Proof of Theorem cvbrt
| Step | Hyp | Ref
| Expression |
| 1 | | eleq1 1149 |
. . . . 5
⊢ (y =
A → (y ∈ Cℋ ↔ A ∈ Cℋ )) |
| 2 | 1 | anbi1d 469 |
. . . 4
⊢ (y =
A → ((y ∈ Cℋ ∧ z ∈ Cℋ ) ↔
(A ∈ Cℋ ∧
z ∈ Cℋ
))) |
| 3 | | psseq1 1559 |
. . . . 5
⊢ (y =
A → (y ⊂ z ↔
A ⊂ z)) |
| 4 | | psseq1 1559 |
. . . . . . . 8
⊢ (y =
A → (y ⊂ x ↔
A ⊂ x)) |
| 5 | 4 | anbi1d 469 |
. . . . . . 7
⊢ (y =
A → ((y ⊂ x ∧
x ⊂ z) ↔ (A
⊂ x ∧ x ⊂ z))) |
| 6 | 5 | birexdv 1220 |
. . . . . 6
⊢ (y =
A → (∃x ∈ Cℋ (y ⊂ x ∧
x ⊂ z) ↔ ∃x ∈ Cℋ (A ⊂ x ∧
x ⊂ z))) |
| 7 | 6 | negbid 463 |
. . . . 5
⊢ (y =
A → (¬ ∃x ∈ Cℋ (y ⊂ x ∧
x ⊂ z) ↔ ¬ ∃x ∈ Cℋ (A ⊂ x ∧
x ⊂ z))) |
| 8 | 3, 7 | anbi12d 476 |
. . . 4
⊢ (y =
A → ((y ⊂ z ∧
¬ ∃x ∈
Cℋ (y ⊂
x ∧ x ⊂ z))
↔ (A ⊂ z ∧ ¬ ∃x ∈ Cℋ (A ⊂ x ∧
x ⊂ z)))) |
| 9 | 2, 8 | anbi12d 476 |
. . 3
⊢ (y =
A → (((y ∈ Cℋ ∧ z ∈ Cℋ ) ∧
(y ⊂ z ∧ ¬ ∃x ∈ Cℋ (y ⊂ x ∧
x ⊂ z))) ↔ ((A
∈ Cℋ ∧ z
∈ Cℋ ) ∧ (A ⊂ z ∧
¬ ∃x ∈
Cℋ (A ⊂
x ∧ x ⊂ z))))) |
| 10 | | eleq1 1149 |
. . . . 5
⊢ (z =
B → (z ∈ Cℋ ↔ B ∈ Cℋ )) |
| 11 | 10 | anbi2d 468 |
. . . 4
⊢ (z =
B → ((A ∈ Cℋ ∧ z ∈ Cℋ ) ↔
(A ∈ Cℋ ∧
B ∈ Cℋ
))) |
| 12 | | psseq2 1560 |
. . . . 5
⊢ (z =
B → (A ⊂ z ↔
A ⊂ B)) |
| 13 | | psseq2 1560 |
. . . . . . . 8
⊢ (z =
B → (x ⊂ z ↔
x ⊂ B)) |
| 14 | 13 | anbi2d 468 |
. . . . . . 7
⊢ (z =
B → ((A ⊂ x ∧
x ⊂ z) ↔ (A
⊂ x ∧ x ⊂ B))) |
| 15 | 14 | birexdv 1220 |
. . . . . 6
⊢ (z =
B → (∃x ∈ Cℋ (A ⊂ x ∧
x ⊂ z) ↔ ∃x ∈ Cℋ (A ⊂ x ∧
x ⊂ B))) |
| 16 | 15 | negbid 463 |
. . . . 5
⊢ (z =
B → (¬ ∃x ∈ Cℋ (A ⊂ x ∧
x ⊂ z) ↔ ¬ ∃x ∈ Cℋ (A ⊂ x ∧
x ⊂ B))) |
| 17 | 12, 16 | anbi12d 476 |
. . . 4
⊢ (z =
B → ((A ⊂ z ∧
¬ ∃x ∈
Cℋ (A ⊂
x ∧ x ⊂ z))
↔ (A ⊂ B ∧ ¬ ∃x ∈ Cℋ (A ⊂ x ∧
x ⊂ B)))) |
| 18 | 11, 17 | anbi12d 476 |
. . 3
⊢ (z =
B → (((A ∈ Cℋ ∧ z ∈ Cℋ ) ∧
(A ⊂ z ∧ ¬ ∃x ∈ Cℋ (A ⊂ x ∧
x ⊂ z))) ↔ ((A
∈ Cℋ ∧ B
∈ Cℋ ) ∧ (A ⊂ B ∧
¬ ∃x ∈
Cℋ (A ⊂
x ∧ x ⊂ B))))) |
| 19 | | df-cv 5712 |
. . 3
⊢ ⋖ = {〈y, z〉∣((y ∈ Cℋ ∧ z ∈ Cℋ ) ∧
(y ⊂ z ∧ ¬ ∃x ∈ Cℋ (y ⊂ x ∧
x ⊂ z)))} |
| 20 | 9, 18, 19 | brabg 2116 |
. 2
⊢ ((A
∈ Cℋ ∧ B
∈ Cℋ ) → (A ⋖ B
↔ ((A ∈
Cℋ ∧ B ∈
Cℋ ) ∧ (A ⊂
B ∧ ¬ ∃x ∈ Cℋ (A ⊂ x ∧
x ⊂ B))))) |
| 21 | | ibar 487 |
. 2
⊢ ((A
∈ Cℋ ∧ B
∈ Cℋ ) → ((A ⊂ B ∧
¬ ∃x ∈
Cℋ (A ⊂
x ∧ x ⊂ B))
↔ ((A ∈
Cℋ ∧ B ∈
Cℋ ) ∧ (A ⊂
B ∧ ¬ ∃x ∈ Cℋ (A ⊂ x ∧
x ⊂ B))))) |
| 22 | 20, 21 | bitr4d 409 |
1
⊢ ((A
∈ Cℋ ∧ B
∈ Cℋ ) → (A ⋖ B
↔ (A ⊂ B ∧ ¬ ∃x ∈ Cℋ (A ⊂ x ∧
x ⊂ B)))) |