Metamath Proof Explorer


Theorem goldratmolem2

Description: Lemma 2 for determining the value of golden ratio. (Contributed by Ender Ting, 9-May-2026)

Ref Expression
Hypothesis goldra.val ⊢ F = 2 ⁢ cos ⁡ π 5
Assertion goldratmolem2 ⊢ − 1 = F 5 2 - 5 ⁢ F 3 2 + 5 ⁢ F 2

Proof

Step Hyp Ref Expression
1 goldra.val ⊢ F = 2 ⁢ cos ⁡ π 5
2 1 goldracos5teq ⊢ cos ⁡ π = 16 ⁢ F 2 5 - 20 ⁢ F 2 3 + 5 ⁢ F 2
3 cospi ⊢ cos ⁡ π = − 1
4 1 goldrarr ⊢ F ∈ ℝ
5 4 recni ⊢ F ∈ ℂ
6 2cnne0 ⊢ 2 ∈ ℂ ∧ 2 ≠ 0
7 5nn0 ⊢ 5 ∈ ℕ 0
8 expdiv ⊢ F ∈ ℂ ∧ 2 ∈ ℂ ∧ 2 ≠ 0 ∧ 5 ∈ ℕ 0 → F 2 5 = F 5 2 5
9 5 6 7 8 mp3an ⊢ F 2 5 = F 5 2 5
10 9 oveq2i ⊢ 16 ⁢ F 2 5 = 16 ⁢ F 5 2 5
11 expcl ⊢ F ∈ ℂ ∧ 5 ∈ ℕ 0 → F 5 ∈ ℂ
12 5 7 11 mp2an ⊢ F 5 ∈ ℂ
13 2cn ⊢ 2 ∈ ℂ
14 expcl ⊢ 2 ∈ ℂ ∧ 5 ∈ ℕ 0 → 2 5 ∈ ℂ
15 13 7 14 mp2an ⊢ 2 5 ∈ ℂ
16 2ne0 ⊢ 2 ≠ 0
17 5nn ⊢ 5 ∈ ℕ
18 17 nnzi ⊢ 5 ∈ ℤ
19 expne0i ⊢ 2 ∈ ℂ ∧ 2 ≠ 0 ∧ 5 ∈ ℤ → 2 5 ≠ 0
20 13 16 18 19 mp3an ⊢ 2 5 ≠ 0
21 15 20 pm3.2i ⊢ 2 5 ∈ ℂ ∧ 2 5 ≠ 0
22 16nn0 ⊢ 16 ∈ ℕ 0
23 22 nn0cni ⊢ 16 ∈ ℂ
24 1nn0 ⊢ 1 ∈ ℕ 0
25 6nn ⊢ 6 ∈ ℕ
26 24 25 decnncl ⊢ 16 ∈ ℕ
27 26 nnne0i ⊢ 16 ≠ 0
28 23 27 pm3.2i ⊢ 16 ∈ ℂ ∧ 16 ≠ 0
29 divdiv2 ⊢ F 5 ∈ ℂ ∧ 2 5 ∈ ℂ ∧ 2 5 ≠ 0 ∧ 16 ∈ ℂ ∧ 16 ≠ 0 → F 5 2 5 16 = F 5 ⋅ 16 2 5
30 12 21 28 29 mp3an ⊢ F 5 2 5 16 = F 5 ⋅ 16 2 5
31 12 23 mulcomi ⊢ F 5 ⋅ 16 = 16 ⁢ F 5
32 31 oveq1i ⊢ F 5 ⋅ 16 2 5 = 16 ⁢ F 5 2 5
33 23 12 15 20 divassi ⊢ 16 ⁢ F 5 2 5 = 16 ⁢ F 5 2 5
34 30 32 33 3eqtrri ⊢ 16 ⁢ F 5 2 5 = F 5 2 5 16
35 exp1 ⊢ 2 ∈ ℂ → 2 1 = 2
36 13 35 ax-mp ⊢ 2 1 = 2
37 36 eqcomi ⊢ 2 = 2 1
38 4cn ⊢ 4 ∈ ℂ
39 ax-1cn ⊢ 1 ∈ ℂ
40 4p1e5 ⊢ 4 + 1 = 5
41 38 39 40 mvlladdi ⊢ 1 = 5 − 4
42 41 oveq2i ⊢ 2 1 = 2 5 − 4
43 37 42 eqtri ⊢ 2 = 2 5 − 4
44 4z ⊢ 4 ∈ ℤ
45 18 44 pm3.2i ⊢ 5 ∈ ℤ ∧ 4 ∈ ℤ
46 expsub ⊢ 2 ∈ ℂ ∧ 2 ≠ 0 ∧ 5 ∈ ℤ ∧ 4 ∈ ℤ → 2 5 − 4 = 2 5 2 4
