Proof of Theorem nom14
| Step | Hyp | Ref
| Expression |
| 1 | | ax-a2 30 |
. . . . 5
((a ∩ (a ∩ b))
∪ (a⊥ ∩ (a ∩ b))) =
((a⊥ ∩ (a ∩ b))
∪ (a ∩ (a ∩ b))) |
| 2 | | anass 69 |
. . . . . . . 8
((a ∩ a) ∩ b) =
(a ∩ (a ∩ b)) |
| 3 | 2 | ax-r1 34 |
. . . . . . 7
(a ∩ (a ∩ b)) =
((a ∩ a) ∩ b) |
| 4 | | anidm 103 |
. . . . . . . 8
(a ∩ a) = a |
| 5 | 4 | ran 71 |
. . . . . . 7
((a ∩ a) ∩ b) =
(a ∩ b) |
| 6 | 3, 5 | ax-r2 35 |
. . . . . 6
(a ∩ (a ∩ b)) =
(a ∩ b) |
| 7 | 6 | lor 66 |
. . . . 5
((a⊥ ∩ (a ∩ b))
∪ (a ∩ (a ∩ b))) =
((a⊥ ∩ (a ∩ b))
∪ (a ∩ b)) |
| 8 | | lear 153 |
. . . . . 6
(a⊥ ∩ (a ∩ b)) ≤
(a ∩ b) |
| 9 | 8 | df-le2 123 |
. . . . 5
((a⊥ ∩ (a ∩ b))
∪ (a ∩ b)) = (a ∩
b) |
| 10 | 1, 7, 9 | 3tr 62 |
. . . 4
((a ∩ (a ∩ b))
∪ (a⊥ ∩ (a ∩ b))) =
(a ∩ b) |
| 11 | 10 | ax-r5 37 |
. . 3
(((a ∩ (a ∩ b))
∪ (a⊥ ∩ (a ∩ b)))
∪ ((a⊥ ∪ (a ∩ b))
∩ (a ∩ b)⊥ )) = ((a ∩ b) ∪
((a⊥ ∪ (a ∩ b))
∩ (a ∩ b)⊥ )) |
| 12 | | leo 150 |
. . . . 5
(a ∩ b) ≤ ((a
∩ b) ∪ a⊥ ) |
| 13 | | lea 152 |
. . . . . 6
((a⊥ ∪ (a ∩ b))
∩ (a ∩ b)⊥ ) ≤ (a⊥ ∪ (a ∩ b)) |
| 14 | | ax-a2 30 |
. . . . . 6
(a⊥ ∪ (a ∩ b)) =
((a ∩ b) ∪ a⊥ ) |
| 15 | 13, 14 | lbtr 131 |
. . . . 5
((a⊥ ∪ (a ∩ b))
∩ (a ∩ b)⊥ ) ≤ ((a ∩ b) ∪
a⊥ ) |
| 16 | 12, 15 | lel2or 162 |
. . . 4
((a ∩ b) ∪ ((a⊥ ∪ (a ∩ b))
∩ (a ∩ b)⊥ )) ≤ ((a ∩ b) ∪
a⊥ ) |
| 17 | | leo 150 |
. . . . . 6
a⊥ ≤ (a⊥ ∪ (a ∩ b)) |
| 18 | | lea 152 |
. . . . . . 7
(a ∩ b) ≤ a |
| 19 | 18 | lecon 146 |
. . . . . 6
a⊥ ≤ (a ∩ b)⊥ |
| 20 | 17, 19 | ler2an 165 |
. . . . 5
a⊥ ≤ ((a⊥ ∪ (a ∩ b))
∩ (a ∩ b)⊥ ) |
| 21 | 20 | lelor 158 |
. . . 4
((a ∩ b) ∪ a⊥ ) ≤ ((a ∩ b) ∪
((a⊥ ∪ (a ∩ b))
∩ (a ∩ b)⊥ )) |
| 22 | 16, 21 | lebi 137 |
. . 3
((a ∩ b) ∪ ((a⊥ ∪ (a ∩ b))
∩ (a ∩ b)⊥ )) = ((a ∩ b) ∪
a⊥ ) |
| 23 | | ax-a2 30 |
. . 3
((a ∩ b) ∪ a⊥ ) = (a⊥ ∪ (a ∩ b)) |
| 24 | 11, 22, 23 | 3tr 62 |
. 2
(((a ∩ (a ∩ b))
∪ (a⊥ ∩ (a ∩ b)))
∪ ((a⊥ ∪ (a ∩ b))
∩ (a ∩ b)⊥ )) = (a⊥ ∪ (a ∩ b)) |
| 25 | | df-i4 46 |
. 2
(a →4 (a ∩ b)) =
(((a ∩ (a ∩ b))
∪ (a⊥ ∩ (a ∩ b)))
∪ ((a⊥ ∪ (a ∩ b))
∩ (a ∩ b)⊥ )) |
| 26 | | df-i1 43 |
. 2
(a →1 b) = (a⊥ ∪ (a ∩ b)) |
| 27 | 24, 25, 26 | 3tr1 60 |
1
(a →4 (a ∩ b)) =
(a →1 b) |