Proof of Theorem neg3antlem1
| Step | Hyp | Ref
| Expression |
| 1 | | leo 158 |
. . 3
(a ∩ c) ≤ ((a
∩ c) ∪ (a⊥ ∩ c)) |
| 2 | | neg3ant.1 |
. . . . . 6
(a →3 c) = (b
→3 c) |
| 3 | 2 | ran 78 |
. . . . 5
((a →3 c) ∩ c) =
((b →3 c) ∩ c) |
| 4 | | u3lemab 612 |
. . . . 5
((a →3 c) ∩ c) =
((a ∩ c) ∪ (a⊥ ∩ c)) |
| 5 | | u3lemab 612 |
. . . . 5
((b →3 c) ∩ c) =
((b ∩ c) ∪ (b⊥ ∩ c)) |
| 6 | 3, 4, 5 | 3tr2 64 |
. . . 4
((a ∩ c) ∪ (a⊥ ∩ c)) = ((b ∩
c) ∪ (b⊥ ∩ c)) |
| 7 | | u1lemab 610 |
. . . . 5
((b →1 c) ∩ c) =
((b ∩ c) ∪ (b⊥ ∩ c)) |
| 8 | 7 | ax-r1 35 |
. . . 4
((b ∩ c) ∪ (b⊥ ∩ c)) = ((b
→1 c) ∩ c) |
| 9 | 6, 8 | ax-r2 36 |
. . 3
((a ∩ c) ∪ (a⊥ ∩ c)) = ((b
→1 c) ∩ c) |
| 10 | 1, 9 | lbtr 139 |
. 2
(a ∩ c) ≤ ((b
→1 c) ∩ c) |
| 11 | | lea 160 |
. 2
((b →1 c) ∩ c) ≤
(b →1 c) |
| 12 | 10, 11 | letr 137 |
1
(a ∩ c) ≤ (b
→1 c) |