Metamath Proof Explorer


Theorem leweon

Description: Lexicographical order is a well-ordering of On X. On . Proposition 7.56(1) of TakeutiZaring p. 54. Note that unlike r0weon , this order isnot set-like, as the preimage of <. 1o , (/) >. is the proper class ( { (/) } X. On ) . (Contributed by Mario Carneiro, 9-Mar-2013)

Ref Expression
Hypothesis leweon.1 ⊢ L = 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
Assertion leweon ⊢ L We On × On

Proof

Step Hyp Ref Expression
1 leweon.1 ⊢ L = 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
2 epweon ⊢ E We On
3 fvex ⊢ 1 st ⁡ y ∈ V
4 3 epeli ⊢ 1 st ⁡ x E 1 st ⁡ y ↔ 1 st ⁡ x ∈ 1 st ⁡ y
5 fvex ⊢ 2 nd ⁡ y ∈ V
6 5 epeli ⊢ 2 nd ⁡ x E 2 nd ⁡ y ↔ 2 nd ⁡ x ∈ 2 nd ⁡ y
7 6 anbi2i ⊢ 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x E 2 nd ⁡ y ↔ 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x ∈ 2 nd ⁡ y
8 4 7 orbi12i ⊢ 1 st ⁡ x E 1 st ⁡ y ∨ 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x E 2 nd ⁡ y ↔ 1 st ⁡ x ∈ 1 st ⁡ y ∨ 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x ∈ 2 nd ⁡ y
9 8 anbi2i ⊢ x ∈ On × On ∧ y ∈ On × On ∧ 1 st ⁡ x E 1 st ⁡ y ∨ 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x E 2 nd ⁡ 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
10 9 opabbii ⊢ x y | x ∈ On × On ∧ y ∈ On × On ∧ 1 st ⁡ x E 1 st ⁡ y ∨ 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x E 2 nd ⁡ y = 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
11 1 10 eqtr4i ⊢ L = x y | x ∈ On × On ∧ y ∈ On × On ∧ 1 st ⁡ x E 1 st ⁡ y ∨ 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x E 2 nd ⁡ y
12 11 wexp ⊢ E We On ∧ E We On → L We On × On
13 2 2 12 mp2an ⊢ L We On × On