Metamath Proof Explorer


Theorem pcaddlem

Description: Lemma for pcadd . The original numbers A and B have been decomposed using the prime count function as ( P ^ M ) x. ( R / S ) where R , S are both not divisible by P and M = ( P pCnt A ) , and similarly for B . (Contributed by Mario Carneiro, 9-Sep-2014)

Ref Expression
Hypotheses pcaddlem.1 ⊢ φ → P ∈ ℙ
pcaddlem.2 ⊢ φ → A = P M ⁢ R S
pcaddlem.3 ⊢ φ → B = P N ⁢ T U
pcaddlem.4 ⊢ φ → N ∈ ℤ ≥ M
pcaddlem.5 ⊢ φ → R ∈ ℤ ∧ ¬ P ∥ R
pcaddlem.6 ⊢ φ → S ∈ ℕ ∧ ¬ P ∥ S
pcaddlem.7 ⊢ φ → T ∈ ℤ ∧ ¬ P ∥ T
pcaddlem.8 ⊢ φ → U ∈ ℕ ∧ ¬ P ∥ U
Assertion pcaddlem ⊢ φ → M ≤ P pCnt A + B

Proof

Step Hyp Ref Expression
1 pcaddlem.1 ⊢ φ → P ∈ ℙ
2 pcaddlem.2 ⊢ φ → A = P M ⁢ R S
3 pcaddlem.3 ⊢ φ → B = P N ⁢ T U
4 pcaddlem.4 ⊢ φ → N ∈ ℤ ≥ M
5 pcaddlem.5 ⊢ φ → R ∈ ℤ ∧ ¬ P ∥ R
6 pcaddlem.6 ⊢ φ → S ∈ ℕ ∧ ¬ P ∥ S
7 pcaddlem.7 ⊢ φ → T ∈ ℤ ∧ ¬ P ∥ T
8 pcaddlem.8 ⊢ φ → U ∈ ℕ ∧ ¬ P ∥ U
9 oveq2 ⊢ A + B = 0 → P pCnt A + B = P pCnt 0
10 9 breq2d ⊢ A + B = 0 → M ≤ P pCnt A + B ↔ M ≤ P pCnt 0
11 eluzel2 ⊢ N ∈ ℤ ≥ M → M ∈ ℤ
12 4 11 syl ⊢ φ → M ∈ ℤ
13 12 zred ⊢ φ → M ∈ ℝ
14 13 adantr ⊢ φ ∧ A + B ≠ 0 → M ∈ ℝ
15 prmnn ⊢ P ∈ ℙ → P ∈ ℕ
16 1 15 syl ⊢ φ → P ∈ ℕ
17 16 nncnd ⊢ φ → P ∈ ℂ
18 16 nnne0d ⊢ φ → P ≠ 0
19 eluzelz ⊢ N ∈ ℤ ≥ M → N ∈ ℤ
20 4 19 syl ⊢ φ → N ∈ ℤ
21 20 12 zsubcld ⊢ φ → N − M ∈ ℤ
22 17 18 21 expclzd ⊢ φ → P N − M ∈ ℂ
23 7 simpld ⊢ φ → T ∈ ℤ
24 23 zcnd ⊢ φ → T ∈ ℂ
25 8 simpld ⊢ φ → U ∈ ℕ
26 25 nncnd ⊢ φ → U ∈ ℂ
27 25 nnne0d ⊢ φ → U ≠ 0
28 22 24 26 27 divassd ⊢ φ → P N − M ⁢ T U = P N − M ⁢ T U
29 28 oveq2d ⊢ φ → R S + P N − M ⁢ T U = R S + P N − M ⁢ T U
30 5 simpld ⊢ φ → R ∈ ℤ
31 30 zcnd ⊢ φ → R ∈ ℂ
32 6 simpld ⊢ φ → S ∈ ℕ
33 32 nncnd ⊢ φ → S ∈ ℂ
