Metamath Proof Explorer


Theorem dquartlem1

Description: Lemma for dquart . (Contributed by Mario Carneiro, 6-May-2015)

Ref Expression
Hypotheses dquart.b ⊢ φ → B ∈ ℂ
dquart.c ⊢ φ → C ∈ ℂ
dquart.x ⊢ φ → X ∈ ℂ
dquart.s ⊢ φ → S ∈ ℂ
dquart.m ⊢ φ → M = 2 ⁢ S 2
dquart.m0 ⊢ φ → M ≠ 0
dquart.i ⊢ φ → I ∈ ℂ
dquart.i2 ⊢ φ → I 2 = − S 2 - B 2 + C 4 S
Assertion dquartlem1 ⊢ φ → X 2 + M + B 2 + M 2 ⁢ X − C 4 S = 0 ↔ X = - S + I ∨ X = - S - I

Proof

Step Hyp Ref Expression
1 dquart.b ⊢ φ → B ∈ ℂ
2 dquart.c ⊢ φ → C ∈ ℂ
3 dquart.x ⊢ φ → X ∈ ℂ
4 dquart.s ⊢ φ → S ∈ ℂ
5 dquart.m ⊢ φ → M = 2 ⁢ S 2
6 dquart.m0 ⊢ φ → M ≠ 0
7 dquart.i ⊢ φ → I ∈ ℂ
8 dquart.i2 ⊢ φ → I 2 = − S 2 - B 2 + C 4 S
9 3 sqcld ⊢ φ → X 2 ∈ ℂ
10 2cn ⊢ 2 ∈ ℂ
11 mulcl ⊢ 2 ∈ ℂ ∧ S ∈ ℂ → 2 ⁢ S ∈ ℂ
12 10 4 11 sylancr ⊢ φ → 2 ⁢ S ∈ ℂ
13 12 sqcld ⊢ φ → 2 ⁢ S 2 ∈ ℂ
14 5 13 eqeltrd ⊢ φ → M ∈ ℂ
15 14 1 addcld ⊢ φ → M + B ∈ ℂ
16 15 halfcld ⊢ φ → M + B 2 ∈ ℂ
17 9 16 addcld ⊢ φ → X 2 + M + B 2 ∈ ℂ
18 14 halfcld ⊢ φ → M 2 ∈ ℂ
19 18 3 mulcld ⊢ φ → M 2 ⁢ X ∈ ℂ
20 4cn ⊢ 4 ∈ ℂ
21 20 a1i ⊢ φ → 4 ∈ ℂ
22 4ne0 ⊢ 4 ≠ 0
23 22 a1i ⊢ φ → 4 ≠ 0
24 2 21 23 divcld ⊢ φ → C 4 ∈ ℂ
25 19 24 subcld ⊢ φ → M 2 ⁢ X − C 4 ∈ ℂ
26 5 6 eqnetrrd ⊢ φ → 2 ⁢ S 2 ≠ 0
27 sqne0 ⊢ 2 ⁢ S ∈ ℂ → 2 ⁢ S 2 ≠ 0 ↔ 2 ⁢ S ≠ 0
28 12 27 syl ⊢ φ → 2 ⁢ S 2 ≠ 0 ↔ 2 ⁢ S ≠ 0
29 26 28 mpbid ⊢ φ → 2 ⁢ S ≠ 0
30 mulne0b ⊢ 2 ∈ ℂ ∧ S ∈ ℂ → 2 ≠ 0 ∧ S ≠ 0 ↔ 2 ⁢ S ≠ 0
31 10 4 30 sylancr ⊢ φ → 2 ≠ 0 ∧ S ≠ 0 ↔ 2 ⁢ S ≠ 0
32 29 31 mpbird ⊢ φ → 2 ≠ 0 ∧ S ≠ 0
33 32 simprd ⊢ φ → S ≠ 0
34 25 4 33 divcld ⊢ φ → M 2 ⁢ X − C 4 S ∈ ℂ
35 17 34 addcld ⊢ φ → X 2 + M + B 2 + M 2 ⁢ X − C 4 S ∈ ℂ
36 10 a1i ⊢ φ → 2 ∈ ℂ
37 2ne0 ⊢ 2 ≠ 0
38 37 a1i ⊢ φ → 2 ≠ 0
39 35 36 38 diveq0ad ⊢ φ → X 2 + M + B 2 + M 2 ⁢ X − C 4 S 2 = 0 ↔ X 2 + M + B 2 + M 2 ⁢ X − C 4 S = 0
40 9 16 34 addassd ⊢ φ → X 2 + M + B 2 + M 2 ⁢ X − C 4 S = X 2 + M + B 2 + M 2 ⁢ X − C 4 S
