Proof of Theorem mdi
| Step | Hyp | Ref
| Expression |
| 1 | | mdbr 5726 |
. . . . 5
⊢ ((A
∈ Cℋ ∧ B
∈ Cℋ ) → (A Mℋ B ↔ ∀x ∈ Cℋ (x ⊆ B
→ ((x ∨ℋ A) ∩ B) =
(x ∨ℋ (A ∩ B))))) |
| 2 | | sseq1 1521 |
. . . . . . 7
⊢ (x =
C → (x ⊆ B
↔ C ⊆ B)) |
| 3 | | opreq1 3006 |
. . . . . . . . 9
⊢ (x =
C → (x ∨ℋ A) = (C
∨ℋ A)) |
| 4 | 3 | ineq1d 1644 |
. . . . . . . 8
⊢ (x =
C → ((x ∨ℋ A) ∩ B) =
((C ∨ℋ A) ∩ B)) |
| 5 | | opreq1 3006 |
. . . . . . . 8
⊢ (x =
C → (x ∨ℋ (A ∩ B)) =
(C ∨ℋ (A ∩ B))) |
| 6 | 4, 5 | cleq12d 1115 |
. . . . . . 7
⊢ (x =
C → (((x ∨ℋ A) ∩ B) =
(x ∨ℋ (A ∩ B))
↔ ((C ∨ℋ A) ∩ B) =
(C ∨ℋ (A ∩ B)))) |
| 7 | 2, 6 | imbi12d 474 |
. . . . . 6
⊢ (x =
C → ((x ⊆ B
→ ((x ∨ℋ A) ∩ B) =
(x ∨ℋ (A ∩ B)))
↔ (C ⊆ B → ((C
∨ℋ A) ∩ B) = (C
∨ℋ (A ∩ B))))) |
| 8 | 7 | rcla4v 1402 |
. . . . 5
⊢ (∀x ∈ Cℋ (x ⊆ B
→ ((x ∨ℋ A) ∩ B) =
(x ∨ℋ (A ∩ B)))
→ (C ∈ Cℋ
→ (C ⊆ B → ((C
∨ℋ A) ∩ B) = (C
∨ℋ (A ∩ B))))) |
| 9 | 1, 8 | syl6bi 187 |
. . . 4
⊢ ((A
∈ Cℋ ∧ B
∈ Cℋ ) → (A Mℋ B → (C
∈ Cℋ → (C
⊆ B → ((C ∨ℋ A) ∩ B) =
(C ∨ℋ (A ∩ B)))))) |
| 10 | 9 | com23 32 |
. . 3
⊢ ((A
∈ Cℋ ∧ B
∈ Cℋ ) → (C ∈ Cℋ → (A Mℋ B → (C
⊆ B → ((C ∨ℋ A) ∩ B) =
(C ∨ℋ (A ∩ B)))))) |
| 11 | 10 | imp 277 |
. 2
⊢ (((A
∈ Cℋ ∧ B
∈ Cℋ ) ∧ C
∈ Cℋ ) → (A Mℋ B → (C
⊆ B → ((C ∨ℋ A) ∩ B) =
(C ∨ℋ (A ∩ B))))) |
| 12 | 11 | 3impa 609 |
1
⊢ ((A
∈ Cℋ ∧ B
∈ Cℋ ∧ C
∈ Cℋ ) → (A Mℋ B → (C
⊆ B → ((C ∨ℋ A) ∩ B) =
(C ∨ℋ (A ∩ B))))) |