Metamath Proof Explorer


Theorem odd2np1

Description: An integer is odd iff it is one plus twice another integer. (Contributed by Scott Fenton, 3-Apr-2014) (Revised by Mario Carneiro, 19-Apr-2014)

Ref Expression
Assertion odd2np1 ⊢ N ∈ ℤ → ¬ 2 ∥ N ↔ ∃ n ∈ ℤ 2 ⁢ n + 1 = N

Proof

Step Hyp Ref Expression
1 2z ⊢ 2 ∈ ℤ
2 divides ⊢ 2 ∈ ℤ ∧ N ∈ ℤ → 2 ∥ N ↔ ∃ k ∈ ℤ k ⋅ 2 = N
3 1 2 mpan ⊢ N ∈ ℤ → 2 ∥ N ↔ ∃ k ∈ ℤ k ⋅ 2 = N
4 3 notbid ⊢ N ∈ ℤ → ¬ 2 ∥ N ↔ ¬ ∃ k ∈ ℤ k ⋅ 2 = N
5 elznn0 ⊢ N ∈ ℤ ↔ N ∈ ℝ ∧ N ∈ ℕ 0 ∨ − N ∈ ℕ 0
6 odd2np1lem ⊢ N ∈ ℕ 0 → ∃ n ∈ ℤ 2 ⁢ n + 1 = N ∨ ∃ k ∈ ℤ k ⋅ 2 = N
7 6 adantl ⊢ N ∈ ℝ ∧ N ∈ ℕ 0 → ∃ n ∈ ℤ 2 ⁢ n + 1 = N ∨ ∃ k ∈ ℤ k ⋅ 2 = N
8 peano2z ⊢ x ∈ ℤ → x + 1 ∈ ℤ
9 znegcl ⊢ x + 1 ∈ ℤ → − x + 1 ∈ ℤ
10 8 9 syl ⊢ x ∈ ℤ → − x + 1 ∈ ℤ
11 10 ad2antlr ⊢ N ∈ ℝ ∧ x ∈ ℤ ∧ 2 ⁢ x + 1 = − N → − x + 1 ∈ ℤ
12 zcn ⊢ x ∈ ℤ → x ∈ ℂ
13 2cn ⊢ 2 ∈ ℂ
14 mulcl ⊢ 2 ∈ ℂ ∧ x ∈ ℂ → 2 ⁢ x ∈ ℂ
15 13 14 mpan ⊢ x ∈ ℂ → 2 ⁢ x ∈ ℂ
16 peano2cn ⊢ 2 ⁢ x ∈ ℂ → 2 ⁢ x + 1 ∈ ℂ
17 15 16 syl ⊢ x ∈ ℂ → 2 ⁢ x + 1 ∈ ℂ
18 12 17 syl ⊢ x ∈ ℤ → 2 ⁢ x + 1 ∈ ℂ
19 18 adantl ⊢ N ∈ ℝ ∧ x ∈ ℤ → 2 ⁢ x + 1 ∈ ℂ
20 simpl ⊢ N ∈ ℝ ∧ x ∈ ℤ → N ∈ ℝ
21 20 recnd ⊢ N ∈ ℝ ∧ x ∈ ℤ → N ∈ ℂ
22 negcon2 ⊢ 2 ⁢ x + 1 ∈ ℂ ∧ N ∈ ℂ → 2 ⁢ x + 1 = − N ↔ N = − 2 ⁢ x + 1
23 19 21 22 syl2anc ⊢ N ∈ ℝ ∧ x ∈ ℤ → 2 ⁢ x + 1 = − N ↔ N = − 2 ⁢ x + 1
24 eqcom ⊢ N = − 2 ⁢ x + 1 ↔ − 2 ⁢ x + 1 = N
25 13 12 14 sylancr ⊢ x ∈ ℤ → 2 ⁢ x ∈ ℂ
26 ax-1cn ⊢ 1 ∈ ℂ
27 13 26 mulcli ⊢ 2 ⋅ 1 ∈ ℂ
28 addsubass ⊢ 2 ⁢ x ∈ ℂ ∧ 2 ⋅ 1 ∈ ℂ ∧ 1 ∈ ℂ → 2 ⁢ x + 2 ⋅ 1 - 1 = 2 ⁢ x + 2 ⋅ 1 - 1
29 27 26 28 mp3an23 ⊢ 2 ⁢ x ∈ ℂ → 2 ⁢ x + 2 ⋅ 1 - 1 = 2 ⁢ x + 2 ⋅ 1 - 1
