Proof of Theorem 3vcom
| Step | Hyp | Ref
| Expression |
| 1 | | oran3 85 |
. . . . 5
((a⊥ →1
c)⊥ ∪ (b⊥ →1 c)⊥ ) = ((a⊥ →1 c) ∩ (b⊥ →1 c))⊥ |
| 2 | 1 | ax-r1 34 |
. . . 4
((a⊥ →1
c) ∩ (b⊥ →1 c))⊥ = ((a⊥ →1 c)⊥ ∪ (b⊥ →1 c)⊥ ) |
| 3 | | u1lem9ab 761 |
. . . . . 6
(a⊥ →1
c)⊥ ≤ (a →1 c) |
| 4 | | u1lem9ab 761 |
. . . . . 6
(b⊥ →1
c)⊥ ≤ (b →1 c) |
| 5 | 3, 4 | le2or 160 |
. . . . 5
((a⊥ →1
c)⊥ ∪ (b⊥ →1 c)⊥ ) ≤ ((a →1 c) ∪ (b
→1 c)) |
| 6 | 5 | lecom 172 |
. . . 4
((a⊥ →1
c)⊥ ∪ (b⊥ →1 c)⊥ ) C ((a →1 c) ∪ (b
→1 c)) |
| 7 | 2, 6 | bctr 173 |
. . 3
((a⊥ →1
c) ∩ (b⊥ →1 c))⊥ C ((a →1 c) ∪ (b
→1 c)) |
| 8 | 7 | comcom6 441 |
. 2
((a⊥ →1
c) ∩ (b⊥ →1 c)) C ((a
→1 c) ∪ (b →1 c)) |
| 9 | 8 | comcom 435 |
1
((a →1 c) ∪ (b
→1 c)) C ((a⊥ →1 c) ∩ (b⊥ →1 c)) |