Proof of Theorem reuun2
Step | Hyp | Ref
| Expression |
1 | | df-rex 2621 |
. . 3
⊢ (∃x ∈ B φ ↔ ∃x(x ∈ B ∧ φ)) |
2 | | euor2 2272 |
. . 3
⊢ (¬ ∃x(x ∈ B ∧ φ) → (∃!x((x ∈ B ∧ φ) ∨
(x ∈
A ∧ φ)) ↔ ∃!x(x ∈ A ∧ φ))) |
3 | 1, 2 | sylnbi 297 |
. 2
⊢ (¬ ∃x ∈ B φ → (∃!x((x ∈ B ∧ φ) ∨
(x ∈
A ∧ φ)) ↔ ∃!x(x ∈ A ∧ φ))) |
4 | | df-reu 2622 |
. . 3
⊢ (∃!x ∈ (A ∪
B)φ
↔ ∃!x(x ∈ (A ∪
B) ∧ φ)) |
5 | | elun 3221 |
. . . . . 6
⊢ (x ∈ (A ∪ B)
↔ (x ∈ A ∨ x ∈ B)) |
6 | 5 | anbi1i 676 |
. . . . 5
⊢ ((x ∈ (A ∪ B) ∧ φ) ↔
((x ∈
A ∨
x ∈
B) ∧ φ)) |
7 | | andir 838 |
. . . . . 6
⊢ (((x ∈ A ∨ x ∈ B) ∧ φ) ↔ ((x ∈ A ∧ φ) ∨
(x ∈
B ∧ φ))) |
8 | | orcom 376 |
. . . . . 6
⊢ (((x ∈ A ∧ φ) ∨
(x ∈
B ∧ φ)) ↔ ((x ∈ B ∧ φ) ∨
(x ∈
A ∧ φ))) |
9 | 7, 8 | bitri 240 |
. . . . 5
⊢ (((x ∈ A ∨ x ∈ B) ∧ φ) ↔ ((x ∈ B ∧ φ) ∨
(x ∈
A ∧ φ))) |
10 | 6, 9 | bitri 240 |
. . . 4
⊢ ((x ∈ (A ∪ B) ∧ φ) ↔
((x ∈
B ∧ φ) ∨
(x ∈
A ∧ φ))) |
11 | 10 | eubii 2213 |
. . 3
⊢ (∃!x(x ∈ (A ∪ B) ∧ φ) ↔
∃!x((x ∈ B ∧ φ) ∨ (x ∈ A ∧ φ))) |
12 | 4, 11 | bitri 240 |
. 2
⊢ (∃!x ∈ (A ∪
B)φ
↔ ∃!x((x ∈ B ∧ φ) ∨ (x ∈ A ∧ φ))) |
13 | | df-reu 2622 |
. 2
⊢ (∃!x ∈ A φ ↔ ∃!x(x ∈ A ∧ φ)) |
14 | 3, 12, 13 | 3bitr4g 279 |
1
⊢ (¬ ∃x ∈ B φ → (∃!x ∈ (A ∪
B)φ
↔ ∃!x ∈ A φ)) |