30 25 29 syl ⊢ x ∈ ℤ → 2 ⁢ x + 2 ⋅ 1 - 1 = 2 ⁢ x + 2 ⋅ 1 - 1
31 2t1e2 ⊢ 2 ⋅ 1 = 2
32 31 oveq1i ⊢ 2 ⋅ 1 − 1 = 2 − 1
33 2m1e1 ⊢ 2 − 1 = 1
34 32 33 eqtri ⊢ 2 ⋅ 1 − 1 = 1
35 34 oveq2i ⊢ 2 ⁢ x + 2 ⋅ 1 - 1 = 2 ⁢ x + 1
36 30 35 eqtr2di ⊢ x ∈ ℤ → 2 ⁢ x + 1 = 2 ⁢ x + 2 ⋅ 1 - 1
37 adddi ⊢ 2 ∈ ℂ ∧ x ∈ ℂ ∧ 1 ∈ ℂ → 2 ⁢ x + 1 = 2 ⁢ x + 2 ⋅ 1
38 13 26 37 mp3an13 ⊢ x ∈ ℂ → 2 ⁢ x + 1 = 2 ⁢ x + 2 ⋅ 1
39 12 38 syl ⊢ x ∈ ℤ → 2 ⁢ x + 1 = 2 ⁢ x + 2 ⋅ 1
40 39 oveq1d ⊢ x ∈ ℤ → 2 ⁢ x + 1 − 1 = 2 ⁢ x + 2 ⋅ 1 - 1
41 36 40 eqtr4d ⊢ x ∈ ℤ → 2 ⁢ x + 1 = 2 ⁢ x + 1 − 1
42 41 negeqd ⊢ x ∈ ℤ → − 2 ⁢ x + 1 = − 2 ⁢ x + 1 − 1
43 8 zcnd ⊢ x ∈ ℤ → x + 1 ∈ ℂ
44 mulneg2 ⊢ 2 ∈ ℂ ∧ x + 1 ∈ ℂ → 2 ⁢ − x + 1 = − 2 ⁢ x + 1
45 13 43 44 sylancr ⊢ x ∈ ℤ → 2 ⁢ − x + 1 = − 2 ⁢ x + 1
46 45 oveq1d ⊢ x ∈ ℤ → 2 ⁢ − x + 1 + 1 = - 2 ⁢ x + 1 + 1
47 mulcl ⊢ 2 ∈ ℂ ∧ x + 1 ∈ ℂ → 2 ⁢ x + 1 ∈ ℂ
48 13 43 47 sylancr ⊢ x ∈ ℤ → 2 ⁢ x + 1 ∈ ℂ
49 negsubdi ⊢ 2 ⁢ x + 1 ∈ ℂ ∧ 1 ∈ ℂ → − 2 ⁢ x + 1 − 1 = - 2 ⁢ x + 1 + 1
50 48 26 49 sylancl ⊢ x ∈ ℤ → − 2 ⁢ x + 1 − 1 = - 2 ⁢ x + 1 + 1
51 46 50 eqtr4d ⊢ x ∈ ℤ → 2 ⁢ − x + 1 + 1 = − 2 ⁢ x + 1 − 1
52 42 51 eqtr4d ⊢ x ∈ ℤ → − 2 ⁢ x + 1 = 2 ⁢ − x + 1 + 1
53 52 adantl ⊢ N ∈ ℝ ∧ x ∈ ℤ → − 2 ⁢ x + 1 = 2 ⁢ − x + 1 + 1
54 53 eqeq1d ⊢ N ∈ ℝ ∧ x ∈ ℤ → − 2 ⁢ x + 1 = N ↔ 2 ⁢ − x + 1 + 1 = N
55 24 54 bitrid ⊢ N ∈ ℝ ∧ x ∈ ℤ → N = − 2 ⁢ x + 1 ↔ 2 ⁢ − x + 1 + 1 = N
56 23 55 bitrd ⊢ N ∈ ℝ ∧ x ∈ ℤ → 2 ⁢ x + 1 = − N ↔ 2 ⁢ − x + 1 + 1 = N
57 56 biimpa ⊢ N ∈ ℝ ∧ x ∈ ℤ ∧ 2 ⁢ x + 1 = − N → 2 ⁢ − x + 1 + 1 = N
58 oveq2 ⊢ n = − x + 1 → 2 ⁢ n = 2 ⁢ − x + 1
