Proof of Theorem gomaex3lem6
Step | Hyp | Ref
| Expression |
1 | | gomaex3lem5.1 |
. . 3
a ≤ b⊥ |
2 | | gomaex3lem5.2 |
. . 3
b ≤ c⊥ |
3 | | gomaex3lem5.3 |
. . 3
c ≤ d⊥ |
4 | | gomaex3lem5.5 |
. . 3
e ≤ f⊥ |
5 | | gomaex3lem5.6 |
. . 3
f ≤ a⊥ |
6 | | gomaex3lem5.8 |
. . 3
(((i →2 g) ∩ (g
→2 y)) ∩ (((y →2 w) ∩ (w
→2 n)) ∩ ((n →2 k) ∩ (k
→2 i)))) ≤ (g →2 i) |
7 | | gomaex3lem5.9 |
. . 3
p = ((a ∪ b)
→1 (d ∪ e)⊥
)⊥ |
8 | | gomaex3lem5.10 |
. . 3
q = ((e ∪ f)
→1 (b ∪ c)⊥
)⊥ |
9 | | gomaex3lem5.11 |
. . 3
r = ((p⊥ →1 q)⊥ ∩ (c ∪ d)) |
10 | | gomaex3lem5.12 |
. . 3
g = a |
11 | | gomaex3lem5.13 |
. . 3
h = b |
12 | | gomaex3lem5.14 |
. . 3
i = c |
13 | | gomaex3lem5.15 |
. . 3
j = (c ∪ d)⊥ |
14 | | gomaex3lem5.16 |
. . 3
k = r |
15 | | gomaex3lem5.17 |
. . 3
m = (p⊥ →1 q) |
16 | | gomaex3lem5.18 |
. . 3
n = (p⊥ →1 q)⊥ |
17 | | gomaex3lem5.19 |
. . 3
u = (p⊥ ∩ q) |
18 | | gomaex3lem5.20 |
. . 3
w = q⊥ |
19 | | gomaex3lem5.21 |
. . 3
x = q |
20 | | gomaex3lem5.22 |
. . 3
y = (e ∪ f)⊥ |
21 | | gomaex3lem5.23 |
. . 3
z = f |
22 | 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,
11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21 | gomaex3lem5 918 |
. 2
(((g ∪ h) ∩ (i
∪ j)) ∩ (((k ∪ m) ∩
(n ∪ u)) ∩ ((w
∪ x) ∩ (y ∪ z))))
≤ (h ∪ i) |
23 | 10, 11 | 2or 72 |
. . . 4
(g ∪ h) = (a ∪
b) |
24 | 12, 13 | 2or 72 |
. . . 4
(i ∪ j) = (c ∪
(c ∪ d)⊥ ) |
25 | 23, 24 | 2an 79 |
. . 3
((g ∪ h) ∩ (i
∪ j)) = ((a ∪ b) ∩
(c ∪ (c ∪ d)⊥ )) |
26 | 14, 15 | 2or 72 |
. . . . 5
(k ∪ m) = (r ∪
(p⊥ →1
q)) |
27 | 16, 17 | 2or 72 |
. . . . 5
(n ∪ u) = ((p⊥ →1 q)⊥ ∪ (p⊥ ∩ q)) |
28 | 26, 27 | 2an 79 |
. . . 4
((k ∪ m) ∩ (n
∪ u)) = ((r ∪ (p⊥ →1 q)) ∩ ((p⊥ →1 q)⊥ ∪ (p⊥ ∩ q))) |
29 | 18, 19 | 2or 72 |
. . . . 5
(w ∪ x) = (q⊥ ∪ q) |
30 | 20, 21 | 2or 72 |
. . . . 5
(y ∪ z) = ((e ∪
f)⊥ ∪ f) |
31 | 29, 30 | 2an 79 |
. . . 4
((w ∪ x) ∩ (y
∪ z)) = ((q⊥ ∪ q) ∩ ((e
∪ f)⊥ ∪ f)) |
32 | 28, 31 | 2an 79 |
. . 3
(((k ∪ m) ∩ (n
∪ u)) ∩ ((w ∪ x) ∩
(y ∪ z))) = (((r
∪ (p⊥ →1
q)) ∩ ((p⊥ →1 q)⊥ ∪ (p⊥ ∩ q))) ∩ ((q⊥ ∪ q) ∩ ((e
∪ f)⊥ ∪ f))) |
33 | 25, 32 | 2an 79 |
. 2
(((g ∪ h) ∩ (i
∪ j)) ∩ (((k ∪ m) ∩
(n ∪ u)) ∩ ((w
∪ x) ∩ (y ∪ z)))) =
(((a ∪ b) ∩ (c
∪ (c ∪ d)⊥ )) ∩ (((r ∪ (p⊥ →1 q)) ∩ ((p⊥ →1 q)⊥ ∪ (p⊥ ∩ q))) ∩ ((q⊥ ∪ q) ∩ ((e
∪ f)⊥ ∪ f)))) |
34 | 11, 12 | 2or 72 |
. 2
(h ∪ i) = (b ∪
c) |
35 | 22, 33, 34 | le3tr2 141 |
1
(((a ∪ b) ∩ (c
∪ (c ∪ d)⊥ )) ∩ (((r ∪ (p⊥ →1 q)) ∩ ((p⊥ →1 q)⊥ ∪ (p⊥ ∩ q))) ∩ ((q⊥ ∪ q) ∩ ((e
∪ f)⊥ ∪ f)))) ≤ (b
∪ c) |