Proof of Theorem i3orlem5
| Step | Hyp | Ref
| Expression |
| 1 | | leo 158 |
. 2
((a ∪ c)⊥ ∩ (b ∪ c)⊥ ) ≤ (((a ∪ c)⊥ ∩ (b ∪ c)⊥ ) ∪ (((a ∪ c)⊥ ∪ (b ∪ c))
∩ ((a ∪ c) ∪ ((a
∪ c)⊥ ∩ (b ∪ c))))) |
| 2 | | anandir 115 |
. . 3
((a⊥ ∩ b⊥ ) ∩ c⊥ ) = ((a⊥ ∩ c⊥ ) ∩ (b⊥ ∩ c⊥ )) |
| 3 | | oran 87 |
. . . . . 6
(a ∪ c) = (a⊥ ∩ c⊥
)⊥ |
| 4 | 3 | con2 67 |
. . . . 5
(a ∪ c)⊥ = (a⊥ ∩ c⊥ ) |
| 5 | 4 | ax-r1 35 |
. . . 4
(a⊥ ∩ c⊥ ) = (a ∪ c)⊥ |
| 6 | | oran 87 |
. . . . . 6
(b ∪ c) = (b⊥ ∩ c⊥
)⊥ |
| 7 | 6 | con2 67 |
. . . . 5
(b ∪ c)⊥ = (b⊥ ∩ c⊥ ) |
| 8 | 7 | ax-r1 35 |
. . . 4
(b⊥ ∩ c⊥ ) = (b ∪ c)⊥ |
| 9 | 5, 8 | 2an 79 |
. . 3
((a⊥ ∩ c⊥ ) ∩ (b⊥ ∩ c⊥ )) = ((a ∪ c)⊥ ∩ (b ∪ c)⊥ ) |
| 10 | 2, 9 | ax-r2 36 |
. 2
((a⊥ ∩ b⊥ ) ∩ c⊥ ) = ((a ∪ c)⊥ ∩ (b ∪ c)⊥ ) |
| 11 | | df2i3 498 |
. 2
((a ∪ c) →3 (b ∪ c)) =
(((a ∪ c)⊥ ∩ (b ∪ c)⊥ ) ∪ (((a ∪ c)⊥ ∪ (b ∪ c))
∩ ((a ∪ c) ∪ ((a
∪ c)⊥ ∩ (b ∪ c))))) |
| 12 | 1, 10, 11 | le3tr1 140 |
1
((a⊥ ∩ b⊥ ) ∩ c⊥ ) ≤ ((a ∪ c)
→3 (b ∪ c)) |