Metamath Proof Explorer


Theorem pntrlog2bndlem6a

Description: Lemma for pntrlog2bndlem6 . (Contributed by Mario Carneiro, 7-Jun-2016)

Ref Expression
Hypotheses pntsval.1 ⊢ S = a ∈ ℝ ⟼ ∑ i = 1 a Λ ⁡ i ⁢ log ⁡ i + ψ ⁡ a i
pntrlog2bnd.r ⊢ R = a ∈ ℝ + ⟼ ψ ⁡ a − a
pntrlog2bnd.t ⊢ T = a ∈ ℝ ⟼ if a ∈ ℝ + a ⁢ log ⁡ a 0
pntrlog2bndlem5.1 ⊢ φ → B ∈ ℝ +
pntrlog2bndlem5.2 ⊢ φ → ∀ y ∈ ℝ + R ⁡ y y ≤ B
pntrlog2bndlem6.1 ⊢ φ → A ∈ ℝ
pntrlog2bndlem6.2 ⊢ φ → 1 ≤ A
Assertion pntrlog2bndlem6a ⊢ φ ∧ x ∈ 1 +∞ → 1 … x = 1 … x A ∪ x A + 1 … x

Proof

Step Hyp Ref Expression
1 pntsval.1 ⊢ S = a ∈ ℝ ⟼ ∑ i = 1 a Λ ⁡ i ⁢ log ⁡ i + ψ ⁡ a i
2 pntrlog2bnd.r ⊢ R = a ∈ ℝ + ⟼ ψ ⁡ a − a
3 pntrlog2bnd.t ⊢ T = a ∈ ℝ ⟼ if a ∈ ℝ + a ⁢ log ⁡ a 0
4 pntrlog2bndlem5.1 ⊢ φ → B ∈ ℝ +
5 pntrlog2bndlem5.2 ⊢ φ → ∀ y ∈ ℝ + R ⁡ y y ≤ B
6 pntrlog2bndlem6.1 ⊢ φ → A ∈ ℝ
7 pntrlog2bndlem6.2 ⊢ φ → 1 ≤ A
8 elioore ⊢ x ∈ 1 +∞ → x ∈ ℝ
9 8 adantl ⊢ φ ∧ x ∈ 1 +∞ → x ∈ ℝ
10 1rp ⊢ 1 ∈ ℝ +
11 10 a1i ⊢ φ ∧ x ∈ 1 +∞ → 1 ∈ ℝ +
12 11 rpred ⊢ φ ∧ x ∈ 1 +∞ → 1 ∈ ℝ
13 eliooord ⊢ x ∈ 1 +∞ → 1 < x ∧ x < +∞
14 13 adantl ⊢ φ ∧ x ∈ 1 +∞ → 1 < x ∧ x < +∞
15 14 simpld ⊢ φ ∧ x ∈ 1 +∞ → 1 < x
16 12 9 15 ltled ⊢ φ ∧ x ∈ 1 +∞ → 1 ≤ x
17 9 11 16 rpgecld ⊢ φ ∧ x ∈ 1 +∞ → x ∈ ℝ +
18 10 a1i ⊢ φ → 1 ∈ ℝ +
19 6 18 7 rpgecld ⊢ φ → A ∈ ℝ +
20 19 adantr ⊢ φ ∧ x ∈ 1 +∞ → A ∈ ℝ +
21 17 20 rpdivcld ⊢ φ ∧ x ∈ 1 +∞ → x A ∈ ℝ +
22 21 rprege0d ⊢ φ ∧ x ∈ 1 +∞ → x A ∈ ℝ ∧ 0 ≤ x A
23 flge0nn0 ⊢ x A ∈ ℝ ∧ 0 ≤ x A → x A ∈ ℕ 0
24 nn0p1nn ⊢ x A ∈ ℕ 0 → x A + 1 ∈ ℕ
25 22 23 24 3syl ⊢ φ ∧ x ∈ 1 +∞ → x A + 1 ∈ ℕ
26 nnuz ⊢ ℕ = ℤ ≥ 1
27 25 26 eleqtrdi ⊢ φ ∧ x ∈ 1 +∞ → x A + 1 ∈ ℤ ≥ 1
28 21 rpred ⊢ φ ∧ x ∈ 1 +∞ → x A ∈ ℝ
29 17 rpge0d ⊢ φ ∧ x ∈ 1 +∞ → 0 ≤ x
30 7 adantr ⊢ φ ∧ x ∈ 1 +∞ → 1 ≤ A
31 11 20 9 29 30 lediv2ad ⊢ φ ∧ x ∈ 1 +∞ → x A ≤ x 1
32 9 recnd ⊢ φ ∧ x ∈ 1 +∞ → x ∈ ℂ
33 32 div1d ⊢ φ ∧ x ∈ 1 +∞ → x 1 = x
34 31 33 breqtrd ⊢ φ ∧ x ∈ 1 +∞ → x A ≤ x
35 flword2 ⊢ x A ∈ ℝ ∧ x ∈ ℝ ∧ x A ≤ x → x ∈ ℤ ≥ x A
36 28 9 34 35 syl3anc ⊢ φ ∧ x ∈ 1 +∞ → x ∈ ℤ ≥ x A
37 fzsplit2 ⊢ x A + 1 ∈ ℤ ≥ 1 ∧ x ∈ ℤ ≥ x A → 1 … x = 1 … x A ∪ x A + 1 … x
38 27 36 37 syl2anc ⊢ φ ∧ x ∈ 1 +∞ → 1 … x = 1 … x A ∪ x A + 1 … x