Proof of Theorem oa3-1to5
| Step | Hyp | Ref
| Expression |
| 1 | | leid 140 |
. . . . 5
(b →1 c) ≤ (b
→1 c) |
| 2 | | oa3-1to5.1 |
. . . . 5
((a →1 c) ∩ ((a
∩ b) ∪ ((a →1 c) ∩ (b
→1 c)))) ≤ (b →1 c) |
| 3 | 1, 2 | lel2or 162 |
. . . 4
((b →1 c) ∪ ((a
→1 c) ∩ ((a ∩ b) ∪
((a →1 c) ∩ (b
→1 c))))) ≤ (b →1 c) |
| 4 | 3 | lelan 159 |
. . 3
(c ∩ ((b →1 c) ∪ ((a
→1 c) ∩ ((a ∩ b) ∪
((a →1 c) ∩ (b
→1 c)))))) ≤ (c ∩ (b
→1 c)) |
| 5 | | ax-a1 29 |
. . . . . . . 8
b = b⊥ ⊥ |
| 6 | 5 | ran 71 |
. . . . . . 7
(b ∩ c) = (b⊥ ⊥ ∩
c) |
| 7 | 6 | ax-r5 37 |
. . . . . 6
((b ∩ c) ∪ (b⊥ ∩ c)) = ((b⊥ ⊥ ∩
c) ∪ (b⊥ ∩ c)) |
| 8 | | ax-a2 30 |
. . . . . 6
((b⊥ ⊥
∩ c) ∪ (b⊥ ∩ c)) = ((b⊥ ∩ c) ∪ (b⊥ ⊥ ∩
c)) |
| 9 | 7, 8 | ax-r2 35 |
. . . . 5
((b ∩ c) ∪ (b⊥ ∩ c)) = ((b⊥ ∩ c) ∪ (b⊥ ⊥ ∩
c)) |
| 10 | | u1lemab 592 |
. . . . 5
((b →1 c) ∩ c) =
((b ∩ c) ∪ (b⊥ ∩ c)) |
| 11 | | u1lemab 592 |
. . . . 5
((b⊥ →1
c) ∩ c) = ((b⊥ ∩ c) ∪ (b⊥ ⊥ ∩
c)) |
| 12 | 9, 10, 11 | 3tr1 60 |
. . . 4
((b →1 c) ∩ c) =
((b⊥ →1
c) ∩ c) |
| 13 | | ancom 68 |
. . . 4
(c ∩ (b →1 c)) = ((b
→1 c) ∩ c) |
| 14 | | ancom 68 |
. . . 4
(c ∩ (b⊥ →1 c)) = ((b⊥ →1 c) ∩ c) |
| 15 | 12, 13, 14 | 3tr1 60 |
. . 3
(c ∩ (b →1 c)) = (c ∩
(b⊥ →1
c)) |
| 16 | 4, 15 | lbtr 131 |
. 2
(c ∩ ((b →1 c) ∪ ((a
→1 c) ∩ ((a ∩ b) ∪
((a →1 c) ∩ (b
→1 c)))))) ≤ (c ∩ (b⊥ →1 c)) |
| 17 | | lear 153 |
. 2
(c ∩ (b⊥ →1 c)) ≤ (b⊥ →1 c) |
| 18 | 16, 17 | letr 129 |
1
(c ∩ ((b →1 c) ∪ ((a
→1 c) ∩ ((a ∩ b) ∪
((a →1 c) ∩ (b
→1 c)))))) ≤ (b⊥ →1 c) |