Proof of Theorem u5lemana
Step | Hyp | Ref
| Expression |
1 | | df-i5 48 |
. . 3
(a →5 b) = (((a ∩
b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) |
2 | 1 | ran 78 |
. 2
((a →5 b) ∩ a⊥ ) = ((((a ∩ b) ∪
(a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∩ a⊥ ) |
3 | | comanr1 464 |
. . . . . 6
a C (a ∩ b) |
4 | 3 | comcom3 454 |
. . . . 5
a⊥ C
(a ∩ b) |
5 | | comanr1 464 |
. . . . 5
a⊥ C
(a⊥ ∩ b) |
6 | 4, 5 | com2or 483 |
. . . 4
a⊥ C
((a ∩ b) ∪ (a⊥ ∩ b)) |
7 | | comanr1 464 |
. . . 4
a⊥ C
(a⊥ ∩ b⊥ ) |
8 | 6, 7 | fh1r 473 |
. . 3
((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∩ a⊥ ) = ((((a ∩ b) ∪
(a⊥ ∩ b)) ∩ a⊥ ) ∪ ((a⊥ ∩ b⊥ ) ∩ a⊥ )) |
9 | 4, 5 | fh1r 473 |
. . . . 5
(((a ∩ b) ∪ (a⊥ ∩ b)) ∩ a⊥ ) = (((a ∩ b) ∩
a⊥ ) ∪ ((a⊥ ∩ b) ∩ a⊥ )) |
10 | | ax-a2 31 |
. . . . . 6
(((a ∩ b) ∩ a⊥ ) ∪ ((a⊥ ∩ b) ∩ a⊥ )) = (((a⊥ ∩ b) ∩ a⊥ ) ∪ ((a ∩ b) ∩
a⊥ )) |
11 | | an32 83 |
. . . . . . . . 9
((a⊥ ∩ b) ∩ a⊥ ) = ((a⊥ ∩ a⊥ ) ∩ b) |
12 | | anidm 111 |
. . . . . . . . . 10
(a⊥ ∩ a⊥ ) = a⊥ |
13 | 12 | ran 78 |
. . . . . . . . 9
((a⊥ ∩ a⊥ ) ∩ b) = (a⊥ ∩ b) |
14 | 11, 13 | ax-r2 36 |
. . . . . . . 8
((a⊥ ∩ b) ∩ a⊥ ) = (a⊥ ∩ b) |
15 | | an32 83 |
. . . . . . . . 9
((a ∩ b) ∩ a⊥ ) = ((a ∩ a⊥ ) ∩ b) |
16 | | ancom 74 |
. . . . . . . . . 10
((a ∩ a⊥ ) ∩ b) = (b ∩
(a ∩ a⊥ )) |
17 | | dff 101 |
. . . . . . . . . . . . 13
0 = (a ∩ a⊥ ) |
18 | 17 | lan 77 |
. . . . . . . . . . . 12
(b ∩ 0) = (b ∩ (a ∩
a⊥ )) |
19 | 18 | ax-r1 35 |
. . . . . . . . . . 11
(b ∩ (a ∩ a⊥ )) = (b ∩ 0) |
20 | | an0 108 |
. . . . . . . . . . 11
(b ∩ 0) = 0 |
21 | 19, 20 | ax-r2 36 |
. . . . . . . . . 10
(b ∩ (a ∩ a⊥ )) = 0 |
22 | 16, 21 | ax-r2 36 |
. . . . . . . . 9
((a ∩ a⊥ ) ∩ b) = 0 |
23 | 15, 22 | ax-r2 36 |
. . . . . . . 8
((a ∩ b) ∩ a⊥ ) = 0 |
24 | 14, 23 | 2or 72 |
. . . . . . 7
(((a⊥ ∩
b) ∩ a⊥ ) ∪ ((a ∩ b) ∩
a⊥ )) = ((a⊥ ∩ b) ∪ 0) |
25 | | or0 102 |
. . . . . . 7
((a⊥ ∩ b) ∪ 0) = (a⊥ ∩ b) |
26 | 24, 25 | ax-r2 36 |
. . . . . 6
(((a⊥ ∩
b) ∩ a⊥ ) ∪ ((a ∩ b) ∩
a⊥ )) = (a⊥ ∩ b) |
27 | 10, 26 | ax-r2 36 |
. . . . 5
(((a ∩ b) ∩ a⊥ ) ∪ ((a⊥ ∩ b) ∩ a⊥ )) = (a⊥ ∩ b) |
28 | 9, 27 | ax-r2 36 |
. . . 4
(((a ∩ b) ∪ (a⊥ ∩ b)) ∩ a⊥ ) = (a⊥ ∩ b) |
29 | | an32 83 |
. . . . 5
((a⊥ ∩ b⊥ ) ∩ a⊥ ) = ((a⊥ ∩ a⊥ ) ∩ b⊥ ) |
30 | 12 | ran 78 |
. . . . 5
((a⊥ ∩ a⊥ ) ∩ b⊥ ) = (a⊥ ∩ b⊥ ) |
31 | 29, 30 | ax-r2 36 |
. . . 4
((a⊥ ∩ b⊥ ) ∩ a⊥ ) = (a⊥ ∩ b⊥ ) |
32 | 28, 31 | 2or 72 |
. . 3
((((a ∩ b) ∪ (a⊥ ∩ b)) ∩ a⊥ ) ∪ ((a⊥ ∩ b⊥ ) ∩ a⊥ )) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) |
33 | 8, 32 | ax-r2 36 |
. 2
((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∩ a⊥ ) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) |
34 | 2, 33 | ax-r2 36 |
1
((a →5 b) ∩ a⊥ ) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) |