Metamath Proof Explorer


Theorem abelthlem4

Description: Lemma for abelth . (Contributed by Mario Carneiro, 31-Mar-2015)

Ref Expression
Hypotheses abelth.1 ⊢ φ → A : ℕ 0 ⟶ ℂ
abelth.2 ⊢ φ → seq 0 + A ∈ dom ⁡ ⇝
abelth.3 ⊢ φ → M ∈ ℝ
abelth.4 ⊢ φ → 0 ≤ M
abelth.5 ⊢ S = z ∈ ℂ | 1 − z ≤ M ⁢ 1 − z
abelth.6 ⊢ F = x ∈ S ⟼ ∑ n ∈ ℕ 0 A ⁡ n ⁢ x n
Assertion abelthlem4 ⊢ φ → F : S ⟶ ℂ

Proof

Step Hyp Ref Expression
1 abelth.1 ⊢ φ → A : ℕ 0 ⟶ ℂ
2 abelth.2 ⊢ φ → seq 0 + A ∈ dom ⁡ ⇝
3 abelth.3 ⊢ φ → M ∈ ℝ
4 abelth.4 ⊢ φ → 0 ≤ M
5 abelth.5 ⊢ S = z ∈ ℂ | 1 − z ≤ M ⁢ 1 − z
6 abelth.6 ⊢ F = x ∈ S ⟼ ∑ n ∈ ℕ 0 A ⁡ n ⁢ x n
7 nn0uz ⊢ ℕ 0 = ℤ ≥ 0
8 0zd ⊢ φ ∧ x ∈ S → 0 ∈ ℤ
9 fveq2 ⊢ m = n → A ⁡ m = A ⁡ n
10 oveq2 ⊢ m = n → x m = x n
11 9 10 oveq12d ⊢ m = n → A ⁡ m ⁢ x m = A ⁡ n ⁢ x n
12 eqid ⊢ m ∈ ℕ 0 ⟼ A ⁡ m ⁢ x m = m ∈ ℕ 0 ⟼ A ⁡ m ⁢ x m
13 ovex ⊢ A ⁡ n ⁢ x n ∈ V
14 11 12 13 fvmpt ⊢ n ∈ ℕ 0 → m ∈ ℕ 0 ⟼ A ⁡ m ⁢ x m ⁡ n = A ⁡ n ⁢ x n
15 14 adantl ⊢ φ ∧ x ∈ S ∧ n ∈ ℕ 0 → m ∈ ℕ 0 ⟼ A ⁡ m ⁢ x m ⁡ n = A ⁡ n ⁢ x n
16 1 adantr ⊢ φ ∧ x ∈ S → A : ℕ 0 ⟶ ℂ
17 16 ffvelcdmda ⊢ φ ∧ x ∈ S ∧ n ∈ ℕ 0 → A ⁡ n ∈ ℂ
18 5 ssrab3 ⊢ S ⊆ ℂ
19 18 a1i ⊢ φ → S ⊆ ℂ
20 19 sselda ⊢ φ ∧ x ∈ S → x ∈ ℂ
21 expcl ⊢ x ∈ ℂ ∧ n ∈ ℕ 0 → x n ∈ ℂ
22 20 21 sylan ⊢ φ ∧ x ∈ S ∧ n ∈ ℕ 0 → x n ∈ ℂ
23 17 22 mulcld ⊢ φ ∧ x ∈ S ∧ n ∈ ℕ 0 → A ⁡ n ⁢ x n ∈ ℂ
24 1 2 3 4 5 abelthlem3 ⊢ φ ∧ x ∈ S → seq 0 + m ∈ ℕ 0 ⟼ A ⁡ m ⁢ x m ∈ dom ⁡ ⇝
25 7 8 15 23 24 isumcl ⊢ φ ∧ x ∈ S → ∑ n ∈ ℕ 0 A ⁡ n ⁢ x n ∈ ℂ
26 25 6 fmptd ⊢ φ → F : S ⟶ ℂ