Metamath Proof Explorer


Theorem 3cubeslem2

Description: Lemma for 3cubes . Used to show that the denominators in 3cubeslem4 are nonzero. (Contributed by Igor Ieskov, 22-Jan-2024)

Ref Expression
Hypothesis 3cubeslem1.a ⊢ φ → A ∈ ℚ
Assertion 3cubeslem2 ⊢ φ → ¬ 3 3 ⁢ A 2 + 3 2 ⁢ A + 3 = 0

Proof

Step Hyp Ref Expression
1 3cubeslem1.a ⊢ φ → A ∈ ℚ
2 3re ⊢ 3 ∈ ℝ
3 2 a1i ⊢ φ → 3 ∈ ℝ
4 3 recnd ⊢ φ → 3 ∈ ℂ
5 4 mullidd ⊢ φ → 1 ⋅ 3 = 3
6 5 oveq2d ⊢ φ → 3 2 ⁢ A 2 + 3 ⁢ A ⋅ 3 + 1 ⋅ 3 = 3 2 ⁢ A 2 + 3 ⁢ A ⋅ 3 + 3
7 4 sqcld ⊢ φ → 3 2 ∈ ℂ
8 qre ⊢ A ∈ ℚ → A ∈ ℝ
9 1 8 syl ⊢ φ → A ∈ ℝ
10 9 resqcld ⊢ φ → A 2 ∈ ℝ
11 10 recnd ⊢ φ → A 2 ∈ ℂ
12 7 11 mulcld ⊢ φ → 3 2 ⁢ A 2 ∈ ℂ
13 9 recnd ⊢ φ → A ∈ ℂ
14 4 13 mulcld ⊢ φ → 3 ⁢ A ∈ ℂ
15 12 14 addcld ⊢ φ → 3 2 ⁢ A 2 + 3 ⁢ A ∈ ℂ
16 1cnd ⊢ φ → 1 ∈ ℂ
17 15 16 4 adddird ⊢ φ → 3 2 ⁢ A 2 + 3 ⁢ A + 1 ⋅ 3 = 3 2 ⁢ A 2 + 3 ⁢ A ⋅ 3 + 1 ⋅ 3
18 4 13 4 mulassd ⊢ φ → 3 ⁢ A ⋅ 3 = 3 ⁢ A ⋅ 3
19 18 oveq2d ⊢ φ → 3 2 ⁢ A 2 ⋅ 3 + 3 ⁢ A ⋅ 3 = 3 2 ⁢ A 2 ⋅ 3 + 3 ⁢ A ⋅ 3
20 19 oveq1d ⊢ φ → 3 2 ⁢ A 2 ⋅ 3 + 3 ⁢ A ⋅ 3 + 3 = 3 2 ⁢ A 2 ⋅ 3 + 3 ⁢ A ⋅ 3 + 3
21 12 14 4 adddird ⊢ φ → 3 2 ⁢ A 2 + 3 ⁢ A ⋅ 3 = 3 2 ⁢ A 2 ⋅ 3 + 3 ⁢ A ⋅ 3
22 21 oveq1d ⊢ φ → 3 2 ⁢ A 2 + 3 ⁢ A ⋅ 3 + 3 = 3 2 ⁢ A 2 ⋅ 3 + 3 ⁢ A ⋅ 3 + 3
23 4 4 13 mulassd ⊢ φ → 3 ⋅ 3 ⁢ A = 3 ⁢ 3 ⁢ A
24 23 oveq2d ⊢ φ → 3 2 ⁢ A 2 ⋅ 3 + 3 ⋅ 3 ⁢ A = 3 2 ⁢ A 2 ⋅ 3 + 3 ⁢ 3 ⁢ A
25 24 oveq1d ⊢ φ → 3 2 ⁢ A 2 ⋅ 3 + 3 ⋅ 3 ⁢ A + 3 = 3 2 ⁢ A 2 ⋅ 3 + 3 ⁢ 3 ⁢ A + 3
26 11 4 mulcomd ⊢ φ → A 2 ⋅ 3 = 3 ⁢ A 2
27 26 oveq2d ⊢ φ → 3 2 ⁢ A 2 ⋅ 3 = 3 2 ⁢ 3 ⁢ A 2
