Metamath Proof Explorer


Theorem 0dgr

Description: A constant function has degree 0. (Contributed by Mario Carneiro, 24-Jul-2014)

Ref Expression
Assertion 0dgr ⊢ A ∈ ℂ → deg ⁡ ℂ × A = 0

Proof

Step Hyp Ref Expression
1 ssid ⊢ ℂ ⊆ ℂ
2 plyconst ⊢ ℂ ⊆ ℂ ∧ A ∈ ℂ → ℂ × A ∈ Poly ⁡ ℂ
3 1 2 mpan ⊢ A ∈ ℂ → ℂ × A ∈ Poly ⁡ ℂ
4 0nn0 ⊢ 0 ∈ ℕ 0
5 4 a1i ⊢ A ∈ ℂ → 0 ∈ ℕ 0
6 simpl ⊢ A ∈ ℂ ∧ k ∈ 0 … 0 → A ∈ ℂ
7 fconstmpt ⊢ ℂ × A = z ∈ ℂ ⟼ A
8 0z ⊢ 0 ∈ ℤ
9 exp0 ⊢ z ∈ ℂ → z 0 = 1
10 9 oveq2d ⊢ z ∈ ℂ → A ⁢ z 0 = A ⋅ 1
11 mulrid ⊢ A ∈ ℂ → A ⋅ 1 = A
12 10 11 sylan9eqr ⊢ A ∈ ℂ ∧ z ∈ ℂ → A ⁢ z 0 = A
13 simpl ⊢ A ∈ ℂ ∧ z ∈ ℂ → A ∈ ℂ
14 12 13 eqeltrd ⊢ A ∈ ℂ ∧ z ∈ ℂ → A ⁢ z 0 ∈ ℂ
15 oveq2 ⊢ k = 0 → z k = z 0
16 15 oveq2d ⊢ k = 0 → A ⁢ z k = A ⁢ z 0
17 16 fsum1 ⊢ 0 ∈ ℤ ∧ A ⁢ z 0 ∈ ℂ → ∑ k = 0 0 A ⁢ z k = A ⁢ z 0
18 8 14 17 sylancr ⊢ A ∈ ℂ ∧ z ∈ ℂ → ∑ k = 0 0 A ⁢ z k = A ⁢ z 0
19 18 12 eqtrd ⊢ A ∈ ℂ ∧ z ∈ ℂ → ∑ k = 0 0 A ⁢ z k = A
20 19 mpteq2dva ⊢ A ∈ ℂ → z ∈ ℂ ⟼ ∑ k = 0 0 A ⁢ z k = z ∈ ℂ ⟼ A
21 7 20 eqtr4id ⊢ A ∈ ℂ → ℂ × A = z ∈ ℂ ⟼ ∑ k = 0 0 A ⁢ z k
22 3 5 6 21 dgrle ⊢ A ∈ ℂ → deg ⁡ ℂ × A ≤ 0
23 dgrcl ⊢ ℂ × A ∈ Poly ⁡ ℂ → deg ⁡ ℂ × A ∈ ℕ 0
24 nn0le0eq0 ⊢ deg ⁡ ℂ × A ∈ ℕ 0 → deg ⁡ ℂ × A ≤ 0 ↔ deg ⁡ ℂ × A = 0
25 3 23 24 3syl ⊢ A ∈ ℂ → deg ⁡ ℂ × A ≤ 0 ↔ deg ⁡ ℂ × A = 0
26 22 25 mpbid ⊢ A ∈ ℂ → deg ⁡ ℂ × A = 0