Proof of Theorem oa3-1to5
Step | Hyp | Ref
| Expression |
1 | | leid 148 |
. . . . 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 170 |
. . . 4
((b →1 c) ∪ ((a
→1 c) ∩ ((a ∩ b) ∪
((a →1 c) ∩ (b
→1 c))))) ≤ (b →1 c) |
4 | 3 | lelan 167 |
. . 3
(c ∩ ((b →1 c) ∪ ((a
→1 c) ∩ ((a ∩ b) ∪
((a →1 c) ∩ (b
→1 c)))))) ≤ (c ∩ (b
→1 c)) |
5 | | ax-a1 30 |
. . . . . . . 8
b = b⊥
⊥ |
6 | 5 | ran 78 |
. . . . . . 7
(b ∩ c) = (b⊥ ⊥ ∩
c) |
7 | 6 | ax-r5 38 |
. . . . . 6
((b ∩ c) ∪ (b⊥ ∩ c)) = ((b⊥ ⊥ ∩
c) ∪ (b⊥ ∩ c)) |
8 | | ax-a2 31 |
. . . . . 6
((b⊥
⊥ ∩ c) ∪ (b⊥ ∩ c)) = ((b⊥ ∩ c) ∪ (b⊥ ⊥ ∩
c)) |
9 | 7, 8 | ax-r2 36 |
. . . . 5
((b ∩ c) ∪ (b⊥ ∩ c)) = ((b⊥ ∩ c) ∪ (b⊥ ⊥ ∩
c)) |
10 | | u1lemab 610 |
. . . . 5
((b →1 c) ∩ c) =
((b ∩ c) ∪ (b⊥ ∩ c)) |
11 | | u1lemab 610 |
. . . . 5
((b⊥ →1
c) ∩ c) = ((b⊥ ∩ c) ∪ (b⊥ ⊥ ∩
c)) |
12 | 9, 10, 11 | 3tr1 63 |
. . . 4
((b →1 c) ∩ c) =
((b⊥ →1
c) ∩ c) |
13 | | ancom 74 |
. . . 4
(c ∩ (b →1 c)) = ((b
→1 c) ∩ c) |
14 | | ancom 74 |
. . . 4
(c ∩ (b⊥ →1 c)) = ((b⊥ →1 c) ∩ c) |
15 | 12, 13, 14 | 3tr1 63 |
. . 3
(c ∩ (b →1 c)) = (c ∩
(b⊥ →1
c)) |
16 | 4, 15 | lbtr 139 |
. 2
(c ∩ ((b →1 c) ∪ ((a
→1 c) ∩ ((a ∩ b) ∪
((a →1 c) ∩ (b
→1 c)))))) ≤ (c ∩ (b⊥ →1 c)) |
17 | | lear 161 |
. 2
(c ∩ (b⊥ →1 c)) ≤ (b⊥ →1 c) |
18 | 16, 17 | letr 137 |
1
(c ∩ ((b →1 c) ∪ ((a
→1 c) ∩ ((a ∩ b) ∪
((a →1 c) ∩ (b
→1 c)))))) ≤ (b⊥ →1 c) |