Detailed syntax breakdown of Definition df-st
Step | Hyp | Ref
| Expression |
1 | | cst 28897 |
. 2
class
States |
2 | | chba 28854 |
. . . . . 6
class
ℋ |
3 | | vf |
. . . . . . 7
setvar 𝑓 |
4 | 3 | cv 1541 |
. . . . . 6
class 𝑓 |
5 | 2, 4 | cfv 6339 |
. . . . 5
class (𝑓‘
ℋ) |
6 | | c1 10616 |
. . . . 5
class
1 |
7 | 5, 6 | wceq 1542 |
. . . 4
wff (𝑓‘ ℋ) =
1 |
8 | | vx |
. . . . . . . . 9
setvar 𝑥 |
9 | 8 | cv 1541 |
. . . . . . . 8
class 𝑥 |
10 | | vy |
. . . . . . . . . 10
setvar 𝑦 |
11 | 10 | cv 1541 |
. . . . . . . . 9
class 𝑦 |
12 | | cort 28865 |
. . . . . . . . 9
class
⊥ |
13 | 11, 12 | cfv 6339 |
. . . . . . . 8
class
(⊥‘𝑦) |
14 | 9, 13 | wss 3843 |
. . . . . . 7
wff 𝑥 ⊆ (⊥‘𝑦) |
15 | | chj 28868 |
. . . . . . . . . 10
class
∨ℋ |
16 | 9, 11, 15 | co 7170 |
. . . . . . . . 9
class (𝑥 ∨ℋ 𝑦) |
17 | 16, 4 | cfv 6339 |
. . . . . . . 8
class (𝑓‘(𝑥 ∨ℋ 𝑦)) |
18 | 9, 4 | cfv 6339 |
. . . . . . . . 9
class (𝑓‘𝑥) |
19 | 11, 4 | cfv 6339 |
. . . . . . . . 9
class (𝑓‘𝑦) |
20 | | caddc 10618 |
. . . . . . . . 9
class
+ |
21 | 18, 19, 20 | co 7170 |
. . . . . . . 8
class ((𝑓‘𝑥) + (𝑓‘𝑦)) |
22 | 17, 21 | wceq 1542 |
. . . . . . 7
wff (𝑓‘(𝑥 ∨ℋ 𝑦)) = ((𝑓‘𝑥) + (𝑓‘𝑦)) |
23 | 14, 22 | wi 4 |
. . . . . 6
wff (𝑥 ⊆ (⊥‘𝑦) → (𝑓‘(𝑥 ∨ℋ 𝑦)) = ((𝑓‘𝑥) + (𝑓‘𝑦))) |
24 | | cch 28864 |
. . . . . 6
class
Cℋ |
25 | 23, 10, 24 | wral 3053 |
. . . . 5
wff
∀𝑦 ∈
Cℋ (𝑥 ⊆ (⊥‘𝑦) → (𝑓‘(𝑥 ∨ℋ 𝑦)) = ((𝑓‘𝑥) + (𝑓‘𝑦))) |
26 | 25, 8, 24 | wral 3053 |
. . . 4
wff
∀𝑥 ∈
Cℋ ∀𝑦 ∈ Cℋ
(𝑥 ⊆
(⊥‘𝑦) →
(𝑓‘(𝑥 ∨ℋ 𝑦)) = ((𝑓‘𝑥) + (𝑓‘𝑦))) |
27 | 7, 26 | wa 399 |
. . 3
wff ((𝑓‘ ℋ) = 1 ∧
∀𝑥 ∈
Cℋ ∀𝑦 ∈ Cℋ
(𝑥 ⊆
(⊥‘𝑦) →
(𝑓‘(𝑥 ∨ℋ 𝑦)) = ((𝑓‘𝑥) + (𝑓‘𝑦)))) |
28 | | cc0 10615 |
. . . . 5
class
0 |
29 | | cicc 12824 |
. . . . 5
class
[,] |
30 | 28, 6, 29 | co 7170 |
. . . 4
class
(0[,]1) |
31 | | cmap 8437 |
. . . 4
class
↑m |
32 | 30, 24, 31 | co 7170 |
. . 3
class ((0[,]1)
↑m Cℋ ) |
33 | 27, 3, 32 | crab 3057 |
. 2
class {𝑓 ∈ ((0[,]1)
↑m Cℋ ) ∣ ((𝑓‘ ℋ) = 1 ∧
∀𝑥 ∈
Cℋ ∀𝑦 ∈ Cℋ
(𝑥 ⊆
(⊥‘𝑦) →
(𝑓‘(𝑥 ∨ℋ 𝑦)) = ((𝑓‘𝑥) + (𝑓‘𝑦))))} |
34 | 1, 33 | wceq 1542 |
1
wff States =
{𝑓 ∈ ((0[,]1)
↑m Cℋ ) ∣ ((𝑓‘ ℋ) = 1 ∧
∀𝑥 ∈
Cℋ ∀𝑦 ∈ Cℋ
(𝑥 ⊆
(⊥‘𝑦) →
(𝑓‘(𝑥 ∨ℋ 𝑦)) = ((𝑓‘𝑥) + (𝑓‘𝑦))))} |