Metamath Proof Explorer


Theorem kur14lem9

Description: Lemma for kur14 . Since the set T is closed under closure and complement, it contains the minimal set S as a subset, so S also has at most 1 4 elements. (Indeed S = T , and it's not hard to prove this, but we don't need it for this proof.) (Contributed by Mario Carneiro, 11-Feb-2015)

Ref Expression
Hypotheses kur14lem.j JTop
kur14lem.x X=J
kur14lem.k K=clsJ
kur14lem.i I=intJ
kur14lem.a AX
kur14lem.b B=XKA
kur14lem.c C=KXA
kur14lem.d D=IKA
kur14lem.t T=AXAKABCIAKBDKIAICKDIKBKICIKIA
kur14lem.s S=x𝒫𝒫X|AxyxXyKyx
Assertion kur14lem9 SFinS14

Proof

Step Hyp Ref Expression
1 kur14lem.j JTop
2 kur14lem.x X=J
3 kur14lem.k K=clsJ
4 kur14lem.i I=intJ
5 kur14lem.a AX
6 kur14lem.b B=XKA
7 kur14lem.c C=KXA
8 kur14lem.d D=IKA
9 kur14lem.t T=AXAKABCIAKBDKIAICKDIKBKICIKIA
10 kur14lem.s S=x𝒫𝒫X|AxyxXyKyx
11 vex sV
12 11 elintrab sx𝒫𝒫X|AxyxXyKyxx𝒫𝒫XAxyxXyKyxsx
13 ssun1 AXAKAAXAKABCIA
14 ssun1 AXAKABCIAAXAKABCIAKBDKIA
15 ssun1 AXAKABCIAKBDKIAAXAKABCIAKBDKIAICKDIKBKICIKIA
16 15 9 sseqtrri AXAKABCIAKBDKIAT
17 14 16 sstri AXAKABCIAT
18 13 17 sstri AXAKAT
19 2 topopn JTopXJ
20 1 19 ax-mp XJ
21 20 elexi XV
22 21 5 ssexi AV
23 22 tpid1 AAXAKA
24 18 23 sselii AT
25 1 2 3 4 5 6 7 8 9 kur14lem7 yTyXXyKyT
26 25 simprd yTXyKyT
27 26 rgen yTXyKyT
28 25 simpld yTyX
29 21 elpw2 y𝒫XyX
30 28 29 sylibr yTy𝒫X
31 30 ssriv T𝒫X
32 21 pwex 𝒫XV
33 32 elpw2 T𝒫𝒫XT𝒫X
34 31 33 mpbir T𝒫𝒫X
35 eleq2 x=TAxAT
36 sseq2 x=TXyKyxXyKyT
37 36 raleqbi1dv x=TyxXyKyxyTXyKyT
38 35 37 anbi12d x=TAxyxXyKyxATyTXyKyT
39 eleq2 x=TsxsT
40 38 39 imbi12d x=TAxyxXyKyxsxATyTXyKyTsT
41 40 rspccv x𝒫𝒫XAxyxXyKyxsxT𝒫𝒫XATyTXyKyTsT
42 34 41 mpi x𝒫𝒫XAxyxXyKyxsxATyTXyKyTsT
43 24 27 42 mp2ani x𝒫𝒫XAxyxXyKyxsxsT
44 12 43 sylbi sx𝒫𝒫X|AxyxXyKyxsT
45 44 ssriv x𝒫𝒫X|AxyxXyKyxT
46 10 45 eqsstri ST
47 1 2 3 4 5 6 7 8 9 kur14lem8 TFinT14
48 1nn0 10
49 4nn0 40
50 48 49 deccl 140
51 46 47 50 hashsslei SFinS14