Metamath Proof Explorer


Theorem goldpolyfactor

Description: Factorization of a polynomial which has golden ratio among its roots, done by term-by-term-by-term multiplying and summing a few shorter polynomials. (Contributed by Ender Ting, 24-Jul-2026)

Ref Expression
Hypothesis goldpolyfactor.1 ⊢ F ∈ ℂ
Assertion goldpolyfactor ⊢ F 2 - F - 1 ⁢ F 2 - F - 1 ⁢ F + 2 = F 5 − 5 ⁢ F 3 + 5 ⁢ F + 2

Proof

Step Hyp Ref Expression
1 goldpolyfactor.1 ⊢ F ∈ ℂ
2 1 sqcli ⊢ F 2 ∈ ℂ
3 2 1 subcli ⊢ F 2 − F ∈ ℂ
4 ax-1cn ⊢ 1 ∈ ℂ
5 3 4 subcli ⊢ F 2 - F - 1 ∈ ℂ
6 5 3 4 subdii ⊢ F 2 - F - 1 ⁢ F 2 - F - 1 = F 2 - F - 1 ⁢ F 2 − F − F 2 - F - 1 ⋅ 1
7 5 2 1 subdii ⊢ F 2 - F - 1 ⁢ F 2 − F = F 2 - F - 1 ⁢ F 2 − F 2 - F - 1 ⁢ F
8 3 4 2 subdiri ⊢ F 2 - F - 1 ⁢ F 2 = F 2 − F ⁢ F 2 − 1 ⁢ F 2
9 2 1 2 subdiri ⊢ F 2 − F ⁢ F 2 = F 2 ⁢ F 2 − F ⁢ F 2
10 2nn0 ⊢ 2 ∈ ℕ 0
11 expadd ⊢ F ∈ ℂ ∧ 2 ∈ ℕ 0 ∧ 2 ∈ ℕ 0 → F 2 + 2 = F 2 ⁢ F 2
12 1 10 10 11 mp3an ⊢ F 2 + 2 = F 2 ⁢ F 2
13 2p2e4 ⊢ 2 + 2 = 4
14 13 oveq2i ⊢ F 2 + 2 = F 4
15 12 14 eqtr3i ⊢ F 2 ⁢ F 2 = F 4
16 1 2 mulcomi ⊢ F ⁢ F 2 = F 2 ⁢ F
17 df-3 ⊢ 3 = 2 + 1
18 17 oveq2i ⊢ F 3 = F 2 + 1
19 expp1 ⊢ F ∈ ℂ ∧ 2 ∈ ℕ 0 → F 2 + 1 = F 2 ⁢ F
20 1 10 19 mp2an ⊢ F 2 + 1 = F 2 ⁢ F
21 18 20 eqtr2i ⊢ F 2 ⁢ F = F 3
22 16 21 eqtri ⊢ F ⁢ F 2 = F 3
23 15 22 oveq12i ⊢ F 2 ⁢ F 2 − F ⁢ F 2 = F 4 − F 3
24 9 23 eqtri ⊢ F 2 − F ⁢ F 2 = F 4 − F 3
25 2 mullidi ⊢ 1 ⁢ F 2 = F 2
26 24 25 oveq12i ⊢ F 2 − F ⁢ F 2 − 1 ⁢ F 2 = F 4 - F 3 - F 2
27 8 26 eqtri ⊢ F 2 - F - 1 ⁢ F 2 = F 4 - F 3 - F 2
