Metamath Proof Explorer


Theorem quartlem1

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

Ref Expression
Hypotheses quartlem1.p ⊢ φ → P ∈ ℂ
quartlem1.q ⊢ φ → Q ∈ ℂ
quartlem1.r ⊢ φ → R ∈ ℂ
quartlem1.u ⊢ φ → U = P 2 + 12 ⁢ R
quartlem1.v ⊢ φ → V = − 2 ⁢ P 3 - 27 ⁢ Q 2 + 72 ⁢ P ⁢ R
Assertion quartlem1 ⊢ φ → U = 2 ⁢ P 2 − 3 ⁢ P 2 − 4 ⁢ R ∧ V = 2 ⁢ 2 ⁢ P 3 - 9 ⁢ 2 ⁢ P ⁢ P 2 − 4 ⁢ R + 27 ⁢ − Q 2

Proof

Step Hyp Ref Expression
1 quartlem1.p ⊢ φ → P ∈ ℂ
2 quartlem1.q ⊢ φ → Q ∈ ℂ
3 quartlem1.r ⊢ φ → R ∈ ℂ
4 quartlem1.u ⊢ φ → U = P 2 + 12 ⁢ R
5 quartlem1.v ⊢ φ → V = − 2 ⁢ P 3 - 27 ⁢ Q 2 + 72 ⁢ P ⁢ R
6 2cn ⊢ 2 ∈ ℂ
7 sqmul ⊢ 2 ∈ ℂ ∧ P ∈ ℂ → 2 ⁢ P 2 = 2 2 ⁢ P 2
8 6 1 7 sylancr ⊢ φ → 2 ⁢ P 2 = 2 2 ⁢ P 2
9 sq2 ⊢ 2 2 = 4
10 9 oveq1i ⊢ 2 2 ⁢ P 2 = 4 ⁢ P 2
11 8 10 eqtrdi ⊢ φ → 2 ⁢ P 2 = 4 ⁢ P 2
12 11 oveq1d ⊢ φ → 2 ⁢ P 2 − 3 ⁢ P 2 = 4 ⁢ P 2 − 3 ⁢ P 2
13 4cn ⊢ 4 ∈ ℂ
14 13 a1i ⊢ φ → 4 ∈ ℂ
15 3cn ⊢ 3 ∈ ℂ
16 15 a1i ⊢ φ → 3 ∈ ℂ
17 1 sqcld ⊢ φ → P 2 ∈ ℂ
18 14 16 17 subdird ⊢ φ → 4 − 3 ⁢ P 2 = 4 ⁢ P 2 − 3 ⁢ P 2
19 12 18 eqtr4d ⊢ φ → 2 ⁢ P 2 − 3 ⁢ P 2 = 4 − 3 ⁢ P 2
20 ax-1cn ⊢ 1 ∈ ℂ
21 3p1e4 ⊢ 3 + 1 = 4
22 13 15 20 21 subaddrii ⊢ 4 − 3 = 1
23 22 oveq1i ⊢ 4 − 3 ⁢ P 2 = 1 ⁢ P 2
24 mullid ⊢ P 2 ∈ ℂ → 1 ⁢ P 2 = P 2
25 23 24 eqtrid ⊢ P 2 ∈ ℂ → 4 − 3 ⁢ P 2 = P 2
26 17 25 syl ⊢ φ → 4 − 3 ⁢ P 2 = P 2
27 19 26 eqtr2d ⊢ φ → P 2 = 2 ⁢ P 2 − 3 ⁢ P 2
28 27 oveq1d ⊢ φ → P 2 + 12 ⁢ R = 2 ⁢ P 2 - 3 ⁢ P 2 + 12 ⁢ R
29 mulcl ⊢ 2 ∈ ℂ ∧ P ∈ ℂ → 2 ⁢ P ∈ ℂ
30 6 1 29 sylancr ⊢ φ → 2 ⁢ P ∈ ℂ
31 30 sqcld ⊢ φ → 2 ⁢ P 2 ∈ ℂ
32 mulcl ⊢ 3 ∈ ℂ ∧ P 2 ∈ ℂ → 3 ⁢ P 2 ∈ ℂ
33 15 17 32 sylancr ⊢ φ → 3 ⁢ P 2 ∈ ℂ
34 1nn0 ⊢ 1 ∈ ℕ 0
35 2nn ⊢ 2 ∈ ℕ
36 34 35 decnncl ⊢ 12 ∈ ℕ
