Metamath Proof Explorer


Theorem bcthlem1

Description: Lemma for bcth . Substitutions for the function F . (Contributed by Mario Carneiro, 9-Jan-2014)

Ref Expression
Hypotheses bcth.2 J = MetOpen D
bcthlem.4 φ D CMet X
bcthlem.5 F = k , z X × + x r | x X r + r < 1 k cls J x ball D r ball D z M k
Assertion bcthlem1 φ A B X × + C A F B C X × + 2 nd C < 1 A cls J ball D C ball D B M A

Proof

Step Hyp Ref Expression
1 bcth.2 J = MetOpen D
2 bcthlem.4 φ D CMet X
3 bcthlem.5 F = k , z X × + x r | x X r + r < 1 k cls J x ball D r ball D z M k
4 opabssxp x r | x X r + r < 1 A cls J x ball D r ball D B M A X × +
5 elfvdm D CMet X X dom CMet
6 2 5 syl φ X dom CMet
7 reex V
8 rpssre +
9 7 8 ssexi + V
10 xpexg X dom CMet + V X × + V
11 6 9 10 sylancl φ X × + V
12 ssexg x r | x X r + r < 1 A cls J x ball D r ball D B M A X × + X × + V x r | x X r + r < 1 A cls J x ball D r ball D B M A V
13 4 11 12 sylancr φ x r | x X r + r < 1 A cls J x ball D r ball D B M A V
14 oveq2 k = A 1 k = 1 A
15 14 breq2d k = A r < 1 k r < 1 A
16 fveq2 k = A M k = M A
17 16 difeq2d k = A ball D z M k = ball D z M A
18 17 sseq2d k = A cls J x ball D r ball D z M k cls J x ball D r ball D z M A
19 15 18 anbi12d k = A r < 1 k cls J x ball D r ball D z M k r < 1 A cls J x ball D r ball D z M A
20 19 anbi2d k = A x X r + r < 1 k cls J x ball D r ball D z M k x X r + r < 1 A cls J x ball D r ball D z M A
21 20 opabbidv k = A x r | x X r + r < 1 k cls J x ball D r ball D z M k = x r | x X r + r < 1 A cls J x ball D r ball D z M A
22 fveq2 z = B ball D z = ball D B
23 22 difeq1d z = B ball D z M A = ball D B M A
24 23 sseq2d z = B cls J x ball D r ball D z M A cls J x ball D r ball D B M A
25 24 anbi2d z = B r < 1 A cls J x ball D r ball D z M A r < 1 A cls J x ball D r ball D B M A
26 25 anbi2d z = B x X r + r < 1 A cls J x ball D r ball D z M A x X r + r < 1 A cls J x ball D r ball D B M A
27 26 opabbidv z = B x r | x X r + r < 1 A cls J x ball D r ball D z M A = x r | x X r + r < 1 A cls J x ball D r ball D B M A
28 21 27 3 ovmpog A B X × + x r | x X r + r < 1 A cls J x ball D r ball D B M A V A F B = x r | x X r + r < 1 A cls J x ball D r ball D B M A
29 13 28 syl3an3 A B X × + φ A F B = x r | x X r + r < 1 A cls J x ball D r ball D B M A
30 29 3expa A B X × + φ A F B = x r | x X r + r < 1 A cls J x ball D r ball D B M A
31 30 ancoms φ A B X × + A F B = x r | x X r + r < 1 A cls J x ball D r ball D B M A
32 31 eleq2d φ A B X × + C A F B C x r | x X r + r < 1 A cls J x ball D r ball D B M A
33 4 sseli C x r | x X r + r < 1 A cls J x ball D r ball D B M A C X × +
34 simp1 C X × + 2 nd C < 1 A cls J ball D C ball D B M A C X × +
