Proof of Theorem imp3
Step | Hyp | Ref
| Expression |
1 | | df-i1 44 |
. . 3
(b →1 c) = (b⊥ ∪ (b ∩ c)) |
2 | 1 | lan 77 |
. 2
((a →2 b) ∩ (b
→1 c)) = ((a →2 b) ∩ (b⊥ ∪ (b ∩ c))) |
3 | | u2lemc1 681 |
. . . 4
b C (a →2 b) |
4 | 3 | comcom3 454 |
. . 3
b⊥ C
(a →2 b) |
5 | | comanr1 464 |
. . . 4
b C (b ∩ c) |
6 | 5 | comcom3 454 |
. . 3
b⊥ C
(b ∩ c) |
7 | 4, 6 | fh2 470 |
. 2
((a →2 b) ∩ (b⊥ ∪ (b ∩ c))) =
(((a →2 b) ∩ b⊥ ) ∪ ((a →2 b) ∩ (b
∩ c))) |
8 | | u2lemanb 616 |
. . 3
((a →2 b) ∩ b⊥ ) = (a⊥ ∩ b⊥ ) |
9 | | ancom 74 |
. . . 4
((a →2 b) ∩ (b
∩ c)) = ((b ∩ c) ∩
(a →2 b)) |
10 | | lea 160 |
. . . . . 6
(b ∩ c) ≤ b |
11 | | u2lem3 750 |
. . . . . . 7
(b →2 (a →2 b)) = 1 |
12 | 11 | u2lemle2 716 |
. . . . . 6
b ≤ (a →2 b) |
13 | 10, 12 | letr 137 |
. . . . 5
(b ∩ c) ≤ (a
→2 b) |
14 | 13 | df2le2 136 |
. . . 4
((b ∩ c) ∩ (a
→2 b)) = (b ∩ c) |
15 | 9, 14 | ax-r2 36 |
. . 3
((a →2 b) ∩ (b
∩ c)) = (b ∩ c) |
16 | 8, 15 | 2or 72 |
. 2
(((a →2 b) ∩ b⊥ ) ∪ ((a →2 b) ∩ (b
∩ c))) = ((a⊥ ∩ b⊥ ) ∪ (b ∩ c)) |
17 | 2, 7, 16 | 3tr 65 |
1
((a →2 b) ∩ (b
→1 c)) = ((a⊥ ∩ b⊥ ) ∪ (b ∩ c)) |