Metamath Proof Explorer


Theorem dquartlem2

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
dquart.d ⊢ φ → D ∈ ℂ
dquart.3 ⊢ φ → M 3 + 2 ⁢ B ⁢ M 2 + B 2 − 4 ⁢ D ⋅ M + − C 2 = 0
Assertion dquartlem2 ⊢ φ → M + B 2 2 − C 2 4 M = D

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 dquart.d ⊢ φ → D ∈ ℂ
10 dquart.3 ⊢ φ → M 3 + 2 ⁢ B ⁢ M 2 + B 2 − 4 ⁢ D ⋅ M + − C 2 = 0
11 2cn ⊢ 2 ∈ ℂ
12 mulcl ⊢ 2 ∈ ℂ ∧ S ∈ ℂ → 2 ⁢ S ∈ ℂ
13 11 4 12 sylancr ⊢ φ → 2 ⁢ S ∈ ℂ
14 13 sqcld ⊢ φ → 2 ⁢ S 2 ∈ ℂ
15 5 14 eqeltrd ⊢ φ → M ∈ ℂ
16 15 1 addcld ⊢ φ → M + B ∈ ℂ
17 11 a1i ⊢ φ → 2 ∈ ℂ
18 2ne0 ⊢ 2 ≠ 0
19 18 a1i ⊢ φ → 2 ≠ 0
20 16 17 19 sqdivd ⊢ φ → M + B 2 2 = M + B 2 2 2
21 sq2 ⊢ 2 2 = 4
22 21 oveq2i ⊢ M + B 2 2 2 = M + B 2 4
23 20 22 eqtrdi ⊢ φ → M + B 2 2 = M + B 2 4
24 23 oveq1d ⊢ φ → M + B 2 2 − C 2 4 M = M + B 2 4 − C 2 4 M
25 16 sqcld ⊢ φ → M + B 2 ∈ ℂ
26 4cn ⊢ 4 ∈ ℂ
27 26 a1i ⊢ φ → 4 ∈ ℂ
28 4ne0 ⊢ 4 ≠ 0
29 28 a1i ⊢ φ → 4 ≠ 0
30 25 27 29 divcld ⊢ φ → M + B 2 4 ∈ ℂ
31 2 sqcld ⊢ φ → C 2 ∈ ℂ
32 31 27 29 divcld ⊢ φ → C 2 4 ∈ ℂ
33 32 15 6 divcld ⊢ φ → C 2 4 M ∈ ℂ
34 30 33 subcld ⊢ φ → M + B 2 4 − C 2 4 M ∈ ℂ
35 30 33 15 subdird ⊢ φ → M + B 2 4 − C 2 4 M ⋅ M = M + B 2 4 ⋅ M − C 2 4 M ⋅ M
36 25 15 27 29 div23d ⊢ φ → M + B 2 ⋅ M 4 = M + B 2 4 ⋅ M
37 36 eqcomd ⊢ φ → M + B 2 4 ⋅ M = M + B 2 ⋅ M 4
38 32 15 6 divcan1d ⊢ φ → C 2 4 M ⋅ M = C 2 4
39 37 38 oveq12d ⊢ φ → M + B 2 4 ⋅ M − C 2 4 M ⋅ M = M + B 2 ⋅ M 4 − C 2 4
40 binom2 ⊢ M ∈ ℂ ∧ B ∈ ℂ → M + B 2 = M 2 + 2 ⁢ M ⁢ B + B 2
41 15 1 40 syl2anc ⊢ φ → M + B 2 = M 2 + 2 ⁢ M ⁢ B + B 2
42 41 oveq1d ⊢ φ → M + B 2 ⋅ M = M 2 + 2 ⁢ M ⁢ B + B 2 ⋅ M
43 15 sqcld ⊢ φ → M 2 ∈ ℂ
44 15 1 mulcld ⊢ φ → M ⁢ B ∈ ℂ
45 mulcl ⊢ 2 ∈ ℂ ∧ M ⁢ B ∈ ℂ → 2 ⁢ M ⁢ B ∈ ℂ