28 3 4 1 subdiri ⊢ F 2 - F - 1 ⁢ F = F 2 − F ⁢ F − 1 ⁢ F
29 2 1 1 subdiri ⊢ F 2 − F ⁢ F = F 2 ⁢ F − F ⁢ F
30 1 sqvali ⊢ F 2 = F ⁢ F
31 30 eqcomi ⊢ F ⁢ F = F 2
32 21 31 oveq12i ⊢ F 2 ⁢ F − F ⁢ F = F 3 − F 2
33 29 32 eqtri ⊢ F 2 − F ⁢ F = F 3 − F 2
34 1 mullidi ⊢ 1 ⁢ F = F
35 33 34 oveq12i ⊢ F 2 − F ⁢ F − 1 ⁢ F = F 3 - F 2 - F
36 28 35 eqtri ⊢ F 2 - F - 1 ⁢ F = F 3 - F 2 - F
37 27 36 oveq12i ⊢ F 2 - F - 1 ⁢ F 2 − F 2 - F - 1 ⁢ F = F 4 − F 3 - F 2 - F 3 - F 2 - F
38 4nn0 ⊢ 4 ∈ ℕ 0
39 expcl ⊢ F ∈ ℂ ∧ 4 ∈ ℕ 0 → F 4 ∈ ℂ
40 1 38 39 mp2an ⊢ F 4 ∈ ℂ
41 3nn0 ⊢ 3 ∈ ℕ 0
42 expcl ⊢ F ∈ ℂ ∧ 3 ∈ ℕ 0 → F 3 ∈ ℂ
43 1 41 42 mp2an ⊢ F 3 ∈ ℂ
44 40 43 subcli ⊢ F 4 − F 3 ∈ ℂ
45 44 2 subcli ⊢ F 4 - F 3 - F 2 ∈ ℂ
46 43 2 subcli ⊢ F 3 − F 2 ∈ ℂ
47 subsub ⊢ F 4 - F 3 - F 2 ∈ ℂ ∧ F 3 − F 2 ∈ ℂ ∧ F ∈ ℂ → F 4 − F 3 - F 2 - F 3 - F 2 - F = F 4 - F 3 - F 2 - F 3 − F 2 + F
48 45 46 1 47 mp3an ⊢ F 4 − F 3 - F 2 - F 3 - F 2 - F = F 4 - F 3 - F 2 - F 3 − F 2 + F
49 nnncan2 ⊢ F 4 − F 3 ∈ ℂ ∧ F 3 ∈ ℂ ∧ F 2 ∈ ℂ → F 4 − F 3 - F 2 - F 3 − F 2 = F 4 - F 3 - F 3
50 44 43 2 49 mp3an ⊢ F 4 − F 3 - F 2 - F 3 − F 2 = F 4 - F 3 - F 3
51 subsub4 ⊢ F 4 ∈ ℂ ∧ F 3 ∈ ℂ ∧ F 3 ∈ ℂ → F 4 - F 3 - F 3 = F 4 − F 3 + F 3
52 40 43 43 51 mp3an ⊢ F 4 - F 3 - F 3 = F 4 − F 3 + F 3
53 43 2timesi ⊢ 2 ⁢ F 3 = F 3 + F 3
54 53 oveq2i ⊢ F 4 − 2 ⁢ F 3 = F 4 − F 3 + F 3
55 52 54 eqtr4i ⊢ F 4 - F 3 - F 3 = F 4 − 2 ⁢ F 3
56 50 55 eqtri ⊢ F 4 − F 3 - F 2 - F 3 − F 2 = F 4 − 2 ⁢ F 3
57 56 oveq1i ⊢ F 4 - F 3 - F 2 - F 3 − F 2 + F = F 4 - 2 ⁢ F 3 + F