34 22 24 mulcld ⊢ φ → P N − M ⁢ T ∈ ℂ
35 32 nnne0d ⊢ φ → S ≠ 0
36 31 33 34 26 35 27 divadddivd ⊢ φ → R S + P N − M ⁢ T U = R ⁢ U + P N − M ⁢ T ⁢ S S ⁢ U
37 29 36 eqtr3d ⊢ φ → R S + P N − M ⁢ T U = R ⁢ U + P N − M ⁢ T ⁢ S S ⁢ U
38 37 oveq2d ⊢ φ → P pCnt R S + P N − M ⁢ T U = P pCnt R ⁢ U + P N − M ⁢ T ⁢ S S ⁢ U
39 38 adantr ⊢ φ ∧ A + B ≠ 0 → P pCnt R S + P N − M ⁢ T U = P pCnt R ⁢ U + P N − M ⁢ T ⁢ S S ⁢ U
40 1 adantr ⊢ φ ∧ A + B ≠ 0 → P ∈ ℙ
41 25 nnzd ⊢ φ → U ∈ ℤ
42 30 41 zmulcld ⊢ φ → R ⁢ U ∈ ℤ
43 uznn0sub ⊢ N ∈ ℤ ≥ M → N − M ∈ ℕ 0
44 4 43 syl ⊢ φ → N − M ∈ ℕ 0
45 16 44 nnexpcld ⊢ φ → P N − M ∈ ℕ
46 45 nnzd ⊢ φ → P N − M ∈ ℤ
47 46 23 zmulcld ⊢ φ → P N − M ⁢ T ∈ ℤ
48 32 nnzd ⊢ φ → S ∈ ℤ
49 47 48 zmulcld ⊢ φ → P N − M ⁢ T ⁢ S ∈ ℤ
50 42 49 zaddcld ⊢ φ → R ⁢ U + P N − M ⁢ T ⁢ S ∈ ℤ
51 50 adantr ⊢ φ ∧ A + B ≠ 0 → R ⁢ U + P N − M ⁢ T ⁢ S ∈ ℤ
52 17 18 12 expclzd ⊢ φ → P M ∈ ℂ
53 52 mul01d ⊢ φ → P M ⋅ 0 = 0
54 oveq2 ⊢ R S + P N − M ⁢ T U = 0 → P M ⁢ R S + P N − M ⁢ T U = P M ⋅ 0
55 54 eqeq1d ⊢ R S + P N − M ⁢ T U = 0 → P M ⁢ R S + P N − M ⁢ T U = 0 ↔ P M ⋅ 0 = 0
56 53 55 syl5ibrcom ⊢ φ → R S + P N − M ⁢ T U = 0 → P M ⁢ R S + P N − M ⁢ T U = 0
57 56 necon3d ⊢ φ → P M ⁢ R S + P N − M ⁢ T U ≠ 0 → R S + P N − M ⁢ T U ≠ 0
58 31 33 35 divcld ⊢ φ → R S ∈ ℂ
59 24 26 27 divcld ⊢ φ → T U ∈ ℂ
60 22 59 mulcld ⊢ φ → P N − M ⁢ T U ∈ ℂ
61 52 58 60 adddid ⊢ φ → P M ⁢ R S + P N − M ⁢ T U = P M ⁢ R S + P M ⁢ P N − M ⁢ T U
62 12 zcnd ⊢ φ → M ∈ ℂ
63 20 zcnd ⊢ φ → N ∈ ℂ
64 62 63 pncan3d ⊢ φ → M + N - M = N
65 64 oveq2d ⊢ φ → P M + N - M = P N
66 expaddz ⊢ P ∈ ℂ ∧ P ≠ 0 ∧ M ∈ ℤ ∧ N − M ∈ ℤ → P M + N - M = P M ⁢ P N − M
67 17 18 12 21 66 syl22anc ⊢ φ → P M + N - M = P M ⁢ P N − M
68 65 67 eqtr3d ⊢ φ → P N = P M ⁢ P N − M
