Proof of Theorem fiunlem
Step | Hyp | Ref
| Expression |
1 | | vex 3436 |
. . . 4
⊢ 𝑣 ∈ V |
2 | | eqeq1 2742 |
. . . . 5
⊢ (𝑧 = 𝑣 → (𝑧 = 𝐵 ↔ 𝑣 = 𝐵)) |
3 | 2 | rexbidv 3226 |
. . . 4
⊢ (𝑧 = 𝑣 → (∃𝑥 ∈ 𝐴 𝑧 = 𝐵 ↔ ∃𝑥 ∈ 𝐴 𝑣 = 𝐵)) |
4 | 1, 3 | elab 3609 |
. . 3
⊢ (𝑣 ∈ {𝑧 ∣ ∃𝑥 ∈ 𝐴 𝑧 = 𝐵} ↔ ∃𝑥 ∈ 𝐴 𝑣 = 𝐵) |
5 | | fiun.1 |
. . . . . 6
⊢ (𝑥 = 𝑦 → 𝐵 = 𝐶) |
6 | 5 | eqeq2d 2749 |
. . . . 5
⊢ (𝑥 = 𝑦 → (𝑣 = 𝐵 ↔ 𝑣 = 𝐶)) |
7 | 6 | cbvrexvw 3384 |
. . . 4
⊢
(∃𝑥 ∈
𝐴 𝑣 = 𝐵 ↔ ∃𝑦 ∈ 𝐴 𝑣 = 𝐶) |
8 | | r19.29 3184 |
. . . . . . 7
⊢
((∀𝑦 ∈
𝐴 (𝐵 ⊆ 𝐶 ∨ 𝐶 ⊆ 𝐵) ∧ ∃𝑦 ∈ 𝐴 𝑣 = 𝐶) → ∃𝑦 ∈ 𝐴 ((𝐵 ⊆ 𝐶 ∨ 𝐶 ⊆ 𝐵) ∧ 𝑣 = 𝐶)) |
9 | | sseq12 3948 |
. . . . . . . . . . . . 13
⊢ ((𝑢 = 𝐵 ∧ 𝑣 = 𝐶) → (𝑢 ⊆ 𝑣 ↔ 𝐵 ⊆ 𝐶)) |
10 | 9 | ancoms 459 |
. . . . . . . . . . . 12
⊢ ((𝑣 = 𝐶 ∧ 𝑢 = 𝐵) → (𝑢 ⊆ 𝑣 ↔ 𝐵 ⊆ 𝐶)) |
11 | | sseq12 3948 |
. . . . . . . . . . . 12
⊢ ((𝑣 = 𝐶 ∧ 𝑢 = 𝐵) → (𝑣 ⊆ 𝑢 ↔ 𝐶 ⊆ 𝐵)) |
12 | 10, 11 | orbi12d 916 |
. . . . . . . . . . 11
⊢ ((𝑣 = 𝐶 ∧ 𝑢 = 𝐵) → ((𝑢 ⊆ 𝑣 ∨ 𝑣 ⊆ 𝑢) ↔ (𝐵 ⊆ 𝐶 ∨ 𝐶 ⊆ 𝐵))) |
13 | 12 | biimprcd 249 |
. . . . . . . . . 10
⊢ ((𝐵 ⊆ 𝐶 ∨ 𝐶 ⊆ 𝐵) → ((𝑣 = 𝐶 ∧ 𝑢 = 𝐵) → (𝑢 ⊆ 𝑣 ∨ 𝑣 ⊆ 𝑢))) |
14 | 13 | expdimp 453 |
. . . . . . . . 9
⊢ (((𝐵 ⊆ 𝐶 ∨ 𝐶 ⊆ 𝐵) ∧ 𝑣 = 𝐶) → (𝑢 = 𝐵 → (𝑢 ⊆ 𝑣 ∨ 𝑣 ⊆ 𝑢))) |
15 | 14 | rexlimivw 3211 |
. . . . . . . 8
⊢
(∃𝑦 ∈
𝐴 ((𝐵 ⊆ 𝐶 ∨ 𝐶 ⊆ 𝐵) ∧ 𝑣 = 𝐶) → (𝑢 = 𝐵 → (𝑢 ⊆ 𝑣 ∨ 𝑣 ⊆ 𝑢))) |
16 | 15 | imp 407 |
. . . . . . 7
⊢
((∃𝑦 ∈
𝐴 ((𝐵 ⊆ 𝐶 ∨ 𝐶 ⊆ 𝐵) ∧ 𝑣 = 𝐶) ∧ 𝑢 = 𝐵) → (𝑢 ⊆ 𝑣 ∨ 𝑣 ⊆ 𝑢)) |
17 | 8, 16 | sylan 580 |
. . . . . 6
⊢
(((∀𝑦 ∈
𝐴 (𝐵 ⊆ 𝐶 ∨ 𝐶 ⊆ 𝐵) ∧ ∃𝑦 ∈ 𝐴 𝑣 = 𝐶) ∧ 𝑢 = 𝐵) → (𝑢 ⊆ 𝑣 ∨ 𝑣 ⊆ 𝑢)) |
18 | 17 | an32s 649 |
. . . . 5
⊢
(((∀𝑦 ∈
𝐴 (𝐵 ⊆ 𝐶 ∨ 𝐶 ⊆ 𝐵) ∧ 𝑢 = 𝐵) ∧ ∃𝑦 ∈ 𝐴 𝑣 = 𝐶) → (𝑢 ⊆ 𝑣 ∨ 𝑣 ⊆ 𝑢)) |
19 | 18 | adantlll 715 |
. . . 4
⊢ ((((𝐵:𝐷⟶𝑆 ∧ ∀𝑦 ∈ 𝐴 (𝐵 ⊆ 𝐶 ∨ 𝐶 ⊆ 𝐵)) ∧ 𝑢 = 𝐵) ∧ ∃𝑦 ∈ 𝐴 𝑣 = 𝐶) → (𝑢 ⊆ 𝑣 ∨ 𝑣 ⊆ 𝑢)) |
20 | 7, 19 | sylan2b 594 |
. . 3
⊢ ((((𝐵:𝐷⟶𝑆 ∧ ∀𝑦 ∈ 𝐴 (𝐵 ⊆ 𝐶 ∨ 𝐶 ⊆ 𝐵)) ∧ 𝑢 = 𝐵) ∧ ∃𝑥 ∈ 𝐴 𝑣 = 𝐵) → (𝑢 ⊆ 𝑣 ∨ 𝑣 ⊆ 𝑢)) |
21 | 4, 20 | sylan2b 594 |
. 2
⊢ ((((𝐵:𝐷⟶𝑆 ∧ ∀𝑦 ∈ 𝐴 (𝐵 ⊆ 𝐶 ∨ 𝐶 ⊆ 𝐵)) ∧ 𝑢 = 𝐵) ∧ 𝑣 ∈ {𝑧 ∣ ∃𝑥 ∈ 𝐴 𝑧 = 𝐵}) → (𝑢 ⊆ 𝑣 ∨ 𝑣 ⊆ 𝑢)) |
22 | 21 | ralrimiva 3103 |
1
⊢ (((𝐵:𝐷⟶𝑆 ∧ ∀𝑦 ∈ 𝐴 (𝐵 ⊆ 𝐶 ∨ 𝐶 ⊆ 𝐵)) ∧ 𝑢 = 𝐵) → ∀𝑣 ∈ {𝑧 ∣ ∃𝑥 ∈ 𝐴 𝑧 = 𝐵} (𝑢 ⊆ 𝑣 ∨ 𝑣 ⊆ 𝑢)) |