58 48 57 eqtri ⊢ F 4 − F 3 - F 2 - F 3 - F 2 - F = F 4 - 2 ⁢ F 3 + F
59 7 37 58 3eqtri ⊢ F 2 - F - 1 ⁢ F 2 − F = F 4 - 2 ⁢ F 3 + F
60 5 mulridi ⊢ F 2 - F - 1 ⋅ 1 = F 2 - F - 1
61 59 60 oveq12i ⊢ F 2 - F - 1 ⁢ F 2 − F − F 2 - F - 1 ⋅ 1 = F 4 − 2 ⁢ F 3 + F - F 2 - F - 1
62 2cn ⊢ 2 ∈ ℂ
63 62 43 mulcli ⊢ 2 ⁢ F 3 ∈ ℂ
64 40 63 subcli ⊢ F 4 − 2 ⁢ F 3 ∈ ℂ
65 64 1 addcli ⊢ F 4 - 2 ⁢ F 3 + F ∈ ℂ
66 subsub ⊢ F 4 - 2 ⁢ F 3 + F ∈ ℂ ∧ F 2 − F ∈ ℂ ∧ 1 ∈ ℂ → F 4 − 2 ⁢ F 3 + F - F 2 - F - 1 = F 4 - 2 ⁢ F 3 + F - F 2 − F + 1
67 65 3 4 66 mp3an ⊢ F 4 − 2 ⁢ F 3 + F - F 2 - F - 1 = F 4 - 2 ⁢ F 3 + F - F 2 − F + 1
68 64 a1i ⊢ ⊤ → F 4 − 2 ⁢ F 3 ∈ ℂ
69 1 a1i ⊢ ⊤ → F ∈ ℂ
70 2 a1i ⊢ ⊤ → F 2 ∈ ℂ
71 68 69 70 69 addsubsub23 ⊢ ⊤ → F 4 − 2 ⁢ F 3 + F - F 2 − F = F 4 - 2 ⁢ F 3 - F 2 + F + F
72 71 mptru ⊢ F 4 − 2 ⁢ F 3 + F - F 2 − F = F 4 - 2 ⁢ F 3 - F 2 + F + F
73 1 2timesi ⊢ 2 ⁢ F = F + F
74 73 oveq2i ⊢ F 4 − 2 ⁢ F 3 - F 2 + 2 ⁢ F = F 4 - 2 ⁢ F 3 - F 2 + F + F
75 72 74 eqtr4i ⊢ F 4 − 2 ⁢ F 3 + F - F 2 − F = F 4 − 2 ⁢ F 3 - F 2 + 2 ⁢ F
76 75 oveq1i ⊢ F 4 - 2 ⁢ F 3 + F - F 2 − F + 1 = F 4 - 2 ⁢ F 3 - F 2 + 2 ⁢ F + 1
77 67 76 eqtri ⊢ F 4 − 2 ⁢ F 3 + F - F 2 - F - 1 = F 4 - 2 ⁢ F 3 - F 2 + 2 ⁢ F + 1
78 6 61 77 3eqtri ⊢ F 2 - F - 1 ⁢ F 2 - F - 1 = F 4 - 2 ⁢ F 3 - F 2 + 2 ⁢ F + 1
79 78 oveq1i ⊢ F 2 - F - 1 ⁢ F 2 - F - 1 ⁢ F + 2 = F 4 - 2 ⁢ F 3 - F 2 + 2 ⁢ F + 1 ⁢ F + 2
80 64 2 subcli ⊢ F 4 - 2 ⁢ F 3 - F 2 ∈ ℂ
81 62 1 mulcli ⊢ 2 ⁢ F ∈ ℂ
82 80 81 addcli ⊢ F 4 − 2 ⁢ F 3 - F 2 + 2 ⁢ F ∈ ℂ
