Proof of Theorem 4oaiii
Step | Hyp | Ref
| Expression |
1 | | 4oa.1 |
. . . 4
e = (((a ∩ c) ∪
((a →1 d) ∩ (c
→1 d))) ∩ ((b ∩ c) ∪
((b →1 d) ∩ (c
→1 d)))) |
2 | | 4oa.2 |
. . . 4
f = (((a ∩ b) ∪
((a →1 d) ∩ (b
→1 d))) ∪ e) |
3 | 1, 2 | 4oa 1039 |
. . 3
((a →1 d) ∩ f) ≤
(b →1 d) |
4 | | lear 161 |
. . 3
((a →1 d) ∩ f) ≤
f |
5 | 3, 4 | ler2an 173 |
. 2
((a →1 d) ∩ f) ≤
((b →1 d) ∩ f) |
6 | | ancom 74 |
. . . . 5
(((a ∩ c) ∪ ((a
→1 d) ∩ (c →1 d))) ∩ ((b
∩ c) ∪ ((b →1 d) ∩ (c
→1 d)))) = (((b ∩ c) ∪
((b →1 d) ∩ (c
→1 d))) ∩ ((a ∩ c) ∪
((a →1 d) ∩ (c
→1 d)))) |
7 | 1, 6 | ax-r2 36 |
. . . 4
e = (((b ∩ c) ∪
((b →1 d) ∩ (c
→1 d))) ∩ ((a ∩ c) ∪
((a →1 d) ∩ (c
→1 d)))) |
8 | | ancom 74 |
. . . . . . 7
(a ∩ b) = (b ∩
a) |
9 | | ancom 74 |
. . . . . . 7
((a →1 d) ∩ (b
→1 d)) = ((b →1 d) ∩ (a
→1 d)) |
10 | 8, 9 | 2or 72 |
. . . . . 6
((a ∩ b) ∪ ((a
→1 d) ∩ (b →1 d))) = ((b ∩
a) ∪ ((b →1 d) ∩ (a
→1 d))) |
11 | 10 | ax-r5 38 |
. . . . 5
(((a ∩ b) ∪ ((a
→1 d) ∩ (b →1 d))) ∪ e) =
(((b ∩ a) ∪ ((b
→1 d) ∩ (a →1 d))) ∪ e) |
12 | 2, 11 | ax-r2 36 |
. . . 4
f = (((b ∩ a) ∪
((b →1 d) ∩ (a
→1 d))) ∪ e) |
13 | 7, 12 | 4oa 1039 |
. . 3
((b →1 d) ∩ f) ≤
(a →1 d) |
14 | | lear 161 |
. . 3
((b →1 d) ∩ f) ≤
f |
15 | 13, 14 | ler2an 173 |
. 2
((b →1 d) ∩ f) ≤
((a →1 d) ∩ f) |
16 | 5, 15 | lebi 145 |
1
((a →1 d) ∩ f) =
((b →1 d) ∩ f) |