Metamath Proof Explorer


Theorem dcubic2

Description: Reverse direction of dcubic . Given a solution U to the "substitution" quadratic equation X = U - M / U , show that X is in the desired form. (Contributed by Mario Carneiro, 25-Apr-2015)

Ref Expression
Hypotheses dcubic.c ⊢ φ → P ∈ ℂ
dcubic.d ⊢ φ → Q ∈ ℂ
dcubic.x ⊢ φ → X ∈ ℂ
dcubic.t ⊢ φ → T ∈ ℂ
dcubic.3 ⊢ φ → T 3 = G − N
dcubic.g ⊢ φ → G ∈ ℂ
dcubic.2 ⊢ φ → G 2 = N 2 + M 3
dcubic.m ⊢ φ → M = P 3
dcubic.n ⊢ φ → N = Q 2
dcubic.0 ⊢ φ → T ≠ 0
dcubic2.u ⊢ φ → U ∈ ℂ
dcubic2.z ⊢ φ → U ≠ 0
dcubic2.2 ⊢ φ → X = U − M U
dcubic2.x ⊢ φ → X 3 + P ⁢ X + Q = 0
Assertion dcubic2 ⊢ φ → ∃ r ∈ ℂ r 3 = 1 ∧ X = r ⁢ T − M r ⁢ T

Proof

Step Hyp Ref Expression
1 dcubic.c ⊢ φ → P ∈ ℂ
2 dcubic.d ⊢ φ → Q ∈ ℂ
3 dcubic.x ⊢ φ → X ∈ ℂ
4 dcubic.t ⊢ φ → T ∈ ℂ
5 dcubic.3 ⊢ φ → T 3 = G − N
6 dcubic.g ⊢ φ → G ∈ ℂ
7 dcubic.2 ⊢ φ → G 2 = N 2 + M 3
8 dcubic.m ⊢ φ → M = P 3
9 dcubic.n ⊢ φ → N = Q 2
10 dcubic.0 ⊢ φ → T ≠ 0
11 dcubic2.u ⊢ φ → U ∈ ℂ
12 dcubic2.z ⊢ φ → U ≠ 0
13 dcubic2.2 ⊢ φ → X = U − M U
14 dcubic2.x ⊢ φ → X 3 + P ⁢ X + Q = 0
15 11 4 10 divcld ⊢ φ → U T ∈ ℂ
16 15 adantr ⊢ φ ∧ U 3 = G − N → U T ∈ ℂ
17 3nn0 ⊢ 3 ∈ ℕ 0
18 17 a1i ⊢ φ → 3 ∈ ℕ 0
19 11 4 10 18 expdivd ⊢ φ → U T 3 = U 3 T 3
20 19 adantr ⊢ φ ∧ U 3 = G − N → U T 3 = U 3 T 3
21 oveq1 ⊢ U 3 = G − N → U 3 T 3 = G − N T 3
22 5 oveq1d ⊢ φ → T 3 T 3 = G − N T 3
23 expcl ⊢ T ∈ ℂ ∧ 3 ∈ ℕ 0 → T 3 ∈ ℂ
24 4 17 23 sylancl ⊢ φ → T 3 ∈ ℂ
25 3z ⊢ 3 ∈ ℤ
26 25 a1i ⊢ φ → 3 ∈ ℤ
27 4 10 26 expne0d ⊢ φ → T 3 ≠ 0
28 24 27 dividd ⊢ φ → T 3 T 3 = 1
29 22 28 eqtr3d ⊢ φ → G − N T 3 = 1
30 21 29 sylan9eqr ⊢ φ ∧ U 3 = G − N → U 3 T 3 = 1
31 20 30 eqtrd ⊢ φ ∧ U 3 = G − N → U T 3 = 1
32 11 4 10 divcan1d ⊢ φ → U T ⁢ T = U
33 32 oveq2d ⊢ φ → M U T ⁢ T = M U
34 32 33 oveq12d ⊢ φ → U T ⁢ T − M U T ⁢ T = U − M U
