Step | Hyp | Ref
| Expression |
1 | | ssint 4895 |
. . . . . . 7
⊢ (𝐶 ⊆ ∩ {𝑥
∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} ↔ ∀𝑦 ∈ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥}𝐶 ⊆ 𝑦) |
2 | | sseq2 3947 |
. . . . . . . 8
⊢ (𝑥 = 𝑦 → (𝐴 ⊆ 𝑥 ↔ 𝐴 ⊆ 𝑦)) |
3 | 2 | ralrab 3630 |
. . . . . . 7
⊢
(∀𝑦 ∈
{𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥}𝐶 ⊆ 𝑦 ↔ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) |
4 | 1, 3 | bitri 274 |
. . . . . 6
⊢ (𝐶 ⊆ ∩ {𝑥
∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} ↔ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) |
5 | 4 | biimpri 227 |
. . . . 5
⊢
(∀𝑦 ∈
𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦) → 𝐶 ⊆ ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥}) |
6 | 5 | adantl 482 |
. . . 4
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐴 ⊆ 𝐶) ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → 𝐶 ⊆ ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥}) |
7 | | sseq2 3947 |
. . . . . . 7
⊢ (𝑧 = 𝐶 → (𝐴 ⊆ 𝑧 ↔ 𝐴 ⊆ 𝐶)) |
8 | | simpll 764 |
. . . . . . 7
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐴 ⊆ 𝐶) ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → 𝐶 ∈ 𝐵) |
9 | | simplr 766 |
. . . . . . 7
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐴 ⊆ 𝐶) ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → 𝐴 ⊆ 𝐶) |
10 | 7, 8, 9 | elrabd 3626 |
. . . . . 6
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐴 ⊆ 𝐶) ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → 𝐶 ∈ {𝑧 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑧}) |
11 | | sseq2 3947 |
. . . . . . 7
⊢ (𝑧 = 𝑥 → (𝐴 ⊆ 𝑧 ↔ 𝐴 ⊆ 𝑥)) |
12 | 11 | cbvrabv 3426 |
. . . . . 6
⊢ {𝑧 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑧} = {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} |
13 | 10, 12 | eleqtrdi 2849 |
. . . . 5
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐴 ⊆ 𝐶) ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → 𝐶 ∈ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥}) |
14 | | intss1 4894 |
. . . . 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 3938 |
. . 3
⊢ (((𝐶 ∈ 𝐵 ∧ 𝐴 ⊆ 𝐶) ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → 𝐶 = ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥}) |
17 | 16 | expl 458 |
. 2
⊢ (𝐶 ∈ 𝐵 → ((𝐴 ⊆ 𝐶 ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) → 𝐶 = ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥})) |
18 | | ssintub 4897 |
. . . 4
⊢ 𝐴 ⊆ ∩ {𝑥
∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} |
19 | | sseq2 3947 |
. . . 4
⊢ (𝐶 = ∩
{𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} → (𝐴 ⊆ 𝐶 ↔ 𝐴 ⊆ ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥})) |
20 | 18, 19 | mpbiri 257 |
. . 3
⊢ (𝐶 = ∩
{𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} → 𝐴 ⊆ 𝐶) |
21 | | eqimss 3977 |
. . . 4
⊢ (𝐶 = ∩
{𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} → 𝐶 ⊆ ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥}) |
22 | 21, 4 | sylib 217 |
. . 3
⊢ (𝐶 = ∩
{𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} → ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) |
23 | 20, 22 | jca 512 |
. 2
⊢ (𝐶 = ∩
{𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥} → (𝐴 ⊆ 𝐶 ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦))) |
24 | 17, 23 | impbid1 224 |
1
⊢ (𝐶 ∈ 𝐵 → ((𝐴 ⊆ 𝐶 ∧ ∀𝑦 ∈ 𝐵 (𝐴 ⊆ 𝑦 → 𝐶 ⊆ 𝑦)) ↔ 𝐶 = ∩ {𝑥 ∈ 𝐵 ∣ 𝐴 ⊆ 𝑥})) |