59 58 oveq1d ⊢ n = − x + 1 → 2 ⁢ n + 1 = 2 ⁢ − x + 1 + 1
60 59 eqeq1d ⊢ n = − x + 1 → 2 ⁢ n + 1 = N ↔ 2 ⁢ − x + 1 + 1 = N
61 60 rspcev ⊢ − x + 1 ∈ ℤ ∧ 2 ⁢ − x + 1 + 1 = N → ∃ n ∈ ℤ 2 ⁢ n + 1 = N
62 11 57 61 syl2anc ⊢ N ∈ ℝ ∧ x ∈ ℤ ∧ 2 ⁢ x + 1 = − N → ∃ n ∈ ℤ 2 ⁢ n + 1 = N
63 62 rexlimdva2 ⊢ N ∈ ℝ → ∃ x ∈ ℤ 2 ⁢ x + 1 = − N → ∃ n ∈ ℤ 2 ⁢ n + 1 = N
64 znegcl ⊢ y ∈ ℤ → − y ∈ ℤ
65 64 ad2antlr ⊢ N ∈ ℝ ∧ y ∈ ℤ ∧ y ⋅ 2 = − N → − y ∈ ℤ
66 zcn ⊢ y ∈ ℤ → y ∈ ℂ
67 mulcl ⊢ y ∈ ℂ ∧ 2 ∈ ℂ → y ⋅ 2 ∈ ℂ
68 66 13 67 sylancl ⊢ y ∈ ℤ → y ⋅ 2 ∈ ℂ
69 recn ⊢ N ∈ ℝ → N ∈ ℂ
70 negcon2 ⊢ y ⋅ 2 ∈ ℂ ∧ N ∈ ℂ → y ⋅ 2 = − N ↔ N = − y ⋅ 2
71 68 69 70 syl2anr ⊢ N ∈ ℝ ∧ y ∈ ℤ → y ⋅ 2 = − N ↔ N = − y ⋅ 2
72 eqcom ⊢ N = − y ⋅ 2 ↔ − y ⋅ 2 = N
73 mulneg1 ⊢ y ∈ ℂ ∧ 2 ∈ ℂ → − y ⋅ 2 = − y ⋅ 2
74 66 13 73 sylancl ⊢ y ∈ ℤ → − y ⋅ 2 = − y ⋅ 2
75 74 adantl ⊢ N ∈ ℝ ∧ y ∈ ℤ → − y ⋅ 2 = − y ⋅ 2
76 75 eqeq1d ⊢ N ∈ ℝ ∧ y ∈ ℤ → − y ⋅ 2 = N ↔ − y ⋅ 2 = N
77 72 76 bitr4id ⊢ N ∈ ℝ ∧ y ∈ ℤ → N = − y ⋅ 2 ↔ − y ⋅ 2 = N
78 71 77 bitrd ⊢ N ∈ ℝ ∧ y ∈ ℤ → y ⋅ 2 = − N ↔ − y ⋅ 2 = N
79 78 biimpa ⊢ N ∈ ℝ ∧ y ∈ ℤ ∧ y ⋅ 2 = − N → − y ⋅ 2 = N
80 oveq1 ⊢ k = − y → k ⋅ 2 = − y ⋅ 2
81 80 eqeq1d ⊢ k = − y → k ⋅ 2 = N ↔ − y ⋅ 2 = N
82 81 rspcev ⊢ − y ∈ ℤ ∧ − y ⋅ 2 = N → ∃ k ∈ ℤ k ⋅ 2 = N
83 65 79 82 syl2anc ⊢ N ∈ ℝ ∧ y ∈ ℤ ∧ y ⋅ 2 = − N → ∃ k ∈ ℤ k ⋅ 2 = N
84 83 rexlimdva2 ⊢ N ∈ ℝ → ∃ y ∈ ℤ y ⋅ 2 = − N → ∃ k ∈ ℤ k ⋅ 2 = N
85 63 84 orim12d ⊢ N ∈ ℝ → ∃ x ∈ ℤ 2 ⁢ x + 1 = − N ∨ ∃ y ∈ ℤ y ⋅ 2 = − N → ∃ n ∈ ℤ 2 ⁢ n + 1 = N ∨ ∃ k ∈ ℤ k ⋅ 2 = N
86 odd2np1lem ⊢ − N ∈ ℕ 0 → ∃ x ∈ ℤ 2 ⁢ x + 1 = − N ∨ ∃ y ∈ ℤ y ⋅ 2 = − N
