Metamath Proof Explorer


Theorem bposlem3

Description: Lemma for bpos . Since the binomial coefficient does not have any primes in the range ( 2 N / 3 , N ] or ( 2 N , +oo ) by bposlem2 and prmfac1 , respectively, and it does not have any in the range ( N , 2 N ] by hypothesis, the product of the primes up through 2 N / 3 must be sufficient to compose the whole binomial coefficient. (Contributed by Mario Carneiro, 13-Mar-2014)

Ref Expression
Hypotheses bpos.1 φN5
bpos.2 φ¬pN<pp2 N
bpos.3 F=nifnnnpCnt(2 NN)1
bpos.4 K=2 N3
Assertion bposlem3 φseq1×FK=(2 NN)

Proof

Step Hyp Ref Expression
1 bpos.1 φN5
2 bpos.2 φ¬pN<pp2 N
3 bpos.3 F=nifnnnpCnt(2 NN)1
4 bpos.4 K=2 N3
5 simpr φnn
6 5nn 5
7 eluznn 5N5N
8 6 1 7 sylancr φN
9 8 nnnn0d φN0
10 fzctr N0N02 N
11 bccl2 N02 N(2 NN)
12 9 10 11 3syl φ(2 NN)
13 12 adantr φn(2 NN)
14 5 13 pccld φnnpCnt(2 NN)0
15 14 ralrimiva φnnpCnt(2 NN)0
16 15 adantr φpnnpCnt(2 NN)0
17 2nn 2
18 nnmulcl 2N2 N
19 17 8 18 sylancr φ2 N
20 19 nnred φ2 N
21 3nn 3
22 nndivre 2 N32 N3
23 20 21 22 sylancl φ2 N3
24 23 flcld φ2 N3
25 4 24 eqeltrid φK
26 3re 3
27 26 a1i φ3
28 5re 5
29 28 a1i φ5
30 8 nnred φN
31 3lt5 3<5
32 26 28 31 ltleii 35
33 32 a1i φ35
34 eluzle N55N
35 1 34 syl φ5N
36 27 29 30 33 35 letrd φ3N
37 2re 2
38 2pos 0<2
39 37 38 pm3.2i 20<2
40 lemul2 3N20<23N232 N
41 26 39 40 mp3an13 N3N232 N
42 30 41 syl φ3N232 N
43 36 42 mpbid φ232 N
44 3pos 0<3
45 26 44 pm3.2i 30<3
46 lemuldiv 22 N30<3232 N22 N3
47 37 45 46 mp3an13 2 N232 N22 N3
48 20 47 syl φ232 N22 N3
49 43 48 mpbid φ22 N3
50 2z 2
51 flge 2 N3222 N322 N3
52 23 50 51 sylancl φ22 N322 N3
53 49 52 mpbid φ22 N3
54 53 4 breqtrrdi φ2K
55 50 eluz1i K2K2K
56 25 54 55 sylanbrc φK2
57 eluz2nn K2K
58 56 57 syl φK
59 58 adantr φpK
60 simpr φpp
61 oveq1 n=pnpCnt(2 NN)=ppCnt(2 NN)
62 3 16 59 60 61 pcmpt φpppCntseq1×FK=ifpKppCnt(2 NN)0
63 iftrue pKifpKppCnt(2 NN)0=ppCnt(2 NN)
64 63 adantl φppKifpKppCnt(2 NN)0=ppCnt(2 NN)
65 iffalse ¬pKifpKppCnt(2 NN)0=0
66 65 adantl φp¬pKifpKppCnt(2 NN)0=0
67 25 zred φK
68 prmz pp
69 68 zred pp
70 ltnle KpK<p¬pK
71 67 69 70 syl2an φpK<p¬pK
72 71 biimpar φp¬pKK<p
73 8 ad2antrr φpK<ppNN
74 simplr φpK<ppNp
75 37 a1i φpK<ppN2
76 67 ad2antrr φpK<ppNK
77 68 ad2antlr φpK<ppNp
78 77 zred φpK<ppNp
79 54 ad2antrr φpK<ppN2K
