Metamath Proof Explorer

Theorem sdomne0

Description: A class that strictly dominates any set is not empty. (Suggested by SN, 14-Jan-2025.) (Contributed by RP, 14-Jan-2025)

Ref Expression
Assertion sdomne0 
|- ( B ~< A -> A =/= (/) )

Proof

Step Hyp Ref Expression
1 relsdom 
 |-  Rel ~<
2 1 brrelex1i 
 |-  ( B ~< A -> B e. _V )
3 breq1 
 |-  ( B = (/) -> ( B ~< A <-> (/) ~< A ) )
4 3 biimpd 
 |-  ( B = (/) -> ( B ~< A -> (/) ~< A ) )
5 4 a1i 
 |-  ( B e. _V -> ( B = (/) -> ( B ~< A -> (/) ~< A ) ) )
6 0sdomg 
 |-  ( B e. _V -> ( (/) ~< B <-> B =/= (/) ) )
7 sdomtr 
 |-  ( ( (/) ~< B /\ B ~< A ) -> (/) ~< A )
8 7 ex 
 |-  ( (/) ~< B -> ( B ~< A -> (/) ~< A ) )
9 6 8 syl6bir 
 |-  ( B e. _V -> ( B =/= (/) -> ( B ~< A -> (/) ~< A ) ) )
10 5 9 pm2.61dne 
 |-  ( B e. _V -> ( B ~< A -> (/) ~< A ) )
11 2 10 syl 
 |-  ( B ~< A -> ( B ~< A -> (/) ~< A ) )
12 1 brrelex2i 
 |-  ( (/) ~< A -> A e. _V )
13 0sdomg 
 |-  ( A e. _V -> ( (/) ~< A <-> A =/= (/) ) )
14 12 13 syl 
 |-  ( (/) ~< A -> ( (/) ~< A <-> A =/= (/) ) )
15 14 ibi 
 |-  ( (/) ~< A -> A =/= (/) )
16 11 15 syl6 
 |-  ( B ~< A -> ( B ~< A -> A =/= (/) ) )
17 16 pm2.43i 
 |-  ( B ~< A -> A =/= (/) )