Step | Hyp | Ref
| Expression |
1 | | abn0 4295 |
. 2
⊢ ({𝑥 ∣ 𝜑} ≠ ∅ ↔ ∃𝑥𝜑) |
2 | | scott0 9502 |
. . . 4
⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑧 ∈ {𝑥 ∣ 𝜑} ∣ ∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑧) ⊆ (rank‘𝑦)} = ∅) |
3 | | nfcv 2904 |
. . . . . . 7
⊢
Ⅎ𝑧{𝑥 ∣ 𝜑} |
4 | | nfab1 2906 |
. . . . . . 7
⊢
Ⅎ𝑥{𝑥 ∣ 𝜑} |
5 | | nfv 1922 |
. . . . . . . 8
⊢
Ⅎ𝑥(rank‘𝑧) ⊆ (rank‘𝑦) |
6 | 4, 5 | nfralw 3147 |
. . . . . . 7
⊢
Ⅎ𝑥∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑧) ⊆ (rank‘𝑦) |
7 | | nfv 1922 |
. . . . . . 7
⊢
Ⅎ𝑧∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑥) ⊆ (rank‘𝑦) |
8 | | fveq2 6717 |
. . . . . . . . 9
⊢ (𝑧 = 𝑥 → (rank‘𝑧) = (rank‘𝑥)) |
9 | 8 | sseq1d 3932 |
. . . . . . . 8
⊢ (𝑧 = 𝑥 → ((rank‘𝑧) ⊆ (rank‘𝑦) ↔ (rank‘𝑥) ⊆ (rank‘𝑦))) |
10 | 9 | ralbidv 3118 |
. . . . . . 7
⊢ (𝑧 = 𝑥 → (∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑧) ⊆ (rank‘𝑦) ↔ ∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑥) ⊆ (rank‘𝑦))) |
11 | 3, 4, 6, 7, 10 | cbvrabw 3400 |
. . . . . 6
⊢ {𝑧 ∈ {𝑥 ∣ 𝜑} ∣ ∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑧) ⊆ (rank‘𝑦)} = {𝑥 ∈ {𝑥 ∣ 𝜑} ∣ ∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑥) ⊆ (rank‘𝑦)} |
12 | | df-rab 3070 |
. . . . . 6
⊢ {𝑥 ∈ {𝑥 ∣ 𝜑} ∣ ∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑥) ⊆ (rank‘𝑦)} = {𝑥 ∣ (𝑥 ∈ {𝑥 ∣ 𝜑} ∧ ∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑥) ⊆ (rank‘𝑦))} |
13 | | abid 2718 |
. . . . . . . 8
⊢ (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑) |
14 | | df-ral 3066 |
. . . . . . . . 9
⊢
(∀𝑦 ∈
{𝑥 ∣ 𝜑} (rank‘𝑥) ⊆ (rank‘𝑦) ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} → (rank‘𝑥) ⊆ (rank‘𝑦))) |
15 | | df-sbc 3695 |
. . . . . . . . . . 11
⊢
([𝑦 / 𝑥]𝜑 ↔ 𝑦 ∈ {𝑥 ∣ 𝜑}) |
16 | 15 | imbi1i 353 |
. . . . . . . . . 10
⊢
(([𝑦 / 𝑥]𝜑 → (rank‘𝑥) ⊆ (rank‘𝑦)) ↔ (𝑦 ∈ {𝑥 ∣ 𝜑} → (rank‘𝑥) ⊆ (rank‘𝑦))) |
17 | 16 | albii 1827 |
. . . . . . . . 9
⊢
(∀𝑦([𝑦 / 𝑥]𝜑 → (rank‘𝑥) ⊆ (rank‘𝑦)) ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} → (rank‘𝑥) ⊆ (rank‘𝑦))) |
18 | 14, 17 | bitr4i 281 |
. . . . . . . 8
⊢
(∀𝑦 ∈
{𝑥 ∣ 𝜑} (rank‘𝑥) ⊆ (rank‘𝑦) ↔ ∀𝑦([𝑦 / 𝑥]𝜑 → (rank‘𝑥) ⊆ (rank‘𝑦))) |
19 | 13, 18 | anbi12i 630 |
. . . . . . 7
⊢ ((𝑥 ∈ {𝑥 ∣ 𝜑} ∧ ∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑥) ⊆ (rank‘𝑦)) ↔ (𝜑 ∧ ∀𝑦([𝑦 / 𝑥]𝜑 → (rank‘𝑥) ⊆ (rank‘𝑦)))) |
20 | 19 | abbii 2808 |
. . . . . 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 | 3eqtri 2769 |
. . . . 5
⊢ {𝑧 ∈ {𝑥 ∣ 𝜑} ∣ ∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑧) ⊆ (rank‘𝑦)} = {𝑥 ∣ (𝜑 ∧ ∀𝑦([𝑦 / 𝑥]𝜑 → (rank‘𝑥) ⊆ (rank‘𝑦)))} |
22 | 21 | eqeq1i 2742 |
. . . 4
⊢ ({𝑧 ∈ {𝑥 ∣ 𝜑} ∣ ∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑧) ⊆ (rank‘𝑦)} = ∅ ↔ {𝑥 ∣ (𝜑 ∧ ∀𝑦([𝑦 / 𝑥]𝜑 → (rank‘𝑥) ⊆ (rank‘𝑦)))} = ∅) |
23 | 2, 22 | bitri 278 |
. . 3
⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ (𝜑 ∧ ∀𝑦([𝑦 / 𝑥]𝜑 → (rank‘𝑥) ⊆ (rank‘𝑦)))} = ∅) |
24 | 23 | necon3bii 2993 |
. 2
⊢ ({𝑥 ∣ 𝜑} ≠ ∅ ↔ {𝑥 ∣ (𝜑 ∧ ∀𝑦([𝑦 / 𝑥]𝜑 → (rank‘𝑥) ⊆ (rank‘𝑦)))} ≠ ∅) |
25 | 1, 24 | bitr3i 280 |
1
⊢
(∃𝑥𝜑 ↔ {𝑥 ∣ (𝜑 ∧ ∀𝑦([𝑦 / 𝑥]𝜑 → (rank‘𝑥) ⊆ (rank‘𝑦)))} ≠ ∅) |