35 13 34 eqtr4d ⊢ φ → X = U T ⁢ T − M U T ⁢ T
36 35 adantr ⊢ φ ∧ U 3 = G − N → X = U T ⁢ T − M U T ⁢ T
37 oveq1 ⊢ r = U T → r 3 = U T 3
38 37 eqeq1d ⊢ r = U T → r 3 = 1 ↔ U T 3 = 1
39 oveq1 ⊢ r = U T → r ⁢ T = U T ⁢ T
40 39 oveq2d ⊢ r = U T → M r ⁢ T = M U T ⁢ T
41 39 40 oveq12d ⊢ r = U T → r ⁢ T − M r ⁢ T = U T ⁢ T − M U T ⁢ T
42 41 eqeq2d ⊢ r = U T → X = r ⁢ T − M r ⁢ T ↔ X = U T ⁢ T − M U T ⁢ T
43 38 42 anbi12d ⊢ r = U T → r 3 = 1 ∧ X = r ⁢ T − M r ⁢ T ↔ U T 3 = 1 ∧ X = U T ⁢ T − M U T ⁢ T
44 43 rspcev ⊢ U T ∈ ℂ ∧ U T 3 = 1 ∧ X = U T ⁢ T − M U T ⁢ T → ∃ r ∈ ℂ r 3 = 1 ∧ X = r ⁢ T − M r ⁢ T
45 16 31 36 44 syl12anc ⊢ φ ∧ U 3 = G − N → ∃ r ∈ ℂ r 3 = 1 ∧ X = r ⁢ T − M r ⁢ T
46 3cn ⊢ 3 ∈ ℂ
47 46 a1i ⊢ φ → 3 ∈ ℂ
48 3ne0 ⊢ 3 ≠ 0
49 48 a1i ⊢ φ → 3 ≠ 0
50 1 47 49 divcld ⊢ φ → P 3 ∈ ℂ
51 8 50 eqeltrd ⊢ φ → M ∈ ℂ
52 51 11 12 divcld ⊢ φ → M U ∈ ℂ
53 52 negcld ⊢ φ → − M U ∈ ℂ
54 53 4 10 divcld ⊢ φ → − M U T ∈ ℂ
55 54 adantr ⊢ φ ∧ U 3 = − G + N → − M U T ∈ ℂ
56 53 4 10 18 expdivd ⊢ φ → − M U T 3 = − M U 3 T 3
57 51 11 12 divnegd ⊢ φ → − M U = − M U
58 57 oveq1d ⊢ φ → − M U 3 = − M U 3
59 51 negcld ⊢ φ → − M ∈ ℂ
60 59 11 12 18 expdivd ⊢ φ → − M U 3 = − M 3 U 3
61 5 oveq2d ⊢ φ → G + N ⁢ T 3 = G + N ⁢ G − N
62 2 halfcld ⊢ φ → Q 2 ∈ ℂ
63 9 62 eqeltrd ⊢ φ → N ∈ ℂ
64 subsq ⊢ G ∈ ℂ ∧ N ∈ ℂ → G 2 − N 2 = G + N ⁢ G − N
65 6 63 64 syl2anc ⊢ φ → G 2 − N 2 = G + N ⁢ G − N
66 61 65 eqtr4d ⊢ φ → G + N ⁢ T 3 = G 2 − N 2
67 7 oveq1d ⊢ φ → G 2 − N 2 = N 2 + M 3 - N 2
68 63 sqcld ⊢ φ → N 2 ∈ ℂ
69 expcl ⊢ M ∈ ℂ ∧ 3 ∈ ℕ 0 → M 3 ∈ ℂ
70 51 17 69 sylancl ⊢ φ → M 3 ∈ ℂ
71 68 70 pncan2d ⊢ φ → N 2 + M 3 - N 2 = M 3
72 66 67 71 3eqtrd ⊢ φ → G + N ⁢ T 3 = M 3
73 72 negeqd ⊢ φ → − G + N ⁢ T 3 = − M 3
74 6 63 addcld ⊢ φ → G + N ∈ ℂ
75 74 24 mulneg1d ⊢ φ → − G + N ⁢ T 3 = − G + N ⁢ T 3
76 3nn ⊢ 3 ∈ ℕ
