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