Metamath Proof Explorer


Theorem domalom

Description: A class which dominates every natural number is not finite. (Contributed by ML, 14-Dec-2020)

Ref Expression
Assertion domalom ( ∀ 𝑛 ∈ ω 𝑛𝐴 → ¬ 𝐴 ∈ Fin )

Proof

Step Hyp Ref Expression
1 nfra1 𝑛𝑛 ∈ ω 𝑛𝐴
2 breq1 ( 𝑦 = 𝑛 → ( 𝑦𝐴𝑛𝐴 ) )
3 2 imbi2d ( 𝑦 = 𝑛 → ( ( ∀ 𝑛 ∈ ω 𝑛𝐴𝑦𝐴 ) ↔ ( ∀ 𝑛 ∈ ω 𝑛𝐴𝑛𝐴 ) ) )
4 breq1 ( 𝑦 = ∅ → ( 𝑦𝐴 ↔ ∅ ≺ 𝐴 ) )
5 breq1 ( 𝑦 = 𝑧 → ( 𝑦𝐴𝑧𝐴 ) )
6 breq1 ( 𝑦 = suc 𝑧 → ( 𝑦𝐴 ↔ suc 𝑧𝐴 ) )
7 1n0 1o ≠ ∅
8 1onn 1o ∈ ω
9 0sdomg ( 1o ∈ ω → ( ∅ ≺ 1o ↔ 1o ≠ ∅ ) )
10 8 9 ax-mp ( ∅ ≺ 1o ↔ 1o ≠ ∅ )
11 7 10 mpbir ∅ ≺ 1o
12 breq1 ( 𝑛 = 1o → ( 𝑛𝐴 ↔ 1o𝐴 ) )
13 12 rspccv ( ∀ 𝑛 ∈ ω 𝑛𝐴 → ( 1o ∈ ω → 1o𝐴 ) )
14 8 13 mpi ( ∀ 𝑛 ∈ ω 𝑛𝐴 → 1o𝐴 )
15 sdomdomtr ( ( ∅ ≺ 1o ∧ 1o𝐴 ) → ∅ ≺ 𝐴 )
16 11 14 15 sylancr ( ∀ 𝑛 ∈ ω 𝑛𝐴 → ∅ ≺ 𝐴 )
17 peano2 ( 𝑧 ∈ ω → suc 𝑧 ∈ ω )
18 php4 ( suc 𝑧 ∈ ω → suc 𝑧 ≺ suc suc 𝑧 )
19 17 18 syl ( 𝑧 ∈ ω → suc 𝑧 ≺ suc suc 𝑧 )
20 breq1 ( 𝑛 = suc suc 𝑧 → ( 𝑛𝐴 ↔ suc suc 𝑧𝐴 ) )
21 20 rspccv ( ∀ 𝑛 ∈ ω 𝑛𝐴 → ( suc suc 𝑧 ∈ ω → suc suc 𝑧𝐴 ) )
22 peano2 ( suc 𝑧 ∈ ω → suc suc 𝑧 ∈ ω )
23 17 22 syl ( 𝑧 ∈ ω → suc suc 𝑧 ∈ ω )
24 21 23 impel ( ( ∀ 𝑛 ∈ ω 𝑛𝐴𝑧 ∈ ω ) → suc suc 𝑧𝐴 )
25 sdomdomtr ( ( suc 𝑧 ≺ suc suc 𝑧 ∧ suc suc 𝑧𝐴 ) → suc 𝑧𝐴 )
26 19 24 25 syl2an2 ( ( ∀ 𝑛 ∈ ω 𝑛𝐴𝑧 ∈ ω ) → suc 𝑧𝐴 )
27 26 a1d ( ( ∀ 𝑛 ∈ ω 𝑛𝐴𝑧 ∈ ω ) → ( 𝑧𝐴 → suc 𝑧𝐴 ) )
28 27 expcom ( 𝑧 ∈ ω → ( ∀ 𝑛 ∈ ω 𝑛𝐴 → ( 𝑧𝐴 → suc 𝑧𝐴 ) ) )
29 4 5 6 16 28 finds2 ( 𝑦 ∈ ω → ( ∀ 𝑛 ∈ ω 𝑛𝐴𝑦𝐴 ) )
30 3 29 vtoclga ( 𝑛 ∈ ω → ( ∀ 𝑛 ∈ ω 𝑛𝐴𝑛𝐴 ) )
31 30 com12 ( ∀ 𝑛 ∈ ω 𝑛𝐴 → ( 𝑛 ∈ ω → 𝑛𝐴 ) )
32 1 31 ralrimi ( ∀ 𝑛 ∈ ω 𝑛𝐴 → ∀ 𝑛 ∈ ω 𝑛𝐴 )
33 sdomnen ( 𝑛𝐴 → ¬ 𝑛𝐴 )
34 ensym ( 𝐴𝑛𝑛𝐴 )
35 33 34 nsyl ( 𝑛𝐴 → ¬ 𝐴𝑛 )
36 35 ralimi ( ∀ 𝑛 ∈ ω 𝑛𝐴 → ∀ 𝑛 ∈ ω ¬ 𝐴𝑛 )
37 32 36 syl ( ∀ 𝑛 ∈ ω 𝑛𝐴 → ∀ 𝑛 ∈ ω ¬ 𝐴𝑛 )
38 isfi ( 𝐴 ∈ Fin ↔ ∃ 𝑛 ∈ ω 𝐴𝑛 )
39 38 notbii ( ¬ 𝐴 ∈ Fin ↔ ¬ ∃ 𝑛 ∈ ω 𝐴𝑛 )
40 ralnex ( ∀ 𝑛 ∈ ω ¬ 𝐴𝑛 ↔ ¬ ∃ 𝑛 ∈ ω 𝐴𝑛 )
41 39 40 bitr4i ( ¬ 𝐴 ∈ Fin ↔ ∀ 𝑛 ∈ ω ¬ 𝐴𝑛 )
42 37 41 sylibr ( ∀ 𝑛 ∈ ω 𝑛𝐴 → ¬ 𝐴 ∈ Fin )