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 ffvelrni 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