Proof of Theorem cplem2
| Step | Hyp | Ref
| Expression |
| Z | | cleqid 1102 |
. . 3
⊢ {z
∈ B∣∀w ∈ B (rank
‘z) ⊆ (rank ‘w)} = {z ∈
B∣∀w ∈ B (rank
‘z) ⊆ (rank ‘w)} |
| 2 | | cleqid 1102 |
. . 3
⊢ ∪x ∈ A
{z ∈ B∣∀w ∈ B (rank
‘z) ⊆ (rank ‘w)} = ∪x ∈ A
{z ∈ B∣∀w ∈ B (rank
‘z) ⊆ (rank ‘w)} |
| 3 | 1, 2 | cplem1 3545 |
. 2
⊢ ∀x ∈ A
(¬ B = ∅ → ¬ (B ∩ ∪x ∈ A
{z ∈ B∣∀w ∈ B (rank
‘z) ⊆ (rank ‘w)}) = ∅) |
| 4 | | cplem2.1 |
. . . 4
⊢ A
∈ V |
| 5 | | scottex 3541 |
. . . 4
⊢ {z
∈ B∣∀w ∈ B (rank
‘z) ⊆ (rank ‘w)} ∈ V |
| 6 | 4, 5 | iunex 2914 |
. . 3
⊢ ∪x ∈ A
{z ∈ B∣∀w ∈ B (rank
‘z) ⊆ (rank ‘w)} ∈ V |
| 7 | | hbiu1 2012 |
. . . . 5
⊢ (y
∈ ∪x ∈
A {z
∈ B∣∀w ∈ B (rank
‘z) ⊆ (rank ‘w)} → ∀x y ∈ ∪x ∈ A {z ∈
B∣∀w ∈ B (rank
‘z) ⊆ (rank ‘w)}) |
| 8 | 7 | hbeleq 1173 |
. . . 4
⊢ (y =
∪x ∈
A {z
∈ B∣∀w ∈ B (rank
‘z) ⊆ (rank ‘w)} → ∀x y = ∪x ∈ A {z ∈
B∣∀w ∈ B (rank
‘z) ⊆ (rank ‘w)}) |
| 9 | | ineq2 1639 |
. . . . . . 7
⊢ (y =
∪x ∈
A {z
∈ B∣∀w ∈ B (rank
‘z) ⊆ (rank ‘w)} → (B
∩ y) = (B ∩ ∪x ∈ A
{z ∈ B∣∀w ∈ B (rank
‘z) ⊆ (rank ‘w)})) |
| 10 | 9 | cleq1d 1109 |
. . . . . 6
⊢ (y =
∪x ∈
A {z
∈ B∣∀w ∈ B (rank
‘z) ⊆ (rank ‘w)} → ((B
∩ y) = ∅ ↔ (B ∩ ∪x ∈ A
{z ∈ B∣∀w ∈ B (rank
‘z) ⊆ (rank ‘w)}) = ∅)) |
| 11 | 10 | negbid 463 |
. . . . 5
⊢ (y =
∪x ∈
A {z
∈ B∣∀w ∈ B (rank
‘z) ⊆ (rank ‘w)} → (¬ (B ∩ y) =
∅ ↔ ¬ (B ∩ ∪x ∈ A {z ∈
B∣∀w ∈ B (rank
‘z) ⊆ (rank ‘w)}) = ∅)) |
| 12 | 11 | imbi2d 464 |
. . . 4
⊢ (y =
∪x ∈
A {z
∈ B∣∀w ∈ B (rank
‘z) ⊆ (rank ‘w)} → ((¬ B = ∅ → ¬ (B ∩ y) =
∅) ↔ (¬ B = ∅ →
¬ (B ∩ ∪x ∈ A {z ∈
B∣∀w ∈ B (rank
‘z) ⊆ (rank ‘w)}) = ∅))) |
| 13 | 8, 12 | birald 1217 |
. . 3
⊢ (y =
∪x ∈
A {z
∈ B∣∀w ∈ B (rank
‘z) ⊆ (rank ‘w)} → (∀x ∈ A
(¬ B = ∅ → ¬ (B ∩ y) =
∅) ↔ &for |