47 6 45 46 mp2an ⊢ 2 5 − 4 = 2 5 2 4
48 2exp4 ⊢ 2 4 = 16
49 48 oveq2i ⊢ 2 5 2 4 = 2 5 16
50 43 47 49 3eqtri ⊢ 2 = 2 5 16
51 50 eqcomi ⊢ 2 5 16 = 2
52 51 oveq2i ⊢ F 5 2 5 16 = F 5 2
53 10 34 52 3eqtri ⊢ 16 ⁢ F 2 5 = F 5 2
54 3nn0 ⊢ 3 ∈ ℕ 0
55 expdiv ⊢ F ∈ ℂ ∧ 2 ∈ ℂ ∧ 2 ≠ 0 ∧ 3 ∈ ℕ 0 → F 2 3 = F 3 2 3
56 5 6 54 55 mp3an ⊢ F 2 3 = F 3 2 3
57 56 oveq2i ⊢ 20 ⁢ F 2 3 = 20 ⁢ F 3 2 3
58 5t4e20 ⊢ 5 ⋅ 4 = 20
59 58 eqcomi ⊢ 20 = 5 ⋅ 4
60 59 oveq1i ⊢ 20 ⁢ F 3 2 3 = 5 ⋅ 4 ⁢ F 3 2 3
61 5cn ⊢ 5 ∈ ℂ
62 expcl ⊢ F ∈ ℂ ∧ 3 ∈ ℕ 0 → F 3 ∈ ℂ
63 5 54 62 mp2an ⊢ F 3 ∈ ℂ
64 expcl ⊢ 2 ∈ ℂ ∧ 3 ∈ ℕ 0 → 2 3 ∈ ℂ
65 13 54 64 mp2an ⊢ 2 3 ∈ ℂ
66 3z ⊢ 3 ∈ ℤ
67 expne0i ⊢ 2 ∈ ℂ ∧ 2 ≠ 0 ∧ 3 ∈ ℤ → 2 3 ≠ 0
68 13 16 66 67 mp3an ⊢ 2 3 ≠ 0
69 63 65 68 divcli ⊢ F 3 2 3 ∈ ℂ
70 61 38 69 mulassi ⊢ 5 ⋅ 4 ⁢ F 3 2 3 = 5 ⁢ 4 ⁢ F 3 2 3
71 60 70 eqtri ⊢ 20 ⁢ F 3 2 3 = 5 ⁢ 4 ⁢ F 3 2 3
72 65 68 pm3.2i ⊢ 2 3 ∈ ℂ ∧ 2 3 ≠ 0
73 4ne0 ⊢ 4 ≠ 0
74 38 73 pm3.2i ⊢ 4 ∈ ℂ ∧ 4 ≠ 0
75 divdiv2 ⊢ F 3 ∈ ℂ ∧ 2 3 ∈ ℂ ∧ 2 3 ≠ 0 ∧ 4 ∈ ℂ ∧ 4 ≠ 0 → F 3 2 3 4 = F 3 ⋅ 4 2 3
76 63 72 74 75 mp3an ⊢ F 3 2 3 4 = F 3 ⋅ 4 2 3
77 63 38 mulcomi ⊢ F 3 ⋅ 4 = 4 ⁢ F 3
78 77 oveq1i ⊢ F 3 ⋅ 4 2 3 = 4 ⁢ F 3 2 3
79 38 63 65 68 divassi ⊢ 4 ⁢ F 3 2 3 = 4 ⁢ F 3 2 3
80 76 78 79 3eqtrri ⊢ 4 ⁢ F 3 2 3 = F 3 2 3 4
81 4t2e8 ⊢ 4 ⋅ 2 = 8
82 cu2 ⊢ 2 3 = 8
83 82 eqcomi ⊢ 8 = 2 3
84 81 83 eqtri ⊢ 4 ⋅ 2 = 2 3
85 65 38 13 73 divmuli ⊢ 2 3 4 = 2 ↔ 4 ⋅ 2 = 2 3
86 84 85 mpbir ⊢ 2 3 4 = 2
87 86 oveq2i ⊢ F 3 2 3 4 = F 3 2
88 80 87 eqtri ⊢ 4 ⁢ F 3 2 3 = F 3 2
89 88 oveq2i ⊢ 5 ⁢ 4 ⁢ F 3 2 3 = 5 ⁢ F 3 2
90 57 71 89 3eqtri ⊢ 20 ⁢ F 2 3 = 5 ⁢ F 3 2
91 53 90 oveq12i ⊢ 16 ⁢ F 2 5 − 20 ⁢ F 2 3 = F 5 2 − 5 ⁢ F 3 2
92 91 oveq1i ⊢ 16 ⁢ F 2 5 - 20 ⁢ F 2 3 + 5 ⁢ F 2 = F 5 2 - 5 ⁢ F 3 2 + 5 ⁢ F 2
93 2 3 92 3eqtr3i ⊢ − 1 = F 5 2 - 5 ⁢ F 3 2 + 5 ⁢ F 2