Proof of Theorem oa4to6lem1
Step | Hyp | Ref
| Expression |
1 | | leor 159 |
. . . 4
b ≤ (a⊥ ∪ b) |
2 | | comid 187 |
. . . . . . . . 9
a C a |
3 | 2 | comcom3 454 |
. . . . . . . 8
a⊥ C
a |
4 | | oa4to6lem.1 |
. . . . . . . . 9
a⊥ ≤ b |
5 | 4 | lecom 180 |
. . . . . . . 8
a⊥ C
b |
6 | 3, 5 | fh3 471 |
. . . . . . 7
(a⊥ ∪ (a ∩ b)) =
((a⊥ ∪ a) ∩ (a⊥ ∪ b)) |
7 | | ancom 74 |
. . . . . . . 8
(1 ∩ (a⊥ ∪
b)) = ((a⊥ ∪ b) ∩ 1) |
8 | | df-t 41 |
. . . . . . . . . 10
1 = (a ∪ a⊥ ) |
9 | | ax-a2 31 |
. . . . . . . . . 10
(a ∪ a⊥ ) = (a⊥ ∪ a) |
10 | 8, 9 | ax-r2 36 |
. . . . . . . . 9
1 = (a⊥ ∪
a) |
11 | 10 | ran 78 |
. . . . . . . 8
(1 ∩ (a⊥ ∪
b)) = ((a⊥ ∪ a) ∩ (a⊥ ∪ b)) |
12 | | an1 106 |
. . . . . . . 8
((a⊥ ∪ b) ∩ 1) = (a⊥ ∪ b) |
13 | 7, 11, 12 | 3tr2 64 |
. . . . . . 7
((a⊥ ∪ a) ∩ (a⊥ ∪ b)) = (a⊥ ∪ b) |
14 | 6, 13 | ax-r2 36 |
. . . . . 6
(a⊥ ∪ (a ∩ b)) =
(a⊥ ∪ b) |
15 | 14 | ax-r1 35 |
. . . . 5
(a⊥ ∪ b) = (a⊥ ∪ (a ∩ b)) |
16 | | anidm 111 |
. . . . . . . . 9
(a ∩ a) = a |
17 | 16 | ran 78 |
. . . . . . . 8
((a ∩ a) ∩ b) =
(a ∩ b) |
18 | 17 | ax-r1 35 |
. . . . . . 7
(a ∩ b) = ((a ∩
a) ∩ b) |
19 | | anass 76 |
. . . . . . 7
((a ∩ a) ∩ b) =
(a ∩ (a ∩ b)) |
20 | 18, 19 | ax-r2 36 |
. . . . . 6
(a ∩ b) = (a ∩
(a ∩ b)) |
21 | 20 | lor 70 |
. . . . 5
(a⊥ ∪ (a ∩ b)) =
(a⊥ ∪ (a ∩ (a ∩
b))) |
22 | 15, 21 | ax-r2 36 |
. . . 4
(a⊥ ∪ b) = (a⊥ ∪ (a ∩ (a ∩
b))) |
23 | 1, 22 | lbtr 139 |
. . 3
b ≤ (a⊥ ∪ (a ∩ (a ∩
b))) |
24 | | leo 158 |
. . . . . 6
(a ∩ b) ≤ ((a
∩ b) ∪ ((c ∩ d) ∪
(e ∩ f))) |
25 | | ax-a3 32 |
. . . . . . 7
(((a ∩ b) ∪ (c
∩ d)) ∪ (e ∩ f)) =
((a ∩ b) ∪ ((c
∩ d) ∪ (e ∩ f))) |
26 | 25 | ax-r1 35 |
. . . . . 6
((a ∩ b) ∪ ((c
∩ d) ∪ (e ∩ f))) =
(((a ∩ b) ∪ (c
∩ d)) ∪ (e ∩ f)) |
27 | 24, 26 | lbtr 139 |
. . . . 5
(a ∩ b) ≤ (((a
∩ b) ∪ (c ∩ d))
∪ (e ∩ f)) |
28 | 27 | lelan 167 |
. . . 4
(a ∩ (a ∩ b)) ≤
(a ∩ (((a ∩ b) ∪
(c ∩ d)) ∪ (e
∩ f))) |
29 | 28 | lelor 166 |
. . 3
(a⊥ ∪ (a ∩ (a ∩
b))) ≤ (a⊥ ∪ (a ∩ (((a
∩ b) ∪ (c ∩ d))
∪ (e ∩ f)))) |
30 | 23, 29 | letr 137 |
. 2
b ≤ (a⊥ ∪ (a ∩ (((a
∩ b) ∪ (c ∩ d))
∪ (e ∩ f)))) |
31 | | oa4to6lem.4 |
. . . . 5
g = (((a ∩ b) ∪
(c ∩ d)) ∪ (e
∩ f)) |
32 | 31 | ud1lem0a 255 |
. . . 4
(a →1 g) = (a
→1 (((a ∩ b) ∪ (c
∩ d)) ∪ (e ∩ f))) |
33 | | df-i1 44 |
. . . 4
(a →1 (((a ∩ b) ∪
(c ∩ d)) ∪ (e
∩ f))) = (a⊥ ∪ (a ∩ (((a
∩ b) ∪ (c ∩ d))
∪ (e ∩ f)))) |
34 | 32, 33 | ax-r2 36 |
. . 3
(a →1 g) = (a⊥ ∪ (a ∩ (((a
∩ b) ∪ (c ∩ d))
∪ (e ∩ f)))) |
35 | 34 | ax-r1 35 |
. 2
(a⊥ ∪ (a ∩ (((a
∩ b) ∪ (c ∩ d))
∪ (e ∩ f)))) = (a
→1 g) |
36 | 30, 35 | lbtr 139 |
1
b ≤ (a →1 g) |