Metamath Proof Explorer


Theorem ackbij1lem14

Description: Lemma for ackbij1 . (Contributed by Stefan O'Rear, 18-Nov-2014)

Ref Expression
Hypothesis ackbij.f ⊢ F = x ∈ 𝒫 ω ∩ Fin ⟼ card ⁡ ⋃ y ∈ x y × 𝒫 y
Assertion ackbij1lem14 ⊢ A ∈ ω → F ⁡ A = suc ⁡ F ⁡ A

Proof

Step Hyp Ref Expression
1 ackbij.f ⊢ F = x ∈ 𝒫 ω ∩ Fin ⟼ card ⁡ ⋃ y ∈ x y × 𝒫 y
2 1 ackbij1lem8 ⊢ A ∈ ω → F ⁡ A = card ⁡ 𝒫 A
3 pweq ⊢ a = ∅ → 𝒫 a = 𝒫 ∅
4 3 fveq2d ⊢ a = ∅ → card ⁡ 𝒫 a = card ⁡ 𝒫 ∅
5 fveq2 ⊢ a = ∅ → F ⁡ a = F ⁡ ∅
6 suceq ⊢ F ⁡ a = F ⁡ ∅ → suc ⁡ F ⁡ a = suc ⁡ F ⁡ ∅
7 5 6 syl ⊢ a = ∅ → suc ⁡ F ⁡ a = suc ⁡ F ⁡ ∅
8 4 7 eqeq12d ⊢ a = ∅ → card ⁡ 𝒫 a = suc ⁡ F ⁡ a ↔ card ⁡ 𝒫 ∅ = suc ⁡ F ⁡ ∅
9 pweq ⊢ a = b → 𝒫 a = 𝒫 b
10 9 fveq2d ⊢ a = b → card ⁡ 𝒫 a = card ⁡ 𝒫 b
11 fveq2 ⊢ a = b → F ⁡ a = F ⁡ b
12 suceq ⊢ F ⁡ a = F ⁡ b → suc ⁡ F ⁡ a = suc ⁡ F ⁡ b
13 11 12 syl ⊢ a = b → suc ⁡ F ⁡ a = suc ⁡ F ⁡ b
14 10 13 eqeq12d ⊢ a = b → card ⁡ 𝒫 a = suc ⁡ F ⁡ a ↔ card ⁡ 𝒫 b = suc ⁡ F ⁡ b
15 pweq ⊢ a = suc ⁡ b → 𝒫 a = 𝒫 suc ⁡ b
16 15 fveq2d ⊢ a = suc ⁡ b → card ⁡ 𝒫 a = card ⁡ 𝒫 suc ⁡ b
17 fveq2 ⊢ a = suc ⁡ b → F ⁡ a = F ⁡ suc ⁡ b
18 suceq ⊢ F ⁡ a = F ⁡ suc ⁡ b → suc ⁡ F ⁡ a = suc ⁡ F ⁡ suc ⁡ b
19 17 18 syl ⊢ a = suc ⁡ b → suc ⁡ F ⁡ a = suc ⁡ F ⁡ suc ⁡ b
20 16 19 eqeq12d ⊢ a = suc ⁡ b → card ⁡ 𝒫 a = suc ⁡ F ⁡ a ↔ card ⁡ 𝒫 suc ⁡ b = suc ⁡ F ⁡ suc ⁡ b
21 pweq ⊢ a = A → 𝒫 a = 𝒫 A
22 21 fveq2d ⊢ a = A → card ⁡ 𝒫 a = card ⁡ 𝒫 A
23 fveq2 ⊢ a = A → F ⁡ a = F ⁡ A
24 suceq ⊢ F ⁡ a = F ⁡ A → suc ⁡ F ⁡ a = suc ⁡ F ⁡ A
25 23 24 syl ⊢ a = A → suc ⁡ F ⁡ a = suc ⁡ F ⁡ A
