| Step | Hyp | Ref
| Expression |
| 1 | | sseq1 3959 |
. . . . . . 7
⊢ (𝑧 = 𝐶 → (𝑧 ⊆ 𝐴 ↔ 𝐶 ⊆ 𝐴)) |
| 2 | | simpll 776 |
. . . . . . 7
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐶 ⊆ 𝐴) ∧ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) → 𝐶 ∈ 𝐵) |
| 3 | | simplr 778 |
. . . . . . 7
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐶 ⊆ 𝐴) ∧ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) → 𝐶 ⊆ 𝐴) |
| 4 | 1, 2, 3 | elrabd 3651 |
. . . . . 6
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐶 ⊆ 𝐴) ∧ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) → 𝐶 ∈ {𝑧 ∈ 𝐵 ∣ 𝑧 ⊆ 𝐴}) |
| 5 | | sseq1 3959 |
. . . . . . 7
⊢ (𝑧 = 𝑥 → (𝑧 ⊆ 𝐴 ↔ 𝑥 ⊆ 𝐴)) |
| 6 | 5 | cbvrabv 3423 |
. . . . . 6
⊢ {𝑧 ∈ 𝐵 ∣ 𝑧 ⊆ 𝐴} = {𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} |
| 7 | 4, 6 | eleqtrdi 2871 |
. . . . 5
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐶 ⊆ 𝐴) ∧ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) → 𝐶 ∈ {𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴}) |
| 8 | | elssuni 4894 |
. . . . 5
⊢ (𝐶 ∈ {𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} → 𝐶 ⊆ ∪ {𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴}) |
| 9 | 7, 8 | syl 17 |
. . . 4
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐶 ⊆ 𝐴) ∧ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) → 𝐶 ⊆ ∪ {𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴}) |
| 10 | | unissb 4896 |
. . . . . 6
⊢ (∪ {𝑥
∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} ⊆ 𝐶 ↔ ∀𝑦 ∈ {𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴}𝑦 ⊆ 𝐶) |
| 11 | | sseq1 3959 |
. . . . . . 7
⊢ (𝑥 = 𝑦 → (𝑥 ⊆ 𝐴 ↔ 𝑦 ⊆ 𝐴)) |
| 12 | 11 | ralrab 3655 |
. . . . . 6
⊢
(∀𝑦 ∈
{𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴}𝑦 ⊆ 𝐶 ↔ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) |
| 13 | 10, 12 | bitri 277 |
. . . . 5
⊢ (∪ {𝑥
∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} ⊆ 𝐶 ↔ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) |
| 14 | 13 | bilanri 510 |
. . . 4
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐶 ⊆ 𝐴) ∧ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) → ∪
{𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} ⊆ 𝐶) |
| 15 | 9, 14 | eqssd 3951 |
. . 3
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐶 ⊆ 𝐴) ∧ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) → 𝐶 = ∪ {𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴}) |
| 16 | 15 | expl 461 |
. 2
⊢ (𝐶 ∈ 𝐵 → ((𝐶 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) → 𝐶 = ∪ {𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴})) |
| 17 | | unilbss 49399 |
. . . 4
⊢ ∪ {𝑥
∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} ⊆ 𝐴 |
| 18 | | sseq1 3959 |
. . . 4
⊢ (𝐶 = ∪
{𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} → (𝐶 ⊆ 𝐴 ↔ ∪ {𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} ⊆ 𝐴)) |
| 19 | 17, 18 | mpbiri 260 |
. . 3
⊢ (𝐶 = ∪
{𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} → 𝐶 ⊆ 𝐴) |
| 20 | | eqimss2 3993 |
. . . 4
⊢ (𝐶 = ∪
{𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} → ∪ {𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} ⊆ 𝐶) |
| 21 | 20, 13 | sylib 220 |
. . 3
⊢ (𝐶 = ∪
{𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} → ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) |
| 22 | 19, 21 | jca 519 |
. 2
⊢ (𝐶 = ∪
{𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} → (𝐶 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶))) |
| 23 | 16, 22 | impbid1 227 |
1
⊢ (𝐶 ∈ 𝐵 → ((𝐶 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) ↔ 𝐶 = ∪ {𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴})) |