Metamath Proof Explorer


Theorem fodomfib

Description: Equivalence of an onto mapping and dominance for a nonempty finite set. Unlike fodomb for arbitrary sets, this theorem does not require the Axiom of Choice for its proof. (Contributed by NM, 23-Mar-2006)

Ref Expression
Assertion fodomfib A Fin A f f : A onto B B B A

Proof

Step Hyp Ref Expression
1 fof f : A onto B f : A B
2 1 fdmd f : A onto B dom f = A
3 2 eqeq1d f : A onto B dom f = A =
4 dm0rn0 dom f = ran f =
5 forn f : A onto B ran f = B
6 5 eqeq1d f : A onto B ran f = B =
7 4 6 syl5bb f : A onto B dom f = B =
8 3 7 bitr3d f : A onto B A = B =
9 8 necon3bid f : A onto B A B
10 9 biimpac A f : A onto B B
11 10 adantll A Fin A f : A onto B B
12 vex f V
13 12 rnex ran f V
14 5 13 syl6eqelr f : A onto B B V
15 14 adantl A Fin f : A onto B B V
16 0sdomg B V B B
17 15 16 syl A Fin f : A onto B B B
18 17 adantlr A Fin A f : A onto B B B
19 11 18 mpbird A Fin A f : A onto B B
20 19 ex A Fin A f : A onto B B
21 fodomfi A Fin f : A onto B B A
22 21 ex A Fin f : A onto B B A
23 22 adantr A Fin A f : A onto B B A
24 20 23 jcad A Fin A f : A onto B B B A
25 24 exlimdv A Fin A f f : A onto B B B A
26 25 expimpd A Fin A f f : A onto B B B A
27 sdomdomtr B B A A
28 0sdomg A Fin A A
29 27 28 syl5ib A Fin B B A A
30 fodomr B B A f f : A onto B
31 29 30 jca2 A Fin B B A A f f : A onto B
32 26 31 impbid A Fin A f f : A onto B B B A