87 85 86 impel ⊢ N ∈ ℝ ∧ − N ∈ ℕ 0 → ∃ n ∈ ℤ 2 ⁢ n + 1 = N ∨ ∃ k ∈ ℤ k ⋅ 2 = N
88 7 87 jaodan ⊢ N ∈ ℝ ∧ N ∈ ℕ 0 ∨ − N ∈ ℕ 0 → ∃ n ∈ ℤ 2 ⁢ n + 1 = N ∨ ∃ k ∈ ℤ k ⋅ 2 = N
89 5 88 sylbi ⊢ N ∈ ℤ → ∃ n ∈ ℤ 2 ⁢ n + 1 = N ∨ ∃ k ∈ ℤ k ⋅ 2 = N
90 halfnz ⊢ ¬ 1 2 ∈ ℤ
91 reeanv ⊢ ∃ n ∈ ℤ ∃ k ∈ ℤ 2 ⁢ n + 1 = N ∧ k ⋅ 2 = N ↔ ∃ n ∈ ℤ 2 ⁢ n + 1 = N ∧ ∃ k ∈ ℤ k ⋅ 2 = N
92 eqtr3 ⊢ 2 ⁢ n + 1 = N ∧ k ⋅ 2 = N → 2 ⁢ n + 1 = k ⋅ 2
93 zcn ⊢ k ∈ ℤ → k ∈ ℂ
94 mulcom ⊢ k ∈ ℂ ∧ 2 ∈ ℂ → k ⋅ 2 = 2 ⁢ k
95 93 13 94 sylancl ⊢ k ∈ ℤ → k ⋅ 2 = 2 ⁢ k
96 95 eqeq2d ⊢ k ∈ ℤ → 2 ⁢ n + 1 = k ⋅ 2 ↔ 2 ⁢ n + 1 = 2 ⁢ k
97 96 adantl ⊢ n ∈ ℤ ∧ k ∈ ℤ → 2 ⁢ n + 1 = k ⋅ 2 ↔ 2 ⁢ n + 1 = 2 ⁢ k
98 mulcl ⊢ 2 ∈ ℂ ∧ k ∈ ℂ → 2 ⁢ k ∈ ℂ
99 13 93 98 sylancr ⊢ k ∈ ℤ → 2 ⁢ k ∈ ℂ
100 zcn ⊢ n ∈ ℤ → n ∈ ℂ
101 mulcl ⊢ 2 ∈ ℂ ∧ n ∈ ℂ → 2 ⁢ n ∈ ℂ
102 13 100 101 sylancr ⊢ n ∈ ℤ → 2 ⁢ n ∈ ℂ
103 subadd ⊢ 2 ⁢ k ∈ ℂ ∧ 2 ⁢ n ∈ ℂ ∧ 1 ∈ ℂ → 2 ⁢ k − 2 ⁢ n = 1 ↔ 2 ⁢ n + 1 = 2 ⁢ k
104 26 103 mp3an3 ⊢ 2 ⁢ k ∈ ℂ ∧ 2 ⁢ n ∈ ℂ → 2 ⁢ k − 2 ⁢ n = 1 ↔ 2 ⁢ n + 1 = 2 ⁢ k
105 99 102 104 syl2anr ⊢ n ∈ ℤ ∧ k ∈ ℤ → 2 ⁢ k − 2 ⁢ n = 1 ↔ 2 ⁢ n + 1 = 2 ⁢ k
106 subcl ⊢ k ∈ ℂ ∧ n ∈ ℂ → k − n ∈ ℂ
107 2cnne0 ⊢ 2 ∈ ℂ ∧ 2 ≠ 0
108 eqcom ⊢ k − n = 1 2 ↔ 1 2 = k − n
109 divmul ⊢ 1 ∈ ℂ ∧ k − n ∈ ℂ ∧ 2 ∈ ℂ ∧ 2 ≠ 0 → 1 2 = k − n ↔ 2 ⁢ k − n = 1
110 108 109 bitrid ⊢ 1 ∈ ℂ ∧ k − n ∈ ℂ ∧ 2 ∈ ℂ ∧ 2 ≠ 0 → k − n = 1 2 ↔ 2 ⁢ k − n = 1
111 26 107 110 mp3an13 ⊢ k − n ∈ ℂ → k − n = 1 2 ↔ 2 ⁢ k − n = 1
112 106 111 syl ⊢ k ∈ ℂ ∧ n ∈ ℂ → k − n = 1 2 ↔ 2 ⁢ k − n = 1
113 112 ancoms ⊢ n ∈ ℂ ∧ k ∈ ℂ → k − n = 1 2 ↔ 2 ⁢ k − n = 1
114 subdi ⊢ 2 ∈ ℂ ∧ k ∈ ℂ ∧ n ∈ ℂ → 2 ⁢ k − n = 2 ⁢ k − 2 ⁢ n
