Metamath Proof Explorer


Theorem eulerpartlemsv3

Description: Lemma for eulerpart . Value of the sum of a finite partition A (Contributed by Thierry Arnoux, 19-Aug-2018)

Ref Expression
Hypotheses eulerpartlems.r ⊢ R = f | f -1 ℕ ∈ Fin
eulerpartlems.s ⊢ S = f ∈ ℕ 0 ℕ ∩ R ⟼ ∑ k ∈ ℕ f ⁡ k ⁢ k
Assertion eulerpartlemsv3 ⊢ A ∈ ℕ 0 ℕ ∩ R → S ⁡ A = ∑ k = 1 S ⁡ A A ⁡ k ⁢ k

Proof

Step Hyp Ref Expression
1 eulerpartlems.r ⊢ R = f | f -1 ℕ ∈ Fin
2 eulerpartlems.s ⊢ S = f ∈ ℕ 0 ℕ ∩ R ⟼ ∑ k ∈ ℕ f ⁡ k ⁢ k
3 1 2 eulerpartlemsv1 ⊢ A ∈ ℕ 0 ℕ ∩ R → S ⁡ A = ∑ k ∈ ℕ A ⁡ k ⁢ k
4 fzssuz ⊢ 1 … S ⁡ A ⊆ ℤ ≥ 1
5 nnuz ⊢ ℕ = ℤ ≥ 1
6 4 5 sseqtrri ⊢ 1 … S ⁡ A ⊆ ℕ
7 6 a1i ⊢ A ∈ ℕ 0 ℕ ∩ R → 1 … S ⁡ A ⊆ ℕ
8 1 2 eulerpartlemelr ⊢ A ∈ ℕ 0 ℕ ∩ R → A : ℕ ⟶ ℕ 0 ∧ A -1 ℕ ∈ Fin
9 8 simpld ⊢ A ∈ ℕ 0 ℕ ∩ R → A : ℕ ⟶ ℕ 0
10 9 adantr ⊢ A ∈ ℕ 0 ℕ ∩ R ∧ k ∈ 1 … S ⁡ A → A : ℕ ⟶ ℕ 0
11 7 sselda ⊢ A ∈ ℕ 0 ℕ ∩ R ∧ k ∈ 1 … S ⁡ A → k ∈ ℕ
12 10 11 ffvelcdmd ⊢ A ∈ ℕ 0 ℕ ∩ R ∧ k ∈ 1 … S ⁡ A → A ⁡ k ∈ ℕ 0
13 12 nn0cnd ⊢ A ∈ ℕ 0 ℕ ∩ R ∧ k ∈ 1 … S ⁡ A → A ⁡ k ∈ ℂ
14 11 nncnd ⊢ A ∈ ℕ 0 ℕ ∩ R ∧ k ∈ 1 … S ⁡ A → k ∈ ℂ
15 13 14 mulcld ⊢ A ∈ ℕ 0 ℕ ∩ R ∧ k ∈ 1 … S ⁡ A → A ⁡ k ⁢ k ∈ ℂ
16 1 2 eulerpartlems ⊢ A ∈ ℕ 0 ℕ ∩ R ∧ t ∈ ℤ ≥ S ⁡ A + 1 → A ⁡ t = 0
17 16 ralrimiva ⊢ A ∈ ℕ 0 ℕ ∩ R → ∀ t ∈ ℤ ≥ S ⁡ A + 1 A ⁡ t = 0
18 fveqeq2 ⊢ k = t → A ⁡ k = 0 ↔ A ⁡ t = 0
19 18 cbvralvw ⊢ ∀ k ∈ ℤ ≥ S ⁡ A + 1 A ⁡ k = 0 ↔ ∀ t ∈ ℤ ≥ S ⁡ A + 1 A ⁡ t = 0
20 17 19 sylibr ⊢ A ∈ ℕ 0 ℕ ∩ R → ∀ k ∈ ℤ ≥ S ⁡ A + 1 A ⁡ k = 0
21 1 2 eulerpartlemsf ⊢ S : ℕ 0 ℕ ∩ R ⟶ ℕ 0
22 21 ffvelcdmi ⊢ A ∈ ℕ 0 ℕ ∩ R → S ⁡ A ∈ ℕ 0
23 nndiffz1 ⊢ S ⁡ A ∈ ℕ 0 → ℕ ∖ 1 … S ⁡ A = ℤ ≥ S ⁡ A + 1
24 22 23 syl ⊢ A ∈ ℕ 0 ℕ ∩ R → ℕ ∖ 1 … S ⁡ A = ℤ ≥ S ⁡ A + 1
25 20 24 raleqtrrdv ⊢ A ∈ ℕ 0 ℕ ∩ R → ∀ k ∈ ℕ ∖ 1 … S ⁡ A A ⁡ k = 0
26 25 r19.21bi ⊢ A ∈ ℕ 0 ℕ ∩ R ∧ k ∈ ℕ ∖ 1 … S ⁡ A → A ⁡ k = 0
27 26 oveq1d ⊢ A ∈ ℕ 0 ℕ ∩ R ∧ k ∈ ℕ ∖ 1 … S ⁡ A → A ⁡ k ⁢ k = 0 ⋅ k
28 simpr ⊢ A ∈ ℕ 0 ℕ ∩ R ∧ k ∈ ℕ ∖ 1 … S ⁡ A → k ∈ ℕ ∖ 1 … S ⁡ A
29 28 eldifad ⊢ A ∈ ℕ 0 ℕ ∩ R ∧ k ∈ ℕ ∖ 1 … S ⁡ A → k ∈ ℕ
30 29 nncnd ⊢ A ∈ ℕ 0 ℕ ∩ R ∧ k ∈ ℕ ∖ 1 … S ⁡ A → k ∈ ℂ
31 30 mul02d ⊢ A ∈ ℕ 0 ℕ ∩ R ∧ k ∈ ℕ ∖ 1 … S ⁡ A → 0 ⋅ k = 0
32 27 31 eqtrd ⊢ A ∈ ℕ 0 ℕ ∩ R ∧ k ∈ ℕ ∖ 1 … S ⁡ A → A ⁡ k ⁢ k = 0
33 5 eqimssi ⊢ ℕ ⊆ ℤ ≥ 1
34 33 a1i ⊢ A ∈ ℕ 0 ℕ ∩ R → ℕ ⊆ ℤ ≥ 1
35 7 15 32 34 sumss ⊢ A ∈ ℕ 0 ℕ ∩ R → ∑ k = 1 S ⁡ A A ⁡ k ⁢ k = ∑ k ∈ ℕ A ⁡ k ⁢ k
36 3 35 eqtr4d ⊢ A ∈ ℕ 0 ℕ ∩ R → S ⁡ A = ∑ k = 1 S ⁡ A A ⁡ k ⁢ k