28 27 oveq1d ⊢ φ → 3 2 ⁢ A 2 ⋅ 3 + 3 ⋅ 3 ⁢ A = 3 2 ⁢ 3 ⁢ A 2 + 3 ⋅ 3 ⁢ A
29 28 oveq1d ⊢ φ → 3 2 ⁢ A 2 ⋅ 3 + 3 ⋅ 3 ⁢ A + 3 = 3 2 ⁢ 3 ⁢ A 2 + 3 ⋅ 3 ⁢ A + 3
30 7 11 4 mulassd ⊢ φ → 3 2 ⁢ A 2 ⋅ 3 = 3 2 ⁢ A 2 ⋅ 3
31 30 oveq1d ⊢ φ → 3 2 ⁢ A 2 ⋅ 3 + 3 ⋅ 3 ⁢ A = 3 2 ⁢ A 2 ⋅ 3 + 3 ⋅ 3 ⁢ A
32 31 oveq1d ⊢ φ → 3 2 ⁢ A 2 ⋅ 3 + 3 ⋅ 3 ⁢ A + 3 = 3 2 ⁢ A 2 ⋅ 3 + 3 ⋅ 3 ⁢ A + 3
33 df-3 ⊢ 3 = 2 + 1
34 33 a1i ⊢ φ → 3 = 2 + 1
35 34 oveq2d ⊢ φ → 3 3 = 3 2 + 1
36 35 oveq1d ⊢ φ → 3 3 ⁢ A 2 = 3 2 + 1 ⁢ A 2
37 36 oveq1d ⊢ φ → 3 3 ⁢ A 2 + 3 2 ⁢ A = 3 2 + 1 ⁢ A 2 + 3 2 ⁢ A
38 37 oveq1d ⊢ φ → 3 3 ⁢ A 2 + 3 2 ⁢ A + 3 = 3 2 + 1 ⁢ A 2 + 3 2 ⁢ A + 3
39 2nn0 ⊢ 2 ∈ ℕ 0
40 39 a1i ⊢ φ → 2 ∈ ℕ 0
41 4 40 expp1d ⊢ φ → 3 2 + 1 = 3 2 ⋅ 3
42 41 oveq1d ⊢ φ → 3 2 + 1 ⁢ A 2 = 3 2 ⋅ 3 ⁢ A 2
43 42 oveq1d ⊢ φ → 3 2 + 1 ⁢ A 2 + 3 2 ⁢ A = 3 2 ⋅ 3 ⁢ A 2 + 3 2 ⁢ A
44 43 oveq1d ⊢ φ → 3 2 + 1 ⁢ A 2 + 3 2 ⁢ A + 3 = 3 2 ⋅ 3 ⁢ A 2 + 3 2 ⁢ A + 3
45 38 44 eqtrd ⊢ φ → 3 3 ⁢ A 2 + 3 2 ⁢ A + 3 = 3 2 ⋅ 3 ⁢ A 2 + 3 2 ⁢ A + 3
46 4 sqvald ⊢ φ → 3 2 = 3 ⋅ 3
47 46 oveq1d ⊢ φ → 3 2 ⁢ A = 3 ⋅ 3 ⁢ A
48 47 oveq2d ⊢ φ → 3 2 ⋅ 3 ⁢ A 2 + 3 2 ⁢ A = 3 2 ⋅ 3 ⁢ A 2 + 3 ⋅ 3 ⁢ A
49 48 oveq1d ⊢ φ → 3 2 ⋅ 3 ⁢ A 2 + 3 2 ⁢ A + 3 = 3 2 ⋅ 3 ⁢ A 2 + 3 ⋅ 3 ⁢ A + 3
50 45 49 eqtrd ⊢ φ → 3 3 ⁢ A 2 + 3 2 ⁢ A + 3 = 3 2 ⋅ 3 ⁢ A 2 + 3 ⋅ 3 ⁢ A + 3
51 7 4 11 mulassd ⊢ φ → 3 2 ⋅ 3 ⁢ A 2 = 3 2 ⁢ 3 ⁢ A 2
52 51 oveq1d ⊢ φ → 3 2 ⋅ 3 ⁢ A 2 + 3 ⋅ 3 ⁢ A = 3 2 ⁢ 3 ⁢ A 2 + 3 ⋅ 3 ⁢ A
53 52 oveq1d ⊢ φ → 3 2 ⋅ 3 ⁢ A 2 + 3 ⋅ 3 ⁢ A + 3 = 3 2 ⁢ 3 ⁢ A 2 + 3 ⋅ 3 ⁢ A + 3