77 76 a1i ⊢ φ → 3 ∈ ℕ
78 n2dvds3 ⊢ ¬ 2 ∥ 3
79 78 a1i ⊢ φ → ¬ 2 ∥ 3
80 oexpneg ⊢ M ∈ ℂ ∧ 3 ∈ ℕ ∧ ¬ 2 ∥ 3 → − M 3 = − M 3
81 51 77 79 80 syl3anc ⊢ φ → − M 3 = − M 3
82 73 75 81 3eqtr4d ⊢ φ → − G + N ⁢ T 3 = − M 3
83 82 oveq1d ⊢ φ → − G + N ⁢ T 3 U 3 = − M 3 U 3
84 74 negcld ⊢ φ → − G + N ∈ ℂ
85 expcl ⊢ U ∈ ℂ ∧ 3 ∈ ℕ 0 → U 3 ∈ ℂ
86 11 17 85 sylancl ⊢ φ → U 3 ∈ ℂ
87 11 12 26 expne0d ⊢ φ → U 3 ≠ 0
88 84 24 86 87 div23d ⊢ φ → − G + N ⁢ T 3 U 3 = − G + N U 3 ⁢ T 3
89 83 88 eqtr3d ⊢ φ → − M 3 U 3 = − G + N U 3 ⁢ T 3
90 58 60 89 3eqtrd ⊢ φ → − M U 3 = − G + N U 3 ⁢ T 3
91 90 oveq1d ⊢ φ → − M U 3 T 3 = − G + N U 3 ⁢ T 3 T 3
92 84 86 87 divcld ⊢ φ → − G + N U 3 ∈ ℂ
93 92 24 27 divcan4d ⊢ φ → − G + N U 3 ⁢ T 3 T 3 = − G + N U 3
94 56 91 93 3eqtrd ⊢ φ → − M U T 3 = − G + N U 3
95 94 adantr ⊢ φ ∧ U 3 = − G + N → − M U T 3 = − G + N U 3
96 oveq1 ⊢ U 3 = − G + N → U 3 U 3 = − G + N U 3
97 96 eqcomd ⊢ U 3 = − G + N → − G + N U 3 = U 3 U 3
98 86 87 dividd ⊢ φ → U 3 U 3 = 1
99 97 98 sylan9eqr ⊢ φ ∧ U 3 = − G + N → − G + N U 3 = 1
100 95 99 eqtrd ⊢ φ ∧ U 3 = − G + N → − M U T 3 = 1
101 52 11 neg2subd ⊢ φ → - M U - − U = U − M U
102 13 101 eqtr4d ⊢ φ → X = - M U - − U
103 102 adantr ⊢ φ ∧ U 3 = − G + N → X = - M U - − U
104 53 4 10 divcan1d ⊢ φ → − M U T ⁢ T = − M U
105 104 adantr ⊢ φ ∧ U 3 = − G + N → − M U T ⁢ T = − M U
106 51 11 12 divneg2d ⊢ φ → − M U = M − U
107 104 106 eqtrd ⊢ φ → − M U T ⁢ T = M − U
108 107 adantr ⊢ φ ∧ U 3 = − G + N → − M U T ⁢ T = M − U
109 108 oveq2d ⊢ φ ∧ U 3 = − G + N → M − M U T ⁢ T = M M − U
110 51 adantr ⊢ φ ∧ U 3 = − G + N → M ∈ ℂ
111 11 negcld ⊢ φ → − U ∈ ℂ
112 111 adantr ⊢ φ ∧ U 3 = − G + N → − U ∈ ℂ
113 75 73 eqtrd ⊢ φ → − G + N ⁢ T 3 = − M 3
114 113 adantr ⊢ φ ∧ U 3 = − G + N → − G + N ⁢ T 3 = − M 3
115 84 adantr ⊢ φ ∧ U 3 = − G + N → − G + N ∈ ℂ
