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Theorem List for Metamath Proof Explorer - 25201-25300   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremmulogsumlem 25201* Lemma for mulogsum 25202. (Contributed by Mario Carneiro, 14-May-2016.)
(𝑥 ∈ ℝ+ ↦ Σ𝑛 ∈ (1...(⌊‘𝑥))(((μ‘𝑛) / 𝑛) · (Σ𝑚 ∈ (1...(⌊‘(𝑥 / 𝑛)))(1 / 𝑚) − (log‘(𝑥 / 𝑛))))) ∈ 𝑂(1)

Theoremmulogsum 25202* Asymptotic formula for Σ𝑛𝑥, (μ(𝑛) / 𝑛)log(𝑥 / 𝑛) = 𝑂(1). Equation 10.2.6 of [Shapiro], p. 406. (Contributed by Mario Carneiro, 14-May-2016.)
(𝑥 ∈ ℝ+ ↦ Σ𝑛 ∈ (1...(⌊‘𝑥))(((μ‘𝑛) / 𝑛) · (log‘(𝑥 / 𝑛)))) ∈ 𝑂(1)

Theoremlogdivsum 25203* Asymptotic analysis of Σ𝑛𝑥, log𝑛 / 𝑛 = (log𝑥)↑2 / 2 + 𝐿 + 𝑂(log𝑥 / 𝑥). (Contributed by Mario Carneiro, 18-May-2016.)
𝐹 = (𝑦 ∈ ℝ+ ↦ (Σ𝑖 ∈ (1...(⌊‘𝑦))((log‘𝑖) / 𝑖) − (((log‘𝑦)↑2) / 2)))       (𝐹:ℝ+⟶ℝ ∧ 𝐹 ∈ dom ⇝𝑟 ∧ ((𝐹𝑟 𝐿𝐴 ∈ ℝ+ ∧ e ≤ 𝐴) → (abs‘((𝐹𝐴) − 𝐿)) ≤ ((log‘𝐴) / 𝐴)))

Theoremmulog2sumlem1 25204* Asymptotic formula for Σ𝑛𝑥, log(𝑥 / 𝑛) / 𝑛 = (1 / 2)log↑2(𝑥) + γ · log𝑥𝐿 + 𝑂(log𝑥 / 𝑥), with explicit constants. Equation 10.2.7 of [Shapiro], p. 407. (Contributed by Mario Carneiro, 18-May-2016.)
𝐹 = (𝑦 ∈ ℝ+ ↦ (Σ𝑖 ∈ (1...(⌊‘𝑦))((log‘𝑖) / 𝑖) − (((log‘𝑦)↑2) / 2)))    &   (𝜑𝐹𝑟 𝐿)    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑 → e ≤ 𝐴)       (𝜑 → (abs‘(Σ𝑚 ∈ (1...(⌊‘𝐴))((log‘(𝐴 / 𝑚)) / 𝑚) − ((((log‘𝐴)↑2) / 2) + ((γ · (log‘𝐴)) − 𝐿)))) ≤ (2 · ((log‘𝐴) / 𝐴)))

Theoremmulog2sumlem2 25205* Lemma for mulog2sum 25207. (Contributed by Mario Carneiro, 19-May-2016.)
𝐹 = (𝑦 ∈ ℝ+ ↦ (Σ𝑖 ∈ (1...(⌊‘𝑦))((log‘𝑖) / 𝑖) − (((log‘𝑦)↑2) / 2)))    &   (𝜑𝐹𝑟 𝐿)    &   𝑇 = ((((log‘(𝑥 / 𝑛))↑2) / 2) + ((γ · (log‘(𝑥 / 𝑛))) − 𝐿))    &   𝑅 = (((1 / 2) + (γ + (abs‘𝐿))) + Σ𝑚 ∈ (1...2)((log‘(e / 𝑚)) / 𝑚))       (𝜑 → (𝑥 ∈ ℝ+ ↦ (Σ𝑛 ∈ (1...(⌊‘𝑥))(((μ‘𝑛) / 𝑛) · 𝑇) − (log‘𝑥))) ∈ 𝑂(1))

