Proof of Theorem reuun2
| Step | Hyp | Ref
| Expression |
| 1 | | df-rex 1653 |
. . . 4
⊢ (∃x ∈ B φ ↔ ∃x(x ∈ B ⋀ φ)) |
| 2 | 1 | negbii 187 |
. . 3
⊢ (¬ ∃x ∈ B φ ↔ ¬ ∃x(x ∈ B ⋀ φ)) |
| 3 | | euor2 1440 |
. . 3
⊢ (¬ ∃x(x ∈ B ⋀ φ) → (∃!x((x ∈ B ⋀ φ) ⋁
(x ∈
A ⋀
φ)) ↔ ∃!x(x ∈ A ⋀ φ))) |
| 4 | 2, 3 | sylbi 199 |
. 2
⊢ (¬ ∃x ∈ B φ → (∃!x((x ∈ B ⋀ φ) ⋁
(x ∈
A ⋀
φ)) ↔ ∃!x(x ∈ A ⋀ φ))) |
| 5 | | df-reu 1654 |
. . 3
⊢ (∃!x ∈ (A ∪
B)φ
↔ ∃!x(x ∈ (A ∪
B) ⋀
φ)) |
| 6 | | elun 2176 |
. . . . . 6
⊢ (x ∈ (A ∪ B)
↔ (x ∈ A ⋁ x ∈ B)) |
| 7 | 6 | anbi1i 483 |
. . . . 5
⊢ ((x ∈ (A ∪ B) ⋀ φ)
↔ ((x ∈ A ⋁ x ∈ B) ⋀ φ)) |
| 8 | | andir 607 |
. . . . 5
⊢ (((x ∈ A ⋁ x ∈ B) ⋀ φ) ↔ ((x ∈ A ⋀ φ) ⋁
(x ∈
B ⋀
φ))) |
| 9 | | orcom 246 |
. . . . 5
⊢ (((x ∈ A ⋀ φ) ⋁
(x ∈
B ⋀
φ)) ↔ ((x ∈ B ⋀ φ) ⋁
(x ∈
A ⋀
φ))) |
| 10 | 7, 8, 9 | 3bitr 177 |
. . . 4
⊢ ((x ∈ (A ∪ B) ⋀ φ)
↔ ((x ∈ B ⋀ φ) ⋁ (x ∈ A ⋀ φ))) |
| 11 | 10 | eubii 1389 |
. . 3
⊢ (∃!x(x ∈ (A ∪ B) ⋀ φ)
↔ ∃!x((x ∈ B ⋀ φ) ⋁ (x ∈ A ⋀ φ))) |
| 12 | 5, 11 | bitr 173 |
. 2
⊢ (∃!x ∈ (A ∪
B)φ
↔ ∃!x((x ∈ B ⋀ φ) ⋁ (x ∈ A ⋀ φ))) |
| 13 | | df-reu 1654 |
. 2
⊢ (∃!x ∈ A φ ↔ ∃!x(x ∈ A ⋀ φ)) |
| 14 | 4, 12, 13 | 3bitr4g 557 |
1
⊢ (¬ ∃x ∈ B φ → (∃!x ∈ (A ∪
B)φ
↔ ∃!x ∈ A φ)) |