116 24 adantr ⊢ φ ∧ U 3 = − G + N → T 3 ∈ ℂ
117 simpr ⊢ φ ∧ U 3 = − G + N → U 3 = − G + N
118 87 adantr ⊢ φ ∧ U 3 = − G + N → U 3 ≠ 0
119 117 118 eqnetrrd ⊢ φ ∧ U 3 = − G + N → − G + N ≠ 0
120 27 adantr ⊢ φ ∧ U 3 = − G + N → T 3 ≠ 0
121 115 116 119 120 mulne0d ⊢ φ ∧ U 3 = − G + N → − G + N ⁢ T 3 ≠ 0
122 114 121 eqnetrrd ⊢ φ ∧ U 3 = − G + N → − M 3 ≠ 0
123 oveq1 ⊢ M = 0 → M 3 = 0 3
124 0exp ⊢ 3 ∈ ℕ → 0 3 = 0
125 76 124 ax-mp ⊢ 0 3 = 0
126 123 125 eqtrdi ⊢ M = 0 → M 3 = 0
127 126 negeqd ⊢ M = 0 → − M 3 = − 0
128 neg0 ⊢ − 0 = 0
129 127 128 eqtrdi ⊢ M = 0 → − M 3 = 0
130 129 necon3i ⊢ − M 3 ≠ 0 → M ≠ 0
131 122 130 syl ⊢ φ ∧ U 3 = − G + N → M ≠ 0
132 11 12 negne0d ⊢ φ → − U ≠ 0
133 132 adantr ⊢ φ ∧ U 3 = − G + N → − U ≠ 0
134 110 112 131 133 ddcand ⊢ φ ∧ U 3 = − G + N → M M − U = − U
135 109 134 eqtrd ⊢ φ ∧ U 3 = − G + N → M − M U T ⁢ T = − U
136 105 135 oveq12d ⊢ φ ∧ U 3 = − G + N → − M U T ⁢ T − M − M U T ⁢ T = - M U - − U
137 103 136 eqtr4d ⊢ φ ∧ U 3 = − G + N → X = − M U T ⁢ T − M − M U T ⁢ T
138 oveq1 ⊢ r = − M U T → r 3 = − M U T 3
139 138 eqeq1d ⊢ r = − M U T → r 3 = 1 ↔ − M U T 3 = 1
140 oveq1 ⊢ r = − M U T → r ⁢ T = − M U T ⁢ T
141 140 oveq2d ⊢ r = − M U T → M r ⁢ T = M − M U T ⁢ T
142 140 141 oveq12d ⊢ r = − M U T → r ⁢ T − M r ⁢ T = − M U T ⁢ T − M − M U T ⁢ T
143 142 eqeq2d ⊢ r = − M U T → X = r ⁢ T − M r ⁢ T ↔ X = − M U T ⁢ T − M − M U T ⁢ T
144 139 143 anbi12d ⊢ r = − M U T → r 3 = 1 ∧ X = r ⁢ T − M r ⁢ T ↔ − M U T 3 = 1 ∧ X = − M U T ⁢ T − M − M U T ⁢ T
145 144 rspcev ⊢ − M U T ∈ ℂ ∧ − M U T 3 = 1 ∧ X = − M U T ⁢ T − M − M U T ⁢ T → ∃ r ∈ ℂ r 3 = 1 ∧ X = r ⁢ T − M r ⁢ T
146 55 100 137 145 syl12anc ⊢ φ ∧ U 3 = − G + N → ∃ r ∈ ℂ r 3 = 1 ∧ X = r ⁢ T − M r ⁢ T
147 86 sqcld ⊢ φ → U 3 2 ∈ ℂ
148 147 mullidd ⊢ φ → 1 ⁢ U 3 2 = U 3 2
149 2 86 mulcld ⊢ φ → Q ⁢ U 3 ∈ ℂ
