| Step | Hyp | Ref
| Expression |
| 1 | | ssint 4901 |
. . . . . 6
⊢ (𝐶 ⊆ ∩ {𝑥
∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} ↔ ∀𝑦 ∈ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥}𝐶 ⊆ 𝑦) |
| 2 | | sseq2 3948 |
. . . . . . 7
⊢ (𝑥 = 𝑦 → (𝐴 ⊆ 𝑥 ↔ 𝐴 ⊆ 𝑦)) |
| 3 | 2 | ralrab 3642 |
. . . . . 6
⊢
(∀𝑦 ∈
{𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥}𝐶 ⊆ 𝑦 ↔ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) |
| 4 | 1, 3 | bitri 276 |
. . . . 5
⊢ (𝐶 ⊆ ∩ {𝑥
∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} ↔ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) |
| 5 | 4 | bilanri 507 |
. . . 4
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐴 ⊆ 𝐶) ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → 𝐶 ⊆ ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥}) |
| 6 | | sseq2 3948 |
. . . . . . 7
⊢ (𝑧 = 𝐶 → (𝐴 ⊆ 𝑧 ↔ 𝐴 ⊆ 𝐶)) |
| 7 | | simpll 772 |
. . . . . . 7
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐴 ⊆ 𝐶) ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → 𝐶 ∈ 𝐵) |
| 8 | | simplr 774 |
. . . . . . 7
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐴 ⊆ 𝐶) ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → 𝐴 ⊆ 𝐶) |
| 9 | 6, 7, 8 | elrabd 3638 |
. . . . . 6
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐴 ⊆ 𝐶) ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → 𝐶 ∈ {𝑧 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑧}) |
| 10 | | sseq2 3948 |
. . . . . . 7
⊢ (𝑧 = 𝑥 → (𝐴 ⊆ 𝑧 ↔ 𝐴 ⊆ 𝑥)) |
| 11 | 10 | cbvrabv 3402 |
. . . . . 6
⊢ {𝑧 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑧} = {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} |
| 12 | 9, 11 | eleqtrdi 2850 |
. . . . 5
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐴 ⊆ 𝐶) ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → 𝐶 ∈ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥}) |
| 13 | | intss1 4900 |
. . . . 5
⊢ (𝐶 ∈ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} → ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} ⊆ 𝐶) |
| 14 | 12, 13 | syl 17 |
. . . 4
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐴 ⊆ 𝐶) ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} ⊆ 𝐶) |
| 15 | 5, 14 | eqssd 3939 |
. . 3
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐴 ⊆ 𝐶) ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → 𝐶 = ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥}) |
| 16 | 15 | expl 458 |
. 2
⊢ (𝐶 ∈ 𝐵 → ((𝐴 ⊆ 𝐶 ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → 𝐶 = ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥})) |
| 17 | | ssintub 4903 |
. . . 4
⊢ 𝐴 ⊆ ∩ {𝑥
∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} |
| 18 | | sseq2 3948 |
. . . 4
⊢ (𝐶 = ∩
{𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} → (𝐴 ⊆ 𝐶 ↔ 𝐴 ⊆ ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥})) |
| 19 | 17, 18 | mpbiri 259 |
. . 3
⊢ (𝐶 = ∩
{𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} → 𝐴 ⊆ 𝐶) |
| 20 | | eqimss 3980 |
. . . 4
⊢ (𝐶 = ∩
{𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} → 𝐶 ⊆ ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥}) |
| 21 | 20, 4 | sylib 219 |
. . 3
⊢ (𝐶 = ∩
{𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} → ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) |
| 22 | 19, 21 | jca 516 |
. 2
⊢ (𝐶 = ∩
{𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} → (𝐴 ⊆ 𝐶 ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦))) |
| 23 | 16, 22 | impbid1 226 |
1
⊢ (𝐶 ∈ 𝐵 → ((𝐴 ⊆ 𝐶 ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) ↔ 𝐶 = ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥})) |