| Step | Hyp | Ref | Expression | 
|---|
| 1 |  | sseq1 4008 | . . . . . . 7
⊢ (𝑧 = 𝐶 → (𝑧 ⊆ 𝐴 ↔ 𝐶 ⊆ 𝐴)) | 
| 2 |  | simpll 766 | . . . . . . 7
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐶 ⊆ 𝐴) ∧ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) → 𝐶 ∈ 𝐵) | 
| 3 |  | simplr 768 | . . . . . . 7
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐶 ⊆ 𝐴) ∧ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) → 𝐶 ⊆ 𝐴) | 
| 4 | 1, 2, 3 | elrabd 3693 | . . . . . 6
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐶 ⊆ 𝐴) ∧ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) → 𝐶 ∈ {𝑧 ∈ 𝐵 ∣ 𝑧 ⊆ 𝐴}) | 
| 5 |  | sseq1 4008 | . . . . . . 7
⊢ (𝑧 = 𝑥 → (𝑧 ⊆ 𝐴 ↔ 𝑥 ⊆ 𝐴)) | 
| 6 | 5 | cbvrabv 3446 | . . . . . 6
⊢ {𝑧 ∈ 𝐵 ∣ 𝑧 ⊆ 𝐴} = {𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} | 
| 7 | 4, 6 | eleqtrdi 2850 | . . . . 5
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐶 ⊆ 𝐴) ∧ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) → 𝐶 ∈ {𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴}) | 
| 8 |  | elssuni 4936 | . . . . 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 4938 | . . . . . . 7
⊢ (∪ {𝑥
∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} ⊆ 𝐶 ↔ ∀𝑦 ∈ {𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴}𝑦 ⊆ 𝐶) | 
| 11 |  | sseq1 4008 | . . . . . . . 8
⊢ (𝑥 = 𝑦 → (𝑥 ⊆ 𝐴 ↔ 𝑦 ⊆ 𝐴)) | 
| 12 | 11 | ralrab 3698 | . . . . . . 7
⊢
(∀𝑦 ∈
{𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴}𝑦 ⊆ 𝐶 ↔ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) | 
| 13 | 10, 12 | bitri 275 | . . . . . 6
⊢ (∪ {𝑥
∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} ⊆ 𝐶 ↔ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) | 
| 14 | 13 | biimpri 228 | . . . . 5
⊢
(∀𝑦 ∈
𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶) → ∪ {𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} ⊆ 𝐶) | 
| 15 | 14 | adantl 481 | . . . 4
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐶 ⊆ 𝐴) ∧ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) → ∪
{𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} ⊆ 𝐶) | 
| 16 | 9, 15 | eqssd 4000 | . . 3
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐶 ⊆ 𝐴) ∧ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) → 𝐶 = ∪ {𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴}) | 
| 17 | 16 | expl 457 | . 2
⊢ (𝐶 ∈ 𝐵 → ((𝐶 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) → 𝐶 = ∪ {𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴})) | 
| 18 |  | unilbss 48742 | . . . 4
⊢ ∪ {𝑥
∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} ⊆ 𝐴 | 
| 19 |  | sseq1 4008 | . . . 4
⊢ (𝐶 = ∪
{𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} → (𝐶 ⊆ 𝐴 ↔ ∪ {𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} ⊆ 𝐴)) | 
| 20 | 18, 19 | mpbiri 258 | . . 3
⊢ (𝐶 = ∪
{𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} → 𝐶 ⊆ 𝐴) | 
| 21 |  | eqimss2 4042 | . . . 4
⊢ (𝐶 = ∪
{𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} → ∪ {𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} ⊆ 𝐶) | 
| 22 | 21, 13 | sylib 218 | . . 3
⊢ (𝐶 = ∪
{𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} → ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) | 
| 23 | 20, 22 | jca 511 | . 2
⊢ (𝐶 = ∪
{𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴} → (𝐶 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶))) | 
| 24 | 17, 23 | impbid1 225 | 1
⊢ (𝐶 ∈ 𝐵 → ((𝐶 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝐵 (𝑦 ⊆ 𝐴 → 𝑦 ⊆ 𝐶)) ↔ 𝐶 = ∪ {𝑥 ∈ 𝐵 ∣ 𝑥 ⊆ 𝐴})) |