Metamath Proof Explorer


Theorem goldratmolem3

Description: Lemma 3 for determining the value of golden ratio. (Contributed by Ender Ting, 23-Jul-2026)

Ref Expression
Hypothesis goldra.val F = 2 cos π 5
Assertion goldratmolem3 F 5 5 F 3 + 5 F + 2 = 0

Proof

Step Hyp Ref Expression
1 goldra.val F = 2 cos π 5
2 1 goldratmolem2 1 = F 5 2 - 5 F 3 2 + 5 F 2
3 2 oveq1i -1 2 = F 5 2 - 5 F 3 2 + 5 F 2 2
4 2cn 2
5 4 mulm1i -1 2 = 2
6 1 goldrarr F
7 6 recni F
8 5nn0 5 0
9 expcl F 5 0 F 5
10 7 8 9 mp2an F 5
11 2ne0 2 0
12 10 4 11 divcli F 5 2
13 5cn 5
14 3nn0 3 0
15 expcl F 3 0 F 3
16 7 14 15 mp2an F 3
17 16 4 11 divcli F 3 2
18 13 17 mulcli 5 F 3 2
19 12 18 subcli F 5 2 5 F 3 2
20 7 4 11 divcli F 2
21 13 20 mulcli 5 F 2
22 19 21 4 adddiri F 5 2 - 5 F 3 2 + 5 F 2 2 = F 5 2 5 F 3 2 2 + 5 F 2 2
23 12 18 4 subdiri F 5 2 5 F 3 2 2 = F 5 2 2 5 F 3 2 2
24 10 4 11 divcan1i F 5 2 2 = F 5
25 13 17 4 mulassi 5 F 3 2 2 = 5 F 3 2 2
26 16 4 11 divcan1i F 3 2 2 = F 3
27 26 oveq2i 5 F 3 2 2 = 5 F 3
28 25 27 eqtri 5 F 3 2 2 = 5 F 3
29 24 28 oveq12i F 5 2 2 5 F 3 2 2 = F 5 5 F 3
30 23 29 eqtri F 5 2 5 F 3 2 2 = F 5 5 F 3
31 13 20 4 mulassi 5 F 2 2 = 5 F 2 2
32 7 4 11 divcan1i F 2 2 = F
33 32 oveq2i 5 F 2 2 = 5 F
34 31 33 eqtri 5 F 2 2 = 5 F
35 30 34 oveq12i F 5 2 5 F 3 2 2 + 5 F 2 2 = F 5 - 5 F 3 + 5 F
36 22 35 eqtri F 5 2 - 5 F 3 2 + 5 F 2 2 = F 5 - 5 F 3 + 5 F
37 3 5 36 3eqtr3ri F 5 - 5 F 3 + 5 F = 2
38 13 16 mulcli 5 F 3
39 10 38 subcli F 5 5 F 3
40 13 7 mulcli 5 F
41 39 40 addcli F 5 - 5 F 3 + 5 F
42 addeq0 F 5 - 5 F 3 + 5 F 2 F 5 5 F 3 + 5 F + 2 = 0 F 5 - 5 F 3 + 5 F = 2
43 41 4 42 mp2an F 5 5 F 3 + 5 F + 2 = 0 F 5 - 5 F 3 + 5 F = 2
44 37 43 mpbir F 5 5 F 3 + 5 F + 2 = 0