54 50 53 eqtrd ⊢ φ → 3 3 ⁢ A 2 + 3 2 ⁢ A + 3 = 3 2 ⁢ 3 ⁢ A 2 + 3 ⋅ 3 ⁢ A + 3
55 29 32 54 3eqtr4rd ⊢ φ → 3 3 ⁢ A 2 + 3 2 ⁢ A + 3 = 3 2 ⁢ A 2 ⋅ 3 + 3 ⋅ 3 ⁢ A + 3
56 13 4 mulcomd ⊢ φ → A ⋅ 3 = 3 ⁢ A
57 56 oveq2d ⊢ φ → 3 ⁢ A ⋅ 3 = 3 ⁢ 3 ⁢ A
58 57 oveq2d ⊢ φ → 3 2 ⁢ A 2 ⋅ 3 + 3 ⁢ A ⋅ 3 = 3 2 ⁢ A 2 ⋅ 3 + 3 ⁢ 3 ⁢ A
59 58 oveq1d ⊢ φ → 3 2 ⁢ A 2 ⋅ 3 + 3 ⁢ A ⋅ 3 + 3 = 3 2 ⁢ A 2 ⋅ 3 + 3 ⁢ 3 ⁢ A + 3
60 25 55 59 3eqtr4d ⊢ φ → 3 3 ⁢ A 2 + 3 2 ⁢ A + 3 = 3 2 ⁢ A 2 ⋅ 3 + 3 ⁢ A ⋅ 3 + 3
61 20 22 60 3eqtr4rd ⊢ φ → 3 3 ⁢ A 2 + 3 2 ⁢ A + 3 = 3 2 ⁢ A 2 + 3 ⁢ A ⋅ 3 + 3
62 6 17 61 3eqtr4rd ⊢ φ → 3 3 ⁢ A 2 + 3 2 ⁢ A + 3 = 3 2 ⁢ A 2 + 3 ⁢ A + 1 ⋅ 3
63 14 mulridd ⊢ φ → 3 ⁢ A ⋅ 1 = 3 ⁢ A
64 63 oveq2d ⊢ φ → 2 ⁢ 3 ⁢ A ⋅ 1 = 2 ⁢ 3 ⁢ A
65 64 oveq2d ⊢ φ → 3 ⁢ A 2 + 2 ⁢ 3 ⁢ A ⋅ 1 = 3 ⁢ A 2 + 2 ⁢ 3 ⁢ A
66 65 oveq1d ⊢ φ → 3 ⁢ A 2 + 2 ⁢ 3 ⁢ A ⋅ 1 + 1 2 = 3 ⁢ A 2 + 2 ⁢ 3 ⁢ A + 1 2
67 66 oveq1d ⊢ φ → 3 ⁢ A 2 + 2 ⁢ 3 ⁢ A ⋅ 1 + 1 2 - 3 ⁢ A = 3 ⁢ A 2 + 2 ⁢ 3 ⁢ A + 1 2 - 3 ⁢ A
68 14 16 binom2d ⊢ φ → 3 ⁢ A + 1 2 = 3 ⁢ A 2 + 2 ⁢ 3 ⁢ A ⋅ 1 + 1 2
69 68 oveq1d ⊢ φ → 3 ⁢ A + 1 2 − 3 ⁢ A = 3 ⁢ A 2 + 2 ⁢ 3 ⁢ A ⋅ 1 + 1 2 - 3 ⁢ A
70 14 2timesd ⊢ φ → 2 ⁢ 3 ⁢ A = 3 ⁢ A + 3 ⁢ A
71 70 oveq2d ⊢ φ → 3 ⁢ A 2 + 2 ⁢ 3 ⁢ A = 3 ⁢ A 2 + 3 ⁢ A + 3 ⁢ A
72 71 oveq1d ⊢ φ → 3 ⁢ A 2 + 2 ⁢ 3 ⁢ A + 1 = 3 ⁢ A 2 + 3 ⁢ A + 3 ⁢ A + 1
73 72 oveq1d ⊢ φ → 3 ⁢ A 2 + 2 ⁢ 3 ⁢ A + 1 - 3 ⁢ A = 3 ⁢ A 2 + 3 ⁢ A + 3 ⁢ A + 1 - 3 ⁢ A
74 sq1 ⊢ 1 2 = 1
75 74 a1i ⊢ φ → 1 2 = 1
76 75 oveq2d ⊢ φ → 3 ⁢ A 2 + 2 ⁢ 3 ⁢ A + 1 2 = 3 ⁢ A 2 + 2 ⁢ 3 ⁢ A + 1
77 76 oveq1d ⊢ φ → 3 ⁢ A 2 + 2 ⁢ 3 ⁢ A + 1 2 - 3 ⁢ A = 3 ⁢ A 2 + 2 ⁢ 3 ⁢ A + 1 - 3 ⁢ A
