Metamath Proof Explorer


Theorem linply2cfv

Description: The evaluation of a linear polynomial over a subset of the complex numbers with explicit coefficients at a complex number. (Contributed by AV, 31-Oct-2026)

Ref Expression
Hypotheses linplyfv.p ⊢ P = z ∈ ℂ ⟼ ∑ k = 0 1 C ⁡ k ⁢ z k
linplyfv2.c ⊢ C = x ∈ ℕ 0 ⟼ if x = 0 A B
Assertion linply2cfv ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S ∧ X ∈ ℂ → P ⁡ X = A + B ⁢ X

Proof

Step Hyp Ref Expression
1 linplyfv.p ⊢ P = z ∈ ℂ ⟼ ∑ k = 0 1 C ⁡ k ⁢ z k
2 linplyfv2.c ⊢ C = x ∈ ℕ 0 ⟼ if x = 0 A B
3 cnex ⊢ ℂ ∈ V
4 3 ssex ⊢ S ⊆ ℂ → S ∈ V
5 4 3ad2ant1 ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S ∧ X ∈ ℂ → S ∈ V
6 nn0ex ⊢ ℕ 0 ∈ V
7 6 a1i ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S ∧ X ∈ ℂ → ℕ 0 ∈ V
8 ifcl ⊢ A ∈ S ∧ B ∈ S → if x = 0 A B ∈ S
9 8 3ad2ant2 ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S ∧ X ∈ ℂ → if x = 0 A B ∈ S
10 9 adantr ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S ∧ X ∈ ℂ ∧ x ∈ ℕ 0 → if x = 0 A B ∈ S
11 10 2 fmptd ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S ∧ X ∈ ℂ → C : ℕ 0 ⟶ S
12 5 7 11 elmapdd ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S ∧ X ∈ ℂ → C ∈ S ℕ 0
13 1 linplyfv ⊢ S ⊆ ℂ ∧ C ∈ S ℕ 0 ∧ X ∈ ℂ → P ⁡ X = C ⁡ 0 + C ⁡ 1 ⁢ X
14 12 13 syld3an2 ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S ∧ X ∈ ℂ → P ⁡ X = C ⁡ 0 + C ⁡ 1 ⁢ X
15 2 a1i ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S → C = x ∈ ℕ 0 ⟼ if x = 0 A B
16 iftrue ⊢ x = 0 → if x = 0 A B = A
17 16 adantl ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S ∧ x = 0 → if x = 0 A B = A
18 0nn0 ⊢ 0 ∈ ℕ 0
19 18 a1i ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S → 0 ∈ ℕ 0
20 simprl ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S → A ∈ S
21 15 17 19 20 fvmptd ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S → C ⁡ 0 = A
22 ax-1ne0 ⊢ 1 ≠ 0
23 22 neii ⊢ ¬ 1 = 0
24 eqeq1 ⊢ x = 1 → x = 0 ↔ 1 = 0
25 23 24 mtbiri ⊢ x = 1 → ¬ x = 0
26 25 adantl ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S ∧ x = 1 → ¬ x = 0
27 26 iffalsed ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S ∧ x = 1 → if x = 0 A B = B
28 1nn0 ⊢ 1 ∈ ℕ 0
29 28 a1i ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S → 1 ∈ ℕ 0
30 simprr ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S → B ∈ S
31 15 27 29 30 fvmptd ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S → C ⁡ 1 = B
32 31 oveq1d ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S → C ⁡ 1 ⁢ X = B ⁢ X
33 21 32 oveq12d ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S → C ⁡ 0 + C ⁡ 1 ⁢ X = A + B ⁢ X
34 33 3adant3 ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S ∧ X ∈ ℂ → C ⁡ 0 + C ⁡ 1 ⁢ X = A + B ⁢ X
35 14 34 eqtrd ⊢ S ⊆ ℂ ∧ A ∈ S ∧ B ∈ S ∧ X ∈ ℂ → P ⁡ X = A + B ⁢ X