41 40 oveq1d ⊢ φ → X 2 + M + B 2 + M 2 ⁢ X − C 4 S 2 = X 2 + M + B 2 + M 2 ⁢ X − C 4 S 2
42 16 34 addcld ⊢ φ → M + B 2 + M 2 ⁢ X − C 4 S ∈ ℂ
43 9 42 36 38 divdird ⊢ φ → X 2 + M + B 2 + M 2 ⁢ X − C 4 S 2 = X 2 2 + M + B 2 + M 2 ⁢ X − C 4 S 2
44 9 36 38 divrec2d ⊢ φ → X 2 2 = 1 2 ⁢ X 2
45 19 24 4 33 divsubdird ⊢ φ → M 2 ⁢ X − C 4 S = M 2 ⁢ X S − C 4 S
46 18 3 4 33 div23d ⊢ φ → M 2 ⁢ X S = M 2 S ⁢ X
47 4 sqvald ⊢ φ → S 2 = S ⁢ S
48 47 oveq2d ⊢ φ → 2 ⁢ S 2 = 2 ⁢ S ⁢ S
49 sqmul ⊢ 2 ∈ ℂ ∧ S ∈ ℂ → 2 ⁢ S 2 = 2 2 ⁢ S 2
50 10 4 49 sylancr ⊢ φ → 2 ⁢ S 2 = 2 2 ⁢ S 2
51 10 sqvali ⊢ 2 2 = 2 ⋅ 2
52 51 oveq1i ⊢ 2 2 ⁢ S 2 = 2 ⋅ 2 ⁢ S 2
53 50 52 eqtrdi ⊢ φ → 2 ⁢ S 2 = 2 ⋅ 2 ⁢ S 2
54 4 sqcld ⊢ φ → S 2 ∈ ℂ
55 36 36 54 mulassd ⊢ φ → 2 ⋅ 2 ⁢ S 2 = 2 ⁢ 2 ⁢ S 2
56 5 53 55 3eqtrd ⊢ φ → M = 2 ⁢ 2 ⁢ S 2
57 56 oveq1d ⊢ φ → M 2 = 2 ⁢ 2 ⁢ S 2 2
58 mulcl ⊢ 2 ∈ ℂ ∧ S 2 ∈ ℂ → 2 ⁢ S 2 ∈ ℂ
59 10 54 58 sylancr ⊢ φ → 2 ⁢ S 2 ∈ ℂ
60 59 36 38 divcan3d ⊢ φ → 2 ⁢ 2 ⁢ S 2 2 = 2 ⁢ S 2
61 57 60 eqtrd ⊢ φ → M 2 = 2 ⁢ S 2
62 36 4 4 mulassd ⊢ φ → 2 ⁢ S ⁢ S = 2 ⁢ S ⁢ S
63 48 61 62 3eqtr4d ⊢ φ → M 2 = 2 ⁢ S ⁢ S
64 63 oveq1d ⊢ φ → M 2 S = 2 ⁢ S ⁢ S S
65 12 4 33 divcan4d ⊢ φ → 2 ⁢ S ⁢ S S = 2 ⁢ S
66 64 65 eqtrd ⊢ φ → M 2 S = 2 ⁢ S
67 66 oveq1d ⊢ φ → M 2 S ⁢ X = 2 ⁢ S ⁢ X
68 46 67 eqtrd ⊢ φ → M 2 ⁢ X S = 2 ⁢ S ⁢ X
69 68 oveq1d ⊢ φ → M 2 ⁢ X S − C 4 S = 2 ⁢ S ⁢ X − C 4 S
70 45 69 eqtrd ⊢ φ → M 2 ⁢ X − C 4 S = 2 ⁢ S ⁢ X − C 4 S
71 70 oveq2d ⊢ φ → M + B 2 + M 2 ⁢ X − C 4 S = M + B 2 + 2 ⁢ S ⁢ X - C 4 S
72 12 3 mulcld ⊢ φ → 2 ⁢ S ⁢ X ∈ ℂ
73 24 4 33 divcld ⊢ φ → C 4 S ∈ ℂ
74 16 72 73 addsub12d ⊢ φ → M + B 2 + 2 ⁢ S ⁢ X - C 4 S = 2 ⁢ S ⁢ X + M + B 2 - C 4 S
75 71 74 eqtrd ⊢ φ → M + B 2 + M 2 ⁢ X − C 4 S = 2 ⁢ S ⁢ X + M + B 2 - C 4 S
76 75 oveq1d ⊢ φ → M + B 2 + M 2 ⁢ X − C 4 S 2 = 2 ⁢ S ⁢ X + M + B 2 - C 4 S 2
77 16 73 subcld ⊢ φ → M + B 2 − C 4 S ∈ ℂ
