Proof of Theorem mhlemlem2
| Step | Hyp | Ref
| Expression |
| 1 | | ax-a2 30 |
. . . 4
(a ∪ b) = (b ∪
a) |
| 2 | 1 | ax-r5 37 |
. . 3
((a ∪ b) ∪ d) =
((b ∪ a) ∪ d) |
| 3 | | ax-a2 30 |
. . . 4
(c ∪ d) = (d ∪
c) |
| 4 | 3 | lor 66 |
. . 3
(b ∪ (c ∪ d)) =
(b ∪ (d ∪ c)) |
| 5 | 2, 4 | 2an 72 |
. 2
(((a ∪ b) ∪ d)
∩ (b ∪ (c ∪ d))) =
(((b ∪ a) ∪ d)
∩ (b ∪ (d ∪ c))) |
| 6 | | mhlem.1 |
. . . 4
(a ∪ b) ≤ (c ∪
d)⊥ |
| 7 | | ax-a2 30 |
. . . 4
(b ∪ a) = (a ∪
b) |
| 8 | | ax-a2 30 |
. . . . 5
(d ∪ c) = (c ∪
d) |
| 9 | 8 | ax-r4 36 |
. . . 4
(d ∪ c)⊥ = (c ∪ d)⊥ |
| 10 | 6, 7, 9 | le3tr1 132 |
. . 3
(b ∪ a) ≤ (d ∪
c)⊥ |
| 11 | 10 | mhlemlem1 856 |
. 2
(((b ∪ a) ∪ d)
∩ (b ∪ (d ∪ c))) =
(b ∪ d) |
| 12 | 5, 11 | ax-r2 35 |
1
(((a ∪ b) ∪ d)
∩ (b ∪ (c ∪ d))) =
(b ∪ d) |