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 ∈ ℂ
5 4 mulm1i ⊢ -1 ⋅ 2 = − 2
6 1 goldrarr ⊢ F ∈ ℝ
7 6 recni ⊢ F ∈ ℂ
8 5nn0 ⊢ 5 ∈ ℕ 0
9 expcl ⊢ F ∈ ℂ ∧ 5 ∈ ℕ 0 → F 5 ∈ ℂ
10 7 8 9 mp2an ⊢ F 5 ∈ ℂ
11 2ne0 ⊢ 2 ≠ 0
12 10 4 11 divcli ⊢ F 5 2 ∈ ℂ
13 5cn ⊢ 5 ∈ ℂ
14 3nn0 ⊢ 3 ∈ ℕ 0
15 expcl ⊢ F ∈ ℂ ∧ 3 ∈ ℕ 0 → F 3 ∈ ℂ
16 7 14 15 mp2an ⊢ F 3 ∈ ℂ
17 16 4 11 divcli ⊢ F 3 2 ∈ ℂ
18 13 17 mulcli ⊢ 5 ⁢ F 3 2 ∈ ℂ
19 12 18 subcli ⊢ F 5 2 − 5 ⁢ F 3 2 ∈ ℂ
20 7 4 11 divcli ⊢ F 2 ∈ ℂ
21 13 20 mulcli ⊢ 5 ⁢ F 2 ∈ ℂ
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 ∈ ℂ
39 10 38 subcli ⊢ F 5 − 5 ⁢ F 3 ∈ ℂ
40 13 7 mulcli ⊢ 5 ⁢ F ∈ ℂ
41 39 40 addcli ⊢ F 5 - 5 ⁢ F 3 + 5 ⁢ F ∈ ℂ
42 addeq0 ⊢ F 5 - 5 ⁢ F 3 + 5 ⁢ F ∈ ℂ ∧ 2 ∈ ℂ → 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