83 82 4 addcli ⊢ F 4 - 2 ⁢ F 3 - F 2 + 2 ⁢ F + 1 ∈ ℂ
84 83 1 62 adddii ⊢ F 4 - 2 ⁢ F 3 - F 2 + 2 ⁢ F + 1 ⁢ F + 2 = F 4 - 2 ⁢ F 3 - F 2 + 2 ⁢ F + 1 ⁢ F + F 4 - 2 ⁢ F 3 - F 2 + 2 ⁢ F + 1 ⋅ 2
85 82 4 1 adddiri ⊢ F 4 - 2 ⁢ F 3 - F 2 + 2 ⁢ F + 1 ⁢ F = F 4 − 2 ⁢ F 3 - F 2 + 2 ⁢ F ⁢ F + 1 ⁢ F
86 80 81 1 adddiri ⊢ F 4 − 2 ⁢ F 3 - F 2 + 2 ⁢ F ⁢ F = F 4 - 2 ⁢ F 3 - F 2 ⁢ F + 2 ⁢ F ⁢ F
87 64 2 1 subdiri ⊢ F 4 - 2 ⁢ F 3 - F 2 ⁢ F = F 4 − 2 ⁢ F 3 ⁢ F − F 2 ⁢ F
88 40 63 1 subdiri ⊢ F 4 − 2 ⁢ F 3 ⁢ F = F 4 ⁢ F − 2 ⁢ F 3 ⁢ F
89 df-5 ⊢ 5 = 4 + 1
90 89 oveq2i ⊢ F 5 = F 4 + 1
91 expp1 ⊢ F ∈ ℂ ∧ 4 ∈ ℕ 0 → F 4 + 1 = F 4 ⁢ F
92 1 38 91 mp2an ⊢ F 4 + 1 = F 4 ⁢ F
93 90 92 eqtr2i ⊢ F 4 ⁢ F = F 5
94 62 43 1 mulassi ⊢ 2 ⁢ F 3 ⁢ F = 2 ⁢ F 3 ⁢ F
95 df-4 ⊢ 4 = 3 + 1
96 95 oveq2i ⊢ F 4 = F 3 + 1
97 expp1 ⊢ F ∈ ℂ ∧ 3 ∈ ℕ 0 → F 3 + 1 = F 3 ⁢ F
98 1 41 97 mp2an ⊢ F 3 + 1 = F 3 ⁢ F
99 96 98 eqtri ⊢ F 4 = F 3 ⁢ F
100 99 oveq2i ⊢ 2 ⁢ F 4 = 2 ⁢ F 3 ⁢ F
101 94 100 eqtr4i ⊢ 2 ⁢ F 3 ⁢ F = 2 ⁢ F 4
102 93 101 oveq12i ⊢ F 4 ⁢ F − 2 ⁢ F 3 ⁢ F = F 5 − 2 ⁢ F 4
103 88 102 eqtri ⊢ F 4 − 2 ⁢ F 3 ⁢ F = F 5 − 2 ⁢ F 4
104 103 21 oveq12i ⊢ F 4 − 2 ⁢ F 3 ⁢ F − F 2 ⁢ F = F 5 - 2 ⁢ F 4 - F 3
105 87 104 eqtri ⊢ F 4 - 2 ⁢ F 3 - F 2 ⁢ F = F 5 - 2 ⁢ F 4 - F 3
106 62 1 1 mulassi ⊢ 2 ⁢ F ⁢ F = 2 ⁢ F ⁢ F
107 30 oveq2i ⊢ 2 ⁢ F 2 = 2 ⁢ F ⁢ F
108 106 107 eqtr4i ⊢ 2 ⁢ F ⁢ F = 2 ⁢ F 2
109 105 108 oveq12i ⊢ F 4 - 2 ⁢ F 3 - F 2 ⁢ F + 2 ⁢ F ⁢ F = F 5 − 2 ⁢ F 4 - F 3 + 2 ⁢ F 2