26 22 25 eqeq12d ⊢ a = A → card ⁡ 𝒫 a = suc ⁡ F ⁡ a ↔ card ⁡ 𝒫 A = suc ⁡ F ⁡ A
27 df-1o ⊢ 1 𝑜 = suc ⁡ ∅
28 pw0 ⊢ 𝒫 ∅ = ∅
29 28 fveq2i ⊢ card ⁡ 𝒫 ∅ = card ⁡ ∅
30 0ex ⊢ ∅ ∈ V
31 cardsn ⊢ ∅ ∈ V → card ⁡ ∅ = 1 𝑜
32 30 31 ax-mp ⊢ card ⁡ ∅ = 1 𝑜
33 29 32 eqtri ⊢ card ⁡ 𝒫 ∅ = 1 𝑜
34 1 ackbij1lem13 ⊢ F ⁡ ∅ = ∅
35 suceq ⊢ F ⁡ ∅ = ∅ → suc ⁡ F ⁡ ∅ = suc ⁡ ∅
36 34 35 ax-mp ⊢ suc ⁡ F ⁡ ∅ = suc ⁡ ∅
37 27 33 36 3eqtr4i ⊢ card ⁡ 𝒫 ∅ = suc ⁡ F ⁡ ∅
38 oveq2 ⊢ card ⁡ 𝒫 b = suc ⁡ F ⁡ b → card ⁡ 𝒫 b + 𝑜 card ⁡ 𝒫 b = card ⁡ 𝒫 b + 𝑜 suc ⁡ F ⁡ b
39 38 adantl ⊢ b ∈ ω ∧ card ⁡ 𝒫 b = suc ⁡ F ⁡ b → card ⁡ 𝒫 b + 𝑜 card ⁡ 𝒫 b = card ⁡ 𝒫 b + 𝑜 suc ⁡ F ⁡ b
40 ackbij1lem5 ⊢ b ∈ ω → card ⁡ 𝒫 suc ⁡ b = card ⁡ 𝒫 b + 𝑜 card ⁡ 𝒫 b
41 40 adantr ⊢ b ∈ ω ∧ card ⁡ 𝒫 b = suc ⁡ F ⁡ b → card ⁡ 𝒫 suc ⁡ b = card ⁡ 𝒫 b + 𝑜 card ⁡ 𝒫 b
42 df-suc ⊢ suc ⁡ b = b ∪ b
43 42 equncomi ⊢ suc ⁡ b = b ∪ b
44 43 fveq2i ⊢ F ⁡ suc ⁡ b = F ⁡ b ∪ b
45 ackbij1lem4 ⊢ b ∈ ω → b ∈ 𝒫 ω ∩ Fin
46 45 adantr ⊢ b ∈ ω ∧ card ⁡ 𝒫 b = suc ⁡ F ⁡ b → b ∈ 𝒫 ω ∩ Fin
47 ackbij1lem3 ⊢ b ∈ ω → b ∈ 𝒫 ω ∩ Fin
48 47 adantr ⊢ b ∈ ω ∧ card ⁡ 𝒫 b = suc ⁡ F ⁡ b → b ∈ 𝒫 ω ∩ Fin
49 incom ⊢ b ∩ b = b ∩ b
50 nnord ⊢ b ∈ ω → Ord ⁡ b
51 orddisj ⊢ Ord ⁡ b → b ∩ b = ∅
52 50 51 syl ⊢ b ∈ ω → b ∩ b = ∅
53 49 52 eqtrid ⊢ b ∈ ω → b ∩ b = ∅
54 53 adantr ⊢ b ∈ ω ∧ card ⁡ 𝒫 b = suc ⁡ F ⁡ b → b ∩ b = ∅
55 1 ackbij1lem9 ⊢ b ∈ 𝒫 ω ∩ Fin ∧ b ∈ 𝒫 ω ∩ Fin ∧ b ∩ b = ∅ → F ⁡ b ∪ b = F ⁡ b + 𝑜 F ⁡ b
56 46 48 54 55 syl3anc ⊢ b ∈ ω ∧ card ⁡ 𝒫 b = suc ⁡ F ⁡ b → F ⁡ b ∪ b = F ⁡ b + 𝑜 F ⁡ b
57 1 ackbij1lem8 ⊢ b ∈ ω → F ⁡ b = card ⁡ 𝒫 b
