Proof of Theorem ineleq
Step | Hyp | Ref
| Expression |
1 | | orcom 866 |
. . . . 5
⊢ ((𝑥 = 𝑦 ∨ (𝐶 ∩ 𝐷) = ∅) ↔ ((𝐶 ∩ 𝐷) = ∅ ∨ 𝑥 = 𝑦)) |
2 | | df-or 844 |
. . . . 5
⊢ (((𝐶 ∩ 𝐷) = ∅ ∨ 𝑥 = 𝑦) ↔ (¬ (𝐶 ∩ 𝐷) = ∅ → 𝑥 = 𝑦)) |
3 | | neq0 4276 |
. . . . . . . 8
⊢ (¬
(𝐶 ∩ 𝐷) = ∅ ↔ ∃𝑧 𝑧 ∈ (𝐶 ∩ 𝐷)) |
4 | | elin 3899 |
. . . . . . . . 9
⊢ (𝑧 ∈ (𝐶 ∩ 𝐷) ↔ (𝑧 ∈ 𝐶 ∧ 𝑧 ∈ 𝐷)) |
5 | 4 | exbii 1851 |
. . . . . . . 8
⊢
(∃𝑧 𝑧 ∈ (𝐶 ∩ 𝐷) ↔ ∃𝑧(𝑧 ∈ 𝐶 ∧ 𝑧 ∈ 𝐷)) |
6 | 3, 5 | bitri 274 |
. . . . . . 7
⊢ (¬
(𝐶 ∩ 𝐷) = ∅ ↔ ∃𝑧(𝑧 ∈ 𝐶 ∧ 𝑧 ∈ 𝐷)) |
7 | 6 | imbi1i 349 |
. . . . . 6
⊢ ((¬
(𝐶 ∩ 𝐷) = ∅ → 𝑥 = 𝑦) ↔ (∃𝑧(𝑧 ∈ 𝐶 ∧ 𝑧 ∈ 𝐷) → 𝑥 = 𝑦)) |
8 | | 19.23v 1946 |
. . . . . 6
⊢
(∀𝑧((𝑧 ∈ 𝐶 ∧ 𝑧 ∈ 𝐷) → 𝑥 = 𝑦) ↔ (∃𝑧(𝑧 ∈ 𝐶 ∧ 𝑧 ∈ 𝐷) → 𝑥 = 𝑦)) |
9 | 7, 8 | bitr4i 277 |
. . . . 5
⊢ ((¬
(𝐶 ∩ 𝐷) = ∅ → 𝑥 = 𝑦) ↔ ∀𝑧((𝑧 ∈ 𝐶 ∧ 𝑧 ∈ 𝐷) → 𝑥 = 𝑦)) |
10 | 1, 2, 9 | 3bitri 296 |
. . . 4
⊢ ((𝑥 = 𝑦 ∨ (𝐶 ∩ 𝐷) = ∅) ↔ ∀𝑧((𝑧 ∈ 𝐶 ∧ 𝑧 ∈ 𝐷) → 𝑥 = 𝑦)) |
11 | 10 | ralbii 3090 |
. . 3
⊢
(∀𝑦 ∈
𝐵 (𝑥 = 𝑦 ∨ (𝐶 ∩ 𝐷) = ∅) ↔ ∀𝑦 ∈ 𝐵 ∀𝑧((𝑧 ∈ 𝐶 ∧ 𝑧 ∈ 𝐷) → 𝑥 = 𝑦)) |
12 | | ralcom4 3161 |
. . 3
⊢
(∀𝑦 ∈
𝐵 ∀𝑧((𝑧 ∈ 𝐶 ∧ 𝑧 ∈ 𝐷) → 𝑥 = 𝑦) ↔ ∀𝑧∀𝑦 ∈ 𝐵 ((𝑧 ∈ 𝐶 ∧ 𝑧 ∈ 𝐷) → 𝑥 = 𝑦)) |
13 | 11, 12 | bitri 274 |
. 2
⊢
(∀𝑦 ∈
𝐵 (𝑥 = 𝑦 ∨ (𝐶 ∩ 𝐷) = ∅) ↔ ∀𝑧∀𝑦 ∈ 𝐵 ((𝑧 ∈ 𝐶 ∧ 𝑧 ∈ 𝐷) → 𝑥 = 𝑦)) |
14 | 13 | ralbii 3090 |
1
⊢
(∀𝑥 ∈
𝐴 ∀𝑦 ∈ 𝐵 (𝑥 = 𝑦 ∨ (𝐶 ∩ 𝐷) = ∅) ↔ ∀𝑥 ∈ 𝐴 ∀𝑧∀𝑦 ∈ 𝐵 ((𝑧 ∈ 𝐶 ∧ 𝑧 ∈ 𝐷) → 𝑥 = 𝑦)) |