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 φ N 5
bpos.2 φ ¬ p N < p p 2 N
bpos.3 F = n if n n n pCnt ( 2 N N) 1
bpos.4 K = 2 N 3
Assertion bposlem3 φ seq 1 × F K = ( 2 N N)

Proof

Step Hyp Ref Expression
1 bpos.1 φ N 5
2 bpos.2 φ ¬ p N < p p 2 N
3 bpos.3 F = n if n n n pCnt ( 2 N N) 1
4 bpos.4 K = 2 N 3
5 simpr φ n n
6 5nn 5
7 eluznn 5 N 5 N
8 6 1 7 sylancr φ N
9 8 nnnn0d φ N 0
10 fzctr N 0 N 0 2 N
11 bccl2 N 0 2 N ( 2 N N)
12 9 10 11 3syl φ ( 2 N N)
13 12 adantr φ n ( 2 N N)
14 5 13 pccld φ n n pCnt ( 2 N N) 0
15 14 ralrimiva φ n n pCnt ( 2 N N) 0
16 15 adantr φ p n n pCnt ( 2 N N) 0
17 2nn 2
18 nnmulcl 2 N 2 N
19 17 8 18 sylancr φ 2 N
20 19 nnred φ 2 N
21 3nn 3
22 nndivre 2 N 3 2 N 3
23 20 21 22 sylancl φ 2 N 3
24 23 flcld φ 2 N 3
25 4 24 eqeltrid φ K
26 3re 3
27 26 a1i φ 3
28 5re 5
29 28 a1i φ 5
30 8 nnred φ N
31 3lt5 3 < 5
32 26 28 31 ltleii 3 5
33 32 a1i φ 3 5
34 eluzle N 5 5 N
35 1 34 syl φ 5 N
36 27 29 30 33 35 letrd φ 3 N
37 2re 2
38 2pos 0 < 2
39 37 38 pm3.2i 2 0 < 2
40 lemul2 3 N 2 0 < 2 3 N 2 3 2 N
41 26 39 40 mp3an13 N 3 N 2 3 2 N
42 30 41 syl φ 3 N 2 3 2 N
43 36 42 mpbid φ 2 3 2 N
44 3pos 0 < 3
45 26 44 pm3.2i 3 0 < 3
46 lemuldiv 2 2 N 3 0 < 3 2 3 2 N 2 2 N 3
47 37 45 46 mp3an13 2 N 2 3 2 N 2 2 N 3
48 20 47 syl φ 2 3 2 N 2 2 N 3
49 43 48 mpbid φ 2 2 N 3
50 2z 2
51 flge 2 N 3 2 2 2 N 3 2 2 N 3
52 23 50 51 sylancl φ 2 2 N 3 2 2 N 3
53 49 52 mpbid φ 2 2 N 3
54 53 4 breqtrrdi φ 2 K
55 50 eluz1i K 2 K 2 K
56 25 54 55 sylanbrc φ K 2
57 eluz2nn K 2 K
58 56 57 syl φ K
59 58 adantr φ p K
60 simpr φ p p
61 oveq1 n = p n pCnt ( 2 N N) = p pCnt ( 2 N N)
62 3 16 59 60 61 pcmpt φ p p pCnt seq 1 × F K = if p K p pCnt ( 2 N N) 0
63 iftrue p K if p K p pCnt ( 2 N N) 0 = p pCnt ( 2 N N)
64 63 adantl φ p p K if p K p pCnt ( 2 N N) 0 = p pCnt ( 2 N N)
65 iffalse ¬ p K if p K p pCnt ( 2 N N) 0 = 0
66 65 adantl φ p ¬ p K if p K p pCnt ( 2 N N) 0 = 0
67 25 zred φ K
68 prmz p p
69 68 zred p p
70 ltnle K p K < p ¬ p K
71 67 69 70 syl2an φ p K < p ¬ p K
72 71 biimpar φ p ¬ p K K < p
73 8 ad2antrr φ p K < p p N N
74 simplr φ p K < p p N p
75 37 a1i φ p K < p p N 2
76 67 ad2antrr φ p K < p p N K
77 68 ad2antlr φ p K < p p N p
78 77 zred φ p K < p p N p
79 54 ad2antrr φ p K < p p N 2 K
80 simprl φ p K < p p N K < p
81 75 76 78 79 80 lelttrd φ p K < p p N 2 < p
82 4 80 eqbrtrrid φ p K < p p N 2 N 3 < p
83 23 ad2antrr φ p K < p p N 2 N 3
84 fllt 2 N 3 p 2 N 3 < p 2 N 3 < p
85 83 77 84 syl2anc φ p K < p p N 2 N 3 < p 2 N 3 < p
86 82 85 mpbird φ p K < p p N 2 N 3 < p
87 simprr φ p K < p p N p N
88 73 74 81 86 87 bposlem2 φ p K < p p N p pCnt ( 2 N N) = 0
89 88 expr φ p K < p p N p pCnt ( 2 N N) = 0
90 rspe p N < p p 2 N p N < p p 2 N
91 90 adantll φ p N < p p 2 N p N < p p 2 N
92 2 ad2antrr φ p N < p p 2 N ¬ p N < p p 2 N
93 91 92 pm2.21dd φ p N < p p 2 N p pCnt ( 2 N N) = 0
94 93 expr φ p N < p p 2 N p pCnt ( 2 N N) = 0
95 12 nnzd φ ( 2 N N)
96 9 faccld φ N !
