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
5 4 recni F
6 2cnne0 2 2 0
7 5nn0 5 0
8 expdiv F 2 2 0 5 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 5 0 F 5
12 5 7 11 mp2an F 5
13 2cn 2
14 expcl 2 5 0 2 5
15 13 7 14 mp2an 2 5
16 2ne0 2 0
17 5nn 5
18 17 nnzi 5
19 expne0i 2 2 0 5 2 5 0
20 13 16 18 19 mp3an 2 5 0
21 15 20 pm3.2i 2 5 2 5 0
22 16nn0 16 0
23 22 nn0cni 16
24 1nn0 1 0
25 6nn 6
26 24 25 decnncl 16
27 26 nnne0i 16 0
28 23 27 pm3.2i 16 16 0
29 divdiv2 F 5 2 5 2 5 0 16 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 2 1 = 2
36 13 35 ax-mp 2 1 = 2
37 36 eqcomi 2 = 2 1
38 4cn 4
39 ax-1cn 1
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
45 18 44 pm3.2i 5 4
46 expsub 2 2 0 5 4 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 0
55 expdiv F 2 2 0 3 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
62 expcl F 3 0 F 3
63 5 54 62 mp2an F 3
64 expcl 2 3 0 2 3
65 13 54 64 mp2an 2 3
66 3z 3
67 expne0i 2 2 0 3 2 3 0
68 13 16 66 67 mp3an 2 3 0
69 63 65 68 divcli F 3 2 3
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 2 3 0
73 4ne0 4 0
74 38 73 pm3.2i 4 4 0
75 divdiv2 F 3 2 3 2 3 0 4 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