Proof of Theorem distid
Step | Hyp | Ref
| Expression |
1 | | lea 160 |
. . . 4
((a ≡ b) ∩ ((a
≡ c) ∪ (b ≡ c)))
≤ (a ≡ b) |
2 | | mlaconjo 886 |
. . . 4
((a ≡ b) ∩ ((a
≡ c) ∪ (b ≡ c)))
≤ (a ≡ c) |
3 | 1, 2 | ler2an 173 |
. . 3
((a ≡ b) ∩ ((a
≡ c) ∪ (b ≡ c)))
≤ ((a ≡ b) ∩ (a
≡ c)) |
4 | | bicom 96 |
. . . . . 6
(a ≡ b) = (b ≡
a) |
5 | | ax-a2 31 |
. . . . . 6
((a ≡ c) ∪ (b
≡ c)) = ((b ≡ c)
∪ (a ≡ c)) |
6 | 4, 5 | 2an 79 |
. . . . 5
((a ≡ b) ∩ ((a
≡ c) ∪ (b ≡ c))) =
((b ≡ a) ∩ ((b
≡ c) ∪ (a ≡ c))) |
7 | | mlaconjo 886 |
. . . . 5
((b ≡ a) ∩ ((b
≡ c) ∪ (a ≡ c)))
≤ (b ≡ c) |
8 | 6, 7 | bltr 138 |
. . . 4
((a ≡ b) ∩ ((a
≡ c) ∪ (b ≡ c)))
≤ (b ≡ c) |
9 | 1, 8 | ler2an 173 |
. . 3
((a ≡ b) ∩ ((a
≡ c) ∪ (b ≡ c)))
≤ ((a ≡ b) ∩ (b
≡ c)) |
10 | 3, 9 | ler2or 172 |
. 2
((a ≡ b) ∩ ((a
≡ c) ∪ (b ≡ c)))
≤ (((a ≡ b) ∩ (a
≡ c)) ∪ ((a ≡ b)
∩ (b ≡ c))) |
11 | | ledi 174 |
. 2
(((a ≡ b) ∩ (a
≡ c)) ∪ ((a ≡ b)
∩ (b ≡ c))) ≤ ((a
≡ b) ∩ ((a ≡ c)
∪ (b ≡ c))) |
12 | 10, 11 | lebi 145 |
1
((a ≡ b) ∩ ((a
≡ c) ∪ (b ≡ c))) =
(((a ≡ b) ∩ (a
≡ c)) ∪ ((a ≡ b)
∩ (b ≡ c))) |