Proof of Theorem ov2gf
Step | Hyp | Ref
| Expression |
1 | | elex 3185 |
. . 3
⊢ (𝑆 ∈ 𝐻 → 𝑆 ∈ V) |
2 | | ov2gf.a |
. . . 4
⊢
Ⅎ𝑥𝐴 |
3 | | ov2gf.c |
. . . 4
⊢
Ⅎ𝑦𝐴 |
4 | | ov2gf.d |
. . . 4
⊢
Ⅎ𝑦𝐵 |
5 | | ov2gf.1 |
. . . . . 6
⊢
Ⅎ𝑥𝐺 |
6 | 5 | nfel1 2765 |
. . . . 5
⊢
Ⅎ𝑥 𝐺 ∈ V |
7 | | ov2gf.5 |
. . . . . . . 8
⊢ 𝐹 = (𝑥 ∈ 𝐶, 𝑦 ∈ 𝐷 ↦ 𝑅) |
8 | | nfmpt21 6620 |
. . . . . . . 8
⊢
Ⅎ𝑥(𝑥 ∈ 𝐶, 𝑦 ∈ 𝐷 ↦ 𝑅) |
9 | 7, 8 | nfcxfr 2749 |
. . . . . . 7
⊢
Ⅎ𝑥𝐹 |
10 | | nfcv 2751 |
. . . . . . 7
⊢
Ⅎ𝑥𝑦 |
11 | 2, 9, 10 | nfov 6575 |
. . . . . 6
⊢
Ⅎ𝑥(𝐴𝐹𝑦) |
12 | 11, 5 | nfeq 2762 |
. . . . 5
⊢
Ⅎ𝑥(𝐴𝐹𝑦) = 𝐺 |
13 | 6, 12 | nfim 1813 |
. . . 4
⊢
Ⅎ𝑥(𝐺 ∈ V → (𝐴𝐹𝑦) = 𝐺) |
14 | | ov2gf.2 |
. . . . . 6
⊢
Ⅎ𝑦𝑆 |
15 | 14 | nfel1 2765 |
. . . . 5
⊢
Ⅎ𝑦 𝑆 ∈ V |
16 | | nfmpt22 6621 |
. . . . . . . 8
⊢
Ⅎ𝑦(𝑥 ∈ 𝐶, 𝑦 ∈ 𝐷 ↦ 𝑅) |
17 | 7, 16 | nfcxfr 2749 |
. . . . . . 7
⊢
Ⅎ𝑦𝐹 |
18 | 3, 17, 4 | nfov 6575 |
. . . . . 6
⊢
Ⅎ𝑦(𝐴𝐹𝐵) |
19 | 18, 14 | nfeq 2762 |
. . . . 5
⊢
Ⅎ𝑦(𝐴𝐹𝐵) = 𝑆 |
20 | 15, 19 | nfim 1813 |
. . . 4
⊢
Ⅎ𝑦(𝑆 ∈ V → (𝐴𝐹𝐵) = 𝑆) |
21 | | ov2gf.3 |
. . . . . 6
⊢ (𝑥 = 𝐴 → 𝑅 = 𝐺) |
22 | 21 | eleq1d 2672 |
. . . . 5
⊢ (𝑥 = 𝐴 → (𝑅 ∈ V ↔ 𝐺 ∈ V)) |
23 | | oveq1 6556 |
. . . . . 6
⊢ (𝑥 = 𝐴 → (𝑥𝐹𝑦) = (𝐴𝐹𝑦)) |
24 | 23, 21 | eqeq12d 2625 |
. . . . 5
⊢ (𝑥 = 𝐴 → ((𝑥𝐹𝑦) = 𝑅 ↔ (𝐴𝐹𝑦) = 𝐺)) |
25 | 22, 24 | imbi12d 333 |
. . . 4
⊢ (𝑥 = 𝐴 → ((𝑅 ∈ V → (𝑥𝐹𝑦) = 𝑅) ↔ (𝐺 ∈ V → (𝐴𝐹𝑦) = 𝐺))) |
26 | | ov2gf.4 |
. . . . . 6
⊢ (𝑦 = 𝐵 → 𝐺 = 𝑆) |
27 | 26 | eleq1d 2672 |
. . . . 5
⊢ (𝑦 = 𝐵 → (𝐺 ∈ V ↔ 𝑆 ∈ V)) |
28 | | oveq2 6557 |
. . . . . 6
⊢ (𝑦 = 𝐵 → (𝐴𝐹𝑦) = (𝐴𝐹𝐵)) |
29 | 28, 26 | eqeq12d 2625 |
. . . . 5
⊢ (𝑦 = 𝐵 → ((𝐴𝐹𝑦) = 𝐺 ↔ (𝐴𝐹𝐵) = 𝑆)) |
30 | 27, 29 | imbi12d 333 |
. . . 4
⊢ (𝑦 = 𝐵 → ((𝐺 ∈ V → (𝐴𝐹𝑦) = 𝐺) ↔ (𝑆 ∈ V → (𝐴𝐹𝐵) = 𝑆))) |
31 | 7 | ovmpt4g 6681 |
. . . . 5
⊢ ((𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐷 ∧ 𝑅 ∈ V) → (𝑥𝐹𝑦) = 𝑅) |
32 | 31 | 3expia 1259 |
. . . 4
⊢ ((𝑥 ∈ 𝐶 ∧ 𝑦 ∈ 𝐷) → (𝑅 ∈ V → (𝑥𝐹𝑦) = 𝑅)) |
33 | 2, 3, 4, 13, 20, 25, 30, 32 | vtocl2gaf 3246 |
. . 3
⊢ ((𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷) → (𝑆 ∈ V → (𝐴𝐹𝐵) = 𝑆)) |
34 | 1, 33 | syl5 33 |
. 2
⊢ ((𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷) → (𝑆 ∈ 𝐻 → (𝐴𝐹𝐵) = 𝑆)) |
35 | 34 | 3impia 1253 |
1
⊢ ((𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷 ∧ 𝑆 ∈ 𝐻) → (𝐴𝐹𝐵) = 𝑆) |