Proof of Theorem i3n2
Step | Hyp | Ref
| Expression |
1 | | df2i3 498 |
. 2
(a⊥ →3
b⊥ ) = ((a⊥ ⊥ ∩
b⊥ ⊥ )
∪ ((a⊥
⊥ ∪ b⊥ ) ∩ (a⊥ ∪ (a⊥ ⊥ ∩
b⊥ )))) |
2 | | ax-a1 30 |
. . . . 5
a = a⊥
⊥ |
3 | | ax-a1 30 |
. . . . 5
b = b⊥
⊥ |
4 | 2, 3 | 2an 79 |
. . . 4
(a ∩ b) = (a⊥ ⊥ ∩
b⊥ ⊥
) |
5 | 2 | ax-r5 38 |
. . . . 5
(a ∪ b⊥ ) = (a⊥ ⊥ ∪
b⊥ ) |
6 | 2 | ran 78 |
. . . . . 6
(a ∩ b⊥ ) = (a⊥ ⊥ ∩
b⊥ ) |
7 | 6 | lor 70 |
. . . . 5
(a⊥ ∪ (a ∩ b⊥ )) = (a⊥ ∪ (a⊥ ⊥ ∩
b⊥ )) |
8 | 5, 7 | 2an 79 |
. . . 4
((a ∪ b⊥ ) ∩ (a⊥ ∪ (a ∩ b⊥ ))) = ((a⊥ ⊥ ∪
b⊥ ) ∩ (a⊥ ∪ (a⊥ ⊥ ∩
b⊥ ))) |
9 | 4, 8 | 2or 72 |
. . 3
((a ∩ b) ∪ ((a
∪ b⊥ ) ∩ (a⊥ ∪ (a ∩ b⊥ )))) = ((a⊥ ⊥ ∩
b⊥ ⊥ )
∪ ((a⊥
⊥ ∪ b⊥ ) ∩ (a⊥ ∪ (a⊥ ⊥ ∩
b⊥ )))) |
10 | 9 | ax-r1 35 |
. 2
((a⊥
⊥ ∩ b⊥ ⊥ ) ∪
((a⊥ ⊥
∪ b⊥ ) ∩ (a⊥ ∪ (a⊥ ⊥ ∩
b⊥ )))) = ((a ∩ b) ∪
((a ∪ b⊥ ) ∩ (a⊥ ∪ (a ∩ b⊥ )))) |
11 | 1, 10 | ax-r2 36 |
1
(a⊥ →3
b⊥ ) = ((a ∩ b) ∪
((a ∪ b⊥ ) ∩ (a⊥ ∪ (a ∩ b⊥ )))) |