Proof of Theorem 3gencl
Step | Hyp | Ref
| Expression |
1 | | 3gencl.3 |
. . . . 5
⊢ (𝐺 ∈ 𝑆 ↔ ∃𝑧 ∈ 𝑅 𝐶 = 𝐺) |
2 | | df-rex 3069 |
. . . . 5
⊢
(∃𝑧 ∈
𝑅 𝐶 = 𝐺 ↔ ∃𝑧(𝑧 ∈ 𝑅 ∧ 𝐶 = 𝐺)) |
3 | 1, 2 | bitri 274 |
. . . 4
⊢ (𝐺 ∈ 𝑆 ↔ ∃𝑧(𝑧 ∈ 𝑅 ∧ 𝐶 = 𝐺)) |
4 | | 3gencl.6 |
. . . . 5
⊢ (𝐶 = 𝐺 → (𝜒 ↔ 𝜃)) |
5 | 4 | imbi2d 340 |
. . . 4
⊢ (𝐶 = 𝐺 → (((𝐷 ∈ 𝑆 ∧ 𝐹 ∈ 𝑆) → 𝜒) ↔ ((𝐷 ∈ 𝑆 ∧ 𝐹 ∈ 𝑆) → 𝜃))) |
6 | | 3gencl.1 |
. . . . . 6
⊢ (𝐷 ∈ 𝑆 ↔ ∃𝑥 ∈ 𝑅 𝐴 = 𝐷) |
7 | | 3gencl.2 |
. . . . . 6
⊢ (𝐹 ∈ 𝑆 ↔ ∃𝑦 ∈ 𝑅 𝐵 = 𝐹) |
8 | | 3gencl.4 |
. . . . . . 7
⊢ (𝐴 = 𝐷 → (𝜑 ↔ 𝜓)) |
9 | 8 | imbi2d 340 |
. . . . . 6
⊢ (𝐴 = 𝐷 → ((𝑧 ∈ 𝑅 → 𝜑) ↔ (𝑧 ∈ 𝑅 → 𝜓))) |
10 | | 3gencl.5 |
. . . . . . 7
⊢ (𝐵 = 𝐹 → (𝜓 ↔ 𝜒)) |
11 | 10 | imbi2d 340 |
. . . . . 6
⊢ (𝐵 = 𝐹 → ((𝑧 ∈ 𝑅 → 𝜓) ↔ (𝑧 ∈ 𝑅 → 𝜒))) |
12 | | 3gencl.7 |
. . . . . . 7
⊢ ((𝑥 ∈ 𝑅 ∧ 𝑦 ∈ 𝑅 ∧ 𝑧 ∈ 𝑅) → 𝜑) |
13 | 12 | 3expia 1119 |
. . . . . 6
⊢ ((𝑥 ∈ 𝑅 ∧ 𝑦 ∈ 𝑅) → (𝑧 ∈ 𝑅 → 𝜑)) |
14 | 6, 7, 9, 11, 13 | 2gencl 3462 |
. . . . 5
⊢ ((𝐷 ∈ 𝑆 ∧ 𝐹 ∈ 𝑆) → (𝑧 ∈ 𝑅 → 𝜒)) |
15 | 14 | com12 32 |
. . . 4
⊢ (𝑧 ∈ 𝑅 → ((𝐷 ∈ 𝑆 ∧ 𝐹 ∈ 𝑆) → 𝜒)) |
16 | 3, 5, 15 | gencl 3461 |
. . 3
⊢ (𝐺 ∈ 𝑆 → ((𝐷 ∈ 𝑆 ∧ 𝐹 ∈ 𝑆) → 𝜃)) |
17 | 16 | com12 32 |
. 2
⊢ ((𝐷 ∈ 𝑆 ∧ 𝐹 ∈ 𝑆) → (𝐺 ∈ 𝑆 → 𝜃)) |
18 | 17 | 3impia 1115 |
1
⊢ ((𝐷 ∈ 𝑆 ∧ 𝐹 ∈ 𝑆 ∧ 𝐺 ∈ 𝑆) → 𝜃) |