Metamath Proof Explorer


Theorem eulerthlem1

Description: Lemma for eulerth . (Contributed by Mario Carneiro, 8-May-2015)

Ref Expression
Hypotheses eulerth.1 φNAAgcdN=1
eulerth.2 S=y0..^N|ygcdN=1
eulerth.3 T=1ϕN
eulerth.4 φF:T1-1 ontoS
eulerth.5 G=xTAFxmodN
Assertion eulerthlem1 φG:TS

Proof

Step Hyp Ref Expression
1 eulerth.1 φNAAgcdN=1
2 eulerth.2 S=y0..^N|ygcdN=1
3 eulerth.3 T=1ϕN
4 eulerth.4 φF:T1-1 ontoS
5 eulerth.5 G=xTAFxmodN
6 1 simp2d φA
7 6 adantr φxTA
8 f1of F:T1-1 ontoSF:TS
9 4 8 syl φF:TS
10 9 ffvelcdmda φxTFxS
11 oveq1 y=FxygcdN=FxgcdN
12 11 eqeq1d y=FxygcdN=1FxgcdN=1
13 12 2 elrab2 FxSFx0..^NFxgcdN=1
14 10 13 sylib φxTFx0..^NFxgcdN=1
15 14 simpld φxTFx0..^N
16 elfzoelz Fx0..^NFx
17 15 16 syl φxTFx
18 7 17 zmulcld φxTAFx
19 1 simp1d φN
20 19 adantr φxTN
21 zmodfzo AFxNAFxmodN0..^N
22 18 20 21 syl2anc φxTAFxmodN0..^N
23 modgcd AFxNAFxmodNgcdN=AFxgcdN
24 18 20 23 syl2anc φxTAFxmodNgcdN=AFxgcdN
25 19 nnzd φN
26 25 adantr φxTN
27 18 26 gcdcomd φxTAFxgcdN=NgcdAFx
28 25 6 gcdcomd φNgcdA=AgcdN
29 1 simp3d φAgcdN=1
30 28 29 eqtrd φNgcdA=1
31 30 adantr φxTNgcdA=1
32 26 17 gcdcomd φxTNgcdFx=FxgcdN
33 14 simprd φxTFxgcdN=1
34 32 33 eqtrd φxTNgcdFx=1
35 rpmul NAFxNgcdA=1NgcdFx=1NgcdAFx=1
36 26 7 17 35 syl3anc φxTNgcdA=1NgcdFx=1NgcdAFx=1
37 31 34 36 mp2and φxTNgcdAFx=1
38 24 27 37 3eqtrd φxTAFxmodNgcdN=1
39 oveq1 y=AFxmodNygcdN=AFxmodNgcdN
40 39 eqeq1d y=AFxmodNygcdN=1AFxmodNgcdN=1
41 40 2 elrab2 AFxmodNSAFxmodN0..^NAFxmodNgcdN=1
42 22 38 41 sylanbrc φxTAFxmodNS
43 42 5 fmptd φG:TS