69 68 oveq1d ⊢ φ → P N ⁢ T U = P M ⁢ P N − M ⁢ T U
70 52 22 59 mulassd ⊢ φ → P M ⁢ P N − M ⁢ T U = P M ⁢ P N − M ⁢ T U
71 3 69 70 3eqtrd ⊢ φ → B = P M ⁢ P N − M ⁢ T U
72 2 71 oveq12d ⊢ φ → A + B = P M ⁢ R S + P M ⁢ P N − M ⁢ T U
73 61 72 eqtr4d ⊢ φ → P M ⁢ R S + P N − M ⁢ T U = A + B
74 73 neeq1d ⊢ φ → P M ⁢ R S + P N − M ⁢ T U ≠ 0 ↔ A + B ≠ 0
75 37 neeq1d ⊢ φ → R S + P N − M ⁢ T U ≠ 0 ↔ R ⁢ U + P N − M ⁢ T ⁢ S S ⁢ U ≠ 0
76 57 74 75 3imtr3d ⊢ φ → A + B ≠ 0 → R ⁢ U + P N − M ⁢ T ⁢ S S ⁢ U ≠ 0
77 32 25 nnmulcld ⊢ φ → S ⁢ U ∈ ℕ
78 77 nncnd ⊢ φ → S ⁢ U ∈ ℂ
79 77 nnne0d ⊢ φ → S ⁢ U ≠ 0
80 78 79 div0d ⊢ φ → 0 S ⁢ U = 0
81 oveq1 ⊢ R ⁢ U + P N − M ⁢ T ⁢ S = 0 → R ⁢ U + P N − M ⁢ T ⁢ S S ⁢ U = 0 S ⁢ U
82 81 eqeq1d ⊢ R ⁢ U + P N − M ⁢ T ⁢ S = 0 → R ⁢ U + P N − M ⁢ T ⁢ S S ⁢ U = 0 ↔ 0 S ⁢ U = 0
83 80 82 syl5ibrcom ⊢ φ → R ⁢ U + P N − M ⁢ T ⁢ S = 0 → R ⁢ U + P N − M ⁢ T ⁢ S S ⁢ U = 0
84 83 necon3d ⊢ φ → R ⁢ U + P N − M ⁢ T ⁢ S S ⁢ U ≠ 0 → R ⁢ U + P N − M ⁢ T ⁢ S ≠ 0
85 76 84 syld ⊢ φ → A + B ≠ 0 → R ⁢ U + P N − M ⁢ T ⁢ S ≠ 0
86 85 imp ⊢ φ ∧ A + B ≠ 0 → R ⁢ U + P N − M ⁢ T ⁢ S ≠ 0
87 77 adantr ⊢ φ ∧ A + B ≠ 0 → S ⁢ U ∈ ℕ
88 pcdiv ⊢ P ∈ ℙ ∧ R ⁢ U + P N − M ⁢ T ⁢ S ∈ ℤ ∧ R ⁢ U + P N − M ⁢ T ⁢ S ≠ 0 ∧ S ⁢ U ∈ ℕ → P pCnt R ⁢ U + P N − M ⁢ T ⁢ S S ⁢ U = P pCnt R ⁢ U + P N − M ⁢ T ⁢ S − P pCnt S ⁢ U
89 40 51 86 87 88 syl121anc ⊢ φ ∧ A + B ≠ 0 → P pCnt R ⁢ U + P N − M ⁢ T ⁢ S S ⁢ U = P pCnt R ⁢ U + P N − M ⁢ T ⁢ S − P pCnt S ⁢ U
90 pcmul ⊢ P ∈ ℙ ∧ S ∈ ℤ ∧ S ≠ 0 ∧ U ∈ ℤ ∧ U ≠ 0 → P pCnt S ⁢ U = P pCnt S + P pCnt U
91 1 48 35 41 27 90 syl122anc ⊢ φ → P pCnt S ⁢ U = P pCnt S + P pCnt U
92 6 simprd ⊢ φ → ¬ P ∥ S
93 pceq0 ⊢ P ∈ ℙ ∧ S ∈ ℕ → P pCnt S = 0 ↔ ¬ P ∥ S
94 1 32 93 syl2anc ⊢ φ → P pCnt S = 0 ↔ ¬ P ∥ S
