Metamath Proof Explorer


Theorem aks4d1p1p1

Description: Exponential law for finite products, special case. (Contributed by metakunt, 22-Jul-2024)

Ref Expression
Hypotheses aks4d1p1p1.1 φ A +
aks4d1p1p1.2 φ N
Assertion aks4d1p1p1 φ k = 1 N A k = A k = 1 N k

Proof

Step Hyp Ref Expression
1 aks4d1p1p1.1 φ A +
2 aks4d1p1p1.2 φ N
3 1 rpcnd φ A
4 3 adantr φ k 1 N A
5 1 rpne0d φ A 0
6 5 adantr φ k 1 N A 0
7 elfzelz k 1 N k
8 7 zcnd k 1 N k
9 8 adantl φ k 1 N k
10 4 6 9 3jca φ k 1 N A A 0 k
11 cxpef A A 0 k A k = e k log A
12 10 11 syl φ k 1 N A k = e k log A
13 12 prodeq2dv φ k = 1 N A k = k = 1 N e k log A
14 eqid 1 = 1
15 nnuz = 1
16 2 15 eleqtrdi φ N 1
17 eluzelcn k 1 k
18 17 adantl φ k 1 k
19 3 adantr φ k 1 A
20 5 adantr φ k 1 A 0
21 19 20 logcld φ k 1 log A
22 18 21 mulcld φ k 1 k log A
23 14 16 22 fprodefsum φ k = 1 N e k log A = e k = 1 N k log A
24 fzfid φ 1 N Fin
25 3 5 logcld φ log A
26 24 25 9 fsummulc1 φ k = 1 N k log A = k = 1 N k log A
27 26 eqcomd φ k = 1 N k log A = k = 1 N k log A
28 27 fveq2d φ e k = 1 N k log A = e k = 1 N k log A
29 24 9 fsumcl φ k = 1 N k
30 3 5 29 cxpefd φ A k = 1 N k = e k = 1 N k log A
31 30 eqcomd φ e k = 1 N k log A = A k = 1 N k
32 28 31 eqtrd φ e k = 1 N k log A = A k = 1 N k
33 23 32 eqtrd φ k = 1 N e k log A = A k = 1 N k
34 13 33 eqtrd φ k = 1 N A k = A k = 1 N k