37 36 nncni ⊢ 12 ∈ ℂ
38 mulcl ⊢ 12 ∈ ℂ ∧ R ∈ ℂ → 12 ⁢ R ∈ ℂ
39 37 3 38 sylancr ⊢ φ → 12 ⁢ R ∈ ℂ
40 31 33 39 subsubd ⊢ φ → 2 ⁢ P 2 − 3 ⁢ P 2 − 12 ⁢ R = 2 ⁢ P 2 - 3 ⁢ P 2 + 12 ⁢ R
41 28 40 eqtr4d ⊢ φ → P 2 + 12 ⁢ R = 2 ⁢ P 2 − 3 ⁢ P 2 − 12 ⁢ R
42 mulcl ⊢ 4 ∈ ℂ ∧ R ∈ ℂ → 4 ⁢ R ∈ ℂ
43 13 3 42 sylancr ⊢ φ → 4 ⁢ R ∈ ℂ
44 16 17 43 subdid ⊢ φ → 3 ⁢ P 2 − 4 ⁢ R = 3 ⁢ P 2 − 3 ⁢ 4 ⁢ R
45 4t3e12 ⊢ 4 ⋅ 3 = 12
46 13 15 45 mulcomli ⊢ 3 ⋅ 4 = 12
47 46 oveq1i ⊢ 3 ⋅ 4 ⁢ R = 12 ⁢ R
48 16 14 3 mulassd ⊢ φ → 3 ⋅ 4 ⁢ R = 3 ⁢ 4 ⁢ R
49 47 48 eqtr3id ⊢ φ → 12 ⁢ R = 3 ⁢ 4 ⁢ R
50 49 oveq2d ⊢ φ → 3 ⁢ P 2 − 12 ⁢ R = 3 ⁢ P 2 − 3 ⁢ 4 ⁢ R
51 44 50 eqtr4d ⊢ φ → 3 ⁢ P 2 − 4 ⁢ R = 3 ⁢ P 2 − 12 ⁢ R
52 51 oveq2d ⊢ φ → 2 ⁢ P 2 − 3 ⁢ P 2 − 4 ⁢ R = 2 ⁢ P 2 − 3 ⁢ P 2 − 12 ⁢ R
53 41 4 52 3eqtr4d ⊢ φ → U = 2 ⁢ P 2 − 3 ⁢ P 2 − 4 ⁢ R
54 6 a1i ⊢ φ → 2 ∈ ℂ
55 3nn0 ⊢ 3 ∈ ℕ 0
56 55 a1i ⊢ φ → 3 ∈ ℕ 0
57 54 1 56 mulexpd ⊢ φ → 2 ⁢ P 3 = 2 3 ⁢ P 3
58 cu2 ⊢ 2 3 = 8
59 58 oveq1i ⊢ 2 3 ⁢ P 3 = 8 ⁢ P 3
60 57 59 eqtrdi ⊢ φ → 2 ⁢ P 3 = 8 ⁢ P 3
61 60 oveq2d ⊢ φ → 2 ⁢ 2 ⁢ P 3 = 2 ⁢ 8 ⁢ P 3
62 8cn ⊢ 8 ∈ ℂ
63 62 a1i ⊢ φ → 8 ∈ ℂ
64 expcl ⊢ P ∈ ℂ ∧ 3 ∈ ℕ 0 → P 3 ∈ ℂ
65 1 55 64 sylancl ⊢ φ → P 3 ∈ ℂ
66 54 63 65 mul12d ⊢ φ → 2 ⁢ 8 ⁢ P 3 = 8 ⁢ 2 ⁢ P 3
67 61 66 eqtrd ⊢ φ → 2 ⁢ 2 ⁢ P 3 = 8 ⁢ 2 ⁢ P 3
68 9cn ⊢ 9 ∈ ℂ
69 68 a1i ⊢ φ → 9 ∈ ℂ
70 mulcl ⊢ 2 ∈ ℂ ∧ P 3 ∈ ℂ → 2 ⁢ P 3 ∈ ℂ
71 6 65 70 sylancr ⊢ φ → 2 ⁢ P 3 ∈ ℂ
72 1 3 mulcld ⊢ φ → P ⁢ R ∈ ℂ
73 mulcl ⊢ 8 ∈ ℂ ∧ P ⁢ R ∈ ℂ → 8 ⁢ P ⁢ R ∈ ℂ
74 62 72 73 sylancr ⊢ φ → 8 ⁢ P ⁢ R ∈ ℂ
75 69 71 74 subdid ⊢ φ → 9 ⁢ 2 ⁢ P 3 − 8 ⁢ P ⁢ R = 9 ⁢ 2 ⁢ P 3 − 9 ⁢ 8 ⁢ P ⁢ R
76 30 17 43 subdid ⊢ φ → 2 ⁢ P ⁢ P 2 − 4 ⁢ R = 2 ⁢ P ⁢ P 2 − 2 ⁢ P ⁢ 4 ⁢ R
