Metamath Proof Explorer


Theorem sgmmul

Description: The divisor function for fixed parameter A is a multiplicative function. (Contributed by Mario Carneiro, 2-Jul-2015)

Ref Expression
Assertion sgmmul A M N M gcd N = 1 A σ M N = A σ M A σ N

Proof

Step Hyp Ref Expression
1 simpr1 A M N M gcd N = 1 M
2 simpr2 A M N M gcd N = 1 N
3 simpr3 A M N M gcd N = 1 M gcd N = 1
4 eqid x | x M = x | x M
5 eqid x | x N = x | x N
6 eqid x | x M N = x | x M N
7 ssrab2 x | x M
8 simpr A M N M gcd N = 1 j x | x M j x | x M
9 7 8 sselid A M N M gcd N = 1 j x | x M j
10 9 nncnd A M N M gcd N = 1 j x | x M j
11 simpll A M N M gcd N = 1 j x | x M A
12 10 11 cxpcld A M N M gcd N = 1 j x | x M j A
13 ssrab2 x | x N
14 simpr A M N M gcd N = 1 k x | x N k x | x N
15 13 14 sselid A M N M gcd N = 1 k x | x N k
16 15 nncnd A M N M gcd N = 1 k x | x N k
17 simpll A M N M gcd N = 1 k x | x N A
18 16 17 cxpcld A M N M gcd N = 1 k x | x N k A
19 9 adantrr A M N M gcd N = 1 j x | x M k x | x N j
20 19 nnred A M N M gcd N = 1 j x | x M k x | x N j
21 19 nnnn0d A M N M gcd N = 1 j x | x M k x | x N j 0
22 21 nn0ge0d A M N M gcd N = 1 j x | x M k x | x N 0 j
23 15 adantrl A M N M gcd N = 1 j x | x M k x | x N k
24 23 nnred A M N M gcd N = 1 j x | x M k x | x N k
25 23 nnnn0d A M N M gcd N = 1 j x | x M k x | x N k 0
26 25 nn0ge0d A M N M gcd N = 1 j x | x M k x | x N 0 k
27 simpll A M N M gcd N = 1 j x | x M k x | x N A
28 20 22 24 26 27 mulcxpd A M N M gcd N = 1 j x | x M k x | x N j k A = j A k A
29 28 eqcomd A M N M gcd N = 1 j x | x M k x | x N j A k A = j k A
30 oveq1 i = j k i A = j k A
31 1 2 3 4 5 6 12 18 29 30 fsumdvdsmul A M N M gcd N = 1 j x | x M j A k x | x N k A = i x | x M N i A
32 sgmval A M A σ M = j x | x M j A
33 1 32 syldan A M N M gcd N = 1 A σ M = j x | x M j A
34 sgmval A N A σ N = k x | x N k A
35 2 34 syldan A M N M gcd N = 1 A σ N = k x | x N k A
36 33 35 oveq12d A M N M gcd N = 1 A σ M A σ N = j x | x M j A k x | x N k A
37 1 2 nnmulcld A M N M gcd N = 1 M N
38 sgmval A M N A σ M N = i x | x M N i A
39 37 38 syldan A M N M gcd N = 1 A σ M N = i x | x M N i A
40 31 36 39 3eqtr4rd A M N M gcd N = 1 A σ M N = A σ M A σ N