58 57 adantr ⊢ b ∈ ω ∧ card ⁡ 𝒫 b = suc ⁡ F ⁡ b → F ⁡ b = card ⁡ 𝒫 b
59 58 oveq1d ⊢ b ∈ ω ∧ card ⁡ 𝒫 b = suc ⁡ F ⁡ b → F ⁡ b + 𝑜 F ⁡ b = card ⁡ 𝒫 b + 𝑜 F ⁡ b
60 56 59 eqtrd ⊢ b ∈ ω ∧ card ⁡ 𝒫 b = suc ⁡ F ⁡ b → F ⁡ b ∪ b = card ⁡ 𝒫 b + 𝑜 F ⁡ b
61 44 60 eqtrid ⊢ b ∈ ω ∧ card ⁡ 𝒫 b = suc ⁡ F ⁡ b → F ⁡ suc ⁡ b = card ⁡ 𝒫 b + 𝑜 F ⁡ b
62 suceq ⊢ F ⁡ suc ⁡ b = card ⁡ 𝒫 b + 𝑜 F ⁡ b → suc ⁡ F ⁡ suc ⁡ b = suc ⁡ card ⁡ 𝒫 b + 𝑜 F ⁡ b
63 61 62 syl ⊢ b ∈ ω ∧ card ⁡ 𝒫 b = suc ⁡ F ⁡ b → suc ⁡ F ⁡ suc ⁡ b = suc ⁡ card ⁡ 𝒫 b + 𝑜 F ⁡ b
64 nnfi ⊢ b ∈ ω → b ∈ Fin
65 pwfi ⊢ b ∈ Fin ↔ 𝒫 b ∈ Fin
66 64 65 sylib ⊢ b ∈ ω → 𝒫 b ∈ Fin
67 66 adantr ⊢ b ∈ ω ∧ card ⁡ 𝒫 b = suc ⁡ F ⁡ b → 𝒫 b ∈ Fin
68 ficardom ⊢ 𝒫 b ∈ Fin → card ⁡ 𝒫 b ∈ ω
69 67 68 syl ⊢ b ∈ ω ∧ card ⁡ 𝒫 b = suc ⁡ F ⁡ b → card ⁡ 𝒫 b ∈ ω
70 1 ackbij1lem10 ⊢ F : 𝒫 ω ∩ Fin ⟶ ω
71 70 ffvelcdmi ⊢ b ∈ 𝒫 ω ∩ Fin → F ⁡ b ∈ ω
72 48 71 syl ⊢ b ∈ ω ∧ card ⁡ 𝒫 b = suc ⁡ F ⁡ b → F ⁡ b ∈ ω
73 nnasuc ⊢ card ⁡ 𝒫 b ∈ ω ∧ F ⁡ b ∈ ω → card ⁡ 𝒫 b + 𝑜 suc ⁡ F ⁡ b = suc ⁡ card ⁡ 𝒫 b + 𝑜 F ⁡ b
74 69 72 73 syl2anc ⊢ b ∈ ω ∧ card ⁡ 𝒫 b = suc ⁡ F ⁡ b → card ⁡ 𝒫 b + 𝑜 suc ⁡ F ⁡ b = suc ⁡ card ⁡ 𝒫 b + 𝑜 F ⁡ b
75 63 74 eqtr4d ⊢ b ∈ ω ∧ card ⁡ 𝒫 b = suc ⁡ F ⁡ b → suc ⁡ F ⁡ suc ⁡ b = card ⁡ 𝒫 b + 𝑜 suc ⁡ F ⁡ b
76 39 41 75 3eqtr4d ⊢ b ∈ ω ∧ card ⁡ 𝒫 b = suc ⁡ F ⁡ b → card ⁡ 𝒫 suc ⁡ b = suc ⁡ F ⁡ suc ⁡ b
77 76 ex ⊢ b ∈ ω → card ⁡ 𝒫 b = suc ⁡ F ⁡ b → card ⁡ 𝒫 suc ⁡ b = suc ⁡ F ⁡ suc ⁡ b
78 8 14 20 26 37 77 finds ⊢ A ∈ ω → card ⁡ 𝒫 A = suc ⁡ F ⁡ A
79 2 78 eqtrd ⊢ A ∈ ω → F ⁡ A = suc ⁡ F ⁡ A