110 86 109 eqtri ⊢ F 4 − 2 ⁢ F 3 - F 2 + 2 ⁢ F ⁢ F = F 5 − 2 ⁢ F 4 - F 3 + 2 ⁢ F 2
111 110 34 oveq12i ⊢ F 4 − 2 ⁢ F 3 - F 2 + 2 ⁢ F ⁢ F + 1 ⁢ F = F 5 - 2 ⁢ F 4 - F 3 + 2 ⁢ F 2 + F
112 85 111 eqtri ⊢ F 4 - 2 ⁢ F 3 - F 2 + 2 ⁢ F + 1 ⁢ F = F 5 - 2 ⁢ F 4 - F 3 + 2 ⁢ F 2 + F
113 82 4 62 adddiri ⊢ F 4 - 2 ⁢ F 3 - F 2 + 2 ⁢ F + 1 ⋅ 2 = F 4 − 2 ⁢ F 3 - F 2 + 2 ⁢ F ⋅ 2 + 1 ⋅ 2
114 80 81 62 adddiri ⊢ F 4 − 2 ⁢ F 3 - F 2 + 2 ⁢ F ⋅ 2 = F 4 - 2 ⁢ F 3 - F 2 ⋅ 2 + 2 ⁢ F ⋅ 2
115 64 2 62 subdiri ⊢ F 4 - 2 ⁢ F 3 - F 2 ⋅ 2 = F 4 − 2 ⁢ F 3 ⋅ 2 − F 2 ⋅ 2
116 40 63 62 subdiri ⊢ F 4 − 2 ⁢ F 3 ⋅ 2 = F 4 ⋅ 2 − 2 ⁢ F 3 ⋅ 2
117 40 62 mulcomi ⊢ F 4 ⋅ 2 = 2 ⁢ F 4
118 62 43 62 mul32i ⊢ 2 ⁢ F 3 ⋅ 2 = 2 ⋅ 2 ⁢ F 3
119 2t2e4 ⊢ 2 ⋅ 2 = 4
120 119 oveq1i ⊢ 2 ⋅ 2 ⁢ F 3 = 4 ⁢ F 3
121 118 120 eqtri ⊢ 2 ⁢ F 3 ⋅ 2 = 4 ⁢ F 3
122 117 121 oveq12i ⊢ F 4 ⋅ 2 − 2 ⁢ F 3 ⋅ 2 = 2 ⁢ F 4 − 4 ⁢ F 3
123 116 122 eqtri ⊢ F 4 − 2 ⁢ F 3 ⋅ 2 = 2 ⁢ F 4 − 4 ⁢ F 3
124 2 62 mulcomi ⊢ F 2 ⋅ 2 = 2 ⁢ F 2
125 123 124 oveq12i ⊢ F 4 − 2 ⁢ F 3 ⋅ 2 − F 2 ⋅ 2 = 2 ⁢ F 4 - 4 ⁢ F 3 - 2 ⁢ F 2
126 115 125 eqtri ⊢ F 4 - 2 ⁢ F 3 - F 2 ⋅ 2 = 2 ⁢ F 4 - 4 ⁢ F 3 - 2 ⁢ F 2
127 62 1 62 mul32i ⊢ 2 ⁢ F ⋅ 2 = 2 ⋅ 2 ⁢ F
128 119 oveq1i ⊢ 2 ⋅ 2 ⁢ F = 4 ⁢ F
129 127 128 eqtri ⊢ 2 ⁢ F ⋅ 2 = 4 ⁢ F
130 126 129 oveq12i ⊢ F 4 - 2 ⁢ F 3 - F 2 ⋅ 2 + 2 ⁢ F ⋅ 2 = 2 ⁢ F 4 − 4 ⁢ F 3 - 2 ⁢ F 2 + 4 ⁢ F
131 114 130 eqtri ⊢ F 4 − 2 ⁢ F 3 - F 2 + 2 ⁢ F ⋅ 2 = 2 ⁢ F 4 − 4 ⁢ F 3 - 2 ⁢ F 2 + 4 ⁢ F
132 62 mullidi ⊢ 1 ⋅ 2 = 2
133 131 132 oveq12i ⊢ F 4 − 2 ⁢ F 3 - F 2 + 2 ⁢ F ⋅ 2 + 1 ⋅ 2 = 2 ⁢ F 4 - 4 ⁢ F 3 - 2 ⁢ F 2 + 4 ⁢ F + 2