35 1st2nd2 C X × + C = 1 st C 2 nd C
36 35 eleq1d C X × + C x r | x X r + r < 1 A cls J x ball D r ball D B M A 1 st C 2 nd C x r | x X r + r < 1 A cls J x ball D r ball D B M A
37 fvex 1 st C V
38 fvex 2 nd C V
39 eleq1 x = 1 st C x X 1 st C X
40 eleq1 r = 2 nd C r + 2 nd C +
41 39 40 bi2anan9 x = 1 st C r = 2 nd C x X r + 1 st C X 2 nd C +
42 simpr x = 1 st C r = 2 nd C r = 2 nd C
43 42 breq1d x = 1 st C r = 2 nd C r < 1 A 2 nd C < 1 A
44 oveq12 x = 1 st C r = 2 nd C x ball D r = 1 st C ball D 2 nd C
45 44 fveq2d x = 1 st C r = 2 nd C cls J x ball D r = cls J 1 st C ball D 2 nd C
46 45 sseq1d x = 1 st C r = 2 nd C cls J x ball D r ball D B M A cls J 1 st C ball D 2 nd C ball D B M A
47 43 46 anbi12d x = 1 st C r = 2 nd C r < 1 A cls J x ball D r ball D B M A 2 nd C < 1 A cls J 1 st C ball D 2 nd C ball D B M A
48 41 47 anbi12d x = 1 st C r = 2 nd C x X r + r < 1 A cls J x ball D r ball D B M A 1 st C X 2 nd C + 2 nd C < 1 A cls J 1 st C ball D 2 nd C ball D B M A
49 37 38 48 opelopaba 1 st C 2 nd C x r | x X r + r < 1 A cls J x ball D r ball D B M A 1 st C X 2 nd C + 2 nd C < 1 A cls J 1 st C ball D 2 nd C ball D B M A
50 36 49 bitrdi C X × + C x r | x X r + r < 1 A cls J x ball D r ball D B M A 1 st C X 2 nd C + 2 nd C < 1 A cls J 1 st C ball D 2 nd C ball D B M A
51 35 eleq1d C X × + C X × + 1 st C 2 nd C X × +
52 opelxp 1 st C 2 nd C X × + 1 st C X 2 nd C +
53 51 52 bitr2di C X × + 1 st C X 2 nd C + C X × +
54 df-ov 1 st C ball D 2 nd C = ball D 1 st C 2 nd C
55 35 fveq2d C X × + ball D C = ball D 1 st C 2 nd C
56 54 55 eqtr4id C X × + 1 st C ball D 2 nd C = ball D C
57 56 fveq2d C X × + cls J 1 st C ball D 2 nd C = cls J ball D C
58 57 sseq1d C X × + cls J 1 st C ball D 2 nd C ball D B M A cls J ball D C ball D B M A
59 58 anbi2d C X × + 2 nd C < 1 A cls J 1 st C ball D 2 nd C ball D B M A 2 nd C < 1 A cls J ball D C ball D B M A
60 53 59 anbi12d C X × + 1 st C X 2 nd C + 2 nd C < 1 A cls J 1 st C ball D 2 nd C ball D B M A C X × + 2 nd C < 1 A cls J ball D C ball D B M A
61 3anass C X × + 2 nd C < 1 A cls J ball D C ball D B M A C X × + 2 nd C < 1 A cls J ball D C ball D B M A
62 60 61 bitr4di C X × + 1 st C X 2 nd C + 2 nd C < 1 A cls J 1 st C ball D 2 nd C ball D B M A C X × + 2 nd C < 1 A cls J ball D C ball D B M A
63 50 62 bitrd C X × + C x r | x X r + r < 1 A cls J x ball D r ball D B M A C X × + 2 nd C < 1 A cls J ball D C ball D B M A
64 33 34 63 pm5.21nii C x r | x X r + r < 1 A cls J x ball D r ball D B M A C X × + 2 nd C < 1 A cls J ball D C ball D B M A
65 32 64 bitrdi φ A B X × + C A F B C X × + 2 nd C < 1 A cls J ball D C ball D B M A