Metamath Proof Explorer


Theorem infxpen

Description: Every infinite ordinal is equinumerous to its Cartesian square. Proposition 10.39 of TakeutiZaring p. 94, whose proof we follow closely. The key idea is to show that the relation R is a well-ordering of ( On X. On ) with the additional property that R -initial segments of ( x X. x ) (where x is a limit ordinal) are of cardinality at most x . (Contributed by Mario Carneiro, 9-Mar-2013) (Revised by Mario Carneiro, 26-Jun-2015)

Ref Expression
Assertion infxpen A On ω A A × A A

Proof

Step Hyp Ref Expression
1 eqid 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 = 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 eleq1w s = z s On × On z On × On
3 eleq1w t = w t On × On w On × On
4 2 3 bi2anan9 s = z t = w s On × On t On × On z On × On w On × On
5 fveq2 s = z 1 st s = 1 st z
6 fveq2 s = z 2 nd s = 2 nd z
7 5 6 uneq12d s = z 1 st s 2 nd s = 1 st z 2 nd z
8 7 adantr s = z t = w 1 st s 2 nd s = 1 st z 2 nd z
9 fveq2 t = w 1 st t = 1 st w
10 fveq2 t = w 2 nd t = 2 nd w
11 9 10 uneq12d t = w 1 st t 2 nd t = 1 st w 2 nd w
12 11 adantl s = z t = w 1 st t 2 nd t = 1 st w 2 nd w
13 8 12 eleq12d s = z t = w 1 st s 2 nd s 1 st t 2 nd t 1 st z 2 nd z 1 st w 2 nd w
14 7 11 eqeqan12d s = z t = w 1 st s 2 nd s = 1 st t 2 nd t 1 st z 2 nd z = 1 st w 2 nd w
15 breq12 s = z t = w s 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 t z 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 w
16 14 15 anbi12d s = z t = w 1 st s 2 nd s = 1 st t 2 nd t s 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 t 1 st z 2 nd z = 1 st w 2 nd w z 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 w
17 13 16 orbi12d s = z t = w 1 st s 2 nd s 1 st t 2 nd t 1 st s 2 nd s = 1 st t 2 nd t s 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 t 1 st z 2 nd z 1 st w 2 nd w 1 st z 2 nd z = 1 st w 2 nd w z 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 w
18 4 17 anbi12d s = z t = w s On × On t On × On 1 st s 2 nd s 1 st t 2 nd t 1 st s 2 nd s = 1 st t 2 nd t s 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 t z On × On w On × On 1 st z 2 nd z 1 st w 2 nd w 1 st z 2 nd z = 1 st w 2 nd w z 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 w
19 18 cbvopabv s t | s On × On t On × On 1 st s 2 nd s 1 st t 2 nd t 1 st s 2 nd s = 1 st t 2 nd t s 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 t = z w | z On × On w On × On 1 st z 2 nd z 1 st w 2 nd w 1 st z 2 nd z = 1 st w 2 nd w z 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 w
20 eqid s t | s On × On t On × On 1 st s 2 nd s 1 st t 2 nd t 1 st s 2 nd s = 1 st t 2 nd t s 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 t a × a × a × a = s t | s On × On t On × On 1 st s 2 nd s 1 st t 2 nd t 1 st s 2 nd s = 1 st t 2 nd t s 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 t a × a × a × a
21 biid a On m a ω m m × m m ω a m a m a a On m a ω m m × m m ω a m a m a
22 eqid 1 st w 2 nd w = 1 st w 2 nd w
23 eqid OrdIso s t | s On × On t On × On 1 st s 2 nd s 1 st t 2 nd t 1 st s 2 nd s = 1 st t 2 nd t s 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 t a × a × a × a a × a = OrdIso s t | s On × On t On × On 1 st s 2 nd s 1 st t 2 nd t 1 st s 2 nd s = 1 st t 2 nd t s 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 t a × a × a × a a × a
24 1 19 20 21 22 23 infxpenlem A On ω A A × A A