Metamath Proof Explorer


Theorem reumodprminv

Description: For any prime number and for any positive integer less than this prime number, there is a unique modular inverse of this positive integer. (Contributed by Alexander van der Vekens, 12-May-2018)

Ref Expression
Assertion reumodprminv PN1..^P∃!i1P1NimodP=1

Proof

Step Hyp Ref Expression
1 simpl PN1..^PP
2 elfzoelz N1..^PN
3 2 adantl PN1..^PN
4 prmnn PP
5 prmz PP
6 fzoval P1..^P=1P1
7 5 6 syl P1..^P=1P1
8 7 eleq2d PN1..^PN1P1
9 8 biimpa PN1..^PN1P1
10 fzm1ndvds PN1P1¬PN
11 4 9 10 syl2an2r PN1..^P¬PN
12 eqid NP2modP=NP2modP
13 12 modprminv PN¬PNNP2modP1P1NNP2modPmodP=1
14 13 simpld PN¬PNNP2modP1P1
15 13 simprd PN¬PNNNP2modPmodP=1
16 1eluzge0 10
17 fzss1 101P10P1
18 16 17 mp1i P1P10P1
19 18 sseld Ps1P1s0P1
20 19 3ad2ant1 PN¬PNs1P1s0P1
21 20 imdistani PN¬PNs1P1PN¬PNs0P1
22 12 modprminveq PN¬PNs0P1NsmodP=1s=NP2modP
23 22 biimpa PN¬PNs0P1NsmodP=1s=NP2modP
24 23 eqcomd PN¬PNs0P1NsmodP=1NP2modP=s
25 24 expr PN¬PNs0P1NsmodP=1NP2modP=s
26 21 25 syl PN¬PNs1P1NsmodP=1NP2modP=s
27 26 ralrimiva PN¬PNs1P1NsmodP=1NP2modP=s
28 14 15 27 jca32 PN¬PNNP2modP1P1NNP2modPmodP=1s1P1NsmodP=1NP2modP=s
29 1 3 11 28 syl3anc PN1..^PNP2modP1P1NNP2modPmodP=1s1P1NsmodP=1NP2modP=s
30 oveq2 i=NP2modPNi=NNP2modP
31 30 oveq1d i=NP2modPNimodP=NNP2modPmodP
32 31 eqeq1d i=NP2modPNimodP=1NNP2modPmodP=1
33 eqeq1 i=NP2modPi=sNP2modP=s
34 33 imbi2d i=NP2modPNsmodP=1i=sNsmodP=1NP2modP=s
35 34 ralbidv i=NP2modPs1P1NsmodP=1i=ss1P1NsmodP=1NP2modP=s
36 32 35 anbi12d i=NP2modPNimodP=1s1P1NsmodP=1i=sNNP2modPmodP=1s1P1NsmodP=1NP2modP=s
37 36 rspcev NP2modP1P1NNP2modPmodP=1s1P1NsmodP=1NP2modP=si1P1NimodP=1s1P1NsmodP=1i=s
38 29 37 syl PN1..^Pi1P1NimodP=1s1P1NsmodP=1i=s
39 oveq2 i=sNi=Ns
40 39 oveq1d i=sNimodP=NsmodP
41 40 eqeq1d i=sNimodP=1NsmodP=1
42 41 reu8 ∃!i1P1NimodP=1i1P1NimodP=1s1P1NsmodP=1i=s
43 38 42 sylibr PN1..^P∃!i1P1NimodP=1