Metamath Proof Explorer


Theorem sdomel

Description: For ordinals, strict dominance implies membership. (Contributed by Mario Carneiro, 13-Jan-2013)

Ref Expression
Assertion sdomel
|- ( ( A e. On /\ B e. On ) -> ( A ~< B -> A e. B ) )

Proof

Step Hyp Ref Expression
1 ssdomg
 |-  ( A e. On -> ( B C_ A -> B ~<_ A ) )
2 1 adantl
 |-  ( ( B e. On /\ A e. On ) -> ( B C_ A -> B ~<_ A ) )
3 ontri1
 |-  ( ( B e. On /\ A e. On ) -> ( B C_ A <-> -. A e. B ) )
4 domtriord
 |-  ( ( B e. On /\ A e. On ) -> ( B ~<_ A <-> -. A ~< B ) )
5 2 3 4 3imtr3d
 |-  ( ( B e. On /\ A e. On ) -> ( -. A e. B -> -. A ~< B ) )
6 5 con4d
 |-  ( ( B e. On /\ A e. On ) -> ( A ~< B -> A e. B ) )
7 6 ancoms
 |-  ( ( A e. On /\ B e. On ) -> ( A ~< B -> A e. B ) )