Proof of Theorem eqop
Step | Hyp | Ref
| Expression |
1 | | dfcleq 2347 |
. 2
⊢ (A = 〈B, C〉 ↔ ∀z(z ∈ A ↔ z ∈ 〈B, C〉)) |
2 | | df-op 4567 |
. . . . . . 7
⊢ 〈B, C〉 = ({z ∣ ∃t ∈ B z = Phi t} ∪ {z
∣ ∃t ∈ C z = ( Phi t ∪ {0c})}) |
3 | 2 | eleq2i 2417 |
. . . . . 6
⊢ (z ∈ 〈B, C〉 ↔ z ∈ ({z ∣ ∃t ∈ B z = Phi t} ∪ {z
∣ ∃t ∈ C z = ( Phi t ∪ {0c})})) |
4 | | elun 3221 |
. . . . . 6
⊢ (z ∈ ({z ∣ ∃t ∈ B z = Phi t} ∪ {z
∣ ∃t ∈ C z = ( Phi t ∪ {0c})}) ↔ (z ∈ {z ∣ ∃t ∈ B z = Phi t} ∨ z ∈ {z ∣ ∃t ∈ C z = ( Phi t ∪ {0c})})) |
5 | 3, 4 | bitri 240 |
. . . . 5
⊢ (z ∈ 〈B, C〉 ↔
(z ∈
{z ∣
∃t ∈ B z = Phi t} ∨ z ∈ {z ∣ ∃t ∈ C z = ( Phi t ∪ {0c})})) |
6 | | abid 2341 |
. . . . . 6
⊢ (z ∈ {z ∣ ∃t ∈ B z = Phi t} ↔ ∃t ∈ B z = Phi t) |
7 | | abid 2341 |
. . . . . 6
⊢ (z ∈ {z ∣ ∃t ∈ C z = ( Phi t ∪ {0c})} ↔ ∃t ∈ C z = ( Phi t ∪ {0c})) |
8 | 6, 7 | orbi12i 507 |
. . . . 5
⊢ ((z ∈ {z ∣ ∃t ∈ B z = Phi t} ∨ z ∈ {z ∣ ∃t ∈ C z = ( Phi t ∪ {0c})}) ↔ (∃t ∈ B z = Phi t ∨ ∃t ∈ C z = ( Phi t ∪ {0c}))) |
9 | 5, 8 | bitri 240 |
. . . 4
⊢ (z ∈ 〈B, C〉 ↔ (∃t ∈ B z = Phi t ∨ ∃t ∈ C z = ( Phi t ∪ {0c}))) |
10 | 9 | bibi2i 304 |
. . 3
⊢ ((z ∈ A ↔ z ∈ 〈B, C〉) ↔ (z
∈ A
↔ (∃t ∈ B z = Phi t ∨ ∃t ∈ C z = ( Phi t ∪
{0c})))) |
11 | 10 | albii 1566 |
. 2
⊢ (∀z(z ∈ A ↔ z ∈ 〈B, C〉) ↔ ∀z(z ∈ A ↔ (∃t ∈ B z = Phi t ∨ ∃t ∈ C z = ( Phi t ∪ {0c})))) |
12 | 1, 11 | bitri 240 |
1
⊢ (A = 〈B, C〉 ↔ ∀z(z ∈ A ↔ (∃t ∈ B z = Phi t ∨ ∃t ∈ C z = ( Phi t ∪ {0c})))) |