80 simprl φpK<ppNK<p
81 75 76 78 79 80 lelttrd φpK<ppN2<p
82 4 80 eqbrtrrid φpK<ppN2 N3<p
83 23 ad2antrr φpK<ppN2 N3
84 fllt 2 N3p2 N3<p2 N3<p
85 83 77 84 syl2anc φpK<ppN2 N3<p2 N3<p
86 82 85 mpbird φpK<ppN2 N3<p
87 simprr φpK<ppNpN
88 73 74 81 86 87 bposlem2 φpK<ppNppCnt(2 NN)=0
89 88 expr φpK<ppNppCnt(2 NN)=0
90 rspe pN<pp2 NpN<pp2 N
91 90 adantll φpN<pp2 NpN<pp2 N
92 2 ad2antrr φpN<pp2 N¬pN<pp2 N
93 91 92 pm2.21dd φpN<pp2 NppCnt(2 NN)=0
94 93 expr φpN<pp2 NppCnt(2 NN)=0
95 12 nnzd φ(2 NN)
96 9 faccld φN!
97 96 96 nnmulcld φN!N!
98 97 nnzd φN!N!
99 dvdsmul1 (2 NN)N!N!(2 NN)(2 NN)N!N!
100 95 98 99 syl2anc φ(2 NN)(2 NN)N!N!
101 bcctr N0(2 NN)=2 N!N!N!
102 9 101 syl φ(2 NN)=2 N!N!N!
103 102 oveq1d φ(2 NN)N!N!=2 N!N!N!N!N!
104 19 nnnn0d φ2 N0
105 104 faccld φ2 N!
106 105 nncnd φ2 N!
107 97 nncnd φN!N!
108 97 nnne0d φN!N!0
109 106 107 108 divcan1d φ2 N!N!N!N!N!=2 N!
110 103 109 eqtrd φ(2 NN)N!N!=2 N!
111 100 110 breqtrd φ(2 NN)2 N!
112 111 adantr φp(2 NN)2 N!
113 68 adantl φpp
114 95 adantr φp(2 NN)
115 105 nnzd φ2 N!
116 115 adantr φp2 N!
117 dvdstr p(2 NN)2 N!p(2 NN)(2 NN)2 N!p2 N!
118 113 114 116 117 syl3anc φpp(2 NN)(2 NN)2 N!p2 N!
119 112 118 mpan2d φpp(2 NN)p2 N!
120 prmfac1 2 N0pp2 N!p2 N
121 120 3expia 2 N0pp2 N!p2 N
122 104 121 sylan φpp2 N!p2 N
123 119 122 syld φpp(2 NN)p2 N
124 123 con3d φp¬p2 N¬p(2 NN)
125 id pp
126 pceq0 p(2 NN)ppCnt(2 NN)=0¬p(2 NN)
127 125 12 126 syl2anr φpppCnt(2 NN)=0¬p(2 NN)
128 124 127 sylibrd φp¬p2 NppCnt(2 NN)=0
129 128 adantr φpN<p¬p2 NppCnt(2 NN)=0
130 94 129 pm2.61d φpN<pppCnt(2 NN)=0
131 130 ex φpN<pppCnt(2 NN)=0
132 131 adantr φpK<pN<pppCnt(2 NN)=0
133 lelttric pNpNN<p
134 69 30 133 syl2anr φppNN<p
135 134 adantr φpK<ppNN<p
136 89 132 135 mpjaod φpK<pppCnt(2 NN)=0
137 72 136 syldan φp¬pKppCnt(2 NN)=0
138 66 137 eqtr4d φp¬pKifpKppCnt(2 NN)0=ppCnt(2 NN)
139 64 138 pm2.61dan φpifpKppCnt(2 NN)0=ppCnt(2 NN)
140 62 139 eqtrd φpppCntseq1×FK=ppCnt(2 NN)
141 140 ralrimiva φpppCntseq1×FK=ppCnt(2 NN)
142 3 15 pcmptcl φF:seq1×F:
143 142 simprd φseq1×F:
144 143 58 ffvelcdmd φseq1×FK
145 144 nnnn0d φseq1×FK0
146 12 nnnn0d φ(2 NN)0
147 pc11 seq1×FK0(2 NN)0seq1×FK=(2 NN)pppCntseq1×FK=ppCnt(2 NN)
148 145 146 147 syl2anc φseq1×FK=(2 NN)pppCntseq1×FK=ppCnt(2 NN)
149 141 148 mpbird φseq1×FK=(2 NN)