Metamath Proof Explorer


Theorem ablfac1lem

Description: Lemma for ablfac1b . Satisfy the assumptions of ablfacrp. (Contributed by Mario Carneiro, 26-Apr-2016)

Ref Expression
Hypotheses ablfac1.b B=BaseG
ablfac1.o O=odG
ablfac1.s S=pAxB|OxpppCntB
ablfac1.g φGAbel
ablfac1.f φBFin
ablfac1.1 φA
ablfac1.m M=PPpCntB
ablfac1.n N=BM
Assertion ablfac1lem φPAMNMgcdN=1B= M N

Proof

Step Hyp Ref Expression
1 ablfac1.b B=BaseG
2 ablfac1.o O=odG
3 ablfac1.s S=pAxB|OxpppCntB
4 ablfac1.g φGAbel
5 ablfac1.f φBFin
6 ablfac1.1 φA
7 ablfac1.m M=PPpCntB
8 ablfac1.n N=BM
9 6 sselda φPAP
10 prmnn PP
11 9 10 syl φPAP
12 ablgrp GAbelGGrp
13 1 grpbn0 GGrpB
14 4 12 13 3syl φB
15 hashnncl BFinBB
16 5 15 syl φBB
17 14 16 mpbird φB
18 17 adantr φPAB
19 9 18 pccld φPAPpCntB0
20 11 19 nnexpcld φPAPPpCntB
21 7 20 eqeltrid φPAM
22 pcdvds PBPPpCntBB
23 9 18 22 syl2anc φPAPPpCntBB
24 7 23 eqbrtrid φPAMB
25 nndivdvds BMMBBM
26 18 21 25 syl2anc φPAMBBM
27 24 26 mpbid φPABM
28 8 27 eqeltrid φPAN
29 21 28 jca φPAMN
30 7 oveq1i MgcdN=PPpCntBgcdN
31 pcndvds2 PB¬PBPPpCntB
32 9 18 31 syl2anc φPA¬PBPPpCntB
33 7 oveq2i BM=BPPpCntB
34 8 33 eqtri N=BPPpCntB
35 34 breq2i PNPBPPpCntB
36 32 35 sylnibr φPA¬PN
37 28 nnzd φPAN
38 coprm PN¬PNPgcdN=1
39 9 37 38 syl2anc φPA¬PNPgcdN=1
40 36 39 mpbid φPAPgcdN=1
41 prmz PP
42 9 41 syl φPAP
43 rpexp1i PNPpCntB0PgcdN=1PPpCntBgcdN=1
44 42 37 19 43 syl3anc φPAPgcdN=1PPpCntBgcdN=1
45 40 44 mpd φPAPPpCntBgcdN=1
46 30 45 eqtrid φPAMgcdN=1
47 8 oveq2i M N=MBM
48 18 nncnd φPAB
49 21 nncnd φPAM
50 21 nnne0d φPAM0
51 48 49 50 divcan2d φPAMBM=B
52 47 51 eqtr2id φPAB= M N
53 29 46 52 3jca φPAMNMgcdN=1B= M N