Proof of Theorem dfso3
Step | Hyp | Ref
| Expression |
1 | | ne0i 4265 |
. . . . 5
⊢ (𝑦 ∈ 𝐴 → 𝐴 ≠ ∅) |
2 | | r19.27zv 4433 |
. . . . 5
⊢ (𝐴 ≠ ∅ →
(∀𝑧 ∈ 𝐴 ((¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ∧ (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥)) ↔ (∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ∧ (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥)))) |
3 | 1, 2 | syl 17 |
. . . 4
⊢ (𝑦 ∈ 𝐴 → (∀𝑧 ∈ 𝐴 ((¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ∧ (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥)) ↔ (∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ∧ (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥)))) |
4 | 3 | ralbiia 3089 |
. . 3
⊢
(∀𝑦 ∈
𝐴 ∀𝑧 ∈ 𝐴 ((¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ∧ (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥)) ↔ ∀𝑦 ∈ 𝐴 (∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ∧ (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥))) |
5 | 4 | ralbii 3090 |
. 2
⊢
(∀𝑥 ∈
𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ∧ (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥)) ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ∧ (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥))) |
6 | | df-3an 1087 |
. . . 4
⊢ ((¬
𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧) ∧ (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥)) ↔ ((¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ∧ (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥))) |
7 | 6 | ralbii 3090 |
. . 3
⊢
(∀𝑧 ∈
𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧) ∧ (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥)) ↔ ∀𝑧 ∈ 𝐴 ((¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ∧ (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥))) |
8 | 7 | 2ralbii 3091 |
. 2
⊢
(∀𝑥 ∈
𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧) ∧ (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥)) ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 ((¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ∧ (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥))) |
9 | | df-po 5494 |
. . . 4
⊢ (𝑅 Po 𝐴 ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧))) |
10 | 9 | anbi1i 623 |
. . 3
⊢ ((𝑅 Po 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥)) ↔ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥))) |
11 | | df-so 5495 |
. . 3
⊢ (𝑅 Or 𝐴 ↔ (𝑅 Po 𝐴 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥))) |
12 | | r19.26-2 3095 |
. . 3
⊢
(∀𝑥 ∈
𝐴 ∀𝑦 ∈ 𝐴 (∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ∧ (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥)) ↔ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥))) |
13 | 10, 11, 12 | 3bitr4i 302 |
. 2
⊢ (𝑅 Or 𝐴 ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ∧ (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥))) |
14 | 5, 8, 13 | 3bitr4ri 303 |
1
⊢ (𝑅 Or 𝐴 ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧) ∧ (𝑥𝑅𝑦 ∨ 𝑥 = 𝑦 ∨ 𝑦𝑅𝑥))) |