Metamath Proof Explorer


Theorem plymul02

Description: Product of a polynomial with the zero polynomial. (Contributed by Thierry Arnoux, 26-Sep-2018)

Ref Expression
Assertion plymul02 ⊢ F ∈ Poly ⁡ S → 0 𝑝 × f F = 0 𝑝

Proof

Step Hyp Ref Expression
1 plyf ⊢ F ∈ Poly ⁡ S → F : ℂ ⟶ ℂ
2 1 ffvelcdmda ⊢ F ∈ Poly ⁡ S ∧ x ∈ ℂ → F ⁡ x ∈ ℂ
3 2 mul02d ⊢ F ∈ Poly ⁡ S ∧ x ∈ ℂ → 0 ⋅ F ⁡ x = 0
4 3 mpteq2dva ⊢ F ∈ Poly ⁡ S → x ∈ ℂ ⟼ 0 ⋅ F ⁡ x = x ∈ ℂ ⟼ 0
5 c0ex ⊢ 0 ∈ V
6 5 fconst ⊢ ℂ × 0 : ℂ ⟶ 0
7 df-0p ⊢ 0 𝑝 = ℂ × 0
8 7 feq1i ⊢ 0 𝑝 : ℂ ⟶ 0 ↔ ℂ × 0 : ℂ ⟶ 0
9 6 8 mpbir ⊢ 0 𝑝 : ℂ ⟶ 0
10 ffn ⊢ 0 𝑝 : ℂ ⟶ 0 → 0 𝑝 Fn ℂ
11 9 10 mp1i ⊢ F ∈ Poly ⁡ S → 0 𝑝 Fn ℂ
12 1 ffnd ⊢ F ∈ Poly ⁡ S → F Fn ℂ
13 cnex ⊢ ℂ ∈ V
14 13 a1i ⊢ F ∈ Poly ⁡ S → ℂ ∈ V
15 inidm ⊢ ℂ ∩ ℂ = ℂ
16 0pval ⊢ x ∈ ℂ → 0 𝑝 ⁡ x = 0
17 16 adantl ⊢ F ∈ Poly ⁡ S ∧ x ∈ ℂ → 0 𝑝 ⁡ x = 0
18 eqidd ⊢ F ∈ Poly ⁡ S ∧ x ∈ ℂ → F ⁡ x = F ⁡ x
19 11 12 14 14 15 17 18 offval ⊢ F ∈ Poly ⁡ S → 0 𝑝 × f F = x ∈ ℂ ⟼ 0 ⋅ F ⁡ x
20 fconstmpt ⊢ ℂ × 0 = x ∈ ℂ ⟼ 0
21 7 20 eqtri ⊢ 0 𝑝 = x ∈ ℂ ⟼ 0
22 21 a1i ⊢ F ∈ Poly ⁡ S → 0 𝑝 = x ∈ ℂ ⟼ 0
23 4 19 22 3eqtr4d ⊢ F ∈ Poly ⁡ S → 0 𝑝 × f F = 0 𝑝