46 11 44 45 sylancr ⊢ φ → 2 ⁢ M ⁢ B ∈ ℂ
47 43 46 addcld ⊢ φ → M 2 + 2 ⁢ M ⁢ B ∈ ℂ
48 1 sqcld ⊢ φ → B 2 ∈ ℂ
49 47 48 15 adddird ⊢ φ → M 2 + 2 ⁢ M ⁢ B + B 2 ⋅ M = M 2 + 2 ⁢ M ⁢ B ⋅ M + B 2 ⋅ M
50 43 46 15 adddird ⊢ φ → M 2 + 2 ⁢ M ⁢ B ⋅ M = M 2 ⋅ M + 2 ⁢ M ⁢ B ⋅ M
51 df-3 ⊢ 3 = 2 + 1
52 51 oveq2i ⊢ M 3 = M 2 + 1
53 2nn0 ⊢ 2 ∈ ℕ 0
54 expp1 ⊢ M ∈ ℂ ∧ 2 ∈ ℕ 0 → M 2 + 1 = M 2 ⋅ M
55 15 53 54 sylancl ⊢ φ → M 2 + 1 = M 2 ⋅ M
56 52 55 eqtr2id ⊢ φ → M 2 ⋅ M = M 3
57 mulcl ⊢ 2 ∈ ℂ ∧ B ∈ ℂ → 2 ⁢ B ∈ ℂ
58 11 1 57 sylancr ⊢ φ → 2 ⁢ B ∈ ℂ
59 58 15 15 mulassd ⊢ φ → 2 ⁢ B ⋅ M ⋅ M = 2 ⁢ B ⁢ M ⋅ M
60 17 15 1 mulassd ⊢ φ → 2 ⋅ M ⁢ B = 2 ⁢ M ⁢ B
61 17 15 1 mul32d ⊢ φ → 2 ⋅ M ⁢ B = 2 ⁢ B ⋅ M
62 60 61 eqtr3d ⊢ φ → 2 ⁢ M ⁢ B = 2 ⁢ B ⋅ M
63 62 oveq1d ⊢ φ → 2 ⁢ M ⁢ B ⋅ M = 2 ⁢ B ⋅ M ⋅ M
64 15 sqvald ⊢ φ → M 2 = M ⋅ M
65 64 oveq2d ⊢ φ → 2 ⁢ B ⁢ M 2 = 2 ⁢ B ⁢ M ⋅ M
66 59 63 65 3eqtr4d ⊢ φ → 2 ⁢ M ⁢ B ⋅ M = 2 ⁢ B ⁢ M 2
67 56 66 oveq12d ⊢ φ → M 2 ⋅ M + 2 ⁢ M ⁢ B ⋅ M = M 3 + 2 ⁢ B ⁢ M 2
68 50 67 eqtrd ⊢ φ → M 2 + 2 ⁢ M ⁢ B ⋅ M = M 3 + 2 ⁢ B ⁢ M 2
69 68 oveq1d ⊢ φ → M 2 + 2 ⁢ M ⁢ B ⋅ M + B 2 ⋅ M = M 3 + 2 ⁢ B ⁢ M 2 + B 2 ⋅ M
70 42 49 69 3eqtrd ⊢ φ → M + B 2 ⋅ M = M 3 + 2 ⁢ B ⁢ M 2 + B 2 ⋅ M
71 70 oveq1d ⊢ φ → M + B 2 ⋅ M − 4 ⁢ D ⋅ M = M 3 + 2 ⁢ B ⁢ M 2 + B 2 ⋅ M - 4 ⁢ D ⋅ M
72 3nn0 ⊢ 3 ∈ ℕ 0
73 expcl ⊢ M ∈ ℂ ∧ 3 ∈ ℕ 0 → M 3 ∈ ℂ
74 15 72 73 sylancl ⊢ φ → M 3 ∈ ℂ
75 58 43 mulcld ⊢ φ → 2 ⁢ B ⁢ M 2 ∈ ℂ
76 74 75 addcld ⊢ φ → M 3 + 2 ⁢ B ⁢ M 2 ∈ ℂ
77 48 15 mulcld ⊢ φ → B 2 ⋅ M ∈ ℂ
78 mulcl ⊢ 4 ∈ ℂ ∧ D ∈ ℂ → 4 ⁢ D ∈ ℂ
79 26 9 78 sylancr ⊢ φ → 4 ⁢ D ∈ ℂ
80 79 15 mulcld ⊢ φ → 4 ⁢ D ⋅ M ∈ ℂ
81 76 77 80 addsubassd ⊢ φ → M 3 + 2 ⁢ B ⁢ M 2 + B 2 ⋅ M - 4 ⁢ D ⋅ M = M 3 + 2 ⁢ B ⁢ M 2 + B 2 ⋅ M − 4 ⁢ D ⋅ M