78 72 77 36 38 divdird ⊢ φ → 2 ⁢ S ⁢ X + M + B 2 - C 4 S 2 = 2 ⁢ S ⁢ X 2 + M + B 2 − C 4 S 2
79 36 4 3 mulassd ⊢ φ → 2 ⁢ S ⁢ X = 2 ⁢ S ⁢ X
80 79 oveq1d ⊢ φ → 2 ⁢ S ⁢ X 2 = 2 ⁢ S ⁢ X 2
81 4 3 mulcld ⊢ φ → S ⁢ X ∈ ℂ
82 81 36 38 divcan3d ⊢ φ → 2 ⁢ S ⁢ X 2 = S ⁢ X
83 80 82 eqtrd ⊢ φ → 2 ⁢ S ⁢ X 2 = S ⁢ X
84 54 negcld ⊢ φ → − S 2 ∈ ℂ
85 1 halfcld ⊢ φ → B 2 ∈ ℂ
86 84 85 subcld ⊢ φ → - S 2 - B 2 ∈ ℂ
87 54 86 73 subsub4d ⊢ φ → S 2 - - S 2 - B 2 - C 4 S = S 2 − − S 2 - B 2 + C 4 S
88 14 1 36 38 divdird ⊢ φ → M + B 2 = M 2 + B 2
89 54 2timesd ⊢ φ → 2 ⁢ S 2 = S 2 + S 2
90 61 89 eqtrd ⊢ φ → M 2 = S 2 + S 2
91 90 oveq1d ⊢ φ → M 2 + B 2 = S 2 + S 2 + B 2
92 88 91 eqtrd ⊢ φ → M + B 2 = S 2 + S 2 + B 2
93 54 54 85 addassd ⊢ φ → S 2 + S 2 + B 2 = S 2 + S 2 + B 2
94 54 85 addcld ⊢ φ → S 2 + B 2 ∈ ℂ
95 54 94 subnegd ⊢ φ → S 2 − − S 2 + B 2 = S 2 + S 2 + B 2
96 54 85 negdi2d ⊢ φ → − S 2 + B 2 = - S 2 - B 2
97 96 oveq2d ⊢ φ → S 2 − − S 2 + B 2 = S 2 − - S 2 - B 2
98 95 97 eqtr3d ⊢ φ → S 2 + S 2 + B 2 = S 2 − - S 2 - B 2
99 92 93 98 3eqtrd ⊢ φ → M + B 2 = S 2 − - S 2 - B 2
100 99 oveq1d ⊢ φ → M + B 2 − C 4 S = S 2 - - S 2 - B 2 - C 4 S
101 8 oveq2d ⊢ φ → S 2 − I 2 = S 2 − − S 2 - B 2 + C 4 S
102 87 100 101 3eqtr4d ⊢ φ → M + B 2 − C 4 S = S 2 − I 2
103 102 oveq1d ⊢ φ → M + B 2 − C 4 S 2 = S 2 − I 2 2
104 83 103 oveq12d ⊢ φ → 2 ⁢ S ⁢ X 2 + M + B 2 − C 4 S 2 = S ⁢ X + S 2 − I 2 2
105 76 78 104 3eqtrd ⊢ φ → M + B 2 + M 2 ⁢ X − C 4 S 2 = S ⁢ X + S 2 − I 2 2
106 44 105 oveq12d ⊢ φ → X 2 2 + M + B 2 + M 2 ⁢ X − C 4 S 2 = 1 2 ⁢ X 2 + S ⁢ X + S 2 − I 2 2
107 41 43 106 3eqtrd ⊢ φ → X 2 + M + B 2 + M 2 ⁢ X − C 4 S 2 = 1 2 ⁢ X 2 + S ⁢ X + S 2 − I 2 2
108 107 eqeq1d ⊢ φ → X 2 + M + B 2 + M 2 ⁢ X − C 4 S 2 = 0 ↔ 1 2 ⁢ X 2 + S ⁢ X + S 2 − I 2 2 = 0
109 39 108 bitr3d ⊢ φ → X 2 + M + B 2 + M 2 ⁢ X − C 4 S = 0 ↔ 1 2 ⁢ X 2 + S ⁢ X + S 2 − I 2 2 = 0
110 ax-1cn ⊢ 1 ∈ ℂ
111 halfcl ⊢ 1 ∈ ℂ → 1 2 ∈ ℂ
