Proof of Theorem scott0s
| Step | Hyp | Ref
| Expression |
| 1 | | abn0 1715 |
. 2
⊢ (¬ {x∣φ} =
∅ ↔ ∃xφ) |
| 2 | | scott0 3542 |
. . . 4
⊢ ({x∣φ} =
∅ ↔ {z ∈ {x∣φ}∣∀y ∈ {x∣φ}
(rank ‘z) ⊆ (rank
‘y)} = ∅) |
| 3 | | ax-17 925 |
. . . . . . 7
⊢ (y
∈ {x∣φ} → ∀z y ∈
{x∣φ}) |
| 4 | | hbab1 1095 |
. . . . . . 7
⊢ (y
∈ {x∣φ} → ∀x y ∈
{x∣φ}) |
| 5 | | ax-17 925 |
. . . . . . . 8
⊢ ((rank ‘z) ⊆ (rank ‘y) → ∀x(rank ‘z)
⊆ (rank ‘y)) |
| 6 | 4, 5 | hbral 1236 |
. . . . . . 7
⊢ (∀y ∈ {x∣φ}
(rank ‘z) ⊆ (rank
‘y) → ∀x∀y
∈ {x∣φ} (rank ‘z) ⊆ (rank ‘y)) |
| 7 | | ax-17 925 |
. . . . . . 7
⊢ (∀y ∈ {x∣φ}
(rank ‘x) ⊆ (rank
‘y) → ∀z∀y
∈ {x∣φ} (rank ‘x) ⊆ (rank ‘y)) |
| 8 | | fveq2 2832 |
. . . . . . . . 9
⊢ (z =
x → (rank ‘z) = (rank ‘x)) |
| 9 | 8 | sseq1d 1527 |
. . . . . . . 8
⊢ (z =
x → ((rank ‘z) ⊆ (rank ‘y) ↔ (rank ‘x) ⊆ (rank ‘y))) |
| 10 | 9 | biraldv 1219 |
. . . . . . 7
⊢ (z =
x → (∀y ∈ {x∣φ}
(rank ‘z) ⊆ (rank
‘y) ↔ ∀y ∈ {x∣φ}
(rank ‘x) ⊆ (rank
‘y))) |
| 11 | 3, 4, 6, 7, 10 | cbvrab 1425 |
. . . . . 6
⊢ {z
∈ {x∣φ}∣∀y ∈ {x∣φ}
(rank ‘z) ⊆ (rank
‘y)} = {x ∈ {x∣φ}∣∀y ∈ {x∣φ}
(rank ‘x) ⊆ (rank
‘y)} |
| 12 | | df-rab 1208 |
. . . . . 6
⊢ {x
∈ {x∣φ}∣∀y ∈ {x∣φ}
(rank ‘x) ⊆ (rank
‘y)} = {x∣(x
∈ {x∣φ} ∧ ∀y ∈ {x∣φ}
(rank ‘x) ⊆ (rank
‘y))} |
| 13 | | abid 1094 |
. . . . . . . 8
⊢ (x
∈ {x∣φ} ↔ φ) |
| 14 | | df-ral 1205 |
. . . . . . . . 9
⊢ (∀y ∈ {x∣φ}
(rank ‘x) ⊆ (rank
‘y) ↔ ∀y(y ∈
{x∣φ} → (rank ‘x) ⊆ (rank ‘y))) |
| 15 | | df-clab 1093 |
. . . . . . . . . . 11
⊢ (y
∈ {x∣φ} ↔ [y / x]φ) |
| 16 | 15 | imbi1i 161 |
. . . . . . . . . 10
⊢ ((y
∈ {x∣φ} → (rank ‘x) ⊆ (rank ‘y)) ↔ ([y /
x]φ
→ (rank ‘x) ⊆ (rank
‘y))) |
| 17 | 16 | bial 695 |
. . . . . . . . 9
⊢ (∀y(y ∈
{x∣φ} → (rank ‘x) ⊆ (rank ‘y)) ↔ ∀y([y / x]φ →
(rank ‘x) ⊆ (rank
‘y))) |
| 18 | 14, 17 | bitr 151 |
. . . . . . . 8
⊢ (∀y ∈ {x∣φ}
(rank ‘x) ⊆ (rank
‘y) ↔ ∀y([y / x]φ →
(rank ‘x) ⊆ (rank
‘y))) |
| 19 | 13, 18 | anbi12i 369 |
. . . . . . 7
⊢ ((x
∈ {x∣φ} ∧ ∀y ∈ {x∣φ}
(rank ‘x) ⊆ (rank
‘y)) ↔ (φ ∧ ∀y([y / x]φ →
(rank ‘x) ⊆ (rank
‘y)))) |
| 20 | 19 | biabi 1181 |
. . . . . 6
⊢ {x∣(x
∈ {x∣φ} ∧ ∀y ∈ {x∣φ}
(rank ‘x) ⊆ (rank
‘y))} = {x∣(φ
∧ ∀y([y / x]φ → (rank ‘x) ⊆ (rank ‘y)))} |
| 21 | 11, 12, 20 | 3eqtr 1123 |
. . . . 5
⊢ {z
∈ {x∣φ}∣∀y ∈ {x∣φ}
(rank ‘z) ⊆ (rank
‘y)} = {x∣(φ
∧ ∀y([y / x]φ → (rank ‘x) ⊆ (rank ‘y)))} |
| 22 | 21 | cleq1i 1108 |
. . . 4
⊢ ({z
∈ {x∣φ}∣∀y ∈ {x∣φ}
(rank ‘z) ⊆ (rank
‘y)} = ∅ ↔ {x∣(φ
∧ ∀y([y / x]φ → (rank ‘x) ⊆ (rank ‘y)))} = ∅) |
| 23 | 2, 22 | bitr 151 |
. . 3
⊢ ({x∣φ} =
∅ ↔ {x∣(φ ∧ ∀y([y / x]φ →
(rank ‘x) ⊆ (rank
‘y)))} = ∅) |
| 24 | 23 | negbii 162 |
. 2
⊢ (¬ {x∣φ} =
∅ ↔ ¬ {x∣(φ ∧ ∀y([y / x]φ →
(rank ‘x) ⊆ (rank
‘y)))} = ∅) |
| 25 | 1, 24 | bitr3 153 |
1
⊢ (∃xφ ↔
¬ {x∣(φ ∧ ∀y([y / x]φ →
(rank ‘x) ⊆ (rank
‘y)))} = ∅) |