Metamath Proof Explorer


Theorem gchdomtri

Description: Under certain conditions, a GCH-set can demonstrate trichotomy of dominance. Lemma for gchac . (Contributed by Mario Carneiro, 15-May-2015)

Ref Expression
Assertion gchdomtri AGCHA⊔︀AAB𝒫AABBA

Proof

Step Hyp Ref Expression
1 sdomdom ABAB
2 1 con3i ¬AB¬AB
3 reldom Rel
4 3 brrelex1i B𝒫ABV
5 4 3ad2ant3 AGCHA⊔︀AAB𝒫ABV
6 fidomtri2 BVAFinBA¬AB
7 5 6 sylan AGCHA⊔︀AAB𝒫AAFinBA¬AB
8 2 7 imbitrrid AGCHA⊔︀AAB𝒫AAFin¬ABBA
9 8 orrd AGCHA⊔︀AAB𝒫AAFinABBA
10 simp1 AGCHA⊔︀AAB𝒫AAGCH
11 10 adantr AGCHA⊔︀AAB𝒫A¬AFinAGCH
12 simpr AGCHA⊔︀AAB𝒫A¬AFin¬AFin
13 djudoml AGCHBVAA⊔︀B
14 10 5 13 syl2anc AGCHA⊔︀AAB𝒫AAA⊔︀B
15 14 adantr AGCHA⊔︀AAB𝒫A¬AFinAA⊔︀B
16 djulepw A⊔︀AAB𝒫AA⊔︀B𝒫A
17 16 3adant1 AGCHA⊔︀AAB𝒫AA⊔︀B𝒫A
18 17 adantr AGCHA⊔︀AAB𝒫A¬AFinA⊔︀B𝒫A
19 gchor AGCH¬AFinAA⊔︀BA⊔︀B𝒫AAA⊔︀BA⊔︀B𝒫A
20 11 12 15 18 19 syl22anc AGCHA⊔︀AAB𝒫A¬AFinAA⊔︀BA⊔︀B𝒫A
21 djudoml BVAGCHBB⊔︀A
22 5 10 21 syl2anc AGCHA⊔︀AAB𝒫ABB⊔︀A
23 djucomen BVAGCHB⊔︀AA⊔︀B
24 5 10 23 syl2anc AGCHA⊔︀AAB𝒫AB⊔︀AA⊔︀B
25 domentr BB⊔︀AB⊔︀AA⊔︀BBA⊔︀B
26 22 24 25 syl2anc AGCHA⊔︀AAB𝒫ABA⊔︀B
27 domen2 AA⊔︀BBABA⊔︀B
28 26 27 syl5ibrcom AGCHA⊔︀AAB𝒫AAA⊔︀BBA
29 28 imp AGCHA⊔︀AAB𝒫AAA⊔︀BBA
30 29 olcd AGCHA⊔︀AAB𝒫AAA⊔︀BABBA
31 simpl1 AGCHA⊔︀AAB𝒫AA⊔︀B𝒫AAGCH
32 canth2g AGCHA𝒫A
33 sdomdom A𝒫AA𝒫A
34 31 32 33 3syl AGCHA⊔︀AAB𝒫AA⊔︀B𝒫AA𝒫A
35 simpl2 AGCHA⊔︀AAB𝒫AA⊔︀B𝒫AA⊔︀AA
36 pwen A⊔︀AA𝒫A⊔︀A𝒫A
37 35 36 syl AGCHA⊔︀AAB𝒫AA⊔︀B𝒫A𝒫A⊔︀A𝒫A
38 enen2 A⊔︀B𝒫A𝒫A⊔︀AA⊔︀B𝒫A⊔︀A𝒫A
39 38 adantl AGCHA⊔︀AAB𝒫AA⊔︀B𝒫A𝒫A⊔︀AA⊔︀B𝒫A⊔︀A𝒫A
40 37 39 mpbird AGCHA⊔︀AAB𝒫AA⊔︀B𝒫A𝒫A⊔︀AA⊔︀B
41 endom 𝒫A⊔︀AA⊔︀B𝒫A⊔︀AA⊔︀B
42 pwdjudom 𝒫A⊔︀AA⊔︀B𝒫AB
43 40 41 42 3syl AGCHA⊔︀AAB𝒫AA⊔︀B𝒫A𝒫AB
44 domtr A𝒫A𝒫ABAB
45 34 43 44 syl2anc AGCHA⊔︀AAB𝒫AA⊔︀B𝒫AAB
46 45 orcd AGCHA⊔︀AAB𝒫AA⊔︀B𝒫AABBA
47 30 46 jaodan AGCHA⊔︀AAB𝒫AAA⊔︀BA⊔︀B𝒫AABBA
48 20 47 syldan AGCHA⊔︀AAB𝒫A¬AFinABBA
49 9 48 pm2.61dan AGCHA⊔︀AAB𝒫AABBA