| Step | Hyp | Ref | Expression | 
|---|
| 1 |  | ssint 4964 | . . . . . . 7
⊢ (𝐶 ⊆ ∩ {𝑥
∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} ↔ ∀𝑦 ∈ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥}𝐶 ⊆ 𝑦) | 
| 2 |  | sseq2 4010 | . . . . . . . 8
⊢ (𝑥 = 𝑦 → (𝐴 ⊆ 𝑥 ↔ 𝐴 ⊆ 𝑦)) | 
| 3 | 2 | ralrab 3699 | . . . . . . 7
⊢
(∀𝑦 ∈
{𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥}𝐶 ⊆ 𝑦 ↔ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) | 
| 4 | 1, 3 | bitri 275 | . . . . . 6
⊢ (𝐶 ⊆ ∩ {𝑥
∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} ↔ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) | 
| 5 | 4 | biimpri 228 | . . . . 5
⊢
(∀𝑦 ∈
𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦) → 𝐶 ⊆ ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥}) | 
| 6 | 5 | adantl 481 | . . . 4
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐴 ⊆ 𝐶) ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → 𝐶 ⊆ ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥}) | 
| 7 |  | sseq2 4010 | . . . . . . 7
⊢ (𝑧 = 𝐶 → (𝐴 ⊆ 𝑧 ↔ 𝐴 ⊆ 𝐶)) | 
| 8 |  | simpll 767 | . . . . . . 7
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐴 ⊆ 𝐶) ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → 𝐶 ∈ 𝐵) | 
| 9 |  | simplr 769 | . . . . . . 7
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐴 ⊆ 𝐶) ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → 𝐴 ⊆ 𝐶) | 
| 10 | 7, 8, 9 | elrabd 3694 | . . . . . 6
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐴 ⊆ 𝐶) ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → 𝐶 ∈ {𝑧 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑧}) | 
| 11 |  | sseq2 4010 | . . . . . . 7
⊢ (𝑧 = 𝑥 → (𝐴 ⊆ 𝑧 ↔ 𝐴 ⊆ 𝑥)) | 
| 12 | 11 | cbvrabv 3447 | . . . . . 6
⊢ {𝑧 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑧} = {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} | 
| 13 | 10, 12 | eleqtrdi 2851 | . . . . 5
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐴 ⊆ 𝐶) ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → 𝐶 ∈ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥}) | 
| 14 |  | intss1 4963 | . . . . 5
⊢ (𝐶 ∈ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} → ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} ⊆ 𝐶) | 
| 15 | 13, 14 | syl 17 | . . . 4
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐴 ⊆ 𝐶) ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} ⊆ 𝐶) | 
| 16 | 6, 15 | eqssd 4001 | . . 3
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐴 ⊆ 𝐶) ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → 𝐶 = ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥}) | 
| 17 | 16 | expl 457 | . 2
⊢ (𝐶 ∈ 𝐵 → ((𝐴 ⊆ 𝐶 ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → 𝐶 = ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥})) | 
| 18 |  | ssintub 4966 | . . . 4
⊢ 𝐴 ⊆ ∩ {𝑥
∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} | 
| 19 |  | sseq2 4010 | . . . 4
⊢ (𝐶 = ∩
{𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} → (𝐴 ⊆ 𝐶 ↔ 𝐴 ⊆ ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥})) | 
| 20 | 18, 19 | mpbiri 258 | . . 3
⊢ (𝐶 = ∩
{𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} → 𝐴 ⊆ 𝐶) | 
| 21 |  | eqimss 4042 | . . . 4
⊢ (𝐶 = ∩
{𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} → 𝐶 ⊆ ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥}) | 
| 22 | 21, 4 | sylib 218 | . . 3
⊢ (𝐶 = ∩
{𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} → ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) | 
| 23 | 20, 22 | jca 511 | . 2
⊢ (𝐶 = ∩
{𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} → (𝐴 ⊆ 𝐶 ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦))) | 
| 24 | 17, 23 | impbid1 225 | 1
⊢ (𝐶 ∈ 𝐵 → ((𝐴 ⊆ 𝐶 ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) ↔ 𝐶 = ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥})) |