Detailed syntax breakdown of Definition df-cndprob
Step | Hyp | Ref
| Expression |
1 | | ccprob 32398 |
. 2
class
cprob |
2 | | vp |
. . 3
setvar 𝑝 |
3 | | cprb 32374 |
. . 3
class
Prob |
4 | | va |
. . . 4
setvar 𝑎 |
5 | | vb |
. . . 4
setvar 𝑏 |
6 | 2 | cv 1538 |
. . . . 5
class 𝑝 |
7 | 6 | cdm 5589 |
. . . 4
class dom 𝑝 |
8 | 4 | cv 1538 |
. . . . . . 7
class 𝑎 |
9 | 5 | cv 1538 |
. . . . . . 7
class 𝑏 |
10 | 8, 9 | cin 3886 |
. . . . . 6
class (𝑎 ∩ 𝑏) |
11 | 10, 6 | cfv 6433 |
. . . . 5
class (𝑝‘(𝑎 ∩ 𝑏)) |
12 | 9, 6 | cfv 6433 |
. . . . 5
class (𝑝‘𝑏) |
13 | | cdiv 11632 |
. . . . 5
class
/ |
14 | 11, 12, 13 | co 7275 |
. . . 4
class ((𝑝‘(𝑎 ∩ 𝑏)) / (𝑝‘𝑏)) |
15 | 4, 5, 7, 7, 14 | cmpo 7277 |
. . 3
class (𝑎 ∈ dom 𝑝, 𝑏 ∈ dom 𝑝 ↦ ((𝑝‘(𝑎 ∩ 𝑏)) / (𝑝‘𝑏))) |
16 | 2, 3, 15 | cmpt 5157 |
. 2
class (𝑝 ∈ Prob ↦ (𝑎 ∈ dom 𝑝, 𝑏 ∈ dom 𝑝 ↦ ((𝑝‘(𝑎 ∩ 𝑏)) / (𝑝‘𝑏)))) |
17 | 1, 16 | wceq 1539 |
1
wff cprob =
(𝑝 ∈ Prob ↦
(𝑎 ∈ dom 𝑝, 𝑏 ∈ dom 𝑝 ↦ ((𝑝‘(𝑎 ∩ 𝑏)) / (𝑝‘𝑏)))) |