Proof of Theorem reuuni1
| Step | Hyp | Ref
| Expression |
| 1 | | euuni 1954 |
. . . . . . . 8
⊢ (∃!x(x ∈
A ∧ φ) → ((x ∈ A ∧
φ) ↔ ∪{x∣(x ∈ A ∧
φ)} = x)) |
| 2 | 1 | biimpd 135 |
. . . . . . 7
⊢ (∃!x(x ∈
A ∧ φ) → ((x ∈ A ∧
φ) → ∪{x∣(x ∈ A ∧
φ)} = x)) |
| 3 | 2 | exp3a 292 |
. . . . . 6
⊢ (∃!x(x ∈
A ∧ φ) → (x ∈ A
→ (φ → ∪{x∣(x ∈ A ∧
φ)} = x))) |
| 4 | 3 | com12 13 |
. . . . 5
⊢ (x
∈ A → (∃!x(x ∈
A ∧ φ) → (φ → ∪{x∣(x ∈ A ∧
φ)} = x))) |
| 5 | 4 | imp 277 |
. . . 4
⊢ ((x
∈ A ∧ ∃!x(x ∈
A ∧ φ)) → (φ → ∪{x∣(x ∈ A ∧
φ)} = x)) |
| 6 | | pm3.27 260 |
. . . . . 6
⊢ ((x
∈ A ∧ φ) → φ) |
| 7 | 1, 6 | syl6bir 188 |
. . . . 5
⊢ (∃!x(x ∈
A ∧ φ) → (∪{x∣(x ∈ A ∧
φ)} = x → φ)) |
| 8 | 7 | adantl 305 |
. . . 4
⊢ ((x
∈ A ∧ ∃!x(x ∈
A ∧ φ)) → (∪{x∣(x ∈ A ∧
φ)} = x → φ)) |
| 9 | 5, 8 | impbid 397 |
. . 3
⊢ ((x
∈ A ∧ ∃!x(x ∈
A ∧ φ)) → (φ ↔ ∪{x∣(x ∈ A ∧
φ)} = x)) |
| 10 | | df-rab 1208 |
. . . . 5
⊢ {x
∈ A∣φ} = {x∣(x
∈ A ∧ φ)} |
| 11 | 10 | unieqi 1928 |
. . . 4
⊢ ∪{x ∈ A∣φ} =
∪{x∣(x
∈ A ∧ φ)} |
| 12 | 11 | cleq1i 1108 |
. . 3
⊢ (∪{x ∈ A∣φ} =
x ↔ ∪{x∣(x ∈ A ∧
φ)} = x) |
| 13 | 9, 12 | syl6bbr 416 |
. 2
⊢ ((x
∈ A ∧ ∃!x(x ∈
A ∧ φ)) → (φ ↔ ∪{x ∈ A∣φ} =
x)) |
| 14 | | df-reu 1207 |
. 2
⊢ (∃!x ∈ A φ ↔ ∃!x(x ∈
A ∧ φ)) |
| 15 | 13, 14 | sylan2b 347 |
1
⊢ ((x
∈ A ∧ ∃!x ∈ A φ) → (φ ↔ ∪{x ∈ A∣φ} =
x)) |