Metamath Proof Explorer


Theorem alephcard

Description: Every aleph is a cardinal number. Theorem 65 of Suppes p. 229. (Contributed by NM, 25-Oct-2003) (Revised by Mario Carneiro, 2-Feb-2013)

Ref Expression
Assertion alephcard card A = A

Proof

Step Hyp Ref Expression
1 2fveq3 x = card x = card
2 fveq2 x = x =
3 1 2 eqeq12d x = card x = x card =
4 2fveq3 x = y card x = card y
5 fveq2 x = y x = y
6 4 5 eqeq12d x = y card x = x card y = y
7 2fveq3 x = suc y card x = card suc y
8 fveq2 x = suc y x = suc y
9 7 8 eqeq12d x = suc y card x = x card suc y = suc y
10 2fveq3 x = A card x = card A
11 fveq2 x = A x = A
12 10 11 eqeq12d x = A card x = x card A = A
13 cardom card ω = ω
14 aleph0 = ω
15 14 fveq2i card = card ω
16 13 15 14 3eqtr4i card =
17 harcard card har y = har y
18 alephsuc y On suc y = har y
19 18 fveq2d y On card suc y = card har y
20 17 19 18 3eqtr4a y On card suc y = suc y
21 20 a1d y On card y = y card suc y = suc y
22 cardiun x V y x card y = y card y x y = y x y
23 22 elv y x card y = y card y x y = y x y
24 23 adantl Lim x y x card y = y card y x y = y x y
25 vex x V
26 alephlim x V Lim x x = y x y
27 25 26 mpan Lim x x = y x y
28 27 adantr Lim x y x card y = y x = y x y
29 28 fveq2d Lim x y x card y = y card x = card y x y
30 24 29 28 3eqtr4d Lim x y x card y = y card x = x
31 30 ex Lim x y x card y = y card x = x
32 3 6 9 12 16 21 31 tfinds A On card A = A
33 card0 card =
34 alephfnon Fn On
35 34 fndmi dom = On
36 35 eleq2i A dom A On
37 ndmfv ¬ A dom A =
38 36 37 sylnbir ¬ A On A =
39 38 fveq2d ¬ A On card A = card
40 33 39 38 3eqtr4a ¬ A On card A = A
41 32 40 pm2.61i card A = A