Proof of Theorem i3orlem7
Step | Hyp | Ref
| Expression |
1 | | lea 160 |
. . . . . . 7
(a ∩ b⊥ ) ≤ a |
2 | | leo 158 |
. . . . . . 7
a ≤ (a ∪ b) |
3 | 1, 2 | letr 137 |
. . . . . 6
(a ∩ b⊥ ) ≤ (a ∪ b) |
4 | | leo 158 |
. . . . . 6
(a ∩ b⊥ ) ≤ ((a ∩ b⊥ ) ∪ (a ∩ b)) |
5 | 3, 4 | ler2an 173 |
. . . . 5
(a ∩ b⊥ ) ≤ ((a ∪ b) ∩
((a ∩ b⊥ ) ∪ (a ∩ b))) |
6 | 5 | ler 149 |
. . . 4
(a ∩ b⊥ ) ≤ (((a ∪ b) ∩
((a ∩ b⊥ ) ∪ (a ∩ b)))
∪ ((a ∪ b) ∩ (a⊥ ∩ (a ∪ b⊥ )))) |
7 | | i3n1 249 |
. . . . . . 7
(a⊥ →3
b⊥ ) = (((a ∩ b⊥ ) ∪ (a ∩ b))
∪ (a⊥ ∩ (a ∪ b⊥ ))) |
8 | 7 | lan 77 |
. . . . . 6
((a ∪ b) ∩ (a⊥ →3 b⊥ )) = ((a ∪ b) ∩
(((a ∩ b⊥ ) ∪ (a ∩ b))
∪ (a⊥ ∩ (a ∪ b⊥ )))) |
9 | | comor1 461 |
. . . . . . . . 9
(a ∪ b) C a |
10 | | comor2 462 |
. . . . . . . . . 10
(a ∪ b) C b |
11 | 10 | comcom2 183 |
. . . . . . . . 9
(a ∪ b) C b⊥ |
12 | 9, 11 | com2an 484 |
. . . . . . . 8
(a ∪ b) C (a
∩ b⊥
) |
13 | 9, 10 | com2an 484 |
. . . . . . . 8
(a ∪ b) C (a
∩ b) |
14 | 12, 13 | com2or 483 |
. . . . . . 7
(a ∪ b) C ((a
∩ b⊥ ) ∪ (a ∩ b)) |
15 | 9 | comcom2 183 |
. . . . . . . 8
(a ∪ b) C a⊥ |
16 | 9, 11 | com2or 483 |
. . . . . . . 8
(a ∪ b) C (a
∪ b⊥
) |
17 | 15, 16 | com2an 484 |
. . . . . . 7
(a ∪ b) C (a⊥ ∩ (a ∪ b⊥ )) |
18 | 14, 17 | fh1 469 |
. . . . . 6
((a ∪ b) ∩ (((a
∩ b⊥ ) ∪ (a ∩ b))
∪ (a⊥ ∩ (a ∪ b⊥ )))) = (((a ∪ b) ∩
((a ∩ b⊥ ) ∪ (a ∩ b)))
∪ ((a ∪ b) ∩ (a⊥ ∩ (a ∪ b⊥ )))) |
19 | 8, 18 | ax-r2 36 |
. . . . 5
((a ∪ b) ∩ (a⊥ →3 b⊥ )) = (((a ∪ b) ∩
((a ∩ b⊥ ) ∪ (a ∩ b)))
∪ ((a ∪ b) ∩ (a⊥ ∩ (a ∪ b⊥ )))) |
20 | 19 | ax-r1 35 |
. . . 4
(((a ∪ b) ∩ ((a
∩ b⊥ ) ∪ (a ∩ b)))
∪ ((a ∪ b) ∩ (a⊥ ∩ (a ∪ b⊥ )))) = ((a ∪ b) ∩
(a⊥ →3
b⊥
)) |
21 | 6, 20 | lbtr 139 |
. . 3
(a ∩ b⊥ ) ≤ ((a ∪ b) ∩
(a⊥ →3
b⊥
)) |
22 | 21 | ler 149 |
. 2
(a ∩ b⊥ ) ≤ (((a ∪ b) ∩
(a⊥ →3
b⊥ )) ∪
((a ∪ c) →3 (b ∪ c))) |
23 | | i3orlem6 557 |
. . 3
((a →3 b)⊥ ∪ ((a ∪ c)
→3 (b ∪ c))) = (((a
∪ b) ∩ (a⊥ →3 b⊥ )) ∪ ((a ∪ c)
→3 (b ∪ c))) |
24 | 23 | ax-r1 35 |
. 2
(((a ∪ b) ∩ (a⊥ →3 b⊥ )) ∪ ((a ∪ c)
→3 (b ∪ c))) = ((a
→3 b)⊥
∪ ((a ∪ c) →3 (b ∪ c))) |
25 | 22, 24 | lbtr 139 |
1
(a ∩ b⊥ ) ≤ ((a →3 b)⊥ ∪ ((a ∪ c)
→3 (b ∪ c))) |