Metamath Proof Explorer


Theorem aks4d1p1p1

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

Ref Expression
Hypotheses aks4d1p1p1.1 φA+
aks4d1p1p1.2 φN
Assertion aks4d1p1p1 φk=1NAk=Ak=1Nk

Proof

Step Hyp Ref Expression
1 aks4d1p1p1.1 φA+
2 aks4d1p1p1.2 φN
3 1 rpcnd φA
4 3 adantr φk1NA
5 1 rpne0d φA0
6 5 adantr φk1NA0
7 elfzelz k1Nk
8 7 zcnd k1Nk
9 8 adantl φk1Nk
10 4 6 9 3jca φk1NAA0k
11 cxpef AA0kAk=eklogA
12 10 11 syl φk1NAk=eklogA
13 12 prodeq2dv φk=1NAk=k=1NeklogA
14 eqid 1=1
15 nnuz =1
16 2 15 eleqtrdi φN1
17 eluzelcn k1k
18 17 adantl φk1k
19 3 adantr φk1A
20 5 adantr φk1A0
21 19 20 logcld φk1logA
22 18 21 mulcld φk1klogA
23 14 16 22 fprodefsum φk=1NeklogA=ek=1NklogA
24 fzfid φ1NFin
25 3 5 logcld φlogA
26 24 25 9 fsummulc1 φk=1NklogA=k=1NklogA
27 26 eqcomd φk=1NklogA=k=1NklogA
28 27 fveq2d φek=1NklogA=ek=1NklogA
29 24 9 fsumcl φk=1Nk
30 3 5 29 cxpefd φAk=1Nk=ek=1NklogA
31 30 eqcomd φek=1NklogA=Ak=1Nk
32 28 31 eqtrd φek=1NklogA=Ak=1Nk
33 23 32 eqtrd φk=1NeklogA=Ak=1Nk
34 13 33 eqtrd φk=1NAk=Ak=1Nk