Metamath Proof Explorer


Definition df-lexo

Description: Define the lexicographical ordering of On X. On . Based on Definition 7.55 of TakeutiZaring p. 54. (Contributed by BTernaryTau, 2-Sep-2026)

Ref Expression
Assertion df-lexo Could not format assertion : No typesetting found for |- LexOrd = { <. x , y >. | ( ( x e. ( On X. On ) /\ y e. ( On X. On ) ) /\ ( ( 1st ` x ) e. ( 1st ` y ) \/ ( ( 1st ` x ) = ( 1st ` y ) /\ ( 2nd ` x ) e. ( 2nd ` y ) ) ) ) } with typecode |-

Detailed syntax breakdown

Step Hyp Ref Expression
0 clexo Could not format LexOrd : No typesetting found for class LexOrd with typecode class
1 vx setvar x
2 vy setvar y
3 1 cv setvar x
4 con0 class On
5 4 4 cxp class On × On
6 3 5 wcel wff x ∈ On × On
7 2 cv setvar y
8 7 5 wcel wff y ∈ On × On
9 6 8 wa wff x ∈ On × On ∧ y ∈ On × On
10 c1st class 1 st
11 3 10 cfv class 1 st ⁡ x
12 7 10 cfv class 1 st ⁡ y
13 11 12 wcel wff 1 st ⁡ x ∈ 1 st ⁡ y
14 11 12 wceq wff 1 st ⁡ x = 1 st ⁡ y
15 c2nd class 2 nd
16 3 15 cfv class 2 nd ⁡ x
17 7 15 cfv class 2 nd ⁡ y
18 16 17 wcel wff 2 nd ⁡ x ∈ 2 nd ⁡ y
19 14 18 wa wff 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x ∈ 2 nd ⁡ y
20 13 19 wo wff 1 st ⁡ x ∈ 1 st ⁡ y ∨ 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x ∈ 2 nd ⁡ y
21 9 20 wa wff x ∈ On × On ∧ y ∈ On × On ∧ 1 st ⁡ x ∈ 1 st ⁡ y ∨ 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x ∈ 2 nd ⁡ y
22 21 1 2 copab class x y | x ∈ On × On ∧ y ∈ On × On ∧ 1 st ⁡ x ∈ 1 st ⁡ y ∨ 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x ∈ 2 nd ⁡ y
23 0 22 wceq Could not format LexOrd = { <. x , y >. | ( ( x e. ( On X. On ) /\ y e. ( On X. On ) ) /\ ( ( 1st ` x ) e. ( 1st ` y ) \/ ( ( 1st ` x ) = ( 1st ` y ) /\ ( 2nd ` x ) e. ( 2nd ` y ) ) ) ) } : No typesetting found for wff LexOrd = { <. x , y >. | ( ( x e. ( On X. On ) /\ y e. ( On X. On ) ) /\ ( ( 1st ` x ) e. ( 1st ` y ) \/ ( ( 1st ` x ) = ( 1st ` y ) /\ ( 2nd ` x ) e. ( 2nd ` y ) ) ) ) } with typecode wff