Metamath Proof Explorer


Theorem mule1

Description: The Möbius function takes on values in magnitude at most 1 . (Together with mucl , this implies that it takes a value in { -u 1 , 0 , 1 } for every positive integer.) (Contributed by Mario Carneiro, 22-Sep-2014)

Ref Expression
Assertion mule1 A μ A 1

Proof

Step Hyp Ref Expression
1 muval A μ A = if p p 2 A 0 1 p | p A
2 iftrue p p 2 A if p p 2 A 0 1 p | p A = 0
3 1 2 sylan9eq A p p 2 A μ A = 0
4 3 fveq2d A p p 2 A μ A = 0
5 abs0 0 = 0
6 0le1 0 1
7 5 6 eqbrtri 0 1
8 4 7 eqbrtrdi A p p 2 A μ A 1
9 iffalse ¬ p p 2 A if p p 2 A 0 1 p | p A = 1 p | p A
10 1 9 sylan9eq A ¬ p p 2 A μ A = 1 p | p A
11 10 fveq2d A ¬ p p 2 A μ A = 1 p | p A
12 neg1cn 1
13 prmdvdsfi A p | p A Fin
14 hashcl p | p A Fin p | p A 0
15 13 14 syl A p | p A 0
16 absexp 1 p | p A 0 1 p | p A = 1 p | p A
17 12 15 16 sylancr A 1 p | p A = 1 p | p A
18 ax-1cn 1
19 18 absnegi 1 = 1
20 abs1 1 = 1
21 19 20 eqtri 1 = 1
22 21 oveq1i 1 p | p A = 1 p | p A
23 15 nn0zd A p | p A
24 1exp p | p A 1 p | p A = 1
25 23 24 syl A 1 p | p A = 1
26 22 25 syl5eq A 1 p | p A = 1
27 17 26 eqtrd A 1 p | p A = 1
28 27 adantr A ¬ p p 2 A 1 p | p A = 1
29 11 28 eqtrd A ¬ p p 2 A μ A = 1
30 1le1 1 1
31 29 30 eqbrtrdi A ¬ p p 2 A μ A 1
32 8 31 pm2.61dan A μ A 1