Proof of Theorem ni32
Step | Hyp | Ref
| Expression |
1 | | df2i3 498 |
. . 3
(a →3 b) = ((a⊥ ∩ b⊥ ) ∪ ((a⊥ ∪ b) ∩ (a
∪ (a⊥ ∩ b)))) |
2 | | oran 87 |
. . . 4
((a⊥ ∩ b⊥ ) ∪ ((a⊥ ∪ b) ∩ (a
∪ (a⊥ ∩ b)))) = ((a⊥ ∩ b⊥ )⊥ ∩
((a⊥ ∪ b) ∩ (a
∪ (a⊥ ∩ b)))⊥
)⊥ |
3 | | oran 87 |
. . . . . . 7
(a ∪ b) = (a⊥ ∩ b⊥
)⊥ |
4 | | oran 87 |
. . . . . . . 8
((a ∩ b⊥ ) ∪ (a⊥ ∩ (a ∪ b⊥ ))) = ((a ∩ b⊥ )⊥ ∩
(a⊥ ∩ (a ∪ b⊥ ))⊥
)⊥ |
5 | | anor1 88 |
. . . . . . . . . . . . 13
(a ∩ b⊥ ) = (a⊥ ∪ b)⊥ |
6 | 5 | con2 67 |
. . . . . . . . . . . 12
(a ∩ b⊥ )⊥ = (a⊥ ∪ b) |
7 | 6 | ax-r1 35 |
. . . . . . . . . . 11
(a⊥ ∪ b) = (a ∩
b⊥
)⊥ |
8 | | oran 87 |
. . . . . . . . . . . 12
(a ∪ (a⊥ ∩ b)) = (a⊥ ∩ (a⊥ ∩ b)⊥
)⊥ |
9 | | anor2 89 |
. . . . . . . . . . . . . . 15
(a⊥ ∩ b) = (a ∪
b⊥
)⊥ |
10 | 9 | con2 67 |
. . . . . . . . . . . . . 14
(a⊥ ∩ b)⊥ = (a ∪ b⊥ ) |
11 | 10 | lan 77 |
. . . . . . . . . . . . 13
(a⊥ ∩ (a⊥ ∩ b)⊥ ) = (a⊥ ∩ (a ∪ b⊥ )) |
12 | 11 | ax-r4 37 |
. . . . . . . . . . . 12
(a⊥ ∩ (a⊥ ∩ b)⊥ )⊥ = (a⊥ ∩ (a ∪ b⊥
))⊥ |
13 | 8, 12 | ax-r2 36 |
. . . . . . . . . . 11
(a ∪ (a⊥ ∩ b)) = (a⊥ ∩ (a ∪ b⊥
))⊥ |
14 | 7, 13 | 2an 79 |
. . . . . . . . . 10
((a⊥ ∪ b) ∩ (a
∪ (a⊥ ∩ b))) = ((a ∩
b⊥ )⊥
∩ (a⊥ ∩ (a ∪ b⊥ ))⊥
) |
15 | 14 | ax-r1 35 |
. . . . . . . . 9
((a ∩ b⊥ )⊥ ∩
(a⊥ ∩ (a ∪ b⊥ ))⊥ ) =
((a⊥ ∪ b) ∩ (a
∪ (a⊥ ∩ b))) |
16 | 15 | ax-r4 37 |
. . . . . . . 8
((a ∩ b⊥ )⊥ ∩
(a⊥ ∩ (a ∪ b⊥ ))⊥
)⊥ = ((a⊥
∪ b) ∩ (a ∪ (a⊥ ∩ b)))⊥ |
17 | 4, 16 | ax-r2 36 |
. . . . . . 7
((a ∩ b⊥ ) ∪ (a⊥ ∩ (a ∪ b⊥ ))) = ((a⊥ ∪ b) ∩ (a
∪ (a⊥ ∩ b)))⊥ |
18 | 3, 17 | 2an 79 |
. . . . . 6
((a ∪ b) ∩ ((a
∩ b⊥ ) ∪ (a⊥ ∩ (a ∪ b⊥ )))) = ((a⊥ ∩ b⊥ )⊥ ∩
((a⊥ ∪ b) ∩ (a
∪ (a⊥ ∩ b)))⊥ ) |
19 | 18 | ax-r1 35 |
. . . . 5
((a⊥ ∩ b⊥ )⊥ ∩
((a⊥ ∪ b) ∩ (a
∪ (a⊥ ∩ b)))⊥ ) = ((a ∪ b) ∩
((a ∩ b⊥ ) ∪ (a⊥ ∩ (a ∪ b⊥ )))) |
20 | 19 | ax-r4 37 |
. . . 4
((a⊥ ∩ b⊥ )⊥ ∩
((a⊥ ∪ b) ∩ (a
∪ (a⊥ ∩ b)))⊥ )⊥ =
((a ∪ b) ∩ ((a
∩ b⊥ ) ∪ (a⊥ ∩ (a ∪ b⊥
))))⊥ |
21 | 2, 20 | ax-r2 36 |
. . 3
((a⊥ ∩ b⊥ ) ∪ ((a⊥ ∪ b) ∩ (a
∪ (a⊥ ∩ b)))) = ((a
∪ b) ∩ ((a ∩ b⊥ ) ∪ (a⊥ ∩ (a ∪ b⊥
))))⊥ |
22 | 1, 21 | ax-r2 36 |
. 2
(a →3 b) = ((a ∪
b) ∩ ((a ∩ b⊥ ) ∪ (a⊥ ∩ (a ∪ b⊥
))))⊥ |
23 | 22 | con2 67 |
1
(a →3 b)⊥ = ((a ∪ b) ∩
((a ∩ b⊥ ) ∪ (a⊥ ∩ (a ∪ b⊥ )))) |