Metamath Proof Explorer


Theorem deg1nn0clb

Description: A polynomial is nonzero iff it has definite degree. (Contributed by Stefan O'Rear, 23-Mar-2015)

Ref Expression
Hypotheses deg1z.d ⊢ D = deg 1 ⁡ R
deg1z.p ⊢ P = Poly 1 ⁡ R
deg1z.z ⊢ 0 ˙ = 0 P
deg1nn0cl.b ⊢ B = Base P
Assertion deg1nn0clb ⊢ R ∈ Ring ∧ F ∈ B → F ≠ 0 ˙ ↔ D ⁡ F ∈ ℕ 0

Proof

Step Hyp Ref Expression
1 deg1z.d ⊢ D = deg 1 ⁡ R
2 deg1z.p ⊢ P = Poly 1 ⁡ R
3 deg1z.z ⊢ 0 ˙ = 0 P
4 deg1nn0cl.b ⊢ B = Base P
5 1 2 3 4 deg1nn0cl ⊢ R ∈ Ring ∧ F ∈ B ∧ F ≠ 0 ˙ → D ⁡ F ∈ ℕ 0
6 5 3expia ⊢ R ∈ Ring ∧ F ∈ B → F ≠ 0 ˙ → D ⁡ F ∈ ℕ 0
7 mnfnre ⊢ −∞ ∉ ℝ
8 7 neli ⊢ ¬ −∞ ∈ ℝ
9 nn0re ⊢ −∞ ∈ ℕ 0 → −∞ ∈ ℝ
10 8 9 mto ⊢ ¬ −∞ ∈ ℕ 0
11 1 2 3 deg1z ⊢ R ∈ Ring → D ⁡ 0 ˙ = −∞
12 11 adantr ⊢ R ∈ Ring ∧ F ∈ B → D ⁡ 0 ˙ = −∞
13 12 eleq1d ⊢ R ∈ Ring ∧ F ∈ B → D ⁡ 0 ˙ ∈ ℕ 0 ↔ −∞ ∈ ℕ 0
14 10 13 mtbiri ⊢ R ∈ Ring ∧ F ∈ B → ¬ D ⁡ 0 ˙ ∈ ℕ 0
15 fveq2 ⊢ F = 0 ˙ → D ⁡ F = D ⁡ 0 ˙
16 15 eleq1d ⊢ F = 0 ˙ → D ⁡ F ∈ ℕ 0 ↔ D ⁡ 0 ˙ ∈ ℕ 0
17 16 notbid ⊢ F = 0 ˙ → ¬ D ⁡ F ∈ ℕ 0 ↔ ¬ D ⁡ 0 ˙ ∈ ℕ 0
18 14 17 syl5ibrcom ⊢ R ∈ Ring ∧ F ∈ B → F = 0 ˙ → ¬ D ⁡ F ∈ ℕ 0
19 18 necon2ad ⊢ R ∈ Ring ∧ F ∈ B → D ⁡ F ∈ ℕ 0 → F ≠ 0 ˙
20 6 19 impbid ⊢ R ∈ Ring ∧ F ∈ B → F ≠ 0 ˙ ↔ D ⁡ F ∈ ℕ 0