95 92 94 mpbird ⊢ φ → P pCnt S = 0
96 8 simprd ⊢ φ → ¬ P ∥ U
97 pceq0 ⊢ P ∈ ℙ ∧ U ∈ ℕ → P pCnt U = 0 ↔ ¬ P ∥ U
98 1 25 97 syl2anc ⊢ φ → P pCnt U = 0 ↔ ¬ P ∥ U
99 96 98 mpbird ⊢ φ → P pCnt U = 0
100 95 99 oveq12d ⊢ φ → P pCnt S + P pCnt U = 0 + 0
101 00id ⊢ 0 + 0 = 0
102 100 101 eqtrdi ⊢ φ → P pCnt S + P pCnt U = 0
103 91 102 eqtrd ⊢ φ → P pCnt S ⁢ U = 0
104 103 oveq2d ⊢ φ → P pCnt R ⁢ U + P N − M ⁢ T ⁢ S − P pCnt S ⁢ U = P pCnt R ⁢ U + P N − M ⁢ T ⁢ S − 0
105 104 adantr ⊢ φ ∧ A + B ≠ 0 → P pCnt R ⁢ U + P N − M ⁢ T ⁢ S − P pCnt S ⁢ U = P pCnt R ⁢ U + P N − M ⁢ T ⁢ S − 0
106 pczcl ⊢ P ∈ ℙ ∧ R ⁢ U + P N − M ⁢ T ⁢ S ∈ ℤ ∧ R ⁢ U + P N − M ⁢ T ⁢ S ≠ 0 → P pCnt R ⁢ U + P N − M ⁢ T ⁢ S ∈ ℕ 0
107 40 51 86 106 syl12anc ⊢ φ ∧ A + B ≠ 0 → P pCnt R ⁢ U + P N − M ⁢ T ⁢ S ∈ ℕ 0
108 107 nn0cnd ⊢ φ ∧ A + B ≠ 0 → P pCnt R ⁢ U + P N − M ⁢ T ⁢ S ∈ ℂ
109 108 subid1d ⊢ φ ∧ A + B ≠ 0 → P pCnt R ⁢ U + P N − M ⁢ T ⁢ S − 0 = P pCnt R ⁢ U + P N − M ⁢ T ⁢ S
110 105 109 eqtrd ⊢ φ ∧ A + B ≠ 0 → P pCnt R ⁢ U + P N − M ⁢ T ⁢ S − P pCnt S ⁢ U = P pCnt R ⁢ U + P N − M ⁢ T ⁢ S
111 39 89 110 3eqtrd ⊢ φ ∧ A + B ≠ 0 → P pCnt R S + P N − M ⁢ T U = P pCnt R ⁢ U + P N − M ⁢ T ⁢ S
112 111 107 eqeltrd ⊢ φ ∧ A + B ≠ 0 → P pCnt R S + P N − M ⁢ T U ∈ ℕ 0
113 nn0addge1 ⊢ M ∈ ℝ ∧ P pCnt R S + P N − M ⁢ T U ∈ ℕ 0 → M ≤ M + P pCnt R S + P N − M ⁢ T U
114 14 112 113 syl2anc ⊢ φ ∧ A + B ≠ 0 → M ≤ M + P pCnt R S + P N − M ⁢ T U
115 nnq ⊢ P ∈ ℕ → P ∈ ℚ
116 16 115 syl ⊢ φ → P ∈ ℚ
117 qexpclz ⊢ P ∈ ℚ ∧ P ≠ 0 ∧ M ∈ ℤ → P M ∈ ℚ
118 116 18 12 117 syl3anc ⊢ φ → P M ∈ ℚ
119 118 adantr ⊢ φ ∧ A + B ≠ 0 → P M ∈ ℚ
120 17 18 12 expne0d ⊢ φ → P M ≠ 0
121 120 adantr ⊢ φ ∧ A + B ≠ 0 → P M ≠ 0
122 znq ⊢ R ∈ ℤ ∧ S ∈ ℕ → R S ∈ ℚ
