Metamath Proof Explorer


Theorem phibndlem

Description: Lemma for phibnd . (Contributed by Mario Carneiro, 23-Feb-2014)

Ref Expression
Assertion phibndlem N2x1N|xgcdN=11N1

Proof

Step Hyp Ref Expression
1 eluz2nn N2N
2 fzm1 N1x1Nx1N1x=N
3 nnuz =1
4 2 3 eleq2s Nx1Nx1N1x=N
5 4 biimpa Nx1Nx1N1x=N
6 5 ord Nx1N¬x1N1x=N
7 1 6 sylan N2x1N¬x1N1x=N
8 eluzelz N2N
9 gcdid NNgcdN=N
10 8 9 syl N2NgcdN=N
11 nnre NN
12 nnnn0 NN0
13 12 nn0ge0d N0N
14 11 13 absidd NN=N
15 1 14 syl N2N=N
16 10 15 eqtrd N2NgcdN=N
17 1re 1
18 eluz2gt1 N21<N
19 ltne 11<NN1
20 17 18 19 sylancr N2N1
21 16 20 eqnetrd N2NgcdN1
22 oveq1 x=NxgcdN=NgcdN
23 22 neeq1d x=NxgcdN1NgcdN1
24 21 23 syl5ibrcom N2x=NxgcdN1
25 24 adantr N2x1Nx=NxgcdN1
26 7 25 syld N2x1N¬x1N1xgcdN1
27 26 necon4bd N2x1NxgcdN=1x1N1
28 27 ralrimiva N2x1NxgcdN=1x1N1
29 rabss x1N|xgcdN=11N1x1NxgcdN=1x1N1
30 28 29 sylibr N2x1N|xgcdN=11N1