Metamath Proof Explorer


Theorem evlsvval

Description: Give a formula for the evaluation of a polynomial. (Contributed by SN, 9-Feb-2025)

Ref Expression
Hypotheses evlsvval.q Q = I evalSub S R
evlsvval.p P = I mPoly U
evlsvval.b B = Base P
evlsvval.d D = h 0 I | h -1 Fin
evlsvval.k K = Base S
evlsvval.u U = S 𝑠 R
evlsvval.t T = S 𝑠 K I
evlsvval.m M = mulGrp T
evlsvval.w × ˙ = M
evlsvval.x · ˙ = T
evlsvval.f F = x R K I × x
evlsvval.g G = x I a K I a x
evlsvval.i φ I V
evlsvval.s φ S CRing
evlsvval.r φ R SubRing S
evlsvval.a φ A B
Assertion evlsvval φ Q A = T b D F A b · ˙ M b × ˙ f G

Proof

Step Hyp Ref Expression
1 evlsvval.q Q = I evalSub S R
2 evlsvval.p P = I mPoly U
3 evlsvval.b B = Base P
4 evlsvval.d D = h 0 I | h -1 Fin
5 evlsvval.k K = Base S
6 evlsvval.u U = S 𝑠 R
7 evlsvval.t T = S 𝑠 K I
8 evlsvval.m M = mulGrp T
9 evlsvval.w × ˙ = M
10 evlsvval.x · ˙ = T
11 evlsvval.f F = x R K I × x
12 evlsvval.g G = x I a K I a x
13 evlsvval.i φ I V
14 evlsvval.s φ S CRing
15 evlsvval.r φ R SubRing S
16 evlsvval.a φ A B
17 fveq1 p = A p b = A b
18 17 fveq2d p = A F p b = F A b
19 18 oveq1d p = A F p b · ˙ M b × ˙ f G = F A b · ˙ M b × ˙ f G
20 19 mpteq2dv p = A b D F p b · ˙ M b × ˙ f G = b D F A b · ˙ M b × ˙ f G
21 20 oveq2d p = A T b D F p b · ˙ M b × ˙ f G = T b D F A b · ˙ M b × ˙ f G
22 eqid p B T b D F p b · ˙ M b × ˙ f G = p B T b D F p b · ˙ M b × ˙ f G
23 1 2 3 4 5 6 7 8 9 10 22 11 12 13 14 15 evlsval3 φ Q = p B T b D F p b · ˙ M b × ˙ f G
24 ovexd φ T b D F A b · ˙ M b × ˙ f G V
25 21 23 16 24 fvmptd4 φ Q A = T b D F A b · ˙ M b × ˙ f G