Metamath Proof Explorer


Theorem rpnnen2lem10

Description: Lemma for rpnnen2 . (Contributed by Mario Carneiro, 13-May-2013) (Revised by Mario Carneiro, 30-Apr-2014)

Ref Expression
Hypotheses rpnnen2.1 ⊢ F = x ∈ 𝒫 ℕ ⟼ n ∈ ℕ ⟼ if n ∈ x 1 3 n 0
rpnnen2.2 ⊢ φ → A ⊆ ℕ
rpnnen2.3 ⊢ φ → B ⊆ ℕ
rpnnen2.4 ⊢ φ → m ∈ A ∖ B
rpnnen2.5 ⊢ φ → ∀ n ∈ ℕ n < m → n ∈ A ↔ n ∈ B
rpnnen2.6 ⊢ ψ ↔ ∑ k ∈ ℕ F ⁡ A ⁡ k = ∑ k ∈ ℕ F ⁡ B ⁡ k
Assertion rpnnen2lem10 ⊢ φ ∧ ψ → ∑ k ∈ ℤ ≥ m F ⁡ A ⁡ k = ∑ k ∈ ℤ ≥ m F ⁡ B ⁡ k

Proof

Step Hyp Ref Expression
1 rpnnen2.1 ⊢ F = x ∈ 𝒫 ℕ ⟼ n ∈ ℕ ⟼ if n ∈ x 1 3 n 0
2 rpnnen2.2 ⊢ φ → A ⊆ ℕ
3 rpnnen2.3 ⊢ φ → B ⊆ ℕ
4 rpnnen2.4 ⊢ φ → m ∈ A ∖ B
5 rpnnen2.5 ⊢ φ → ∀ n ∈ ℕ n < m → n ∈ A ↔ n ∈ B
6 rpnnen2.6 ⊢ ψ ↔ ∑ k ∈ ℕ F ⁡ A ⁡ k = ∑ k ∈ ℕ F ⁡ B ⁡ k
7 6 bilani ⊢ φ ∧ ψ → ∑ k ∈ ℕ F ⁡ A ⁡ k = ∑ k ∈ ℕ F ⁡ B ⁡ k
8 eldifi ⊢ m ∈ A ∖ B → m ∈ A
9 ssel2 ⊢ A ⊆ ℕ ∧ m ∈ A → m ∈ ℕ
10 8 9 sylan2 ⊢ A ⊆ ℕ ∧ m ∈ A ∖ B → m ∈ ℕ
11 2 4 10 syl2anc ⊢ φ → m ∈ ℕ
12 1 rpnnen2lem8 ⊢ A ⊆ ℕ ∧ m ∈ ℕ → ∑ k ∈ ℕ F ⁡ A ⁡ k = ∑ k = 1 m − 1 F ⁡ A ⁡ k + ∑ k ∈ ℤ ≥ m F ⁡ A ⁡ k
13 2 11 12 syl2anc ⊢ φ → ∑ k ∈ ℕ F ⁡ A ⁡ k = ∑ k = 1 m − 1 F ⁡ A ⁡ k + ∑ k ∈ ℤ ≥ m F ⁡ A ⁡ k
14 1z ⊢ 1 ∈ ℤ
15 nnz ⊢ m ∈ ℕ → m ∈ ℤ
16 elfzm11 ⊢ 1 ∈ ℤ ∧ m ∈ ℤ → k ∈ 1 … m − 1 ↔ k ∈ ℤ ∧ 1 ≤ k ∧ k < m
17 14 15 16 sylancr ⊢ m ∈ ℕ → k ∈ 1 … m − 1 ↔ k ∈ ℤ ∧ 1 ≤ k ∧ k < m
18 17 biimpa ⊢ m ∈ ℕ ∧ k ∈ 1 … m − 1 → k ∈ ℤ ∧ 1 ≤ k ∧ k < m
19 11 18 sylan ⊢ φ ∧ k ∈ 1 … m − 1 → k ∈ ℤ ∧ 1 ≤ k ∧ k < m