77 54 1 17 mulassd ⊢ φ → 2 ⁢ P ⁢ P 2 = 2 ⁢ P ⁢ P 2
78 1 17 mulcomd ⊢ φ → P ⁢ P 2 = P 2 ⁢ P
79 df-3 ⊢ 3 = 2 + 1
80 79 oveq2i ⊢ P 3 = P 2 + 1
81 2nn0 ⊢ 2 ∈ ℕ 0
82 expp1 ⊢ P ∈ ℂ ∧ 2 ∈ ℕ 0 → P 2 + 1 = P 2 ⁢ P
83 1 81 82 sylancl ⊢ φ → P 2 + 1 = P 2 ⁢ P
84 80 83 eqtrid ⊢ φ → P 3 = P 2 ⁢ P
85 78 84 eqtr4d ⊢ φ → P ⁢ P 2 = P 3
86 85 oveq2d ⊢ φ → 2 ⁢ P ⁢ P 2 = 2 ⁢ P 3
87 77 86 eqtrd ⊢ φ → 2 ⁢ P ⁢ P 2 = 2 ⁢ P 3
88 54 1 14 3 mul4d ⊢ φ → 2 ⁢ P ⁢ 4 ⁢ R = 2 ⋅ 4 ⁢ P ⁢ R
89 2t4e8 ⊢ 2 ⋅ 4 = 8
90 89 oveq1i ⊢ 2 ⋅ 4 ⁢ P ⁢ R = 8 ⁢ P ⁢ R
91 88 90 eqtrdi ⊢ φ → 2 ⁢ P ⁢ 4 ⁢ R = 8 ⁢ P ⁢ R
92 87 91 oveq12d ⊢ φ → 2 ⁢ P ⁢ P 2 − 2 ⁢ P ⁢ 4 ⁢ R = 2 ⁢ P 3 − 8 ⁢ P ⁢ R
93 76 92 eqtrd ⊢ φ → 2 ⁢ P ⁢ P 2 − 4 ⁢ R = 2 ⁢ P 3 − 8 ⁢ P ⁢ R
94 93 oveq2d ⊢ φ → 9 ⁢ 2 ⁢ P ⁢ P 2 − 4 ⁢ R = 9 ⁢ 2 ⁢ P 3 − 8 ⁢ P ⁢ R
95 9t8e72 ⊢ 9 ⋅ 8 = 72
96 95 oveq1i ⊢ 9 ⋅ 8 ⁢ P ⁢ R = 72 ⁢ P ⁢ R
97 69 63 72 mulassd ⊢ φ → 9 ⋅ 8 ⁢ P ⁢ R = 9 ⁢ 8 ⁢ P ⁢ R
98 96 97 eqtr3id ⊢ φ → 72 ⁢ P ⁢ R = 9 ⁢ 8 ⁢ P ⁢ R
99 98 oveq2d ⊢ φ → 9 ⁢ 2 ⁢ P 3 − 72 ⁢ P ⁢ R = 9 ⁢ 2 ⁢ P 3 − 9 ⁢ 8 ⁢ P ⁢ R
100 75 94 99 3eqtr4d ⊢ φ → 9 ⁢ 2 ⁢ P ⁢ P 2 − 4 ⁢ R = 9 ⁢ 2 ⁢ P 3 − 72 ⁢ P ⁢ R
101 67 100 oveq12d ⊢ φ → 2 ⁢ 2 ⁢ P 3 − 9 ⁢ 2 ⁢ P ⁢ P 2 − 4 ⁢ R = 8 ⁢ 2 ⁢ P 3 − 9 ⁢ 2 ⁢ P 3 − 72 ⁢ P ⁢ R
102 mulcl ⊢ 8 ∈ ℂ ∧ 2 ⁢ P 3 ∈ ℂ → 8 ⁢ 2 ⁢ P 3 ∈ ℂ
103 62 71 102 sylancr ⊢ φ → 8 ⁢ 2 ⁢ P 3 ∈ ℂ
104 mulcl ⊢ 9 ∈ ℂ ∧ 2 ⁢ P 3 ∈ ℂ → 9 ⁢ 2 ⁢ P 3 ∈ ℂ
105 68 71 104 sylancr ⊢ φ → 9 ⁢ 2 ⁢ P 3 ∈ ℂ
106 7nn0 ⊢ 7 ∈ ℕ 0
107 106 35 decnncl ⊢ 72 ∈ ℕ
108 107 nncni ⊢ 72 ∈ ℂ
109 mulcl ⊢ 72 ∈ ℂ ∧ P ⁢ R ∈ ℂ → 72 ⁢ P ⁢ R ∈ ℂ
110 108 72 109 sylancr ⊢ φ → 72 ⁢ P ⁢ R ∈ ℂ
111 103 105 110 subsubd ⊢ φ → 8 ⁢ 2 ⁢ P 3 − 9 ⁢ 2 ⁢ P 3 − 72 ⁢ P ⁢ R = 8 ⁢ 2 ⁢ P 3 - 9 ⁢ 2 ⁢ P 3 + 72 ⁢ P ⁢ R
