Proof of Theorem wcom3i
Step | Hyp | Ref
| Expression |
1 | | anor1 88 |
. . . . . . . . 9
(a ∩ b⊥ ) = (a⊥ ∪ b)⊥ |
2 | 1 | bi1 118 |
. . . . . . . 8
((a ∩ b⊥ ) ≡ (a⊥ ∪ b)⊥ ) = 1 |
3 | 2 | wcon2 208 |
. . . . . . 7
((a ∩ b⊥ )⊥ ≡
(a⊥ ∪ b)) = 1 |
4 | 3 | wran 369 |
. . . . . 6
(((a ∩ b⊥ )⊥ ∩
a) ≡ ((a⊥ ∪ b) ∩ a)) =
1 |
5 | | ancom 74 |
. . . . . . 7
((a⊥ ∪ b) ∩ a) =
(a ∩ (a⊥ ∪ b)) |
6 | 5 | bi1 118 |
. . . . . 6
(((a⊥ ∪
b) ∩ a) ≡ (a
∩ (a⊥ ∪ b))) = 1 |
7 | 4, 6 | wr2 371 |
. . . . 5
(((a ∩ b⊥ )⊥ ∩
a) ≡ (a ∩ (a⊥ ∪ b))) = 1 |
8 | | wcom3i.1 |
. . . . 5
((a ∩ (a⊥ ∪ b)) ≡ (a
∩ b)) = 1 |
9 | 7, 8 | wr2 371 |
. . . 4
(((a ∩ b⊥ )⊥ ∩
a) ≡ (a ∩ b)) =
1 |
10 | 9 | wlor 368 |
. . 3
(((a ∩ b⊥ ) ∪ ((a ∩ b⊥ )⊥ ∩
a)) ≡ ((a ∩ b⊥ ) ∪ (a ∩ b))) =
1 |
11 | | wlea 388 |
. . . 4
((a ∩ b⊥ ) ≤2 a) = 1 |
12 | 11 | wom4 380 |
. . 3
(((a ∩ b⊥ ) ∪ ((a ∩ b⊥ )⊥ ∩
a)) ≡ a) = 1 |
13 | | ax-a2 31 |
. . . 4
((a ∩ b⊥ ) ∪ (a ∩ b)) =
((a ∩ b) ∪ (a
∩ b⊥
)) |
14 | 13 | bi1 118 |
. . 3
(((a ∩ b⊥ ) ∪ (a ∩ b))
≡ ((a ∩ b) ∪ (a
∩ b⊥ ))) =
1 |
15 | 10, 12, 14 | w3tr2 375 |
. 2
(a ≡ ((a ∩ b) ∪
(a ∩ b⊥ ))) = 1 |
16 | 15 | wdf-c1 383 |
1
C (a, b) = 1 |