82 48 79 15 subdird ⊢ φ → B 2 − 4 ⁢ D ⋅ M = B 2 ⋅ M − 4 ⁢ D ⋅ M
83 82 oveq2d ⊢ φ → M 3 + 2 ⁢ B ⁢ M 2 + B 2 − 4 ⁢ D ⋅ M = M 3 + 2 ⁢ B ⁢ M 2 + B 2 ⋅ M − 4 ⁢ D ⋅ M
84 81 83 eqtr4d ⊢ φ → M 3 + 2 ⁢ B ⁢ M 2 + B 2 ⋅ M - 4 ⁢ D ⋅ M = M 3 + 2 ⁢ B ⁢ M 2 + B 2 − 4 ⁢ D ⋅ M
85 48 79 subcld ⊢ φ → B 2 − 4 ⁢ D ∈ ℂ
86 85 15 mulcld ⊢ φ → B 2 − 4 ⁢ D ⋅ M ∈ ℂ
87 76 86 addcld ⊢ φ → M 3 + 2 ⁢ B ⁢ M 2 + B 2 − 4 ⁢ D ⋅ M ∈ ℂ
88 31 negcld ⊢ φ → − C 2 ∈ ℂ
89 76 86 88 addassd ⊢ φ → M 3 + 2 ⁢ B ⁢ M 2 + B 2 − 4 ⁢ D ⋅ M + − C 2 = M 3 + 2 ⁢ B ⁢ M 2 + B 2 − 4 ⁢ D ⋅ M + − C 2
90 87 31 negsubd ⊢ φ → M 3 + 2 ⁢ B ⁢ M 2 + B 2 − 4 ⁢ D ⋅ M + − C 2 = M 3 + 2 ⁢ B ⁢ M 2 + B 2 − 4 ⁢ D ⋅ M - C 2
91 89 90 10 3eqtr3d ⊢ φ → M 3 + 2 ⁢ B ⁢ M 2 + B 2 − 4 ⁢ D ⋅ M - C 2 = 0
92 87 31 91 subeq0d ⊢ φ → M 3 + 2 ⁢ B ⁢ M 2 + B 2 − 4 ⁢ D ⋅ M = C 2
93 71 84 92 3eqtrd ⊢ φ → M + B 2 ⋅ M − 4 ⁢ D ⋅ M = C 2
94 25 15 mulcld ⊢ φ → M + B 2 ⋅ M ∈ ℂ
95 subsub23 ⊢ M + B 2 ⋅ M ∈ ℂ ∧ 4 ⁢ D ⋅ M ∈ ℂ ∧ C 2 ∈ ℂ → M + B 2 ⋅ M − 4 ⁢ D ⋅ M = C 2 ↔ M + B 2 ⋅ M − C 2 = 4 ⁢ D ⋅ M
96 94 80 31 95 syl3anc ⊢ φ → M + B 2 ⋅ M − 4 ⁢ D ⋅ M = C 2 ↔ M + B 2 ⋅ M − C 2 = 4 ⁢ D ⋅ M
97 93 96 mpbid ⊢ φ → M + B 2 ⋅ M − C 2 = 4 ⁢ D ⋅ M
98 27 9 15 mulassd ⊢ φ → 4 ⁢ D ⋅ M = 4 ⁢ D ⋅ M
99 97 98 eqtrd ⊢ φ → M + B 2 ⋅ M − C 2 = 4 ⁢ D ⋅ M
100 99 oveq1d ⊢ φ → M + B 2 ⋅ M − C 2 4 = 4 ⁢ D ⋅ M 4
101 94 31 27 29 divsubdird ⊢ φ → M + B 2 ⋅ M − C 2 4 = M + B 2 ⋅ M 4 − C 2 4
102 9 15 mulcld ⊢ φ → D ⋅ M ∈ ℂ
103 102 27 29 divcan3d ⊢ φ → 4 ⁢ D ⋅ M 4 = D ⋅ M
104 100 101 103 3eqtr3d ⊢ φ → M + B 2 ⋅ M 4 − C 2 4 = D ⋅ M
105 35 39 104 3eqtrd ⊢ φ → M + B 2 4 − C 2 4 M ⋅ M = D ⋅ M
106 34 9 15 6 105 mulcan2ad ⊢ φ → M + B 2 4 − C 2 4 M = D
107 24 106 eqtrd ⊢ φ → M + B 2 2 − C 2 4 M = D