Proof of Theorem resoprab2
Step | Hyp | Ref
| Expression |
1 | | resoprab 7370 |
. 2
⊢
({〈〈𝑥,
𝑦〉, 𝑧〉 ∣ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑)} ↾ (𝐶 × 𝐷)) = {〈〈𝑥, 𝑦〉, 𝑧〉 ∣ ((𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐷) ∧ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑))} |
2 | | anass 468 |
. . . 4
⊢ ((((𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐷) ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐷) ∧ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑))) |
3 | | an4 652 |
. . . . . 6
⊢ (((𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐷) ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)) ↔ ((𝑥 ∈ 𝐶 ∧ 𝑥 ∈ 𝐴) ∧ (𝑦 ∈ 𝐷 ∧ 𝑦 ∈ 𝐵))) |
4 | | ssel 3910 |
. . . . . . . . 9
⊢ (𝐶 ⊆ 𝐴 → (𝑥 ∈ 𝐶 → 𝑥 ∈ 𝐴)) |
5 | 4 | pm4.71d 561 |
. . . . . . . 8
⊢ (𝐶 ⊆ 𝐴 → (𝑥 ∈ 𝐶 ↔ (𝑥 ∈ 𝐶 ∧ 𝑥 ∈ 𝐴))) |
6 | 5 | bicomd 222 |
. . . . . . 7
⊢ (𝐶 ⊆ 𝐴 → ((𝑥 ∈ 𝐶 ∧ 𝑥 ∈ 𝐴) ↔ 𝑥 ∈ 𝐶)) |
7 | | ssel 3910 |
. . . . . . . . 9
⊢ (𝐷 ⊆ 𝐵 → (𝑦 ∈ 𝐷 → 𝑦 ∈ 𝐵)) |
8 | 7 | pm4.71d 561 |
. . . . . . . 8
⊢ (𝐷 ⊆ 𝐵 → (𝑦 ∈ 𝐷 ↔ (𝑦 ∈ 𝐷 ∧ 𝑦 ∈ 𝐵))) |
9 | 8 | bicomd 222 |
. . . . . . 7
⊢ (𝐷 ⊆ 𝐵 → ((𝑦 ∈ 𝐷 ∧ 𝑦 ∈ 𝐵) ↔ 𝑦 ∈ 𝐷)) |
10 | 6, 9 | bi2anan9 635 |
. . . . . 6
⊢ ((𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵) → (((𝑥 ∈ 𝐶 ∧ 𝑥 ∈ 𝐴) ∧ (𝑦 ∈ 𝐷 ∧ 𝑦 ∈ 𝐵)) ↔ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐷))) |
11 | 3, 10 | syl5bb 282 |
. . . . 5
⊢ ((𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵) → (((𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐷) ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)) ↔ (𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐷))) |
12 | 11 | anbi1d 629 |
. . . 4
⊢ ((𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵) → ((((𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐷) ∧ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵)) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐷) ∧ 𝜑))) |
13 | 2, 12 | bitr3id 284 |
. . 3
⊢ ((𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵) → (((𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐷) ∧ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑)) ↔ ((𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐷) ∧ 𝜑))) |
14 | 13 | oprabbidv 7319 |
. 2
⊢ ((𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵) → {〈〈𝑥, 𝑦〉, 𝑧〉 ∣ ((𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐷) ∧ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑))} = {〈〈𝑥, 𝑦〉, 𝑧〉 ∣ ((𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐷) ∧ 𝜑)}) |
15 | 1, 14 | eqtrid 2790 |
1
⊢ ((𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵) → ({〈〈𝑥, 𝑦〉, 𝑧〉 ∣ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑)} ↾ (𝐶 × 𝐷)) = {〈〈𝑥, 𝑦〉, 𝑧〉 ∣ ((𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐷) ∧ 𝜑)}) |