20 19 simp3d ⊢ φ ∧ k ∈ 1 … m − 1 → k < m
21 elfznn ⊢ k ∈ 1 … m − 1 → k ∈ ℕ
22 breq1 ⊢ n = k → n < m ↔ k < m
23 eleq1w ⊢ n = k → n ∈ A ↔ k ∈ A
24 eleq1w ⊢ n = k → n ∈ B ↔ k ∈ B
25 23 24 bibi12d ⊢ n = k → n ∈ A ↔ n ∈ B ↔ k ∈ A ↔ k ∈ B
26 22 25 imbi12d ⊢ n = k → n < m → n ∈ A ↔ n ∈ B ↔ k < m → k ∈ A ↔ k ∈ B
27 26 rspccva ⊢ ∀ n ∈ ℕ n < m → n ∈ A ↔ n ∈ B ∧ k ∈ ℕ → k < m → k ∈ A ↔ k ∈ B
28 5 21 27 syl2an ⊢ φ ∧ k ∈ 1 … m − 1 → k < m → k ∈ A ↔ k ∈ B
29 20 28 mpd ⊢ φ ∧ k ∈ 1 … m − 1 → k ∈ A ↔ k ∈ B
30 29 ifbid ⊢ φ ∧ k ∈ 1 … m − 1 → if k ∈ A 1 3 k 0 = if k ∈ B 1 3 k 0
31 1 rpnnen2lem1 ⊢ A ⊆ ℕ ∧ k ∈ ℕ → F ⁡ A ⁡ k = if k ∈ A 1 3 k 0
32 2 21 31 syl2an ⊢ φ ∧ k ∈ 1 … m − 1 → F ⁡ A ⁡ k = if k ∈ A 1 3 k 0
33 1 rpnnen2lem1 ⊢ B ⊆ ℕ ∧ k ∈ ℕ → F ⁡ B ⁡ k = if k ∈ B 1 3 k 0
34 3 21 33 syl2an ⊢ φ ∧ k ∈ 1 … m − 1 → F ⁡ B ⁡ k = if k ∈ B 1 3 k 0
35 30 32 34 3eqtr4d ⊢ φ ∧ k ∈ 1 … m − 1 → F ⁡ A ⁡ k = F ⁡ B ⁡ k
36 35 sumeq2dv ⊢ φ → ∑ k = 1 m − 1 F ⁡ A ⁡ k = ∑ k = 1 m − 1 F ⁡ B ⁡ k
37 36 oveq1d ⊢ φ → ∑ k = 1 m − 1 F ⁡ A ⁡ k + ∑ k ∈ ℤ ≥ m F ⁡ A ⁡ k = ∑ k = 1 m − 1 F ⁡ B ⁡ k + ∑ k ∈ ℤ ≥ m F ⁡ A ⁡ k
38 13 37 eqtrd ⊢ φ → ∑ k ∈ ℕ F ⁡ A ⁡ k = ∑ k = 1 m − 1 F ⁡ B ⁡ k + ∑ k ∈ ℤ ≥ m F ⁡ A ⁡ k
39 38 adantr ⊢ φ ∧ ψ → ∑ k ∈ ℕ F ⁡ A ⁡ k = ∑ k = 1 m − 1 F ⁡ B ⁡ k + ∑ k ∈ ℤ ≥ m F ⁡ A ⁡ k
40 1 rpnnen2lem8 ⊢ B ⊆ ℕ ∧ m ∈ ℕ → ∑ k ∈ ℕ F ⁡ B ⁡ k = ∑ k = 1 m − 1 F ⁡ B ⁡ k + ∑ k ∈ ℤ ≥ m F ⁡ B ⁡ k
