Proof of Theorem rabxfrd
| Step | Hyp | Ref
| Expression |
| 1 | | rabxfrd.3 |
. . . . . . . . . . 11
⊢ ((𝜑 ∧ 𝑦 ∈ 𝐷) → 𝐴 ∈ 𝐷) |
| 2 | 1 | ex 412 |
. . . . . . . . . 10
⊢ (𝜑 → (𝑦 ∈ 𝐷 → 𝐴 ∈ 𝐷)) |
| 3 | | ibibr 368 |
. . . . . . . . . 10
⊢ ((𝑦 ∈ 𝐷 → 𝐴 ∈ 𝐷) ↔ (𝑦 ∈ 𝐷 → (𝐴 ∈ 𝐷 ↔ 𝑦 ∈ 𝐷))) |
| 4 | 2, 3 | sylib 218 |
. . . . . . . . 9
⊢ (𝜑 → (𝑦 ∈ 𝐷 → (𝐴 ∈ 𝐷 ↔ 𝑦 ∈ 𝐷))) |
| 5 | 4 | imp 406 |
. . . . . . . 8
⊢ ((𝜑 ∧ 𝑦 ∈ 𝐷) → (𝐴 ∈ 𝐷 ↔ 𝑦 ∈ 𝐷)) |
| 6 | 5 | anbi1d 631 |
. . . . . . 7
⊢ ((𝜑 ∧ 𝑦 ∈ 𝐷) → ((𝐴 ∈ 𝐷 ∧ 𝜒) ↔ (𝑦 ∈ 𝐷 ∧ 𝜒))) |
| 7 | | rabxfrd.4 |
. . . . . . . 8
⊢ (𝑥 = 𝐴 → (𝜓 ↔ 𝜒)) |
| 8 | 7 | elrab 3692 |
. . . . . . 7
⊢ (𝐴 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓} ↔ (𝐴 ∈ 𝐷 ∧ 𝜒)) |
| 9 | | rabid 3458 |
. . . . . . 7
⊢ (𝑦 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒} ↔ (𝑦 ∈ 𝐷 ∧ 𝜒)) |
| 10 | 6, 8, 9 | 3bitr4g 314 |
. . . . . 6
⊢ ((𝜑 ∧ 𝑦 ∈ 𝐷) → (𝐴 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓} ↔ 𝑦 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒})) |
| 11 | 10 | rabbidva 3443 |
. . . . 5
⊢ (𝜑 → {𝑦 ∈ 𝐷 ∣ 𝐴 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓}} = {𝑦 ∈ 𝐷 ∣ 𝑦 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒}}) |
| 12 | 11 | eleq2d 2827 |
. . . 4
⊢ (𝜑 → (𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝐴 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓}} ↔ 𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝑦 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒}})) |
| 13 | | rabxfrd.1 |
. . . . 5
⊢
Ⅎ𝑦𝐵 |
| 14 | | nfcv 2905 |
. . . . 5
⊢
Ⅎ𝑦𝐷 |
| 15 | | rabxfrd.2 |
. . . . . 6
⊢
Ⅎ𝑦𝐶 |
| 16 | 15 | nfel1 2922 |
. . . . 5
⊢
Ⅎ𝑦 𝐶 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓} |
| 17 | | rabxfrd.5 |
. . . . . 6
⊢ (𝑦 = 𝐵 → 𝐴 = 𝐶) |
| 18 | 17 | eleq1d 2826 |
. . . . 5
⊢ (𝑦 = 𝐵 → (𝐴 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓} ↔ 𝐶 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓})) |
| 19 | 13, 14, 16, 18 | elrabf 3688 |
. . . 4
⊢ (𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝐴 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓}} ↔ (𝐵 ∈ 𝐷 ∧ 𝐶 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓})) |
| 20 | | nfrab1 3457 |
. . . . . 6
⊢
Ⅎ𝑦{𝑦 ∈ 𝐷 ∣ 𝜒} |
| 21 | 13, 20 | nfel 2920 |
. . . . 5
⊢
Ⅎ𝑦 𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒} |
| 22 | | eleq1 2829 |
. . . . 5
⊢ (𝑦 = 𝐵 → (𝑦 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒} ↔ 𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒})) |
| 23 | 13, 14, 21, 22 | elrabf 3688 |
. . . 4
⊢ (𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝑦 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒}} ↔ (𝐵 ∈ 𝐷 ∧ 𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒})) |
| 24 | 12, 19, 23 | 3bitr3g 313 |
. . 3
⊢ (𝜑 → ((𝐵 ∈ 𝐷 ∧ 𝐶 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓}) ↔ (𝐵 ∈ 𝐷 ∧ 𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒}))) |
| 25 | | pm5.32 573 |
. . 3
⊢ ((𝐵 ∈ 𝐷 → (𝐶 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓} ↔ 𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒})) ↔ ((𝐵 ∈ 𝐷 ∧ 𝐶 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓}) ↔ (𝐵 ∈ 𝐷 ∧ 𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒}))) |
| 26 | 24, 25 | sylibr 234 |
. 2
⊢ (𝜑 → (𝐵 ∈ 𝐷 → (𝐶 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓} ↔ 𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒}))) |
| 27 | 26 | imp 406 |
1
⊢ ((𝜑 ∧ 𝐵 ∈ 𝐷) → (𝐶 ∈ {𝑥 ∈ 𝐷 ∣ 𝜓} ↔ 𝐵 ∈ {𝑦 ∈ 𝐷 ∣ 𝜒})) |