Metamath Proof Explorer


Theorem fin23lem21

Description: Lemma for fin23 . X is not empty. We only need here that t has at least one set in its range besides (/) ; the much stronger hypothesis here will serve as our induction hypothesis though. (Contributed by Stefan O'Rear, 1-Nov-2014) (Revised by Mario Carneiro, 6-May-2015)

Ref Expression
Hypotheses fin23lem.a U = seq ω i ω , u V if t i u = u t i u ran t
fin23lem17.f F = g | a 𝒫 g ω x ω a suc x a x ran a ran a
Assertion fin23lem21 ran t F t : ω 1-1 V ran U

Proof

Step Hyp Ref Expression
1 fin23lem.a U = seq ω i ω , u V if t i u = u t i u ran t
2 fin23lem17.f F = g | a 𝒫 g ω x ω a suc x a x ran a ran a
3 1 2 fin23lem17 ran t F t : ω 1-1 V ran U ran U
4 1 fnseqom U Fn ω
5 fvelrnb U Fn ω ran U ran U a ω U a = ran U
6 4 5 ax-mp ran U ran U a ω U a = ran U
7 id a ω a ω
8 vex t V
9 f1f1orn t : ω 1-1 V t : ω 1-1 onto ran t
10 f1oen3g t V t : ω 1-1 onto ran t ω ran t
11 8 9 10 sylancr t : ω 1-1 V ω ran t
12 ominf ¬ ω Fin
13 ssdif0 ran t ran t =
14 snfi Fin
15 ssfi Fin ran t ran t Fin
16 14 15 mpan ran t ran t Fin
17 enfi ω ran t ω Fin ran t Fin
18 16 17 syl5ibr ω ran t ran t ω Fin
19 13 18 syl5bir ω ran t ran t = ω Fin
20 19 necon3bd ω ran t ¬ ω Fin ran t
21 11 12 20 mpisyl t : ω 1-1 V ran t
22 n0 ran t a a ran t
23 eldifsn a ran t a ran t a
24 elssuni a ran t a ran t
25 ssn0 a ran t a ran t
26 24 25 sylan a ran t a ran t
27 23 26 sylbi a ran t ran t
28 27 exlimiv a a ran t ran t
29 22 28 sylbi ran t ran t
30 21 29 syl t : ω 1-1 V ran t
31 1 fin23lem14 a ω ran t U a
32 7 30 31 syl2anr t : ω 1-1 V a ω U a
33 neeq1 U a = ran U U a ran U
34 32 33 syl5ibcom t : ω 1-1 V a ω U a = ran U ran U
35 34 rexlimdva t : ω 1-1 V a ω U a = ran U ran U
36 6 35 syl5bi t : ω 1-1 V ran U ran U ran U
37 36 adantl ran t F t : ω 1-1 V ran U ran U ran U
38 3 37 mpd ran t F t : ω 1-1 V ran U