Metamath Proof Explorer


Theorem musumsum

Description: Evaluate a collapsing sum over the Möbius function. (Contributed by Mario Carneiro, 4-May-2016)

Ref Expression
Hypotheses musumsum.1 m = 1 B = C
musumsum.2 φ A Fin
musumsum.3 φ A
musumsum.4 φ 1 A
musumsum.5 φ m A B
Assertion musumsum φ m A k n | n m μ k B = C

Proof

Step Hyp Ref Expression
1 musumsum.1 m = 1 B = C
2 musumsum.2 φ A Fin
3 musumsum.3 φ A
4 musumsum.4 φ 1 A
5 musumsum.5 φ m A B
6 3 sselda φ m A m
7 musum m k n | n m μ k = if m = 1 1 0
8 6 7 syl φ m A k n | n m μ k = if m = 1 1 0
9 8 oveq1d φ m A k n | n m μ k B = if m = 1 1 0 B
10 fzfid φ m A 1 m Fin
11 dvdsssfz1 m n | n m 1 m
12 6 11 syl φ m A n | n m 1 m
13 10 12 ssfid φ m A n | n m Fin
14 elrabi k n | n m k
15 mucl k μ k
16 14 15 syl k n | n m μ k
17 16 zcnd k n | n m μ k
18 17 adantl φ m A k n | n m μ k
19 13 5 18 fsummulc1 φ m A k n | n m μ k B = k n | n m μ k B
20 ovif if m = 1 1 0 B = if m = 1 1 B 0 B
21 velsn m 1 m = 1
22 21 bicomi m = 1 m 1
23 22 a1i B m = 1 m 1
24 mulid2 B 1 B = B
25 mul02 B 0 B = 0
26 23 24 25 ifbieq12d B if m = 1 1 B 0 B = if m 1 B 0
27 5 26 syl φ m A if m = 1 1 B 0 B = if m 1 B 0
28 20 27 syl5eq φ m A if m = 1 1 0 B = if m 1 B 0
29 9 19 28 3eqtr3d φ m A k n | n m μ k B = if m 1 B 0
30 29 sumeq2dv φ m A k n | n m μ k B = m A if m 1 B 0
31 4 snssd φ 1 A
32 31 sselda φ m 1 m A
33 32 5 syldan φ m 1 B
34 33 ralrimiva φ m 1 B
35 2 olcd φ A 1 A Fin
36 sumss2 1 A m 1 B A 1 A Fin m 1 B = m A if m 1 B 0
37 31 34 35 36 syl21anc φ m 1 B = m A if m 1 B 0
38 1 eleq1d m = 1 B C
39 5 ralrimiva φ m A B
40 38 39 4 rspcdva φ C
41 1 sumsn 1 A C m 1 B = C
42 4 40 41 syl2anc φ m 1 B = C
43 30 37 42 3eqtr2d φ m A k n | n m μ k B = C