Proof of Theorem k1-5
Step | Hyp | Ref
| Expression |
1 | | k1-5.1 |
. . 3
(x⊥ ∩ (x ∪ c)) =
(((x⊥ ∩ (x ∪ c))
∩ c) ∪ ((x⊥ ∩ (x ∪ c))
∩ c⊥
)) |
2 | | ax-a1 30 |
. . . . 5
c = c⊥
⊥ |
3 | 2 | lor 70 |
. . . 4
(x ∪ c) = (x ∪
c⊥ ⊥
) |
4 | 3 | lan 77 |
. . 3
(x⊥ ∩ (x ∪ c)) =
(x⊥ ∩ (x ∪ c⊥ ⊥
)) |
5 | | orcom 73 |
. . . 4
(((x⊥ ∩
(x ∪ c)) ∩ c)
∪ ((x⊥ ∩ (x ∪ c))
∩ c⊥ )) = (((x⊥ ∩ (x ∪ c))
∩ c⊥ ) ∪
((x⊥ ∩ (x ∪ c))
∩ c)) |
6 | 4 | ran 78 |
. . . . 5
((x⊥ ∩
(x ∪ c)) ∩ c⊥ ) = ((x⊥ ∩ (x ∪ c⊥ ⊥ )) ∩
c⊥ ) |
7 | 4, 2 | 2an 79 |
. . . . 5
((x⊥ ∩
(x ∪ c)) ∩ c) =
((x⊥ ∩ (x ∪ c⊥ ⊥ )) ∩
c⊥ ⊥
) |
8 | 6, 7 | 2or 72 |
. . . 4
(((x⊥ ∩
(x ∪ c)) ∩ c⊥ ) ∪ ((x⊥ ∩ (x ∪ c))
∩ c)) = (((x⊥ ∩ (x ∪ c⊥ ⊥ )) ∩
c⊥ ) ∪ ((x⊥ ∩ (x ∪ c⊥ ⊥ )) ∩
c⊥ ⊥
)) |
9 | 5, 8 | tr 62 |
. . 3
(((x⊥ ∩
(x ∪ c)) ∩ c)
∪ ((x⊥ ∩ (x ∪ c))
∩ c⊥ )) = (((x⊥ ∩ (x ∪ c⊥ ⊥ )) ∩
c⊥ ) ∪ ((x⊥ ∩ (x ∪ c⊥ ⊥ )) ∩
c⊥ ⊥
)) |
10 | 1, 4, 9 | 3tr2 64 |
. 2
(x⊥ ∩ (x ∪ c⊥ ⊥ )) =
(((x⊥ ∩ (x ∪ c⊥ ⊥ )) ∩
c⊥ ) ∪ ((x⊥ ∩ (x ∪ c⊥ ⊥ )) ∩
c⊥ ⊥
)) |
11 | | k1-5.2 |
. 2
x ≤ c⊥ |
12 | 10, 11 | k1-4 359 |
1
(x ∪ (x⊥ ∩ c⊥ )) = c⊥ |