134 113 133 eqtri ⊢ F 4 - 2 ⁢ F 3 - F 2 + 2 ⁢ F + 1 ⋅ 2 = 2 ⁢ F 4 - 4 ⁢ F 3 - 2 ⁢ F 2 + 4 ⁢ F + 2
135 112 134 oveq12i ⊢ F 4 - 2 ⁢ F 3 - F 2 + 2 ⁢ F + 1 ⁢ F + F 4 - 2 ⁢ F 3 - F 2 + 2 ⁢ F + 1 ⋅ 2 = F 5 − 2 ⁢ F 4 - F 3 + 2 ⁢ F 2 + F + 2 ⁢ F 4 − 4 ⁢ F 3 - 2 ⁢ F 2 + 4 ⁢ F + 2
136 5nn0 ⊢ 5 ∈ ℕ 0
137 expcl ⊢ F ∈ ℂ ∧ 5 ∈ ℕ 0 → F 5 ∈ ℂ
138 1 136 137 mp2an ⊢ F 5 ∈ ℂ
139 62 40 mulcli ⊢ 2 ⁢ F 4 ∈ ℂ
140 138 139 subcli ⊢ F 5 − 2 ⁢ F 4 ∈ ℂ
141 140 43 subcli ⊢ F 5 - 2 ⁢ F 4 - F 3 ∈ ℂ
142 62 2 mulcli ⊢ 2 ⁢ F 2 ∈ ℂ
143 141 142 addcli ⊢ F 5 − 2 ⁢ F 4 - F 3 + 2 ⁢ F 2 ∈ ℂ
144 143 1 addcli ⊢ F 5 - 2 ⁢ F 4 - F 3 + 2 ⁢ F 2 + F ∈ ℂ
145 4cn ⊢ 4 ∈ ℂ
146 145 43 mulcli ⊢ 4 ⁢ F 3 ∈ ℂ
147 139 146 subcli ⊢ 2 ⁢ F 4 − 4 ⁢ F 3 ∈ ℂ
148 147 142 subcli ⊢ 2 ⁢ F 4 - 4 ⁢ F 3 - 2 ⁢ F 2 ∈ ℂ
149 145 1 mulcli ⊢ 4 ⁢ F ∈ ℂ
150 148 149 addcli ⊢ 2 ⁢ F 4 − 4 ⁢ F 3 - 2 ⁢ F 2 + 4 ⁢ F ∈ ℂ
151 144 150 62 addassi ⊢ F 5 - 2 ⁢ F 4 - F 3 + 2 ⁢ F 2 + F + 2 ⁢ F 4 − 4 ⁢ F 3 - 2 ⁢ F 2 + 4 ⁢ F + 2 = F 5 − 2 ⁢ F 4 - F 3 + 2 ⁢ F 2 + F + 2 ⁢ F 4 − 4 ⁢ F 3 - 2 ⁢ F 2 + 4 ⁢ F + 2
152 143 1 148 149 add4i ⊢ F 5 − 2 ⁢ F 4 - F 3 + 2 ⁢ F 2 + F + 2 ⁢ F 4 - 4 ⁢ F 3 - 2 ⁢ F 2 + 4 ⁢ F = F 5 − 2 ⁢ F 4 - F 3 + 2 ⁢ F 2 + 2 ⁢ F 4 - 4 ⁢ F 3 - 2 ⁢ F 2 + F + 4 ⁢ F
153 ppncan ⊢ F 5 - 2 ⁢ F 4 - F 3 ∈ ℂ ∧ 2 ⁢ F 2 ∈ ℂ ∧ 2 ⁢ F 4 − 4 ⁢ F 3 ∈ ℂ → F 5 - 2 ⁢ F 4 - F 3 + 2 ⁢ F 2 + 2 ⁢ F 4 - 4 ⁢ F 3 - 2 ⁢ F 2 = F 5 - 2 ⁢ F 4 - F 3 + 2 ⁢ F 4 - 4 ⁢ F 3
154 141 142 147 153 mp3an ⊢ F 5 - 2 ⁢ F 4 - F 3 + 2 ⁢ F 2 + 2 ⁢ F 4 - 4 ⁢ F 3 - 2 ⁢ F 2 = F 5 - 2 ⁢ F 4 - F 3 + 2 ⁢ F 4 - 4 ⁢ F 3
