Metamath Proof Explorer


Theorem deg1ldgn

Description: An index at which a polynomial is zero, cannot be its 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
deg1ldg.y ⊢ Y = 0 R
deg1ldg.a ⊢ A = coe 1 ⁡ F
deg1ldgn.r ⊢ φ → R ∈ Ring
deg1ldgn.f ⊢ φ → F ∈ B
deg1ldgn.x ⊢ φ → X ∈ ℕ 0
deg1ldgn.e ⊢ φ → A ⁡ X = Y
Assertion deg1ldgn ⊢ φ → D ⁡ F ≠ X

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 deg1ldg.y ⊢ Y = 0 R
6 deg1ldg.a ⊢ A = coe 1 ⁡ F
7 deg1ldgn.r ⊢ φ → R ∈ Ring
8 deg1ldgn.f ⊢ φ → F ∈ B
9 deg1ldgn.x ⊢ φ → X ∈ ℕ 0
10 deg1ldgn.e ⊢ φ → A ⁡ X = Y
11 fveq2 ⊢ D ⁡ F = X → A ⁡ D ⁡ F = A ⁡ X
12 11 adantl ⊢ φ ∧ D ⁡ F = X → A ⁡ D ⁡ F = A ⁡ X
13 7 adantr ⊢ φ ∧ D ⁡ F = X → R ∈ Ring
14 8 adantr ⊢ φ ∧ D ⁡ F = X → F ∈ B
15 eleq1a ⊢ X ∈ ℕ 0 → D ⁡ F = X → D ⁡ F ∈ ℕ 0
16 9 15 syl ⊢ φ → D ⁡ F = X → D ⁡ F ∈ ℕ 0
17 16 imp ⊢ φ ∧ D ⁡ F = X → D ⁡ F ∈ ℕ 0
18 1 2 3 4 deg1nn0clb ⊢ R ∈ Ring ∧ F ∈ B → F ≠ 0 ˙ ↔ D ⁡ F ∈ ℕ 0
19 7 8 18 syl2anc ⊢ φ → F ≠ 0 ˙ ↔ D ⁡ F ∈ ℕ 0
20 19 adantr ⊢ φ ∧ D ⁡ F = X → F ≠ 0 ˙ ↔ D ⁡ F ∈ ℕ 0
21 17 20 mpbird ⊢ φ ∧ D ⁡ F = X → F ≠ 0 ˙
22 1 2 3 4 5 6 deg1ldg ⊢ R ∈ Ring ∧ F ∈ B ∧ F ≠ 0 ˙ → A ⁡ D ⁡ F ≠ Y
23 13 14 21 22 syl3anc ⊢ φ ∧ D ⁡ F = X → A ⁡ D ⁡ F ≠ Y
24 12 23 eqnetrrd ⊢ φ ∧ D ⁡ F = X → A ⁡ X ≠ Y
25 24 ex ⊢ φ → D ⁡ F = X → A ⁡ X ≠ Y
26 25 necon2d ⊢ φ → A ⁡ X = Y → D ⁡ F ≠ X
27 10 26 mpd ⊢ φ → D ⁡ F ≠ X