112 110 111 mp1i ⊢ φ → 1 2 ∈ ℂ
113 ax-1ne0 ⊢ 1 ≠ 0
114 110 10 113 37 divne0i ⊢ 1 2 ≠ 0
115 114 a1i ⊢ φ → 1 2 ≠ 0
116 7 sqcld ⊢ φ → I 2 ∈ ℂ
117 54 116 subcld ⊢ φ → S 2 − I 2 ∈ ℂ
118 117 halfcld ⊢ φ → S 2 − I 2 2 ∈ ℂ
119 110 a1i ⊢ φ → 1 ∈ ℂ
120 2cnne0 ⊢ 2 ∈ ℂ ∧ 2 ≠ 0
121 120 a1i ⊢ φ → 2 ∈ ℂ ∧ 2 ≠ 0
122 divmuldiv ⊢ 1 ∈ ℂ ∧ S 2 − I 2 ∈ ℂ ∧ 2 ∈ ℂ ∧ 2 ≠ 0 ∧ 2 ∈ ℂ ∧ 2 ≠ 0 → 1 2 ⁢ S 2 − I 2 2 = 1 ⁢ S 2 − I 2 2 ⋅ 2
123 119 117 121 121 122 syl22anc ⊢ φ → 1 2 ⁢ S 2 − I 2 2 = 1 ⁢ S 2 − I 2 2 ⋅ 2
124 117 mullidd ⊢ φ → 1 ⁢ S 2 − I 2 = S 2 − I 2
125 2t2e4 ⊢ 2 ⋅ 2 = 4
126 125 a1i ⊢ φ → 2 ⋅ 2 = 4
127 124 126 oveq12d ⊢ φ → 1 ⁢ S 2 − I 2 2 ⋅ 2 = S 2 − I 2 4
128 123 127 eqtrd ⊢ φ → 1 2 ⁢ S 2 − I 2 2 = S 2 − I 2 4
129 128 oveq2d ⊢ φ → 4 ⁢ 1 2 ⁢ S 2 − I 2 2 = 4 ⁢ S 2 − I 2 4
130 117 21 23 divcan2d ⊢ φ → 4 ⁢ S 2 − I 2 4 = S 2 − I 2
131 129 130 eqtrd ⊢ φ → 4 ⁢ 1 2 ⁢ S 2 − I 2 2 = S 2 − I 2
132 131 oveq2d ⊢ φ → S 2 − 4 ⁢ 1 2 ⁢ S 2 − I 2 2 = S 2 − S 2 − I 2
133 54 116 nncand ⊢ φ → S 2 − S 2 − I 2 = I 2
134 132 133 eqtr2d ⊢ φ → I 2 = S 2 − 4 ⁢ 1 2 ⁢ S 2 − I 2 2
135 112 115 4 118 3 7 134 quad2 ⊢ φ → 1 2 ⁢ X 2 + S ⁢ X + S 2 − I 2 2 = 0 ↔ X = - S + I 2 ⁢ 1 2 ∨ X = - S - I 2 ⁢ 1 2
136 2thalfe1 ⊢ 2 ⁢ 1 2 = 1
137 136 oveq2i ⊢ - S + I 2 ⁢ 1 2 = - S + I 1
138 4 negcld ⊢ φ → − S ∈ ℂ
139 138 7 addcld ⊢ φ → - S + I ∈ ℂ
140 139 div1d ⊢ φ → - S + I 1 = - S + I
141 137 140 eqtrid ⊢ φ → - S + I 2 ⁢ 1 2 = - S + I
142 141 eqeq2d ⊢ φ → X = - S + I 2 ⁢ 1 2 ↔ X = - S + I
143 136 oveq2i ⊢ - S - I 2 ⁢ 1 2 = - S - I 1
144 138 7 subcld ⊢ φ → - S - I ∈ ℂ
145 144 div1d ⊢ φ → - S - I 1 = - S - I
146 143 145 eqtrid ⊢ φ → - S - I 2 ⁢ 1 2 = - S - I
147 146 eqeq2d ⊢ φ → X = - S - I 2 ⁢ 1 2 ↔ X = - S - I
148 142 147 orbi12d ⊢ φ → X = - S + I 2 ⁢ 1 2 ∨ X = - S - I 2 ⁢ 1 2 ↔ X = - S + I ∨ X = - S - I
149 109 135 148 3bitrd ⊢ φ → X 2 + M + B 2 + M 2 ⁢ X − C 4 S = 0 ↔ X = - S + I ∨ X = - S - I