Metamath Proof Explorer


Theorem deg1le0

Description: A polynomial has nonpositive degree iff it is a constant. (Contributed by Stefan O'Rear, 29-Mar-2015)

Ref Expression
Hypotheses deg1le0.d ⊢ D = deg 1 ⁡ R
deg1le0.p ⊢ P = Poly 1 ⁡ R
deg1le0.b ⊢ B = Base P
deg1le0.a ⊢ A = algSc ⁡ P
Assertion deg1le0 ⊢ R ∈ Ring ∧ F ∈ B → D ⁡ F ≤ 0 ↔ F = A ⁡ coe 1 ⁡ F ⁡ 0

Proof

Step Hyp Ref Expression
1 deg1le0.d ⊢ D = deg 1 ⁡ R
2 deg1le0.p ⊢ P = Poly 1 ⁡ R
3 deg1le0.b ⊢ B = Base P
4 deg1le0.a ⊢ A = algSc ⁡ P
5 eqid ⊢ 1 𝑜 mPoly R = 1 𝑜 mPoly R
6 1 deg1fval ⊢ D = 1 𝑜 mDeg R
7 1on ⊢ 1 𝑜 ∈ On
8 7 a1i ⊢ R ∈ Ring ∧ F ∈ B → 1 𝑜 ∈ On
9 simpl ⊢ R ∈ Ring ∧ F ∈ B → R ∈ Ring
10 2 3 ply1bas ⊢ B = Base 1 𝑜 mPoly R
11 2 4 ply1ascl ⊢ A = algSc ⁡ 1 𝑜 mPoly R
12 simpr ⊢ R ∈ Ring ∧ F ∈ B → F ∈ B
13 5 6 8 9 10 11 12 mdegle0 ⊢ R ∈ Ring ∧ F ∈ B → D ⁡ F ≤ 0 ↔ F = A ⁡ F ⁡ 1 𝑜 × 0
14 0nn0 ⊢ 0 ∈ ℕ 0
15 eqid ⊢ coe 1 ⁡ F = coe 1 ⁡ F
16 15 coe1fv ⊢ F ∈ B ∧ 0 ∈ ℕ 0 → coe 1 ⁡ F ⁡ 0 = F ⁡ 1 𝑜 × 0
17 12 14 16 sylancl ⊢ R ∈ Ring ∧ F ∈ B → coe 1 ⁡ F ⁡ 0 = F ⁡ 1 𝑜 × 0
18 17 fveq2d ⊢ R ∈ Ring ∧ F ∈ B → A ⁡ coe 1 ⁡ F ⁡ 0 = A ⁡ F ⁡ 1 𝑜 × 0
19 18 eqeq2d ⊢ R ∈ Ring ∧ F ∈ B → F = A ⁡ coe 1 ⁡ F ⁡ 0 ↔ F = A ⁡ F ⁡ 1 𝑜 × 0
20 13 19 bitr4d ⊢ R ∈ Ring ∧ F ∈ B → D ⁡ F ≤ 0 ↔ F = A ⁡ coe 1 ⁡ F ⁡ 0