Metamath Proof Explorer


Theorem deg1lt0

Description: A polynomial is zero iff it has negative degree. (Contributed by Stefan O'Rear, 1-Apr-2015)

Ref Expression
Hypotheses deg1z.d 𝐷 = ( deg1𝑅 )
deg1z.p 𝑃 = ( Poly1𝑅 )
deg1z.z 0 = ( 0g𝑃 )
deg1nn0cl.b 𝐵 = ( Base ‘ 𝑃 )
Assertion deg1lt0 ( ( 𝑅 ∈ Ring ∧ 𝐹𝐵 ) → ( ( 𝐷𝐹 ) < 0 ↔ 𝐹 = 0 ) )

Proof

Step Hyp Ref Expression
1 deg1z.d 𝐷 = ( deg1𝑅 )
2 deg1z.p 𝑃 = ( Poly1𝑅 )
3 deg1z.z 0 = ( 0g𝑃 )
4 deg1nn0cl.b 𝐵 = ( Base ‘ 𝑃 )
5 1 2 3 4 deg1nn0cl ( ( 𝑅 ∈ Ring ∧ 𝐹𝐵𝐹0 ) → ( 𝐷𝐹 ) ∈ ℕ0 )
6 nn0nlt0 ( ( 𝐷𝐹 ) ∈ ℕ0 → ¬ ( 𝐷𝐹 ) < 0 )
7 5 6 syl ( ( 𝑅 ∈ Ring ∧ 𝐹𝐵𝐹0 ) → ¬ ( 𝐷𝐹 ) < 0 )
8 7 3expia ( ( 𝑅 ∈ Ring ∧ 𝐹𝐵 ) → ( 𝐹0 → ¬ ( 𝐷𝐹 ) < 0 ) )
9 8 necon4ad ( ( 𝑅 ∈ Ring ∧ 𝐹𝐵 ) → ( ( 𝐷𝐹 ) < 0 → 𝐹 = 0 ) )
10 1 2 3 deg1z ( 𝑅 ∈ Ring → ( 𝐷0 ) = -∞ )
11 mnflt0 -∞ < 0
12 10 11 eqbrtrdi ( 𝑅 ∈ Ring → ( 𝐷0 ) < 0 )
13 12 adantr ( ( 𝑅 ∈ Ring ∧ 𝐹𝐵 ) → ( 𝐷0 ) < 0 )
14 fveq2 ( 𝐹 = 0 → ( 𝐷𝐹 ) = ( 𝐷0 ) )
15 14 breq1d ( 𝐹 = 0 → ( ( 𝐷𝐹 ) < 0 ↔ ( 𝐷0 ) < 0 ) )
16 13 15 syl5ibrcom ( ( 𝑅 ∈ Ring ∧ 𝐹𝐵 ) → ( 𝐹 = 0 → ( 𝐷𝐹 ) < 0 ) )
17 9 16 impbid ( ( 𝑅 ∈ Ring ∧ 𝐹𝐵 ) → ( ( 𝐷𝐹 ) < 0 ↔ 𝐹 = 0 ) )