Metamath Proof Explorer


Theorem psercnlem2

Description: Lemma for psercn . (Contributed by Mario Carneiro, 18-Mar-2015)

Ref Expression
Hypotheses pserf.g G = x n 0 A n x n
pserf.f F = y S j 0 G y j
pserf.a φ A : 0
pserf.r R = sup r | seq 0 + G r dom * <
psercn.s S = abs -1 0 R
psercnlem2.i φ a S M + a < M M < R
Assertion psercnlem2 φ a S a 0 ball abs M 0 ball abs M abs -1 0 M abs -1 0 M S

Proof

Step Hyp Ref Expression
1 pserf.g G = x n 0 A n x n
2 pserf.f F = y S j 0 G y j
3 pserf.a φ A : 0
4 pserf.r R = sup r | seq 0 + G r dom * <
5 psercn.s S = abs -1 0 R
6 psercnlem2.i φ a S M + a < M M < R
7 cnvimass abs -1 0 R dom abs
8 absf abs :
9 8 fdmi dom abs =
10 7 9 sseqtri abs -1 0 R
11 5 10 eqsstri S
12 11 a1i φ S
13 12 sselda φ a S a
14 13 abscld φ a S a
15 13 absge0d φ a S 0 a
16 6 simp2d φ a S a < M
17 0re 0
18 6 simp1d φ a S M +
19 18 rpxrd φ a S M *
20 elico2 0 M * a 0 M a 0 a a < M
21 17 19 20 sylancr φ a S a 0 M a 0 a a < M
22 14 15 16 21 mpbir3and φ a S a 0 M
23 ffn abs : abs Fn
24 elpreima abs Fn a abs -1 0 M a a 0 M
25 8 23 24 mp2b a abs -1 0 M a a 0 M
26 13 22 25 sylanbrc φ a S a abs -1 0 M
27 eqid abs = abs
28 27 cnbl0 M * abs -1 0 M = 0 ball abs M
29 19 28 syl φ a S abs -1 0 M = 0 ball abs M
30 26 29 eleqtrd φ a S a 0 ball abs M
31 icossicc 0 M 0 M
32 imass2 0 M 0 M abs -1 0 M abs -1 0 M
33 31 32 mp1i φ a S abs -1 0 M abs -1 0 M
34 29 33 eqsstrrd φ a S 0 ball abs M abs -1 0 M
35 iccssxr 0 +∞ *
36 1 3 4 radcnvcl φ R 0 +∞
37 36 adantr φ a S R 0 +∞
38 35 37 sselid φ a S R *
39 6 simp3d φ a S M < R
40 df-ico . = u * , v * w * | u w w < v
41 df-icc . = u * , v * w * | u w w v
42 xrlelttr z * M * R * z M M < R z < R
43 40 41 42 ixxss2 R * M < R 0 M 0 R
44 38 39 43 syl2anc φ a S 0 M 0 R
45 imass2 0 M 0 R abs -1 0 M abs -1 0 R
46 44 45 syl φ a S abs -1 0 M abs -1 0 R
47 46 5 sseqtrrdi φ a S abs -1 0 M S
48 30 34 47 3jca φ a S a 0 ball abs M 0 ball abs M abs -1 0 M abs -1 0 M S