Metamath Proof Explorer


Theorem sucdom

Description: Strict dominance of a set over a natural number is the same as dominance over its successor. (Contributed by Mario Carneiro, 12-Jan-2013) Avoid ax-pow . (Revised by BTernaryTau, 4-Dec-2024) (Proof shortened by BJ, 11-Jan-2025)

Ref Expression
Assertion sucdom
|- ( A e. _om -> ( A ~< B <-> suc A ~<_ B ) )

Proof

Step Hyp Ref Expression
1 sucdom2
 |-  ( A ~< B -> suc A ~<_ B )
2 nnfi
 |-  ( A e. _om -> A e. Fin )
3 php4
 |-  ( A e. _om -> A ~< suc A )
4 sdomdomtrfi
 |-  ( ( A e. Fin /\ A ~< suc A /\ suc A ~<_ B ) -> A ~< B )
5 4 3expia
 |-  ( ( A e. Fin /\ A ~< suc A ) -> ( suc A ~<_ B -> A ~< B ) )
6 2 3 5 syl2anc
 |-  ( A e. _om -> ( suc A ~<_ B -> A ~< B ) )
7 1 6 impbid2
 |-  ( A e. _om -> ( A ~< B <-> suc A ~<_ B ) )