Proof of Theorem axdc3lem3
Step | Hyp | Ref
| Expression |
1 | | axdc3lem3.2 |
. . 3
⊢ 𝑆 = {𝑠 ∣ ∃𝑛 ∈ ω (𝑠:suc 𝑛⟶𝐴 ∧ (𝑠‘∅) = 𝐶 ∧ ∀𝑘 ∈ 𝑛 (𝑠‘suc 𝑘) ∈ (𝐹‘(𝑠‘𝑘)))} |
2 | 1 | eleq2i 2824 |
. 2
⊢ (𝐵 ∈ 𝑆 ↔ 𝐵 ∈ {𝑠 ∣ ∃𝑛 ∈ ω (𝑠:suc 𝑛⟶𝐴 ∧ (𝑠‘∅) = 𝐶 ∧ ∀𝑘 ∈ 𝑛 (𝑠‘suc 𝑘) ∈ (𝐹‘(𝑠‘𝑘)))}) |
3 | | axdc3lem3.3 |
. . 3
⊢ 𝐵 ∈ V |
4 | | feq1 6479 |
. . . . 5
⊢ (𝑠 = 𝐵 → (𝑠:suc 𝑛⟶𝐴 ↔ 𝐵:suc 𝑛⟶𝐴)) |
5 | | fveq1 6667 |
. . . . . 6
⊢ (𝑠 = 𝐵 → (𝑠‘∅) = (𝐵‘∅)) |
6 | 5 | eqeq1d 2740 |
. . . . 5
⊢ (𝑠 = 𝐵 → ((𝑠‘∅) = 𝐶 ↔ (𝐵‘∅) = 𝐶)) |
7 | | fveq1 6667 |
. . . . . . 7
⊢ (𝑠 = 𝐵 → (𝑠‘suc 𝑘) = (𝐵‘suc 𝑘)) |
8 | | fveq1 6667 |
. . . . . . . 8
⊢ (𝑠 = 𝐵 → (𝑠‘𝑘) = (𝐵‘𝑘)) |
9 | 8 | fveq2d 6672 |
. . . . . . 7
⊢ (𝑠 = 𝐵 → (𝐹‘(𝑠‘𝑘)) = (𝐹‘(𝐵‘𝑘))) |
10 | 7, 9 | eleq12d 2827 |
. . . . . 6
⊢ (𝑠 = 𝐵 → ((𝑠‘suc 𝑘) ∈ (𝐹‘(𝑠‘𝑘)) ↔ (𝐵‘suc 𝑘) ∈ (𝐹‘(𝐵‘𝑘)))) |
11 | 10 | ralbidv 3109 |
. . . . 5
⊢ (𝑠 = 𝐵 → (∀𝑘 ∈ 𝑛 (𝑠‘suc 𝑘) ∈ (𝐹‘(𝑠‘𝑘)) ↔ ∀𝑘 ∈ 𝑛 (𝐵‘suc 𝑘) ∈ (𝐹‘(𝐵‘𝑘)))) |
12 | 4, 6, 11 | 3anbi123d 1437 |
. . . 4
⊢ (𝑠 = 𝐵 → ((𝑠:suc 𝑛⟶𝐴 ∧ (𝑠‘∅) = 𝐶 ∧ ∀𝑘 ∈ 𝑛 (𝑠‘suc 𝑘) ∈ (𝐹‘(𝑠‘𝑘))) ↔ (𝐵:suc 𝑛⟶𝐴 ∧ (𝐵‘∅) = 𝐶 ∧ ∀𝑘 ∈ 𝑛 (𝐵‘suc 𝑘) ∈ (𝐹‘(𝐵‘𝑘))))) |
13 | 12 | rexbidv 3206 |
. . 3
⊢ (𝑠 = 𝐵 → (∃𝑛 ∈ ω (𝑠:suc 𝑛⟶𝐴 ∧ (𝑠‘∅) = 𝐶 ∧ ∀𝑘 ∈ 𝑛 (𝑠‘suc 𝑘) ∈ (𝐹‘(𝑠‘𝑘))) ↔ ∃𝑛 ∈ ω (𝐵:suc 𝑛⟶𝐴 ∧ (𝐵‘∅) = 𝐶 ∧ ∀𝑘 ∈ 𝑛 (𝐵‘suc 𝑘) ∈ (𝐹‘(𝐵‘𝑘))))) |
14 | 3, 13 | elab 3571 |
. 2
⊢ (𝐵 ∈ {𝑠 ∣ ∃𝑛 ∈ ω (𝑠:suc 𝑛⟶𝐴 ∧ (𝑠‘∅) = 𝐶 ∧ ∀𝑘 ∈ 𝑛 (𝑠‘suc 𝑘) ∈ (𝐹‘(𝑠‘𝑘)))} ↔ ∃𝑛 ∈ ω (𝐵:suc 𝑛⟶𝐴 ∧ (𝐵‘∅) = 𝐶 ∧ ∀𝑘 ∈ 𝑛 (𝐵‘suc 𝑘) ∈ (𝐹‘(𝐵‘𝑘)))) |
15 | | suceq 6231 |
. . . . 5
⊢ (𝑛 = 𝑚 → suc 𝑛 = suc 𝑚) |
16 | 15 | feq2d 6484 |
. . . 4
⊢ (𝑛 = 𝑚 → (𝐵:suc 𝑛⟶𝐴 ↔ 𝐵:suc 𝑚⟶𝐴)) |
17 | | raleq 3309 |
. . . 4
⊢ (𝑛 = 𝑚 → (∀𝑘 ∈ 𝑛 (𝐵‘suc 𝑘) ∈ (𝐹‘(𝐵‘𝑘)) ↔ ∀𝑘 ∈ 𝑚 (𝐵‘suc 𝑘) ∈ (𝐹‘(𝐵‘𝑘)))) |
18 | 16, 17 | 3anbi13d 1439 |
. . 3
⊢ (𝑛 = 𝑚 → ((𝐵:suc 𝑛⟶𝐴 ∧ (𝐵‘∅) = 𝐶 ∧ ∀𝑘 ∈ 𝑛 (𝐵‘suc 𝑘) ∈ (𝐹‘(𝐵‘𝑘))) ↔ (𝐵:suc 𝑚⟶𝐴 ∧ (𝐵‘∅) = 𝐶 ∧ ∀𝑘 ∈ 𝑚 (𝐵‘suc 𝑘) ∈ (𝐹‘(𝐵‘𝑘))))) |
19 | 18 | cbvrexvw 3349 |
. 2
⊢
(∃𝑛 ∈
ω (𝐵:suc 𝑛⟶𝐴 ∧ (𝐵‘∅) = 𝐶 ∧ ∀𝑘 ∈ 𝑛 (𝐵‘suc 𝑘) ∈ (𝐹‘(𝐵‘𝑘))) ↔ ∃𝑚 ∈ ω (𝐵:suc 𝑚⟶𝐴 ∧ (𝐵‘∅) = 𝐶 ∧ ∀𝑘 ∈ 𝑚 (𝐵‘suc 𝑘) ∈ (𝐹‘(𝐵‘𝑘)))) |
20 | 2, 14, 19 | 3bitri 300 |
1
⊢ (𝐵 ∈ 𝑆 ↔ ∃𝑚 ∈ ω (𝐵:suc 𝑚⟶𝐴 ∧ (𝐵‘∅) = 𝐶 ∧ ∀𝑘 ∈ 𝑚 (𝐵‘suc 𝑘) ∈ (𝐹‘(𝐵‘𝑘)))) |