Metamath Proof Explorer


Theorem rplpwr

Description: If A and B are relatively prime, then so are A ^ N and B . (Contributed by Scott Fenton, 12-Apr-2014) (Revised by Mario Carneiro, 19-Apr-2014)

Ref Expression
Assertion rplpwr ABNAgcdB=1ANgcdB=1

Proof

Step Hyp Ref Expression
1 oveq2 k=1Ak=A1
2 1 oveq1d k=1AkgcdB=A1gcdB
3 2 eqeq1d k=1AkgcdB=1A1gcdB=1
4 3 imbi2d k=1ABAgcdB=1AkgcdB=1ABAgcdB=1A1gcdB=1
5 oveq2 k=nAk=An
6 5 oveq1d k=nAkgcdB=AngcdB
7 6 eqeq1d k=nAkgcdB=1AngcdB=1
8 7 imbi2d k=nABAgcdB=1AkgcdB=1ABAgcdB=1AngcdB=1
9 oveq2 k=n+1Ak=An+1
10 9 oveq1d k=n+1AkgcdB=An+1gcdB
11 10 eqeq1d k=n+1AkgcdB=1An+1gcdB=1
12 11 imbi2d k=n+1ABAgcdB=1AkgcdB=1ABAgcdB=1An+1gcdB=1
13 oveq2 k=NAk=AN
14 13 oveq1d k=NAkgcdB=ANgcdB
15 14 eqeq1d k=NAkgcdB=1ANgcdB=1
16 15 imbi2d k=NABAgcdB=1AkgcdB=1ABAgcdB=1ANgcdB=1
17 nncn AA
18 17 exp1d AA1=A
19 18 oveq1d AA1gcdB=AgcdB
20 19 adantr ABA1gcdB=AgcdB
21 20 eqeq1d ABA1gcdB=1AgcdB=1
22 21 biimpar ABAgcdB=1A1gcdB=1
23 df-3an ABnABn
24 simpl1 ABnAgcdB=1A
25 24 nncnd ABnAgcdB=1A
26 simpl3 ABnAgcdB=1n
27 26 nnnn0d ABnAgcdB=1n0
28 25 27 expp1d ABnAgcdB=1An+1=AnA
29 simp1 ABnA
30 nnnn0 nn0
31 30 3ad2ant3 ABnn0
32 29 31 nnexpcld ABnAn
33 32 nnzd ABnAn
34 33 adantr ABnAgcdB=1An
35 34 zcnd ABnAgcdB=1An
36 35 25 mulcomd ABnAgcdB=1AnA=AAn
37 28 36 eqtrd ABnAgcdB=1An+1=AAn
38 37 oveq2d ABnAgcdB=1BgcdAn+1=BgcdAAn
39 simpl2 ABnAgcdB=1B
40 32 adantr ABnAgcdB=1An
41 nnz AA
42 41 3ad2ant1 ABnA
43 nnz BB
44 43 3ad2ant2 ABnB
45 42 44 gcdcomd ABnAgcdB=BgcdA
46 45 eqeq1d ABnAgcdB=1BgcdA=1
47 46 biimpa ABnAgcdB=1BgcdA=1
48 rpmulgcd BAAnBgcdA=1BgcdAAn=BgcdAn
49 39 24 40 47 48 syl31anc ABnAgcdB=1BgcdAAn=BgcdAn
50 38 49 eqtrd ABnAgcdB=1BgcdAn+1=BgcdAn
51 peano2nn nn+1
52 51 3ad2ant3 ABnn+1
53 52 adantr ABnAgcdB=1n+1
54 53 nnnn0d ABnAgcdB=1n+10
55 24 54 nnexpcld ABnAgcdB=1An+1
56 55 nnzd ABnAgcdB=1An+1
57 44 adantr ABnAgcdB=1B
58 56 57 gcdcomd ABnAgcdB=1An+1gcdB=BgcdAn+1
59 34 57 gcdcomd ABnAgcdB=1AngcdB=BgcdAn
60 50 58 59 3eqtr4d ABnAgcdB=1An+1gcdB=AngcdB
61 60 eqeq1d ABnAgcdB=1An+1gcdB=1AngcdB=1
62 61 biimprd ABnAgcdB=1AngcdB=1An+1gcdB=1
63 23 62 sylanbr ABnAgcdB=1AngcdB=1An+1gcdB=1
64 63 an32s ABAgcdB=1nAngcdB=1An+1gcdB=1
65 64 expcom nABAgcdB=1AngcdB=1An+1gcdB=1
66 65 a2d nABAgcdB=1AngcdB=1ABAgcdB=1An+1gcdB=1
67 4 8 12 16 22 66 nnind NABAgcdB=1ANgcdB=1
68 67 expd NABAgcdB=1ANgcdB=1
69 68 com12 ABNAgcdB=1ANgcdB=1
70 69 3impia ABNAgcdB=1ANgcdB=1