Proof of Theorem oa4to6lem3
Step | Hyp | Ref
| Expression |
1 | | leor 159 |
. . . 4
f ≤ (e⊥ ∪ f) |
2 | | comid 187 |
. . . . . . . . 9
e C e |
3 | 2 | comcom3 454 |
. . . . . . . 8
e⊥ C
e |
4 | | oa4to6lem.3 |
. . . . . . . . 9
e⊥ ≤ f |
5 | 4 | lecom 180 |
. . . . . . . 8
e⊥ C
f |
6 | 3, 5 | fh3 471 |
. . . . . . 7
(e⊥ ∪ (e ∩ f)) =
((e⊥ ∪ e) ∩ (e⊥ ∪ f)) |
7 | | ancom 74 |
. . . . . . . 8
(1 ∩ (e⊥ ∪
f)) = ((e⊥ ∪ f) ∩ 1) |
8 | | df-t 41 |
. . . . . . . . . 10
1 = (e ∪ e⊥ ) |
9 | | ax-a2 31 |
. . . . . . . . . 10
(e ∪ e⊥ ) = (e⊥ ∪ e) |
10 | 8, 9 | ax-r2 36 |
. . . . . . . . 9
1 = (e⊥ ∪
e) |
11 | 10 | ran 78 |
. . . . . . . 8
(1 ∩ (e⊥ ∪
f)) = ((e⊥ ∪ e) ∩ (e⊥ ∪ f)) |
12 | | an1 106 |
. . . . . . . 8
((e⊥ ∪ f) ∩ 1) = (e⊥ ∪ f) |
13 | 7, 11, 12 | 3tr2 64 |
. . . . . . 7
((e⊥ ∪ e) ∩ (e⊥ ∪ f)) = (e⊥ ∪ f) |
14 | 6, 13 | ax-r2 36 |
. . . . . 6
(e⊥ ∪ (e ∩ f)) =
(e⊥ ∪ f) |
15 | 14 | ax-r1 35 |
. . . . 5
(e⊥ ∪ f) = (e⊥ ∪ (e ∩ f)) |
16 | | anidm 111 |
. . . . . . . . 9
(e ∩ e) = e |
17 | 16 | ran 78 |
. . . . . . . 8
((e ∩ e) ∩ f) =
(e ∩ f) |
18 | 17 | ax-r1 35 |
. . . . . . 7
(e ∩ f) = ((e ∩
e) ∩ f) |
19 | | anass 76 |
. . . . . . 7
((e ∩ e) ∩ f) =
(e ∩ (e ∩ f)) |
20 | 18, 19 | ax-r2 36 |
. . . . . 6
(e ∩ f) = (e ∩
(e ∩ f)) |
21 | 20 | lor 70 |
. . . . 5
(e⊥ ∪ (e ∩ f)) =
(e⊥ ∪ (e ∩ (e ∩
f))) |
22 | 15, 21 | ax-r2 36 |
. . . 4
(e⊥ ∪ f) = (e⊥ ∪ (e ∩ (e ∩
f))) |
23 | 1, 22 | lbtr 139 |
. . 3
f ≤ (e⊥ ∪ (e ∩ (e ∩
f))) |
24 | | leor 159 |
. . . . 5
(e ∩ f) ≤ (((a
∩ b) ∪ (c ∩ d))
∪ (e ∩ f)) |
25 | 24 | lelan 167 |
. . . 4
(e ∩ (e ∩ f)) ≤
(e ∩ (((a ∩ b) ∪
(c ∩ d)) ∪ (e
∩ f))) |
26 | 25 | lelor 166 |
. . 3
(e⊥ ∪ (e ∩ (e ∩
f))) ≤ (e⊥ ∪ (e ∩ (((a
∩ b) ∪ (c ∩ d))
∪ (e ∩ f)))) |
27 | 23, 26 | letr 137 |
. 2
f ≤ (e⊥ ∪ (e ∩ (((a
∩ b) ∪ (c ∩ d))
∪ (e ∩ f)))) |
28 | | oa4to6lem.4 |
. . . . 5
g = (((a ∩ b) ∪
(c ∩ d)) ∪ (e
∩ f)) |
29 | 28 | ud1lem0a 255 |
. . . 4
(e →1 g) = (e
→1 (((a ∩ b) ∪ (c
∩ d)) ∪ (e ∩ f))) |
30 | | df-i1 44 |
. . . 4
(e →1 (((a ∩ b) ∪
(c ∩ d)) ∪ (e
∩ f))) = (e⊥ ∪ (e ∩ (((a
∩ b) ∪ (c ∩ d))
∪ (e ∩ f)))) |
31 | 29, 30 | ax-r2 36 |
. . 3
(e →1 g) = (e⊥ ∪ (e ∩ (((a
∩ b) ∪ (c ∩ d))
∪ (e ∩ f)))) |
32 | 31 | ax-r1 35 |
. 2
(e⊥ ∪ (e ∩ (((a
∩ b) ∪ (c ∩ d))
∪ (e ∩ f)))) = (e
→1 g) |
33 | 27, 32 | lbtr 139 |
1
f ≤ (e →1 g) |