Metamath Proof Explorer


Theorem brdomg

Description: Dominance relation. (Contributed by NM, 15-Jun-1998) (Proof shortened by BTernaryTau, 29-Nov-2024)

Ref Expression
Assertion brdomg B C A B f f : A 1-1 B

Proof

Step Hyp Ref Expression
1 brdom2g A V B C A B f f : A 1-1 B
2 1 ex A V B C A B f f : A 1-1 B
3 reldom Rel
4 3 brrelex1i A B A V
5 f1f f : A 1-1 B f : A B
6 fdm f : A B dom f = A
7 vex f V
8 7 dmex dom f V
9 6 8 eqeltrrdi f : A B A V
10 5 9 syl f : A 1-1 B A V
11 10 exlimiv f f : A 1-1 B A V
12 4 11 pm5.21ni ¬ A V A B f f : A 1-1 B
13 12 a1d ¬ A V B C A B f f : A 1-1 B
14 2 13 pm2.61i B C A B f f : A 1-1 B