97 96 96 nnmulcld φ N ! N !
98 97 nnzd φ N ! N !
99 dvdsmul1 ( 2 N N) N ! N ! ( 2 N N) ( 2 N N) N ! N !
100 95 98 99 syl2anc φ ( 2 N N) ( 2 N N) N ! N !
101 bcctr N 0 ( 2 N N) = 2 N ! N ! N !
102 9 101 syl φ ( 2 N N) = 2 N ! N ! N !
103 102 oveq1d φ ( 2 N N) N ! N ! = 2 N ! N ! N ! N ! N !
104 19 nnnn0d φ 2 N 0
105 104 faccld φ 2 N !
106 105 nncnd φ 2 N !
107 97 nncnd φ N ! N !
108 97 nnne0d φ N ! N ! 0
109 106 107 108 divcan1d φ 2 N ! N ! N ! N ! N ! = 2 N !
110 103 109 eqtrd φ ( 2 N N) N ! N ! = 2 N !
111 100 110 breqtrd φ ( 2 N N) 2 N !
112 111 adantr φ p ( 2 N N) 2 N !
113 68 adantl φ p p
114 95 adantr φ p ( 2 N N)
115 105 nnzd φ 2 N !
116 115 adantr φ p 2 N !
117 dvdstr p ( 2 N N) 2 N ! p ( 2 N N) ( 2 N N) 2 N ! p 2 N !
118 113 114 116 117 syl3anc φ p p ( 2 N N) ( 2 N N) 2 N ! p 2 N !
119 112 118 mpan2d φ p p ( 2 N N) p 2 N !
120 prmfac1 2 N 0 p p 2 N ! p 2 N
121 120 3expia 2 N 0 p p 2 N ! p 2 N
122 104 121 sylan φ p p 2 N ! p 2 N
123 119 122 syld φ p p ( 2 N N) p 2 N
124 123 con3d φ p ¬ p 2 N ¬ p ( 2 N N)
125 id p p
126 pceq0 p ( 2 N N) p pCnt ( 2 N N) = 0 ¬ p ( 2 N N)
127 125 12 126 syl2anr φ p p pCnt ( 2 N N) = 0 ¬ p ( 2 N N)
128 124 127 sylibrd φ p ¬ p 2 N p pCnt ( 2 N N) = 0
129 128 adantr φ p N < p ¬ p 2 N p pCnt ( 2 N N) = 0
130 94 129 pm2.61d φ p N < p p pCnt ( 2 N N) = 0
131 130 ex φ p N < p p pCnt ( 2 N N) = 0
132 131 adantr φ p K < p N < p p pCnt ( 2 N N) = 0
133 lelttric p N p N N < p
134 69 30 133 syl2anr φ p p N N < p
135 134 adantr φ p K < p p N N < p
136 89 132 135 mpjaod φ p K < p p pCnt ( 2 N N) = 0
137 72 136 syldan φ p ¬ p K p pCnt ( 2 N N) = 0
138 66 137 eqtr4d φ p ¬ p K if p K p pCnt ( 2 N N) 0 = p pCnt ( 2 N N)
139 64 138 pm2.61dan φ p if p K p pCnt ( 2 N N) 0 = p pCnt ( 2 N N)
140 62 139 eqtrd φ p p pCnt seq 1 × F K = p pCnt ( 2 N N)
141 140 ralrimiva φ p p pCnt seq 1 × F K = p pCnt ( 2 N N)
142 3 15 pcmptcl φ F : seq 1 × F :
143 142 simprd φ seq 1 × F :
144 143 58 ffvelrnd φ seq 1 × F K
145 144 nnnn0d φ seq 1 × F K 0
146 12 nnnn0d φ ( 2 N N) 0
147 pc11 seq 1 × F K 0 ( 2 N N) 0 seq 1 × F K = ( 2 N N) p p pCnt seq 1 × F K = p pCnt ( 2 N N)
148 145 146 147 syl2anc φ seq 1 × F K = ( 2 N N) p p pCnt seq 1 × F K = p pCnt ( 2 N N)
149 141 148 mpbird φ seq 1 × F K = ( 2 N N)