Proof of Theorem 3gencl
Step | Hyp | Ref
| Expression |
1 | | 3gencl.3 |
. . . . 5
⊢ (G ∈ S ↔ ∃z ∈ R C = G) |
2 | | df-rex 2621 |
. . . . 5
⊢ (∃z ∈ R C = G ↔
∃z(z ∈ R ∧ C = G)) |
3 | 1, 2 | bitri 240 |
. . . 4
⊢ (G ∈ S ↔ ∃z(z ∈ R ∧ C = G)) |
4 | | 3gencl.6 |
. . . . 5
⊢ (C = G →
(χ ↔ θ)) |
5 | 4 | imbi2d 307 |
. . . 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 307 |
. . . . . 6
⊢ (A = D →
((z ∈
R → φ) ↔ (z ∈ R → ψ))) |
10 | | 3gencl.5 |
. . . . . . 7
⊢ (B = F →
(ψ ↔ χ)) |
11 | 10 | imbi2d 307 |
. . . . . 6
⊢ (B = F →
((z ∈
R → ψ) ↔ (z ∈ R → χ))) |
12 | | 3gencl.7 |
. . . . . . 7
⊢ ((x ∈ R ∧ y ∈ R ∧ z ∈ R) → φ) |
13 | 12 | 3expia 1153 |
. . . . . 6
⊢ ((x ∈ R ∧ y ∈ R) → (z
∈ R
→ φ)) |
14 | 6, 7, 9, 11, 13 | 2gencl 2889 |
. . . . 5
⊢ ((D ∈ S ∧ F ∈ S) → (z
∈ R
→ χ)) |
15 | 14 | com12 27 |
. . . 4
⊢ (z ∈ R → ((D
∈ S ∧ F ∈ S) →
χ)) |
16 | 3, 5, 15 | gencl 2888 |
. . 3
⊢ (G ∈ S → ((D
∈ S ∧ F ∈ S) →
θ)) |
17 | 16 | com12 27 |
. 2
⊢ ((D ∈ S ∧ F ∈ S) → (G
∈ S
→ θ)) |
18 | 17 | 3impia 1148 |
1
⊢ ((D ∈ S ∧ F ∈ S ∧ G ∈ S) → θ) |