Metamath Proof Explorer


Theorem cardlim

Description: An infinite cardinal is a limit ordinal. Equivalent to Exercise 4 of TakeutiZaring p. 91. (Contributed by Mario Carneiro, 13-Jan-2013)

Ref Expression
Assertion cardlim ω card A Lim card A

Proof

Step Hyp Ref Expression
1 sseq2 card A = suc x ω card A ω suc x
2 1 biimpd card A = suc x ω card A ω suc x
3 limom Lim ω
4 limsssuc Lim ω ω x ω suc x
5 3 4 ax-mp ω x ω suc x
6 infensuc x On ω x x suc x
7 6 ex x On ω x x suc x
8 5 7 syl5bir x On ω suc x x suc x
9 2 8 sylan9r x On card A = suc x ω card A x suc x
10 breq2 card A = suc x x card A x suc x
11 10 adantl x On card A = suc x x card A x suc x
12 9 11 sylibrd x On card A = suc x ω card A x card A
13 12 ex x On card A = suc x ω card A x card A
14 13 com3r ω card A x On card A = suc x x card A
15 14 imp ω card A x On card A = suc x x card A
16 vex x V
17 16 sucid x suc x
18 eleq2 card A = suc x x card A x suc x
19 17 18 mpbiri card A = suc x x card A
20 cardidm card card A = card A
21 19 20 eleqtrrdi card A = suc x x card card A
22 cardne x card card A ¬ x card A
23 21 22 syl card A = suc x ¬ x card A
24 23 a1i ω card A x On card A = suc x ¬ x card A
25 15 24 pm2.65d ω card A x On ¬ card A = suc x
26 25 nrexdv ω card A ¬ x On card A = suc x
27 peano1 ω
28 ssel ω card A ω card A
29 27 28 mpi ω card A card A
30 n0i card A ¬ card A =
31 cardon card A On
32 31 onordi Ord card A
33 ordzsl Ord card A card A = x On card A = suc x Lim card A
34 32 33 mpbi card A = x On card A = suc x Lim card A
35 3orass card A = x On card A = suc x Lim card A card A = x On card A = suc x Lim card A
36 34 35 mpbi card A = x On card A = suc x Lim card A
37 36 ori ¬ card A = x On card A = suc x Lim card A
38 29 30 37 3syl ω card A x On card A = suc x Lim card A
39 38 ord ω card A ¬ x On card A = suc x Lim card A
40 26 39 mpd ω card A Lim card A
41 limomss Lim card A ω card A
42 40 41 impbii ω card A Lim card A