155 140 139 43 146 addsub4i ⊢ F 5 − 2 ⁢ F 4 + 2 ⁢ F 4 - F 3 + 4 ⁢ F 3 = F 5 - 2 ⁢ F 4 - F 3 + 2 ⁢ F 4 - 4 ⁢ F 3
156 npcan ⊢ F 5 ∈ ℂ ∧ 2 ⁢ F 4 ∈ ℂ → F 5 - 2 ⁢ F 4 + 2 ⁢ F 4 = F 5
157 138 139 156 mp2an ⊢ F 5 - 2 ⁢ F 4 + 2 ⁢ F 4 = F 5
158 145 4 89 comraddi ⊢ 5 = 1 + 4
159 158 oveq1i ⊢ 5 ⁢ F 3 = 1 + 4 ⁢ F 3
160 4 145 43 adddiri ⊢ 1 + 4 ⁢ F 3 = 1 ⁢ F 3 + 4 ⁢ F 3
161 43 mullidi ⊢ 1 ⁢ F 3 = F 3
162 161 oveq1i ⊢ 1 ⁢ F 3 + 4 ⁢ F 3 = F 3 + 4 ⁢ F 3
163 159 160 162 3eqtrri ⊢ F 3 + 4 ⁢ F 3 = 5 ⁢ F 3
164 157 163 oveq12i ⊢ F 5 − 2 ⁢ F 4 + 2 ⁢ F 4 - F 3 + 4 ⁢ F 3 = F 5 − 5 ⁢ F 3
165 154 155 164 3eqtr2i ⊢ F 5 - 2 ⁢ F 4 - F 3 + 2 ⁢ F 2 + 2 ⁢ F 4 - 4 ⁢ F 3 - 2 ⁢ F 2 = F 5 − 5 ⁢ F 3
166 4 145 1 adddiri ⊢ 1 + 4 ⁢ F = 1 ⁢ F + 4 ⁢ F
167 158 oveq1i ⊢ 5 ⁢ F = 1 + 4 ⁢ F
168 34 eqcomi ⊢ F = 1 ⁢ F
169 168 oveq1i ⊢ F + 4 ⁢ F = 1 ⁢ F + 4 ⁢ F
170 166 167 169 3eqtr4ri ⊢ F + 4 ⁢ F = 5 ⁢ F
171 165 170 oveq12i ⊢ F 5 − 2 ⁢ F 4 - F 3 + 2 ⁢ F 2 + 2 ⁢ F 4 - 4 ⁢ F 3 - 2 ⁢ F 2 + F + 4 ⁢ F = F 5 - 5 ⁢ F 3 + 5 ⁢ F
172 152 171 eqtri ⊢ F 5 − 2 ⁢ F 4 - F 3 + 2 ⁢ F 2 + F + 2 ⁢ F 4 - 4 ⁢ F 3 - 2 ⁢ F 2 + 4 ⁢ F = F 5 - 5 ⁢ F 3 + 5 ⁢ F
173 172 oveq1i ⊢ F 5 - 2 ⁢ F 4 - F 3 + 2 ⁢ F 2 + F + 2 ⁢ F 4 − 4 ⁢ F 3 - 2 ⁢ F 2 + 4 ⁢ F + 2 = F 5 − 5 ⁢ F 3 + 5 ⁢ F + 2
174 135 151 173 3eqtr2i ⊢ F 4 - 2 ⁢ F 3 - F 2 + 2 ⁢ F + 1 ⁢ F + F 4 - 2 ⁢ F 3 - F 2 + 2 ⁢ F + 1 ⋅ 2 = F 5 − 5 ⁢ F 3 + 5 ⁢ F + 2
175 79 84 174 3eqtri ⊢ F 2 - F - 1 ⁢ F 2 - F - 1 ⁢ F + 2 = F 5 − 5 ⁢ F 3 + 5 ⁢ F + 2