41 3 11 40 syl2anc ⊢ φ → ∑ k ∈ ℕ F ⁡ B ⁡ k = ∑ k = 1 m − 1 F ⁡ B ⁡ k + ∑ k ∈ ℤ ≥ m F ⁡ B ⁡ k
42 41 adantr ⊢ φ ∧ ψ → ∑ k ∈ ℕ F ⁡ B ⁡ k = ∑ k = 1 m − 1 F ⁡ B ⁡ k + ∑ k ∈ ℤ ≥ m F ⁡ B ⁡ k
43 7 39 42 3eqtr3d ⊢ φ ∧ ψ → ∑ k = 1 m − 1 F ⁡ B ⁡ k + ∑ k ∈ ℤ ≥ m F ⁡ A ⁡ k = ∑ k = 1 m − 1 F ⁡ B ⁡ k + ∑ k ∈ ℤ ≥ m F ⁡ B ⁡ k
44 1 rpnnen2lem6 ⊢ A ⊆ ℕ ∧ m ∈ ℕ → ∑ k ∈ ℤ ≥ m F ⁡ A ⁡ k ∈ ℝ
45 2 11 44 syl2anc ⊢ φ → ∑ k ∈ ℤ ≥ m F ⁡ A ⁡ k ∈ ℝ
46 1 rpnnen2lem6 ⊢ B ⊆ ℕ ∧ m ∈ ℕ → ∑ k ∈ ℤ ≥ m F ⁡ B ⁡ k ∈ ℝ
47 3 11 46 syl2anc ⊢ φ → ∑ k ∈ ℤ ≥ m F ⁡ B ⁡ k ∈ ℝ
48 fzfid ⊢ φ → 1 … m − 1 ∈ Fin
49 1 rpnnen2lem2 ⊢ B ⊆ ℕ → F ⁡ B : ℕ ⟶ ℝ
50 3 49 syl ⊢ φ → F ⁡ B : ℕ ⟶ ℝ
51 ffvelcdm ⊢ F ⁡ B : ℕ ⟶ ℝ ∧ k ∈ ℕ → F ⁡ B ⁡ k ∈ ℝ
52 50 21 51 syl2an ⊢ φ ∧ k ∈ 1 … m − 1 → F ⁡ B ⁡ k ∈ ℝ
53 48 52 fsumrecl ⊢ φ → ∑ k = 1 m − 1 F ⁡ B ⁡ k ∈ ℝ
54 readdcan ⊢ ∑ k ∈ ℤ ≥ m F ⁡ A ⁡ k ∈ ℝ ∧ ∑ k ∈ ℤ ≥ m F ⁡ B ⁡ k ∈ ℝ ∧ ∑ k = 1 m − 1 F ⁡ B ⁡ k ∈ ℝ → ∑ k = 1 m − 1 F ⁡ B ⁡ k + ∑ k ∈ ℤ ≥ m F ⁡ A ⁡ k = ∑ k = 1 m − 1 F ⁡ B ⁡ k + ∑ k ∈ ℤ ≥ m F ⁡ B ⁡ k ↔ ∑ k ∈ ℤ ≥ m F ⁡ A ⁡ k = ∑ k ∈ ℤ ≥ m F ⁡ B ⁡ k
55 45 47 53 54 syl3anc ⊢ φ → ∑ k = 1 m − 1 F ⁡ B ⁡ k + ∑ k ∈ ℤ ≥ m F ⁡ A ⁡ k = ∑ k = 1 m − 1 F ⁡ B ⁡ k + ∑ k ∈ ℤ ≥ m F ⁡ B ⁡ k ↔ ∑ k ∈ ℤ ≥ m F ⁡ A ⁡ k = ∑ k ∈ ℤ ≥ m F ⁡ B ⁡ k
56 55 adantr ⊢ φ ∧ ψ → ∑ k = 1 m − 1 F ⁡ B ⁡ k + ∑ k ∈ ℤ ≥ m F ⁡ A ⁡ k = ∑ k = 1 m − 1 F ⁡ B ⁡ k + ∑ k ∈ ℤ ≥ m F ⁡ B ⁡ k ↔ ∑ k ∈ ℤ ≥ m F ⁡ A ⁡ k = ∑ k ∈ ℤ ≥ m F ⁡ B ⁡ k
57 43 56 mpbid ⊢ φ ∧ ψ → ∑ k ∈ ℤ ≥ m F ⁡ A ⁡ k = ∑ k ∈ ℤ ≥ m F ⁡ B ⁡ k