Proof of Theorem k1-6
Step | Hyp | Ref
| Expression |
1 | | anor3 90 |
. . . . 5
((x ∩ c)⊥ ∩ (x ∩ c⊥ )⊥ ) =
((x ∩ c) ∪ (x
∩ c⊥
))⊥ |
2 | 1 | cm 61 |
. . . 4
((x ∩ c) ∪ (x
∩ c⊥
))⊥ = ((x ∩ c)⊥ ∩ (x ∩ c⊥ )⊥
) |
3 | | k1-6.1 |
. . . . 5
x = ((x ∩ c) ∪
(x ∩ c⊥ )) |
4 | 3 | con4 69 |
. . . 4
x⊥ = ((x ∩ c) ∪
(x ∩ c⊥
))⊥ |
5 | | oran3 93 |
. . . . 5
(x⊥ ∪ c⊥ ) = (x ∩ c)⊥ |
6 | | oran2 92 |
. . . . 5
(x⊥ ∪ c) = (x ∩
c⊥
)⊥ |
7 | 5, 6 | 2an 79 |
. . . 4
((x⊥ ∪ c⊥ ) ∩ (x⊥ ∪ c)) = ((x ∩
c)⊥ ∩ (x ∩ c⊥ )⊥
) |
8 | 2, 4, 7 | 3tr1 63 |
. . 3
x⊥ = ((x⊥ ∪ c⊥ ) ∩ (x⊥ ∪ c)) |
9 | 8 | ran 78 |
. 2
(x⊥ ∩ c) = (((x⊥ ∪ c⊥ ) ∩ (x⊥ ∪ c)) ∩ c) |
10 | | anass 76 |
. 2
(((x⊥ ∪
c⊥ ) ∩ (x⊥ ∪ c)) ∩ c) =
((x⊥ ∪ c⊥ ) ∩ ((x⊥ ∪ c) ∩ c)) |
11 | | ancom 74 |
. . . 4
((x⊥ ∪ c) ∩ c) =
(c ∩ (x⊥ ∪ c)) |
12 | | ax-a2 31 |
. . . . 5
(x⊥ ∪ c) = (c ∪
x⊥ ) |
13 | 12 | lan 77 |
. . . 4
(c ∩ (x⊥ ∪ c)) = (c ∩
(c ∪ x⊥ )) |
14 | | anabs 121 |
. . . 4
(c ∩ (c ∪ x⊥ )) = c |
15 | 11, 13, 14 | 3tr 65 |
. . 3
((x⊥ ∪ c) ∩ c) =
c |
16 | 15 | lan 77 |
. 2
((x⊥ ∪ c⊥ ) ∩ ((x⊥ ∪ c) ∩ c)) =
((x⊥ ∪ c⊥ ) ∩ c) |
17 | 9, 10, 16 | 3tr 65 |
1
(x⊥ ∩ c) = ((x⊥ ∪ c⊥ ) ∩ c) |