78 14 16 addcomd ⊢ φ → 3 ⁢ A + 1 = 1 + 3 ⁢ A
79 78 oveq2d ⊢ φ → 3 ⁢ A 2 + 3 ⁢ A + 3 ⁢ A + 1 = 3 ⁢ A 2 + 3 ⁢ A + 1 + 3 ⁢ A
80 79 oveq1d ⊢ φ → 3 ⁢ A 2 + 3 ⁢ A + 3 ⁢ A + 1 - 3 ⁢ A = 3 ⁢ A 2 + 3 ⁢ A + 1 + 3 ⁢ A - 3 ⁢ A
81 4 13 sqmuld ⊢ φ → 3 ⁢ A 2 = 3 2 ⁢ A 2
82 81 12 eqeltrd ⊢ φ → 3 ⁢ A 2 ∈ ℂ
83 82 14 addcld ⊢ φ → 3 ⁢ A 2 + 3 ⁢ A ∈ ℂ
84 83 14 16 addassd ⊢ φ → 3 ⁢ A 2 + 3 ⁢ A + 3 ⁢ A + 1 = 3 ⁢ A 2 + 3 ⁢ A + 3 ⁢ A + 1
85 84 oveq1d ⊢ φ → 3 ⁢ A 2 + 3 ⁢ A + 3 ⁢ A + 1 - 3 ⁢ A = 3 ⁢ A 2 + 3 ⁢ A + 3 ⁢ A + 1 - 3 ⁢ A
86 15 16 addcld ⊢ φ → 3 2 ⁢ A 2 + 3 ⁢ A + 1 ∈ ℂ
87 86 14 14 addsubassd ⊢ φ → 3 2 ⁢ A 2 + 3 ⁢ A + 1 + 3 ⁢ A - 3 ⁢ A = 3 2 ⁢ A 2 + 3 ⁢ A + 1 + 3 ⁢ A − 3 ⁢ A
88 81 oveq1d ⊢ φ → 3 ⁢ A 2 + 3 ⁢ A = 3 2 ⁢ A 2 + 3 ⁢ A
89 88 oveq1d ⊢ φ → 3 ⁢ A 2 + 3 ⁢ A + 1 = 3 2 ⁢ A 2 + 3 ⁢ A + 1
90 89 oveq1d ⊢ φ → 3 ⁢ A 2 + 3 ⁢ A + 1 + 3 ⁢ A = 3 2 ⁢ A 2 + 3 ⁢ A + 1 + 3 ⁢ A
91 90 oveq1d ⊢ φ → 3 ⁢ A 2 + 3 ⁢ A + 1 + 3 ⁢ A - 3 ⁢ A = 3 2 ⁢ A 2 + 3 ⁢ A + 1 + 3 ⁢ A - 3 ⁢ A
92 14 subidd ⊢ φ → 3 ⁢ A − 3 ⁢ A = 0
93 92 oveq2d ⊢ φ → 3 2 ⁢ A 2 + 3 ⁢ A + 1 + 3 ⁢ A − 3 ⁢ A = 3 2 ⁢ A 2 + 3 ⁢ A + 1 + 0
94 86 addridd ⊢ φ → 3 2 ⁢ A 2 + 3 ⁢ A + 1 + 0 = 3 2 ⁢ A 2 + 3 ⁢ A + 1
95 93 94 eqtr2d ⊢ φ → 3 2 ⁢ A 2 + 3 ⁢ A + 1 = 3 2 ⁢ A 2 + 3 ⁢ A + 1 + 3 ⁢ A − 3 ⁢ A
96 87 91 95 3eqtr4rd ⊢ φ → 3 2 ⁢ A 2 + 3 ⁢ A + 1 = 3 ⁢ A 2 + 3 ⁢ A + 1 + 3 ⁢ A - 3 ⁢ A
97 83 16 14 addassd ⊢ φ → 3 ⁢ A 2 + 3 ⁢ A + 1 + 3 ⁢ A = 3 ⁢ A 2 + 3 ⁢ A + 1 + 3 ⁢ A
98 97 oveq1d ⊢ φ → 3 ⁢ A 2 + 3 ⁢ A + 1 + 3 ⁢ A - 3 ⁢ A = 3 ⁢ A 2 + 3 ⁢ A + 1 + 3 ⁢ A - 3 ⁢ A
