Proof of Theorem ka4lem
Step | Hyp | Ref
| Expression |
1 | | df-a 40 |
. . . 4
(a ∩ b) = (a⊥ ∪ b⊥
)⊥ |
2 | 1 | con2 67 |
. . 3
(a ∩ b)⊥ = (a⊥ ∪ b⊥ ) |
3 | | df-a 40 |
. . . . 5
(a ∩ c) = (a⊥ ∪ c⊥
)⊥ |
4 | | df-a 40 |
. . . . 5
(b ∩ c) = (b⊥ ∪ c⊥
)⊥ |
5 | 3, 4 | 2bi 99 |
. . . 4
((a ∩ c) ≡ (b
∩ c)) = ((a⊥ ∪ c⊥ )⊥ ≡
(b⊥ ∪ c⊥ )⊥
) |
6 | | conb 122 |
. . . . 5
((a⊥ ∪ c⊥ ) ≡ (b⊥ ∪ c⊥ )) = ((a⊥ ∪ c⊥ )⊥ ≡
(b⊥ ∪ c⊥ )⊥
) |
7 | 6 | ax-r1 35 |
. . . 4
((a⊥ ∪ c⊥ )⊥ ≡
(b⊥ ∪ c⊥ )⊥ ) =
((a⊥ ∪ c⊥ ) ≡ (b⊥ ∪ c⊥ )) |
8 | 5, 7 | ax-r2 36 |
. . 3
((a ∩ c) ≡ (b
∩ c)) = ((a⊥ ∪ c⊥ ) ≡ (b⊥ ∪ c⊥ )) |
9 | 2, 8 | 2or 72 |
. 2
((a ∩ b)⊥ ∪ ((a ∩ c)
≡ (b ∩ c))) = ((a⊥ ∪ b⊥ ) ∪ ((a⊥ ∪ c⊥ ) ≡ (b⊥ ∪ c⊥ ))) |
10 | | ka4lemo 228 |
. 2
((a⊥ ∪ b⊥ ) ∪ ((a⊥ ∪ c⊥ ) ≡ (b⊥ ∪ c⊥ ))) = 1 |
11 | 9, 10 | ax-r2 36 |
1
((a ∩ b)⊥ ∪ ((a ∩ c)
≡ (b ∩ c))) = 1 |