Proof of Theorem anass
| Step | Hyp | Ref
| Expression |
| 1 | | ax-a3 31 |
. . . 4
((a⊥ ∪ b⊥ ) ∪ c⊥ ) = (a⊥ ∪ (b⊥ ∪ c⊥ )) |
| 2 | | df-a 39 |
. . . . . 6
(a ∩ b) = (a⊥ ∪ b⊥ )⊥ |
| 3 | 2 | con2 64 |
. . . . 5
(a ∩ b)⊥ = (a⊥ ∪ b⊥ ) |
| 4 | 3 | ax-r5 37 |
. . . 4
((a ∩ b)⊥ ∪ c⊥ ) = ((a⊥ ∪ b⊥ ) ∪ c⊥ ) |
| 5 | | df-a 39 |
. . . . . 6
(b ∩ c) = (b⊥ ∪ c⊥ )⊥ |
| 6 | 5 | con2 64 |
. . . . 5
(b ∩ c)⊥ = (b⊥ ∪ c⊥ ) |
| 7 | 6 | lor 66 |
. . . 4
(a⊥ ∪ (b ∩ c)⊥ ) = (a⊥ ∪ (b⊥ ∪ c⊥ )) |
| 8 | 1, 4, 7 | 3tr1 60 |
. . 3
((a ∩ b)⊥ ∪ c⊥ ) = (a⊥ ∪ (b ∩ c)⊥ ) |
| 9 | 8 | ax-r4 36 |
. 2
((a ∩ b)⊥ ∪ c⊥ )⊥ = (a⊥ ∪ (b ∩ c)⊥ )⊥ |
| 10 | | df-a 39 |
. 2
((a ∩ b) ∩ c) =
((a ∩ b)⊥ ∪ c⊥ )⊥ |
| 11 | | df-a 39 |
. 2
(a ∩ (b ∩ c)) =
(a⊥ ∪ (b ∩ c)⊥ )⊥ |
| 12 | 9, 10, 11 | 3tr1 60 |
1
((a ∩ b) ∩ c) =
(a ∩ (b ∩ c)) |