B -> ( f : A -1-1-> y <-> f : A -1-1-> B ) )
4 3 exbidv
 |-  ( y = B -> ( E. f f : A -1-1-> y <-> E. f f : A -1-1-> B ) )
5 df-dom
 |-  ~<_ = { <. x , y >. | E. f f : x -1-1-> y }
6 2 4 5 brabg
 |-  ( ( A e. V /\ B e. W ) -> ( A ~<_ B <-> E. f f : A -1-1-> B ) )