112 105 103 negsubdi2d ⊢ φ → − 9 ⁢ 2 ⁢ P 3 − 8 ⁢ 2 ⁢ P 3 = 8 ⁢ 2 ⁢ P 3 − 9 ⁢ 2 ⁢ P 3
113 69 63 71 subdird ⊢ φ → 9 − 8 ⁢ 2 ⁢ P 3 = 9 ⁢ 2 ⁢ P 3 − 8 ⁢ 2 ⁢ P 3
114 8p1e9 ⊢ 8 + 1 = 9
115 68 62 20 114 subaddrii ⊢ 9 − 8 = 1
116 115 oveq1i ⊢ 9 − 8 ⁢ 2 ⁢ P 3 = 1 ⁢ 2 ⁢ P 3
117 71 mullidd ⊢ φ → 1 ⁢ 2 ⁢ P 3 = 2 ⁢ P 3
118 116 117 eqtrid ⊢ φ → 9 − 8 ⁢ 2 ⁢ P 3 = 2 ⁢ P 3
119 113 118 eqtr3d ⊢ φ → 9 ⁢ 2 ⁢ P 3 − 8 ⁢ 2 ⁢ P 3 = 2 ⁢ P 3
120 119 negeqd ⊢ φ → − 9 ⁢ 2 ⁢ P 3 − 8 ⁢ 2 ⁢ P 3 = − 2 ⁢ P 3
121 112 120 eqtr3d ⊢ φ → 8 ⁢ 2 ⁢ P 3 − 9 ⁢ 2 ⁢ P 3 = − 2 ⁢ P 3
122 121 oveq1d ⊢ φ → 8 ⁢ 2 ⁢ P 3 - 9 ⁢ 2 ⁢ P 3 + 72 ⁢ P ⁢ R = - 2 ⁢ P 3 + 72 ⁢ P ⁢ R
123 101 111 122 3eqtrd ⊢ φ → 2 ⁢ 2 ⁢ P 3 − 9 ⁢ 2 ⁢ P ⁢ P 2 − 4 ⁢ R = - 2 ⁢ P 3 + 72 ⁢ P ⁢ R
124 7nn ⊢ 7 ∈ ℕ
125 81 124 decnncl ⊢ 27 ∈ ℕ
126 125 nncni ⊢ 27 ∈ ℂ
127 2 sqcld ⊢ φ → Q 2 ∈ ℂ
128 mulneg2 ⊢ 27 ∈ ℂ ∧ Q 2 ∈ ℂ → 27 ⁢ − Q 2 = − 27 ⁢ Q 2
129 126 127 128 sylancr ⊢ φ → 27 ⁢ − Q 2 = − 27 ⁢ Q 2
130 123 129 oveq12d ⊢ φ → 2 ⁢ 2 ⁢ P 3 - 9 ⁢ 2 ⁢ P ⁢ P 2 − 4 ⁢ R + 27 ⁢ − Q 2 = − 2 ⁢ P 3 + 72 ⁢ P ⁢ R + − 27 ⁢ Q 2
131 71 negcld ⊢ φ → − 2 ⁢ P 3 ∈ ℂ
132 mulcl ⊢ 27 ∈ ℂ ∧ Q 2 ∈ ℂ → 27 ⁢ Q 2 ∈ ℂ
133 126 127 132 sylancr ⊢ φ → 27 ⁢ Q 2 ∈ ℂ
134 131 110 133 addsubd ⊢ φ → − 2 ⁢ P 3 + 72 ⁢ P ⁢ R - 27 ⁢ Q 2 = − 2 ⁢ P 3 - 27 ⁢ Q 2 + 72 ⁢ P ⁢ R
135 131 110 addcld ⊢ φ → - 2 ⁢ P 3 + 72 ⁢ P ⁢ R ∈ ℂ
136 135 133 negsubd ⊢ φ → − 2 ⁢ P 3 + 72 ⁢ P ⁢ R + − 27 ⁢ Q 2 = − 2 ⁢ P 3 + 72 ⁢ P ⁢ R - 27 ⁢ Q 2
137 134 136 5 3eqtr4d ⊢ φ → − 2 ⁢ P 3 + 72 ⁢ P ⁢ R + − 27 ⁢ Q 2 = V
138 130 137 eqtr2d ⊢ φ → V = 2 ⁢ 2 ⁢ P 3 - 9 ⁢ 2 ⁢ P ⁢ P 2 − 4 ⁢ R + 27 ⁢ − Q 2
139 53 138 jca ⊢ φ → U = 2 ⁢ P 2 − 3 ⁢ P 2 − 4 ⁢ R ∧ V = 2 ⁢ 2 ⁢ P 3 - 9 ⁢ 2 ⁢ P ⁢ P 2 − 4 ⁢ R + 27 ⁢ − Q 2