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 1nn0 1 0
23 6nn0 6 0
24 22 23 deccl 16 0
25 24 nn0cni 16
26 6nn 6
27 22 26 decnncl 16
28 27 nnne0i 16 0
29 25 28 pm3.2i 16 16 0
30 divdiv2 F 5 2 5 2 5 0 16 16 0 F 5 2 5 16 = F 5 16 2 5
31 12 21 29 30 mp3an F 5 2 5 16 = F 5 16 2 5
32 12 25 mulcomi F 5 16 = 16 F 5
33 32 oveq1i F 5 16 2 5 = 16 F 5 2 5
34 25 12 15 20 divassi 16 F 5 2 5 = 16 F 5 2 5
35 31 33 34 3eqtrri 16 F 5 2 5 = F 5 2 5 16
36 exp1 2 2 1 = 2
37 13 36 ax-mp 2 1 = 2
38 37 eqcomi 2 = 2 1
39 4cn 4
40 ax-1cn 1
41 4p1e5 4 + 1 = 5
42 39 40 41 mvlladdi 1 = 5 4
43 42 oveq2i 2 1 = 2 5 4
44 38 43 eqtri 2 = 2 5 4
45 4z 4
46 18 45 pm3.2i 5 4
47 expsub 2 2 0 5 4 2 5 4 = 2 5 2 4
48 6 46 47 mp2an 2 5 4 = 2 5 2 4
49 2exp4 2 4 = 16
50 49 oveq2i 2 5 2 4 = 2 5 16
51 44 48 50 3eqtri 2 = 2 5 16
52 51 eqcomi 2 5 16 = 2
53 52 oveq2i F 5 2 5 16 = F 5 2
54 10 35 53 3eqtri 16 F 2 5 = F 5 2
55 3nn0 3 0
56 expdiv F 2 2 0 3 0 F 2 3 = F 3 2 3
57 5 6 55 56 mp3an F 2 3 = F 3 2 3
58 57 oveq2i 20 F 2 3 = 20 F 3 2 3
59 5t4e20 5 4 = 20
60 59 eqcomi 20 = 5 4
61 60 oveq1i 20 F 3 2 3 = 5 4 F 3 2 3
62 5cn 5
63 expcl F 3 0 F 3
64 5 55 63 mp2an F 3
65 expcl 2 3 0 2 3
66 13 55 65 mp2an 2 3
67 3z 3
68 expne0i 2 2 0 3 2 3 0
69 13 16 67 68 mp3an 2 3 0
70 64 66 69 divcli F 3 2 3
71 62 39 70 mulassi 5 4 F 3 2 3 = 5 4 F 3 2 3
72 61 71 eqtri 20 F 3 2 3 = 5 4 F 3 2 3
73 66 69 pm3.2i 2 3 2 3 0
74 4ne0 4 0
75 39 74 pm3.2i 4 4 0
76 divdiv2 F 3 2 3 2 3 0 4 4 0 F 3 2 3 4 = F 3 4 2 3
77 64 73 75 76 mp3an F 3 2 3 4 = F 3 4 2 3
78 64 39 mulcomi F 3 4 = 4 F 3
79 78 oveq1i F 3 4 2 3 = 4 F 3 2 3
80 39 64 66 69 divassi 4 F 3 2 3 = 4 F 3 2 3
81 77 79 80 3eqtrri 4 F 3 2 3 = F 3 2 3 4
82 4t2e8 4 2 = 8
83 cu2 2 3 = 8
84 83 eqcomi 8 = 2 3
85 82 84 eqtri 4 2 = 2 3
86 66 39 13 74 divmuli 2 3 4 = 2 4 2 = 2 3
87 85 86 mpbir 2 3 4 = 2
88 87 oveq2i F 3 2 3 4 = F 3 2
89 81 88 eqtri 4 F 3 2 3 = F 3 2
90 89 oveq2i 5 4 F 3 2 3 = 5 F 3 2
91 58 72 90 3eqtri 20 F 2 3 = 5 F 3 2
92 54 91 oveq12i 16 F 2 5 20 F 2 3 = F 5 2 5 F 3 2
93 92 oveq1i 16 F 2 5 - 20 F 2 3 + 5 F 2 = F 5 2 - 5 F 3 2 + 5 F 2
94 2 3 93 3eqtr3i 1 = F 5 2 - 5 F 3 2 + 5 F 2