Metamath Proof Explorer


Theorem pcmpt2

Description: Dividing two prime count maps yields a number with all dividing primes confined to an interval. (Contributed by Mario Carneiro, 14-Mar-2014)

Ref Expression
Hypotheses pcmpt.1 F=nifnnA1
pcmpt.2 φnA0
pcmpt.3 φN
pcmpt.4 φP
pcmpt.5 n=PA=B
pcmpt2.6 φMN
Assertion pcmpt2 φPpCntseq1×FMseq1×FN=ifPM¬PNB0

Proof

Step Hyp Ref Expression
1 pcmpt.1 F=nifnnA1
2 pcmpt.2 φnA0
3 pcmpt.3 φN
4 pcmpt.4 φP
5 pcmpt.5 n=PA=B
6 pcmpt2.6 φMN
7 1 2 pcmptcl φF:seq1×F:
8 7 simprd φseq1×F:
9 eluznn NMNM
10 3 6 9 syl2anc φM
11 8 10 ffvelcdmd φseq1×FM
12 11 nnzd φseq1×FM
13 11 nnne0d φseq1×FM0
14 8 3 ffvelcdmd φseq1×FN
15 pcdiv Pseq1×FMseq1×FM0seq1×FNPpCntseq1×FMseq1×FN=PpCntseq1×FMPpCntseq1×FN
16 4 12 13 14 15 syl121anc φPpCntseq1×FMseq1×FN=PpCntseq1×FMPpCntseq1×FN
17 1 2 10 4 5 pcmpt φPpCntseq1×FM=ifPMB0
18 1 2 3 4 5 pcmpt φPpCntseq1×FN=ifPNB0
19 17 18 oveq12d φPpCntseq1×FMPpCntseq1×FN=ifPMB0ifPNB0
20 5 eleq1d n=PA0B0
21 20 2 4 rspcdva φB0
22 21 nn0cnd φB
23 22 subidd φBB=0
24 23 adantr φPNBB=0
25 prmnn PP
26 4 25 syl φP
27 26 nnred φP
28 27 adantr φPNP
29 3 nnred φN
30 29 adantr φPNN
31 10 nnred φM
32 31 adantr φPNM
33 simpr φPNPN
34 eluzle MNNM
35 6 34 syl φNM
36 35 adantr φPNNM
37 28 30 32 33 36 letrd φPNPM
38 37 iftrued φPNifPMB0=B
39 iftrue PNifPNB0=B
40 39 adantl φPNifPNB0=B
41 38 40 oveq12d φPNifPMB0ifPNB0=BB
42 simpr PM¬PN¬PN
43 42 33 nsyl3 φPN¬PM¬PN
44 43 iffalsed φPNifPM¬PNB0=0
45 24 41 44 3eqtr4d φPNifPMB0ifPNB0=ifPM¬PNB0
46 iffalse ¬PNifPNB0=0
47 46 oveq2d ¬PNifPMB0ifPNB0=ifPMB00
48 0cn 0
49 ifcl B0ifPMB0
50 22 48 49 sylancl φifPMB0
51 50 subid1d φifPMB00=ifPMB0
52 47 51 sylan9eqr φ¬PNifPMB0ifPNB0=ifPMB0
53 simpr φ¬PN¬PN
54 53 biantrud φ¬PNPMPM¬PN
55 54 ifbid φ¬PNifPMB0=ifPM¬PNB0
56 52 55 eqtrd φ¬PNifPMB0ifPNB0=ifPM¬PNB0
57 45 56 pm2.61dan φifPMB0ifPNB0=ifPM¬PNB0
58 16 19 57 3eqtrd φPpCntseq1×FMseq1×FN=ifPM¬PNB0