Proof of Theorem oelim
| Step | Hyp | Ref
| Expression |
| 1 | | rdglim2a 2988 |
. . . 4
⊢ ((B
∈ On ∧ Lim B) →
(rec({〈y, z〉∣z
= (y ·o A)}, 1o) ‘B) = ∪x ∈ B
(rec({〈y, z〉∣z
= (y ·o A)}, 1o) ‘x)) |
| 2 | 1 | ad2antlr 321 |
. . 3
⊢ (((A
∈ On ∧ (B ∈ On ∧ Lim
B)) ∧ ∅ ∈ A) → (rec({〈y, z〉∣z
= (y ·o A)}, 1o) ‘B) = ∪x ∈ B
(rec({〈y, z〉∣z
= (y ·o A)}, 1o) ‘x)) |
| 3 | | oevn0 3123 |
. . . . 5
⊢ (((A
∈ On ∧ B ∈ On) ∧ ∅
∈ A) → (A ↑o B) = (rec({〈y, z〉∣z
= (y ·o A)}, 1o) ‘B)) |
| 4 | | oevn0 3123 |
. . . . . . . . . 10
⊢ (((A
∈ On ∧ x ∈ On) ∧ ∅
∈ A) → (A ↑o x) = (rec({〈y, z〉∣z
= (y ·o A)}, 1o) ‘x)) |
| 5 | | onelon 2223 |
. . . . . . . . . 10
⊢ ((B
∈ On ∧ x ∈ B) → x
∈ On) |
| 6 | 4, 5 | sylan12 355 |
. . . . . . . . 9
⊢ (((A
∈ On ∧ (B ∈ On ∧ x ∈ B))
∧ ∅ ∈ A) → (A ↑o x) = (rec({〈y, z〉∣z
= (y ·o A)}, 1o) ‘x)) |
| 7 | 6 | exp42 300 |
. . . . . . . 8
⊢ (A
∈ On → (B ∈ On →
(x ∈ B → (∅ ∈ A → (A
↑o x) =
(rec({〈y, z〉∣z
= (y ·o A)}, 1o) ‘x))))) |
| 8 | 7 | com34 36 |
. . . . . . 7
⊢ (A
∈ On → (B ∈ On →
(∅ ∈ A → (x ∈ B
→ (A ↑o
x) = (rec({〈y, z〉∣z
= (y ·o A)}, 1o) ‘x))))) |
| 9 | 8 | imp31 280 |
. . . . . 6
⊢ (((A
∈ On ∧ B ∈ On) ∧ ∅
∈ A) → (x ∈ B
→ (A ↑o
x) = (rec({〈y, z〉∣z
= (y ·o A)}, 1o) ‘x))) |
| 10 | 9 | iuneq2dv 2010 |
. . . . 5
⊢ (((A
∈ On ∧ B ∈ On) ∧ ∅
∈ A) → ∪x ∈ B (A
↑o x) = ∪x ∈ B (rec({〈y,
z〉∣z = (y
·o A)},
1o) ‘x)) |
| 11 | 3, 10 | cleq12d 1115 |
. . . 4
⊢ (((A
∈ On ∧ B ∈ On) ∧ ∅
∈ A) → ((A ↑o B) = ∪x ∈ B
(A ↑o x) ↔ (rec({〈y, z〉∣z
= (y ·o A)}, 1o) ‘B) = ∪x ∈ B
(rec({〈y, z〉∣z
= (y ·o A)}, 1o) ‘x))) |
| 12 | 11 | adantlrr 315 |
. . 3
⊢ (((A
∈ On ∧ (B ∈ On ∧ Lim
B)) ∧ ∅ ∈ A) → ((A
↑o B) = ∪x ∈ B (A
↑o x) ↔
(rec({〈y, z〉∣z
= (y ·o A)}, 1o) ‘B) = ∪x ∈ B
(rec({〈y, z〉∣z
= (y ·o A)}, 1o) ‘x))) |
| 13 | 2, 12 | mpbird 171 |
. 2
⊢ (((A
∈ On ∧ (B ∈ On ∧ Lim
B)) ∧ ∅ ∈ A) → (A
↑o B) = ∪x ∈ B (A
↑o x)) |
| 14 | | limelon 2286 |
. . 3
⊢ ((B
∈ C ∧ Lim B) → B
∈ On) |
| 15 | | pm3.27 260 |
. . 3
⊢ ((B
∈ C ∧ Lim B) → Lim B) |
| 16 | 14, 15 | jca 236 |
. 2
⊢ ((B
∈ C ∧ Lim B) → (B
∈ On ∧ Lim B)) |
| 17 | 13, 16 | sylan12 355 |
1
⊢ (((A
∈ On ∧ (B ∈ C ∧ Lim B))
∧ ∅ ∈ A) → (A ↑o B) = ∪x ∈ B
(A ↑o x)) |