Metamath Proof Explorer


Theorem evlvvval

Description: Give a formula for the evaluation of a polynomial given assignments from variables to values. (Contributed by SN, 5-Mar-2025)

Ref Expression
Hypotheses evlvvval.q ⊢ Q = I eval R
evlvvval.p ⊢ P = I mPoly R
evlvvval.b ⊢ B = Base P
evlvvval.d ⊢ D = h ∈ ℕ 0 I | h -1 ℕ ∈ Fin
evlvvval.k ⊢ K = Base R
evlvvval.m ⊢ M = mulGrp R
evlvvval.w ⊢ × ˙ = ⋅ M
evlvvval.x ⊢ · ˙ = ⋅ R
evlvvval.i ⊢ φ → I ∈ V
evlvvval.r ⊢ φ → R ∈ CRing
evlvvval.f ⊢ φ → F ∈ B
evlvvval.a ⊢ φ → A ∈ K I
Assertion evlvvval ⊢ φ → Q ⁡ F ⁡ A = ∑ R b ∈ D F ⁡ b · ˙ ∑ M i ∈ I b ⁡ i × ˙ A ⁡ i

Proof

Step Hyp Ref Expression
1 evlvvval.q ⊢ Q = I eval R
2 evlvvval.p ⊢ P = I mPoly R
3 evlvvval.b ⊢ B = Base P
4 evlvvval.d ⊢ D = h ∈ ℕ 0 I | h -1 ℕ ∈ Fin
5 evlvvval.k ⊢ K = Base R
6 evlvvval.m ⊢ M = mulGrp R
7 evlvvval.w ⊢ × ˙ = ⋅ M
8 evlvvval.x ⊢ · ˙ = ⋅ R
9 evlvvval.i ⊢ φ → I ∈ V
10 evlvvval.r ⊢ φ → R ∈ CRing
11 evlvvval.f ⊢ φ → F ∈ B
12 evlvvval.a ⊢ φ → A ∈ K I
13 eqid ⊢ I evalSub R ⁡ Base R = I evalSub R ⁡ Base R
14 eqid ⊢ I mPoly R ↾ 𝑠 Base R = I mPoly R ↾ 𝑠 Base R
15 eqid ⊢ R ↾ 𝑠 Base R = R ↾ 𝑠 Base R
16 eqid ⊢ Base I mPoly R ↾ 𝑠 Base R = Base I mPoly R ↾ 𝑠 Base R
17 10 crngringd ⊢ φ → R ∈ Ring
18 eqid ⊢ Base R = Base R
19 18 subrgid ⊢ R ∈ Ring → Base R ∈ SubRing ⁡ R
20 17 19 syl ⊢ φ → Base R ∈ SubRing ⁡ R
21 18 ressid ⊢ R ∈ CRing → R ↾ 𝑠 Base R = R
22 10 21 syl ⊢ φ → R ↾ 𝑠 Base R = R
23 22 oveq2d ⊢ φ → I mPoly R ↾ 𝑠 Base R = I mPoly R
24 23 2 eqtr4di ⊢ φ → I mPoly R ↾ 𝑠 Base R = P
25 24 fveq2d ⊢ φ → Base I mPoly R ↾ 𝑠 Base R = Base P
26 25 3 eqtr4di ⊢ φ → Base I mPoly R ↾ 𝑠 Base R = B
27 11 26 eleqtrrd ⊢ φ → F ∈ Base I mPoly R ↾ 𝑠 Base R
28 13 1 14 15 16 9 10 20 27 evlsevl ⊢ φ → I evalSub R ⁡ Base R ⁡ F = Q ⁡ F
29 28 fveq1d ⊢ φ → I evalSub R ⁡ Base R ⁡ F ⁡ A = Q ⁡ F ⁡ A
30 13 14 16 15 4 5 6 7 8 9 10 20 27 12 evlsvvval ⊢ φ → I evalSub R ⁡ Base R ⁡ F ⁡ A = ∑ R b ∈ D F ⁡ b · ˙ ∑ M i ∈ I b ⁡ i × ˙ A ⁡ i
31 29 30 eqtr3d ⊢ φ → Q ⁡ F ⁡ A = ∑ R b ∈ D F ⁡ b · ˙ ∑ M i ∈ I b ⁡ i × ˙ A ⁡ i