150 149 70 negsubd ⊢ φ → Q ⁢ U 3 + − M 3 = Q ⁢ U 3 − M 3
151 148 150 oveq12d ⊢ φ → 1 ⁢ U 3 2 + Q ⁢ U 3 + − M 3 = U 3 2 + Q ⁢ U 3 - M 3
152 1 2 3 4 5 6 7 8 9 10 11 12 13 dcubic1lem ⊢ φ → X 3 + P ⁢ X + Q = 0 ↔ U 3 2 + Q ⁢ U 3 - M 3 = 0
153 14 152 mpbid ⊢ φ → U 3 2 + Q ⁢ U 3 - M 3 = 0
154 151 153 eqtrd ⊢ φ → 1 ⁢ U 3 2 + Q ⁢ U 3 + − M 3 = 0
155 1cnd ⊢ φ → 1 ∈ ℂ
156 ax-1ne0 ⊢ 1 ≠ 0
157 156 a1i ⊢ φ → 1 ≠ 0
158 70 negcld ⊢ φ → − M 3 ∈ ℂ
159 2cn ⊢ 2 ∈ ℂ
160 mulcl ⊢ 2 ∈ ℂ ∧ G ∈ ℂ → 2 ⁢ G ∈ ℂ
161 159 6 160 sylancr ⊢ φ → 2 ⁢ G ∈ ℂ
162 sqmul ⊢ 2 ∈ ℂ ∧ G ∈ ℂ → 2 ⁢ G 2 = 2 2 ⁢ G 2
163 159 6 162 sylancr ⊢ φ → 2 ⁢ G 2 = 2 2 ⁢ G 2
164 7 oveq2d ⊢ φ → 2 2 ⁢ G 2 = 2 2 ⁢ N 2 + M 3
165 159 sqcli ⊢ 2 2 ∈ ℂ
166 mulcl ⊢ 2 2 ∈ ℂ ∧ N 2 ∈ ℂ → 2 2 ⁢ N 2 ∈ ℂ
167 165 68 166 sylancr ⊢ φ → 2 2 ⁢ N 2 ∈ ℂ
168 mulcl ⊢ 2 2 ∈ ℂ ∧ M 3 ∈ ℂ → 2 2 ⁢ M 3 ∈ ℂ
169 165 70 168 sylancr ⊢ φ → 2 2 ⁢ M 3 ∈ ℂ
170 167 169 subnegd ⊢ φ → 2 2 ⁢ N 2 − − 2 2 ⁢ M 3 = 2 2 ⁢ N 2 + 2 2 ⁢ M 3
171 9 oveq2d ⊢ φ → 2 ⋅ N = 2 ⁢ Q 2
172 159 a1i ⊢ φ → 2 ∈ ℂ
173 2ne0 ⊢ 2 ≠ 0
174 173 a1i ⊢ φ → 2 ≠ 0
175 2 172 174 divcan2d ⊢ φ → 2 ⁢ Q 2 = Q
176 171 175 eqtrd ⊢ φ → 2 ⋅ N = Q
177 176 oveq1d ⊢ φ → 2 ⋅ N 2 = Q 2
178 sqmul ⊢ 2 ∈ ℂ ∧ N ∈ ℂ → 2 ⋅ N 2 = 2 2 ⁢ N 2
179 159 63 178 sylancr ⊢ φ → 2 ⋅ N 2 = 2 2 ⁢ N 2
180 177 179 eqtr3d ⊢ φ → Q 2 = 2 2 ⁢ N 2
181 158 mullidd ⊢ φ → 1 ⁢ − M 3 = − M 3
182 181 oveq2d ⊢ φ → 4 ⁢ 1 ⁢ − M 3 = 4 ⁢ − M 3
183 4cn ⊢ 4 ∈ ℂ
184 mulneg2 ⊢ 4 ∈ ℂ ∧ M 3 ∈ ℂ → 4 ⁢ − M 3 = − 4 ⁢ M 3
185 183 70 184 sylancr ⊢ φ → 4 ⁢ − M 3 = − 4 ⁢ M 3
186 182 185 eqtrd ⊢ φ → 4 ⁢ 1 ⁢ − M 3 = − 4 ⁢ M 3
187 sq2 ⊢ 2 2 = 4
188 187 oveq1i ⊢ 2 2 ⁢ M 3 = 4 ⁢ M 3
189 188 negeqi ⊢ − 2 2 ⁢ M 3 = − 4 ⁢ M 3
190 186 189 eqtr4di ⊢ φ → 4 ⁢ 1 ⁢ − M 3 = − 2 2 ⁢ M 3
191 180 190 oveq12d ⊢ φ → Q 2 − 4 ⁢ 1 ⁢ − M 3 = 2 2 ⁢ N 2 − − 2 2 ⁢ M 3
192 165 a1i ⊢ φ → 2 2 ∈ ℂ
