Detailed syntax breakdown of Definition df-r0
| Step | Hyp | Ref
| Expression |
| 1 | | cr0 35789 |
. 2
class
𝑅0 |
| 2 | | vx |
. . . . . . 7
setvar 𝑥 |
| 3 | 2 | cv 1569 |
. . . . . 6
class 𝑥 |
| 4 | | con0 6351 |
. . . . . . 7
class
On |
| 5 | 4, 4 | cxp 5645 |
. . . . . 6
class (On
× On) |
| 6 | 3, 5 | wcel 2145 |
. . . . 5
wff 𝑥 ∈ (On ×
On) |
| 7 | | vy |
. . . . . . 7
setvar 𝑦 |
| 8 | 7 | cv 1569 |
. . . . . 6
class 𝑦 |
| 9 | 8, 5 | wcel 2145 |
. . . . 5
wff 𝑦 ∈ (On ×
On) |
| 10 | 6, 9 | wa 401 |
. . . 4
wff (𝑥 ∈ (On × On) ∧
𝑦 ∈ (On ×
On)) |
| 11 | | c1st 7982 |
. . . . . . . 8
class
1st |
| 12 | 3, 11 | cfv 6527 |
. . . . . . 7
class
(1st ‘𝑥) |
| 13 | | c2nd 7983 |
. . . . . . . 8
class
2nd |
| 14 | 3, 13 | cfv 6527 |
. . . . . . 7
class
(2nd ‘𝑥) |
| 15 | 12, 14 | cun 3896 |
. . . . . 6
class
((1st ‘𝑥) ∪ (2nd ‘𝑥)) |
| 16 | 8, 11 | cfv 6527 |
. . . . . . 7
class
(1st ‘𝑦) |
| 17 | 8, 13 | cfv 6527 |
. . . . . . 7
class
(2nd ‘𝑦) |
| 18 | 16, 17 | cun 3896 |
. . . . . 6
class
((1st ‘𝑦) ∪ (2nd ‘𝑦)) |
| 19 | 15, 18 | wcel 2145 |
. . . . 5
wff
((1st ‘𝑥) ∪ (2nd ‘𝑥)) ∈ ((1st
‘𝑦) ∪
(2nd ‘𝑦)) |
| 20 | 15, 18 | wceq 1570 |
. . . . . 6
wff
((1st ‘𝑥) ∪ (2nd ‘𝑥)) = ((1st
‘𝑦) ∪
(2nd ‘𝑦)) |
| 21 | | clexo 35788 |
. . . . . . 7
class
LexOrd |
| 22 | 3, 8, 21 | wbr 5102 |
. . . . . 6
wff 𝑥LexOrd𝑦 |
| 23 | 20, 22 | wa 401 |
. . . . 5
wff
(((1st ‘𝑥) ∪ (2nd ‘𝑥)) = ((1st
‘𝑦) ∪
(2nd ‘𝑦))
∧ 𝑥LexOrd𝑦) |
| 24 | 19, 23 | wo 861 |
. . . 4
wff
(((1st ‘𝑥) ∪ (2nd ‘𝑥)) ∈ ((1st
‘𝑦) ∪
(2nd ‘𝑦))
∨ (((1st ‘𝑥) ∪ (2nd ‘𝑥)) = ((1st
‘𝑦) ∪
(2nd ‘𝑦))
∧ 𝑥LexOrd𝑦)) |
| 25 | 10, 24 | wa 401 |
. . 3
wff ((𝑥 ∈ (On × On) ∧
𝑦 ∈ (On × On))
∧ (((1st ‘𝑥) ∪ (2nd ‘𝑥)) ∈ ((1st
‘𝑦) ∪
(2nd ‘𝑦))
∨ (((1st ‘𝑥) ∪ (2nd ‘𝑥)) = ((1st
‘𝑦) ∪
(2nd ‘𝑦))
∧ 𝑥LexOrd𝑦))) |
| 26 | 25, 2, 7 | copab 5166 |
. 2
class
{〈𝑥, 𝑦〉 ∣ ((𝑥 ∈ (On × On) ∧
𝑦 ∈ (On × On))
∧ (((1st ‘𝑥) ∪ (2nd ‘𝑥)) ∈ ((1st
‘𝑦) ∪
(2nd ‘𝑦))
∨ (((1st ‘𝑥) ∪ (2nd ‘𝑥)) = ((1st
‘𝑦) ∪
(2nd ‘𝑦))
∧ 𝑥LexOrd𝑦)))} |
| 27 | 1, 26 | wceq 1570 |
1
wff
𝑅0 = {〈𝑥, 𝑦〉 ∣ ((𝑥 ∈ (On × On) ∧ 𝑦 ∈ (On × On)) ∧
(((1st ‘𝑥)
∪ (2nd ‘𝑥)) ∈ ((1st ‘𝑦) ∪ (2nd
‘𝑦)) ∨
(((1st ‘𝑥)
∪ (2nd ‘𝑥)) = ((1st ‘𝑦) ∪ (2nd
‘𝑦)) ∧ 𝑥LexOrd𝑦)))} |