115 13 114 mp3an1 ⊢ k ∈ ℂ ∧ n ∈ ℂ → 2 ⁢ k − n = 2 ⁢ k − 2 ⁢ n
116 115 ancoms ⊢ n ∈ ℂ ∧ k ∈ ℂ → 2 ⁢ k − n = 2 ⁢ k − 2 ⁢ n
117 116 eqeq1d ⊢ n ∈ ℂ ∧ k ∈ ℂ → 2 ⁢ k − n = 1 ↔ 2 ⁢ k − 2 ⁢ n = 1
118 113 117 bitrd ⊢ n ∈ ℂ ∧ k ∈ ℂ → k − n = 1 2 ↔ 2 ⁢ k − 2 ⁢ n = 1
119 100 93 118 syl2an ⊢ n ∈ ℤ ∧ k ∈ ℤ → k − n = 1 2 ↔ 2 ⁢ k − 2 ⁢ n = 1
120 zsubcl ⊢ k ∈ ℤ ∧ n ∈ ℤ → k − n ∈ ℤ
121 eleq1 ⊢ k − n = 1 2 → k − n ∈ ℤ ↔ 1 2 ∈ ℤ
122 120 121 syl5ibcom ⊢ k ∈ ℤ ∧ n ∈ ℤ → k − n = 1 2 → 1 2 ∈ ℤ
123 122 ancoms ⊢ n ∈ ℤ ∧ k ∈ ℤ → k − n = 1 2 → 1 2 ∈ ℤ
124 119 123 sylbird ⊢ n ∈ ℤ ∧ k ∈ ℤ → 2 ⁢ k − 2 ⁢ n = 1 → 1 2 ∈ ℤ
125 105 124 sylbird ⊢ n ∈ ℤ ∧ k ∈ ℤ → 2 ⁢ n + 1 = 2 ⁢ k → 1 2 ∈ ℤ
126 97 125 sylbid ⊢ n ∈ ℤ ∧ k ∈ ℤ → 2 ⁢ n + 1 = k ⋅ 2 → 1 2 ∈ ℤ
127 92 126 syl5 ⊢ n ∈ ℤ ∧ k ∈ ℤ → 2 ⁢ n + 1 = N ∧ k ⋅ 2 = N → 1 2 ∈ ℤ
128 127 rexlimivv ⊢ ∃ n ∈ ℤ ∃ k ∈ ℤ 2 ⁢ n + 1 = N ∧ k ⋅ 2 = N → 1 2 ∈ ℤ
129 91 128 sylbir ⊢ ∃ n ∈ ℤ 2 ⁢ n + 1 = N ∧ ∃ k ∈ ℤ k ⋅ 2 = N → 1 2 ∈ ℤ
130 90 129 mto ⊢ ¬ ∃ n ∈ ℤ 2 ⁢ n + 1 = N ∧ ∃ k ∈ ℤ k ⋅ 2 = N
131 pm5.17 ⊢ ∃ n ∈ ℤ 2 ⁢ n + 1 = N ∨ ∃ k ∈ ℤ k ⋅ 2 = N ∧ ¬ ∃ n ∈ ℤ 2 ⁢ n + 1 = N ∧ ∃ k ∈ ℤ k ⋅ 2 = N ↔ ∃ n ∈ ℤ 2 ⁢ n + 1 = N ↔ ¬ ∃ k ∈ ℤ k ⋅ 2 = N
132 bicom ⊢ ∃ n ∈ ℤ 2 ⁢ n + 1 = N ↔ ¬ ∃ k ∈ ℤ k ⋅ 2 = N ↔ ¬ ∃ k ∈ ℤ k ⋅ 2 = N ↔ ∃ n ∈ ℤ 2 ⁢ n + 1 = N
133 131 132 bitri ⊢ ∃ n ∈ ℤ 2 ⁢ n + 1 = N ∨ ∃ k ∈ ℤ k ⋅ 2 = N ∧ ¬ ∃ n ∈ ℤ 2 ⁢ n + 1 = N ∧ ∃ k ∈ ℤ k ⋅ 2 = N ↔ ¬ ∃ k ∈ ℤ k ⋅ 2 = N ↔ ∃ n ∈ ℤ 2 ⁢ n + 1 = N
134 89 130 133 sylanblc ⊢ N ∈ ℤ → ¬ ∃ k ∈ ℤ k ⋅ 2 = N ↔ ∃ n ∈ ℤ 2 ⁢ n + 1 = N
135 4 134 bitrd ⊢ N ∈ ℤ → ¬ 2 ∥ N ↔ ∃ n ∈ ℤ 2 ⁢ n + 1 = N