Metamath Proof Explorer


Theorem lcmineqlem19

Description: Dividing implies inequality for lcm inequality lemma. (Contributed by metakunt, 12-May-2024)

Ref Expression
Hypothesis lcmineqlem19.1 φN
Assertion lcmineqlem19 φN2 N+1(2 NN)lcm_12 N+1

Proof

Step Hyp Ref Expression
1 lcmineqlem19.1 φN
2 2nn 2
3 2 a1i φ2
4 3 1 nnmulcld φ2 N
5 4 peano2nnd φ2 N+1
6 1 nnnn0d φN0
7 1 nnred φN
8 2re 2
9 8 a1i φ2
10 6 nn0ge0d φ0N
11 3 nnge1d φ12
12 7 9 10 11 lemulge12d φN2 N
13 4 6 12 bccl2d φ(2 NN)
14 fz1ssnn 12 N
15 fzfi 12 NFin
16 lcmfnncl 12 N12 NFinlcm_12 N
17 14 15 16 mp2an lcm_12 N
18 17 a1i φlcm_12 N
19 fz1ssnn 12 N+1
20 fzfi 12 N+1Fin
21 lcmfnncl 12 N+112 N+1Finlcm_12 N+1
22 19 20 21 mp2an lcm_12 N+1
23 22 a1i φlcm_12 N+1
24 1 4 12 lcmineqlem16 φN(2 NN)lcm_12 N
25 1 lcmineqlem18 φN+1(2 N+1N+1)=2 N+1(2 NN)
26 1 peano2nnd φN+1
27 9 7 remulcld φ2 N
28 1red φ1
29 7 27 28 12 leadd1dd φN+12 N+1
30 26 5 29 lcmineqlem16 φN+1(2 N+1N+1)lcm_12 N+1
31 25 30 eqbrtrrd φ2 N+1(2 NN)lcm_12 N+1
32 18 nnzd φlcm_12 N
33 5 nnzd φ2 N+1
34 32 33 jca φlcm_12 N2 N+1
35 dvdslcm lcm_12 N2 N+1lcm_12 Nlcm_12 Nlcm2 N+12 N+1lcm_12 Nlcm2 N+1
36 34 35 syl φlcm_12 Nlcm_12 Nlcm2 N+12 N+1lcm_12 Nlcm2 N+1
37 36 simpld φlcm_12 Nlcm_12 Nlcm2 N+1
38 5 lcmfunnnd φlcm_12 N+1=lcm_12 N+1-1lcm2 N+1
39 27 recnd φ2 N
40 1cnd φ1
41 39 40 pncand φ2 N+1-1=2 N
42 41 oveq2d φ12 N+1-1=12 N
43 42 fveq2d φlcm_12 N+1-1=lcm_12 N
44 43 oveq1d φlcm_12 N+1-1lcm2 N+1=lcm_12 Nlcm2 N+1
45 38 44 eqtrd φlcm_12 N+1=lcm_12 Nlcm2 N+1
46 37 45 breqtrrd φlcm_12 Nlcm_12 N+1
47 1 nnzd φN
48 2z 2
49 1z 1
50 gcdaddm 2N1Ngcd1=Ngcd1+2 N
51 48 49 50 mp3an13 NNgcd1=Ngcd1+2 N
52 47 51 syl φNgcd1=Ngcd1+2 N
53 40 39 addcomd φ1+2 N=2 N+1
54 53 oveq2d φNgcd1+2 N=Ngcd2 N+1
55 52 54 eqtrd φNgcd1=Ngcd2 N+1
56 gcd1 NNgcd1=1
57 47 56 syl φNgcd1=1
58 55 57 eqtr3d φNgcd2 N+1=1
59 1 5 13 18 23 24 31 46 58 lcmineqlem14 φN2 N+1(2 NN)lcm_12 N+1