Proof of Theorem nom55
| Step | Hyp | Ref
| Expression |
| 1 | | nom25 318 |
. 2
(b⊥ ≡ (b⊥ ∩ a⊥ )) = (b⊥ →1 a⊥ ) |
| 2 | | conb 122 |
. . 3
((a ∪ b) ≡ b) =
((a ∪ b)⊥ ≡ b⊥ ) |
| 3 | | bicom 96 |
. . 3
((a ∪ b)⊥ ≡ b⊥ ) = (b⊥ ≡ (a ∪ b)⊥ ) |
| 4 | | ancom 74 |
. . . . . 6
(b⊥ ∩ a⊥ ) = (a⊥ ∩ b⊥ ) |
| 5 | | anor3 90 |
. . . . . 6
(a⊥ ∩ b⊥ ) = (a ∪ b)⊥ |
| 6 | 4, 5 | ax-r2 36 |
. . . . 5
(b⊥ ∩ a⊥ ) = (a ∪ b)⊥ |
| 7 | 6 | ax-r1 35 |
. . . 4
(a ∪ b)⊥ = (b⊥ ∩ a⊥ ) |
| 8 | 7 | lbi 97 |
. . 3
(b⊥ ≡ (a ∪ b)⊥ ) = (b⊥ ≡ (b⊥ ∩ a⊥ )) |
| 9 | 2, 3, 8 | 3tr 65 |
. 2
((a ∪ b) ≡ b) =
(b⊥ ≡ (b⊥ ∩ a⊥ )) |
| 10 | | i2i1 267 |
. 2
(a →2 b) = (b⊥ →1 a⊥ ) |
| 11 | 1, 9, 10 | 3tr1 63 |
1
((a ∪ b) ≡ b) =
(a →2 b) |