Proof of Theorem ndisj2
Step | Hyp | Ref
| Expression |
1 | | ndisj2.1 |
. . . 4
⊢ (𝑥 = 𝑦 → 𝐵 = 𝐶) |
2 | 1 | disjor 5054 |
. . 3
⊢
(Disj 𝑥
∈ 𝐴 𝐵 ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥 = 𝑦 ∨ (𝐵 ∩ 𝐶) = ∅)) |
3 | 2 | notbii 320 |
. 2
⊢ (¬
Disj 𝑥 ∈ 𝐴 𝐵 ↔ ¬ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥 = 𝑦 ∨ (𝐵 ∩ 𝐶) = ∅)) |
4 | | rexnal 3169 |
. 2
⊢
(∃𝑥 ∈
𝐴 ¬ ∀𝑦 ∈ 𝐴 (𝑥 = 𝑦 ∨ (𝐵 ∩ 𝐶) = ∅) ↔ ¬ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥 = 𝑦 ∨ (𝐵 ∩ 𝐶) = ∅)) |
5 | | rexnal 3169 |
. . . 4
⊢
(∃𝑦 ∈
𝐴 ¬ (𝑥 = 𝑦 ∨ (𝐵 ∩ 𝐶) = ∅) ↔ ¬ ∀𝑦 ∈ 𝐴 (𝑥 = 𝑦 ∨ (𝐵 ∩ 𝐶) = ∅)) |
6 | | ioran 981 |
. . . . . 6
⊢ (¬
(𝑥 = 𝑦 ∨ (𝐵 ∩ 𝐶) = ∅) ↔ (¬ 𝑥 = 𝑦 ∧ ¬ (𝐵 ∩ 𝐶) = ∅)) |
7 | | df-ne 2944 |
. . . . . . 7
⊢ (𝑥 ≠ 𝑦 ↔ ¬ 𝑥 = 𝑦) |
8 | | df-ne 2944 |
. . . . . . 7
⊢ ((𝐵 ∩ 𝐶) ≠ ∅ ↔ ¬ (𝐵 ∩ 𝐶) = ∅) |
9 | 7, 8 | anbi12i 627 |
. . . . . 6
⊢ ((𝑥 ≠ 𝑦 ∧ (𝐵 ∩ 𝐶) ≠ ∅) ↔ (¬ 𝑥 = 𝑦 ∧ ¬ (𝐵 ∩ 𝐶) = ∅)) |
10 | 6, 9 | bitr4i 277 |
. . . . 5
⊢ (¬
(𝑥 = 𝑦 ∨ (𝐵 ∩ 𝐶) = ∅) ↔ (𝑥 ≠ 𝑦 ∧ (𝐵 ∩ 𝐶) ≠ ∅)) |
11 | 10 | rexbii 3181 |
. . . 4
⊢
(∃𝑦 ∈
𝐴 ¬ (𝑥 = 𝑦 ∨ (𝐵 ∩ 𝐶) = ∅) ↔ ∃𝑦 ∈ 𝐴 (𝑥 ≠ 𝑦 ∧ (𝐵 ∩ 𝐶) ≠ ∅)) |
12 | 5, 11 | bitr3i 276 |
. . 3
⊢ (¬
∀𝑦 ∈ 𝐴 (𝑥 = 𝑦 ∨ (𝐵 ∩ 𝐶) = ∅) ↔ ∃𝑦 ∈ 𝐴 (𝑥 ≠ 𝑦 ∧ (𝐵 ∩ 𝐶) ≠ ∅)) |
13 | 12 | rexbii 3181 |
. 2
⊢
(∃𝑥 ∈
𝐴 ¬ ∀𝑦 ∈ 𝐴 (𝑥 = 𝑦 ∨ (𝐵 ∩ 𝐶) = ∅) ↔ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐴 (𝑥 ≠ 𝑦 ∧ (𝐵 ∩ 𝐶) ≠ ∅)) |
14 | 3, 4, 13 | 3bitr2i 299 |
1
⊢ (¬
Disj 𝑥 ∈ 𝐴 𝐵 ↔ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐴 (𝑥 ≠ 𝑦 ∧ (𝐵 ∩ 𝐶) ≠ ∅)) |