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 imbitrrid ⊢ ω ≈ ran ⁡ t → ran ⁡ t ⊆ ∅ → ω ∈ Fin
19 13 18 biimtrrid ⊢ 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 biimtrid ⊢ 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 ≠ ∅