99 96 98 eqtrd ⊢ φ → 3 2 ⁢ A 2 + 3 ⁢ A + 1 = 3 ⁢ A 2 + 3 ⁢ A + 1 + 3 ⁢ A - 3 ⁢ A
100 80 85 99 3eqtr4rd ⊢ φ → 3 2 ⁢ A 2 + 3 ⁢ A + 1 = 3 ⁢ A 2 + 3 ⁢ A + 3 ⁢ A + 1 - 3 ⁢ A
101 82 14 14 addassd ⊢ φ → 3 ⁢ A 2 + 3 ⁢ A + 3 ⁢ A = 3 ⁢ A 2 + 3 ⁢ A + 3 ⁢ A
102 101 oveq1d ⊢ φ → 3 ⁢ A 2 + 3 ⁢ A + 3 ⁢ A + 1 = 3 ⁢ A 2 + 3 ⁢ A + 3 ⁢ A + 1
103 102 oveq1d ⊢ φ → 3 ⁢ A 2 + 3 ⁢ A + 3 ⁢ A + 1 - 3 ⁢ A = 3 ⁢ A 2 + 3 ⁢ A + 3 ⁢ A + 1 - 3 ⁢ A
104 100 103 eqtrd ⊢ φ → 3 2 ⁢ A 2 + 3 ⁢ A + 1 = 3 ⁢ A 2 + 3 ⁢ A + 3 ⁢ A + 1 - 3 ⁢ A
105 73 77 104 3eqtr4rd ⊢ φ → 3 2 ⁢ A 2 + 3 ⁢ A + 1 = 3 ⁢ A 2 + 2 ⁢ 3 ⁢ A + 1 2 - 3 ⁢ A
106 67 69 105 3eqtr4rd ⊢ φ → 3 2 ⁢ A 2 + 3 ⁢ A + 1 = 3 ⁢ A + 1 2 − 3 ⁢ A
107 106 oveq1d ⊢ φ → 3 2 ⁢ A 2 + 3 ⁢ A + 1 ⋅ 3 = 3 ⁢ A + 1 2 − 3 ⁢ A ⋅ 3
108 62 107 eqtrd ⊢ φ → 3 3 ⁢ A 2 + 3 2 ⁢ A + 3 = 3 ⁢ A + 1 2 − 3 ⁢ A ⋅ 3
109 3 9 remulcld ⊢ φ → 3 ⁢ A ∈ ℝ
110 peano2re ⊢ 3 ⁢ A ∈ ℝ → 3 ⁢ A + 1 ∈ ℝ
111 109 110 syl ⊢ φ → 3 ⁢ A + 1 ∈ ℝ
112 111 resqcld ⊢ φ → 3 ⁢ A + 1 2 ∈ ℝ
113 112 109 resubcld ⊢ φ → 3 ⁢ A + 1 2 − 3 ⁢ A ∈ ℝ
114 113 recnd ⊢ φ → 3 ⁢ A + 1 2 − 3 ⁢ A ∈ ℂ
115 3nn ⊢ 3 ∈ ℕ
116 nnq ⊢ 3 ∈ ℕ → 3 ∈ ℚ
117 115 116 ax-mp ⊢ 3 ∈ ℚ
118 qmulcl ⊢ 3 ∈ ℚ ∧ A ∈ ℚ → 3 ⁢ A ∈ ℚ
119 117 1 118 sylancr ⊢ φ → 3 ⁢ A ∈ ℚ
120 119 3cubeslem1 ⊢ φ → 0 < 3 ⁢ A + 1 2 − 3 ⁢ A
121 120 gt0ne0d ⊢ φ → 3 ⁢ A + 1 2 − 3 ⁢ A ≠ 0
122 3ne0 ⊢ 3 ≠ 0
123 122 a1i ⊢ φ → 3 ≠ 0
124 114 4 121 123 mulne0d ⊢ φ → 3 ⁢ A + 1 2 − 3 ⁢ A ⋅ 3 ≠ 0
125 108 124 eqnetrd ⊢ φ → 3 3 ⁢ A 2 + 3 2 ⁢ A + 3 ≠ 0
126 125 neneqd ⊢ φ → ¬ 3 3 ⁢ A 2 + 3 2 ⁢ A + 3 = 0