123 30 32 122 syl2anc ⊢ φ → R S ∈ ℚ
124 qexpclz ⊢ P ∈ ℚ ∧ P ≠ 0 ∧ N − M ∈ ℤ → P N − M ∈ ℚ
125 116 18 21 124 syl3anc ⊢ φ → P N − M ∈ ℚ
126 znq ⊢ T ∈ ℤ ∧ U ∈ ℕ → T U ∈ ℚ
127 23 25 126 syl2anc ⊢ φ → T U ∈ ℚ
128 qmulcl ⊢ P N − M ∈ ℚ ∧ T U ∈ ℚ → P N − M ⁢ T U ∈ ℚ
129 125 127 128 syl2anc ⊢ φ → P N − M ⁢ T U ∈ ℚ
130 qaddcl ⊢ R S ∈ ℚ ∧ P N − M ⁢ T U ∈ ℚ → R S + P N − M ⁢ T U ∈ ℚ
131 123 129 130 syl2anc ⊢ φ → R S + P N − M ⁢ T U ∈ ℚ
132 131 adantr ⊢ φ ∧ A + B ≠ 0 → R S + P N − M ⁢ T U ∈ ℚ
133 74 57 sylbird ⊢ φ → A + B ≠ 0 → R S + P N − M ⁢ T U ≠ 0
134 133 imp ⊢ φ ∧ A + B ≠ 0 → R S + P N − M ⁢ T U ≠ 0
135 pcqmul ⊢ P ∈ ℙ ∧ P M ∈ ℚ ∧ P M ≠ 0 ∧ R S + P N − M ⁢ T U ∈ ℚ ∧ R S + P N − M ⁢ T U ≠ 0 → P pCnt P M ⁢ R S + P N − M ⁢ T U = P pCnt P M + P pCnt R S + P N − M ⁢ T U
136 40 119 121 132 134 135 syl122anc ⊢ φ ∧ A + B ≠ 0 → P pCnt P M ⁢ R S + P N − M ⁢ T U = P pCnt P M + P pCnt R S + P N − M ⁢ T U
137 73 oveq2d ⊢ φ → P pCnt P M ⁢ R S + P N − M ⁢ T U = P pCnt A + B
138 137 adantr ⊢ φ ∧ A + B ≠ 0 → P pCnt P M ⁢ R S + P N − M ⁢ T U = P pCnt A + B
139 pcid ⊢ P ∈ ℙ ∧ M ∈ ℤ → P pCnt P M = M
140 1 12 139 syl2anc ⊢ φ → P pCnt P M = M
141 140 oveq1d ⊢ φ → P pCnt P M + P pCnt R S + P N − M ⁢ T U = M + P pCnt R S + P N − M ⁢ T U
142 141 adantr ⊢ φ ∧ A + B ≠ 0 → P pCnt P M + P pCnt R S + P N − M ⁢ T U = M + P pCnt R S + P N − M ⁢ T U
143 136 138 142 3eqtr3d ⊢ φ ∧ A + B ≠ 0 → P pCnt A + B = M + P pCnt R S + P N − M ⁢ T U
144 114 143 breqtrrd ⊢ φ ∧ A + B ≠ 0 → M ≤ P pCnt A + B
145 13 rexrd ⊢ φ → M ∈ ℝ *
146 pnfge ⊢ M ∈ ℝ * → M ≤ +∞
147 145 146 syl ⊢ φ → M ≤ +∞
148 pc0 ⊢ P ∈ ℙ → P pCnt 0 = +∞
149 1 148 syl ⊢ φ → P pCnt 0 = +∞
150 147 149 breqtrrd ⊢ φ → M ≤ P pCnt 0
151 10 144 150 pm2.61ne ⊢ φ → M ≤ P pCnt A + B