| Step | Hyp | Ref
| Expression |
| 1 | | abn0 4367 |
. 2
⊢ ({𝑥 ∣ 𝜑} ≠ ∅ ↔ ∃𝑥𝜑) |
| 2 | | scott0 9909 |
. . . 4
⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑧 ∈ {𝑥 ∣ 𝜑} ∣ ∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑧) ⊆ (rank‘𝑦)} = ∅) |
| 3 | | nfcv 2897 |
. . . . . . 7
⊢
Ⅎ𝑧{𝑥 ∣ 𝜑} |
| 4 | | nfab1 2899 |
. . . . . . 7
⊢
Ⅎ𝑥{𝑥 ∣ 𝜑} |
| 5 | | nfv 1913 |
. . . . . . . 8
⊢
Ⅎ𝑥(rank‘𝑧) ⊆ (rank‘𝑦) |
| 6 | 4, 5 | nfralw 3295 |
. . . . . . 7
⊢
Ⅎ𝑥∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑧) ⊆ (rank‘𝑦) |
| 7 | | nfv 1913 |
. . . . . . 7
⊢
Ⅎ𝑧∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑥) ⊆ (rank‘𝑦) |
| 8 | | fveq2 6887 |
. . . . . . . . 9
⊢ (𝑧 = 𝑥 → (rank‘𝑧) = (rank‘𝑥)) |
| 9 | 8 | sseq1d 3997 |
. . . . . . . 8
⊢ (𝑧 = 𝑥 → ((rank‘𝑧) ⊆ (rank‘𝑦) ↔ (rank‘𝑥) ⊆ (rank‘𝑦))) |
| 10 | 9 | ralbidv 3165 |
. . . . . . 7
⊢ (𝑧 = 𝑥 → (∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑧) ⊆ (rank‘𝑦) ↔ ∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑥) ⊆ (rank‘𝑦))) |
| 11 | 3, 4, 6, 7, 10 | cbvrabw 3457 |
. . . . . 6
⊢ {𝑧 ∈ {𝑥 ∣ 𝜑} ∣ ∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑧) ⊆ (rank‘𝑦)} = {𝑥 ∈ {𝑥 ∣ 𝜑} ∣ ∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑥) ⊆ (rank‘𝑦)} |
| 12 | | df-rab 3421 |
. . . . . 6
⊢ {𝑥 ∈ {𝑥 ∣ 𝜑} ∣ ∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑥) ⊆ (rank‘𝑦)} = {𝑥 ∣ (𝑥 ∈ {𝑥 ∣ 𝜑} ∧ ∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑥) ⊆ (rank‘𝑦))} |
| 13 | | abid 2716 |
. . . . . . . 8
⊢ (𝑥 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑) |
| 14 | | df-ral 3051 |
. . . . . . . . 9
⊢
(∀𝑦 ∈
{𝑥 ∣ 𝜑} (rank‘𝑥) ⊆ (rank‘𝑦) ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} → (rank‘𝑥) ⊆ (rank‘𝑦))) |
| 15 | | df-sbc 3773 |
. . . . . . . . . . 11
⊢
([𝑦 / 𝑥]𝜑 ↔ 𝑦 ∈ {𝑥 ∣ 𝜑}) |
| 16 | 15 | imbi1i 349 |
. . . . . . . . . 10
⊢
(([𝑦 / 𝑥]𝜑 → (rank‘𝑥) ⊆ (rank‘𝑦)) ↔ (𝑦 ∈ {𝑥 ∣ 𝜑} → (rank‘𝑥) ⊆ (rank‘𝑦))) |
| 17 | 16 | albii 1818 |
. . . . . . . . 9
⊢
(∀𝑦([𝑦 / 𝑥]𝜑 → (rank‘𝑥) ⊆ (rank‘𝑦)) ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} → (rank‘𝑥) ⊆ (rank‘𝑦))) |
| 18 | 14, 17 | bitr4i 278 |
. . . . . . . 8
⊢
(∀𝑦 ∈
{𝑥 ∣ 𝜑} (rank‘𝑥) ⊆ (rank‘𝑦) ↔ ∀𝑦([𝑦 / 𝑥]𝜑 → (rank‘𝑥) ⊆ (rank‘𝑦))) |
| 19 | 13, 18 | anbi12i 628 |
. . . . . . 7
⊢ ((𝑥 ∈ {𝑥 ∣ 𝜑} ∧ ∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑥) ⊆ (rank‘𝑦)) ↔ (𝜑 ∧ ∀𝑦([𝑦 / 𝑥]𝜑 → (rank‘𝑥) ⊆ (rank‘𝑦)))) |
| 20 | 19 | abbii 2801 |
. . . . . 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 2761 |
. . . . 5
⊢ {𝑧 ∈ {𝑥 ∣ 𝜑} ∣ ∀𝑦 ∈ {𝑥 ∣ 𝜑} (rank‘𝑧) ⊆ (rank‘𝑦)} = {𝑥 ∣ (𝜑 ∧ ∀𝑦([𝑦 / 𝑥]𝜑 → (rank‘𝑥) ⊆ (rank‘𝑦)))} |
| 22 | 21 | eqeq1i 2739 |
. . . 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 275 |
. . 3
⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ (𝜑 ∧ ∀𝑦([𝑦 / 𝑥]𝜑 → (rank‘𝑥) ⊆ (rank‘𝑦)))} = ∅) |
| 24 | 23 | necon3bii 2983 |
. 2
⊢ ({𝑥 ∣ 𝜑} ≠ ∅ ↔ {𝑥 ∣ (𝜑 ∧ ∀𝑦([𝑦 / 𝑥]𝜑 → (rank‘𝑥) ⊆ (rank‘𝑦)))} ≠ ∅) |
| 25 | 1, 24 | bitr3i 277 |
1
⊢
(∃𝑥𝜑 ↔ {𝑥 ∣ (𝜑 ∧ ∀𝑦([𝑦 / 𝑥]𝜑 → (rank‘𝑥) ⊆ (rank‘𝑦)))} ≠ ∅) |