Proof of Theorem mdbr
| 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 | | opreq2 3007 |
. . . . . . . 8
⊢ (y =
A → (x ∨ℋ y) = (x
∨ℋ A)) |
| 4 | 3 | ineq1d 1644 |
. . . . . . 7
⊢ (y =
A → ((x ∨ℋ y) ∩ z) =
((x ∨ℋ A) ∩ z)) |
| 5 | | ineq1 1638 |
. . . . . . . 8
⊢ (y =
A → (y ∩ z) =
(A ∩ z)) |
| 6 | 5 | opreq2d 3013 |
. . . . . . 7
⊢ (y =
A → (x ∨ℋ (y ∩ z)) =
(x ∨ℋ (A ∩ z))) |
| 7 | 4, 6 | cleq12d 1115 |
. . . . . 6
⊢ (y =
A → (((x ∨ℋ y) ∩ z) =
(x ∨ℋ (y ∩ z))
↔ ((x ∨ℋ A) ∩ z) =
(x ∨ℋ (A ∩ z)))) |
| 8 | 7 | imbi2d 464 |
. . . . 5
⊢ (y =
A → ((x ⊆ z
→ ((x ∨ℋ y) ∩ z) =
(x ∨ℋ (y ∩ z)))
↔ (x ⊆ z → ((x
∨ℋ A) ∩ z) = (x
∨ℋ (A ∩ z))))) |
| 9 | 8 | biraldv 1219 |
. . . 4
⊢ (y =
A → (∀x ∈ Cℋ (x ⊆ z
→ ((x ∨ℋ y) ∩ z) =
(x ∨ℋ (y ∩ z)))
↔ ∀x ∈
Cℋ (x ⊆
z → ((x ∨ℋ A) ∩ z) =
(x ∨ℋ (A ∩ z))))) |
| 10 | 2, 9 | anbi12d 476 |
. . 3
⊢ (y =
A → (((y ∈ Cℋ ∧ z ∈ Cℋ ) ∧
∀x ∈ Cℋ
(x ⊆ z → ((x
∨ℋ y) ∩ z) = (x
∨ℋ (y ∩ z)))) ↔ ((A
∈ Cℋ ∧ z
∈ Cℋ ) ∧ ∀x ∈ Cℋ (x ⊆ z
→ ((x ∨ℋ A) ∩ z) =
(x ∨ℋ (A ∩ z)))))) |
| 11 | | eleq1 1149 |
. . . . 5
⊢ (z =
B → (z ∈ Cℋ ↔ B ∈ Cℋ )) |
| 12 | 11 | anbi2d 468 |
. . . 4
⊢ (z =
B → ((A ∈ Cℋ ∧ z ∈ Cℋ ) ↔
(A ∈ Cℋ ∧
B ∈ Cℋ
))) |
| 13 | | sseq2 1522 |
. . . . . 6
⊢ (z =
B → (x ⊆ z
↔ x ⊆ B)) |
| 14 | | ineq2 1639 |
. . . . . . 7
⊢ (z =
B → ((x ∨ℋ A) ∩ z) =
((x ∨ℋ A) ∩ B)) |
| 15 | | ineq2 1639 |
. . . . . . . 8
⊢ (z =
B → (A ∩ z) =
(A ∩ B)) |
| 16 | 15 | opreq2d 3013 |
. . . . . . 7
⊢ (z =
B → (x ∨ℋ (A ∩ z)) =
(x ∨ℋ (A ∩ B))) |
| 17 | 14, 16 | cleq12d 1115 |
. . . . . 6
⊢ (z =
B → (((x ∨ℋ A) ∩ z) =
(x ∨ℋ (A ∩ z))
↔ ((x ∨ℋ A) ∩ B) =
(x ∨ℋ (A ∩ B)))) |
| 18 | 13, 17 | imbi12d 474 |
. . . . 5
⊢ (z =
B → ((x ⊆ z
→ ((x ∨ℋ A) ∩ z) =
(x ∨ℋ (A ∩ z)))
↔ (x ⊆ B → ((x
∨ℋ A) ∩ B) = (x
∨ℋ (A ∩ B))))) |
| 19 | 18 | biraldv 1219 |
. . . 4
⊢ (z =
B → (∀x ∈ Cℋ (x ⊆ z
→ ((x ∨ℋ A) ∩ z) =
(x ∨ℋ (A ∩ z)))
↔ ∀x ∈
Cℋ (x ⊆
B → ((x ∨ℋ A) ∩ B) =
(x ∨ℋ (A ∩ B))))) |
| 20 | 12, 19 | anbi12d 476 |
. . 3
⊢ (z =
B → (((A ∈ Cℋ ∧ z ∈ Cℋ ) ∧
∀x ∈ Cℋ
(x ⊆ z → ((x
∨ℋ A) ∩ z) = (x
∨ℋ (A ∩ z)))) ↔ ((A
∈ Cℋ ∧ B
∈ Cℋ ) ∧ ∀x ∈ Cℋ (x ⊆ B
→ ((x ∨ℋ A) ∩ B) =
(x ∨ℋ (A ∩ B)))))) |
| 21 | | df-md 5713 |
. . 3
⊢ Mℋ =
{〈y, z〉∣((y ∈ Cℋ ∧ z ∈ Cℋ ) ∧
∀x ∈ Cℋ
(x ⊆ z → ((x
∨ℋ y) ∩ z) = (x
∨ℋ (y ∩ z))))} |
| 22 | 10, 20, 21 | brabg 2116 |
. 2
⊢ ((A
∈ Cℋ ∧ B
∈ Cℋ ) → (A Mℋ B ↔ ((A
∈ Cℋ ∧ B
∈ Cℋ ) ∧ ∀x ∈ Cℋ (x ⊆ B
→ ((x ∨ℋ A) ∩ B) =
(x ∨ℋ (A ∩ B)))))) |
| 23 | | ibar 487 |
. 2
⊢ ((A
∈ Cℋ ∧ B
∈ Cℋ ) → (∀x ∈ Cℋ (x ⊆ B
→ ((x ∨ℋ A) ∩ B) =
(x ∨ℋ (A ∩ B)))
↔ ((A ∈
Cℋ ∧ B ∈
Cℋ ) ∧ ∀x ∈ Cℋ (x ⊆ B
→ ((x ∨ℋ A) ∩ B) =
(x ∨ℋ (A ∩ B)))))) |
| 24 | 22, 23 | bitr4d 409 |
1
⊢ ((A
∈ Cℋ ∧ B
∈ Cℋ ) → (A Mℋ B ↔ ∀x ∈ Cℋ (x ⊆ B
→ ((x ∨ℋ A) ∩ B) =
(x ∨ℋ (A ∩ B))))) |