Metamath Proof Explorer


Theorem ply1ascl0

Description: The zero scalar as a polynomial. (Contributed by Thierry Arnoux, 20-Jan-2025)

Ref Expression
Hypotheses ply1ascl0.w ⊢ W = Poly 1 ⁡ R
ply1ascl0.a ⊢ A = algSc ⁡ W
ply1ascl0.o ⊢ O = 0 R
ply1ascl0.1 ⊢ 0 ˙ = 0 W
ply1ascl0.r ⊢ φ → R ∈ Ring
Assertion ply1ascl0 ⊢ φ → A ⁡ O = 0 ˙

Proof

Step Hyp Ref Expression
1 ply1ascl0.w ⊢ W = Poly 1 ⁡ R
2 ply1ascl0.a ⊢ A = algSc ⁡ W
3 ply1ascl0.o ⊢ O = 0 R
4 ply1ascl0.1 ⊢ 0 ˙ = 0 W
5 ply1ascl0.r ⊢ φ → R ∈ Ring
6 1 ply1sca ⊢ R ∈ Ring → R = Scalar ⁡ W
7 5 6 syl ⊢ φ → R = Scalar ⁡ W
8 7 fveq2d ⊢ φ → 0 R = 0 Scalar ⁡ W
9 3 8 eqtrid ⊢ φ → O = 0 Scalar ⁡ W
10 9 fveq2d ⊢ φ → algSc ⁡ W ⁡ O = algSc ⁡ W ⁡ 0 Scalar ⁡ W
11 eqid ⊢ algSc ⁡ W = algSc ⁡ W
12 eqid ⊢ Scalar ⁡ W = Scalar ⁡ W
13 1 ply1lmod ⊢ R ∈ Ring → W ∈ LMod
14 5 13 syl ⊢ φ → W ∈ LMod
15 1 ply1ring ⊢ R ∈ Ring → W ∈ Ring
16 5 15 syl ⊢ φ → W ∈ Ring
17 11 12 14 16 ascl0 ⊢ φ → algSc ⁡ W ⁡ 0 Scalar ⁡ W = 0 W
18 10 17 eqtrd ⊢ φ → algSc ⁡ W ⁡ O = 0 W
19 2 fveq1i ⊢ A ⁡ O = algSc ⁡ W ⁡ O
20 18 19 4 3eqtr4g ⊢ φ → A ⁡ O = 0 ˙