Proof of Theorem iinrab2
| Step | Hyp | Ref
| Expression |
| 1 | | iineq1 4952 |
. . . . . 6
⊢ (𝐴 = ∅ → ∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} = ∩
𝑥 ∈ ∅ {𝑦 ∈ 𝐵 ∣ 𝜑}) |
| 2 | | 0iin 5007 |
. . . . . 6
⊢ ∩ 𝑥 ∈ ∅ {𝑦 ∈ 𝐵 ∣ 𝜑} = V |
| 3 | 1, 2 | eqtrdi 2788 |
. . . . 5
⊢ (𝐴 = ∅ → ∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} = V) |
| 4 | 3 | ineq1d 4160 |
. . . 4
⊢ (𝐴 = ∅ → (∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} ∩ 𝐵) = (V ∩ 𝐵)) |
| 5 | | inv1 4339 |
. . . . 5
⊢ (𝐵 ∩ V) = 𝐵 |
| 6 | 5 | ineqcomi 4152 |
. . . 4
⊢ (V ∩
𝐵) = 𝐵 |
| 7 | 4, 6 | eqtrdi 2788 |
. . 3
⊢ (𝐴 = ∅ → (∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} ∩ 𝐵) = 𝐵) |
| 8 | | rzal 4435 |
. . . 4
⊢ (𝐴 = ∅ → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 𝜑) |
| 9 | | rabid2 3423 |
. . . . 5
⊢ (𝐵 = {𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑} ↔ ∀𝑦 ∈ 𝐵 ∀𝑥 ∈ 𝐴 𝜑) |
| 10 | | ralcom 3266 |
. . . . 5
⊢
(∀𝑦 ∈
𝐵 ∀𝑥 ∈ 𝐴 𝜑 ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 𝜑) |
| 11 | 9, 10 | bitr2i 276 |
. . . 4
⊢
(∀𝑥 ∈
𝐴 ∀𝑦 ∈ 𝐵 𝜑 ↔ 𝐵 = {𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑}) |
| 12 | 8, 11 | sylib 218 |
. . 3
⊢ (𝐴 = ∅ → 𝐵 = {𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑}) |
| 13 | 7, 12 | eqtrd 2772 |
. 2
⊢ (𝐴 = ∅ → (∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} ∩ 𝐵) = {𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑}) |
| 14 | | iinrab 5012 |
. . . 4
⊢ (𝐴 ≠ ∅ → ∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} = {𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑}) |
| 15 | 14 | ineq1d 4160 |
. . 3
⊢ (𝐴 ≠ ∅ → (∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} ∩ 𝐵) = ({𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑} ∩ 𝐵)) |
| 16 | | ssrab2 4021 |
. . . 4
⊢ {𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑} ⊆ 𝐵 |
| 17 | | dfss 3909 |
. . . 4
⊢ ({𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑} ⊆ 𝐵 ↔ {𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑} = ({𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑} ∩ 𝐵)) |
| 18 | 16, 17 | mpbi 230 |
. . 3
⊢ {𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑} = ({𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑} ∩ 𝐵) |
| 19 | 15, 18 | eqtr4di 2790 |
. 2
⊢ (𝐴 ≠ ∅ → (∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} ∩ 𝐵) = {𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑}) |
| 20 | 13, 19 | pm2.61ine 3016 |
1
⊢ (∩ 𝑥 ∈ 𝐴 {𝑦 ∈ 𝐵 ∣ 𝜑} ∩ 𝐵) = {𝑦 ∈ 𝐵 ∣ ∀𝑥 ∈ 𝐴 𝜑} |