Metamath Proof Explorer


Theorem evl1at0

Description: Polynomial evaluation for the 0 scalar. (Contributed by AV, 10-Aug-2019)

Ref Expression
Hypotheses evl1at0.o ⊢ O = eval 1 ⁡ R
evl1at0.p ⊢ P = Poly 1 ⁡ R
evl1at0.0 ⊢ 0 ˙ = 0 R
evl1at0.z ⊢ Z = 0 P
Assertion evl1at0 ⊢ R ∈ CRing → O ⁡ Z ⁡ 0 ˙ = 0 ˙

Proof

Step Hyp Ref Expression
1 evl1at0.o ⊢ O = eval 1 ⁡ R
2 evl1at0.p ⊢ P = Poly 1 ⁡ R
3 evl1at0.0 ⊢ 0 ˙ = 0 R
4 evl1at0.z ⊢ Z = 0 P
5 crngring ⊢ R ∈ CRing → R ∈ Ring
6 eqid ⊢ algSc ⁡ P = algSc ⁡ P
7 2 6 3 4 ply1scl0 ⊢ R ∈ Ring → algSc ⁡ P ⁡ 0 ˙ = Z
8 5 7 syl ⊢ R ∈ CRing → algSc ⁡ P ⁡ 0 ˙ = Z
9 8 eqcomd ⊢ R ∈ CRing → Z = algSc ⁡ P ⁡ 0 ˙
10 9 fveq2d ⊢ R ∈ CRing → O ⁡ Z = O ⁡ algSc ⁡ P ⁡ 0 ˙
11 10 fveq1d ⊢ R ∈ CRing → O ⁡ Z ⁡ 0 ˙ = O ⁡ algSc ⁡ P ⁡ 0 ˙ ⁡ 0 ˙
12 eqid ⊢ Base R = Base R
13 eqid ⊢ Base P = Base P
14 id ⊢ R ∈ CRing → R ∈ CRing
15 ringgrp ⊢ R ∈ Ring → R ∈ Grp
16 12 3 grpidcl ⊢ R ∈ Grp → 0 ˙ ∈ Base R
17 5 15 16 3syl ⊢ R ∈ CRing → 0 ˙ ∈ Base R
18 1 2 12 6 13 14 17 17 evl1scad ⊢ R ∈ CRing → algSc ⁡ P ⁡ 0 ˙ ∈ Base P ∧ O ⁡ algSc ⁡ P ⁡ 0 ˙ ⁡ 0 ˙ = 0 ˙
19 18 simprd ⊢ R ∈ CRing → O ⁡ algSc ⁡ P ⁡ 0 ˙ ⁡ 0 ˙ = 0 ˙
20 11 19 eqtrd ⊢ R ∈ CRing → O ⁡ Z ⁡ 0 ˙ = 0 ˙