193 192 68 70 adddid ⊢ φ → 2 2 ⁢ N 2 + M 3 = 2 2 ⁢ N 2 + 2 2 ⁢ M 3
194 170 191 193 3eqtr4rd ⊢ φ → 2 2 ⁢ N 2 + M 3 = Q 2 − 4 ⁢ 1 ⁢ − M 3
195 163 164 194 3eqtrd ⊢ φ → 2 ⁢ G 2 = Q 2 − 4 ⁢ 1 ⁢ − M 3
196 155 157 2 158 86 161 195 quad2 ⊢ φ → 1 ⁢ U 3 2 + Q ⁢ U 3 + − M 3 = 0 ↔ U 3 = - Q + 2 ⁢ G 2 ⋅ 1 ∨ U 3 = - Q - 2 ⁢ G 2 ⋅ 1
197 154 196 mpbid ⊢ φ → U 3 = - Q + 2 ⁢ G 2 ⋅ 1 ∨ U 3 = - Q - 2 ⁢ G 2 ⋅ 1
198 2t1e2 ⊢ 2 ⋅ 1 = 2
199 198 oveq2i ⊢ - Q + 2 ⁢ G 2 ⋅ 1 = - Q + 2 ⁢ G 2
200 2 negcld ⊢ φ → − Q ∈ ℂ
201 200 161 172 174 divdird ⊢ φ → - Q + 2 ⁢ G 2 = − Q 2 + 2 ⁢ G 2
202 9 negeqd ⊢ φ → − N = − Q 2
203 2 172 174 divnegd ⊢ φ → − Q 2 = − Q 2
204 202 203 eqtr2d ⊢ φ → − Q 2 = − N
205 6 172 174 divcan3d ⊢ φ → 2 ⁢ G 2 = G
206 204 205 oveq12d ⊢ φ → − Q 2 + 2 ⁢ G 2 = - N + G
207 63 negcld ⊢ φ → − N ∈ ℂ
208 207 6 addcomd ⊢ φ → - N + G = G + -N
209 6 63 negsubd ⊢ φ → G + -N = G − N
210 208 209 eqtrd ⊢ φ → - N + G = G − N
211 201 206 210 3eqtrd ⊢ φ → - Q + 2 ⁢ G 2 = G − N
212 199 211 eqtrid ⊢ φ → - Q + 2 ⁢ G 2 ⋅ 1 = G − N
213 212 eqeq2d ⊢ φ → U 3 = - Q + 2 ⁢ G 2 ⋅ 1 ↔ U 3 = G − N
214 198 oveq2i ⊢ - Q - 2 ⁢ G 2 ⋅ 1 = - Q - 2 ⁢ G 2
215 204 205 oveq12d ⊢ φ → − Q 2 − 2 ⁢ G 2 = - N - G
216 200 161 172 174 divsubdird ⊢ φ → - Q - 2 ⁢ G 2 = − Q 2 − 2 ⁢ G 2
217 6 63 addcomd ⊢ φ → G + N = N + G
218 217 negeqd ⊢ φ → − G + N = − N + G
219 63 6 negdi2d ⊢ φ → − N + G = - N - G
220 218 219 eqtrd ⊢ φ → − G + N = - N - G
221 215 216 220 3eqtr4d ⊢ φ → - Q - 2 ⁢ G 2 = − G + N
222 214 221 eqtrid ⊢ φ → - Q - 2 ⁢ G 2 ⋅ 1 = − G + N
223 222 eqeq2d ⊢ φ → U 3 = - Q - 2 ⁢ G 2 ⋅ 1 ↔ U 3 = − G + N
224 213 223 orbi12d ⊢ φ → U 3 = - Q + 2 ⁢ G 2 ⋅ 1 ∨ U 3 = - Q - 2 ⁢ G 2 ⋅ 1 ↔ U 3 = G − N ∨ U 3 = − G + N
225 197 224 mpbid ⊢ φ → U 3 = G − N ∨ U 3 = − G + N
226 45 146 225 mpjaodan ⊢ φ → ∃ r ∈ ℂ r 3 = 1 ∧ X = r ⁢ T − M r ⁢ T