Metamath Proof Explorer


Theorem gch2

Description: It is sufficient to require that all alephs are GCH-sets to ensure the full generalized continuum hypothesis. (The proof uses the Axiom of Regularity.) (Contributed by Mario Carneiro, 15-May-2015)

Ref Expression
Assertion gch2 GCH = V ran GCH

Proof

Step Hyp Ref Expression
1 ssv ran V
2 sseq2 GCH = V ran GCH ran V
3 1 2 mpbiri GCH = V ran GCH
4 cardidm card card x = card x
5 iscard3 card card x = card x card x ω ran
6 4 5 mpbi card x ω ran
7 elun card x ω ran card x ω card x ran
8 6 7 mpbi card x ω card x ran
9 fingch Fin GCH
10 nnfi card x ω card x Fin
11 9 10 sselid card x ω card x GCH
12 11 a1i ran GCH card x ω card x GCH
13 ssel ran GCH card x ran card x GCH
14 12 13 jaod ran GCH card x ω card x ran card x GCH
15 8 14 mpi ran GCH card x GCH
16 vex x V
17 alephon suc x On
18 simpr ran GCH x On x On
19 simpl ran GCH x On ran GCH
20 alephfnon Fn On
21 fnfvelrn Fn On x On x ran
22 20 18 21 sylancr ran GCH x On x ran
23 19 22 sseldd ran GCH x On x GCH
24 suceloni x On suc x On
25 24 adantl ran GCH x On suc x On
26 fnfvelrn Fn On suc x On suc x ran
27 20 25 26 sylancr ran GCH x On suc x ran
28 19 27 sseldd ran GCH x On suc x GCH
29 gchaleph2 x On x GCH suc x GCH suc x 𝒫 x
30 18 23 28 29 syl3anc ran GCH x On suc x 𝒫 x
31 isnumi suc x On suc x 𝒫 x 𝒫 x dom card
32 17 30 31 sylancr ran GCH x On 𝒫 x dom card
33 32 ralrimiva ran GCH x On 𝒫 x dom card
34 dfac12 CHOICE x On 𝒫 x dom card
35 33 34 sylibr ran GCH CHOICE
36 dfac10 CHOICE dom card = V
37 35 36 sylib ran GCH dom card = V
38 16 37 eleqtrrid ran GCH x dom card
39 cardid2 x dom card card x x
40 engch card x x card x GCH x GCH
41 38 39 40 3syl ran GCH card x GCH x GCH
42 15 41 mpbid ran GCH x GCH
43 16 a1i ran GCH x V
44 42 43 2thd ran GCH x GCH x V
45 44 eqrdv ran GCH GCH = V
46 3 45 impbii GCH = V ran GCH