Theoremmulog2sumlem3 25206* Lemma for mulog2sum 25207. (Contributed by Mario Carneiro, 13-May-2016.)
𝐹 = (𝑦 ∈ ℝ+ ↦ (Σ𝑖 ∈ (1...(⌊‘𝑦))((log‘𝑖) / 𝑖) − (((log‘𝑦)↑2) / 2)))    &   (𝜑𝐹𝑟 𝐿)       (𝜑 → (𝑥 ∈ ℝ+ ↦ (Σ𝑛 ∈ (1...(⌊‘𝑥))(((μ‘𝑛) / 𝑛) · ((log‘(𝑥 / 𝑛))↑2)) − (2 · (log‘𝑥)))) ∈ 𝑂(1))

Theoremmulog2sum 25207* Asymptotic formula for Σ𝑛𝑥, (μ(𝑛) / 𝑛)log↑2(𝑥 / 𝑛) = 2log𝑥 + 𝑂(1). Equation 10.2.8 of [Shapiro], p. 407. (Contributed by Mario Carneiro, 19-May-2016.)
(𝑥 ∈ ℝ+ ↦ (Σ𝑛 ∈ (1...(⌊‘𝑥))(((μ‘𝑛) / 𝑛) · ((log‘(𝑥 / 𝑛))↑2)) − (2 · (log‘𝑥)))) ∈ 𝑂(1)

Theoremvmalogdivsum2 25208* The sum Σ𝑛𝑥, Λ(𝑛)log(𝑥 / 𝑛) / 𝑛 is asymptotic to log↑2(𝑥) / 2 + 𝑂(log𝑥). Exercise 9.1.7 of [Shapiro], p. 336. (Contributed by Mario Carneiro, 30-May-2016.)
(𝑥 ∈ (1(,)+∞) ↦ ((Σ𝑛 ∈ (1...(⌊‘𝑥))(((Λ‘𝑛) / 𝑛) · (log‘(𝑥 / 𝑛))) / (log‘𝑥)) − ((log‘𝑥) / 2))) ∈ 𝑂(1)

Theoremvmalogdivsum 25209* The sum Σ𝑛𝑥, Λ(𝑛)log𝑛 / 𝑛 is asymptotic to log↑2(𝑥) / 2 + 𝑂(log𝑥). Exercise 9.1.7 of [Shapiro], p. 336. (Contributed by Mario Carneiro, 30-May-2016.)
(𝑥 ∈ (1(,)+∞) ↦ ((Σ𝑛 ∈ (1...(⌊‘𝑥))(((Λ‘𝑛) / 𝑛) · (log‘𝑛)) / (log‘𝑥)) − ((log‘𝑥) / 2))) ∈ 𝑂(1)

Theorem2vmadivsumlem 25210* Lemma for 2vmadivsum 25211. (Contributed by Mario Carneiro, 30-May-2016.)
(𝜑𝐴 ∈ ℝ+)    &   (𝜑 → ∀𝑦 ∈ (1[,)+∞)(abs‘(Σ𝑖 ∈ (1...(⌊‘𝑦))((Λ‘𝑖) / 𝑖) − (log‘𝑦))) ≤ 𝐴)       (𝜑 → (𝑥 ∈ (1(,)+∞) ↦ ((Σ𝑛 ∈ (1...(⌊‘𝑥))(((Λ‘𝑛) / 𝑛) · Σ𝑚 ∈ (1...(⌊‘(𝑥 / 𝑛)))((Λ‘𝑚) / 𝑚)) / (log‘𝑥)) − ((log‘𝑥) / 2))) ∈ 𝑂(1))

Theorem2vmadivsum 25211* The sum Σ𝑚𝑛𝑥, Λ(𝑚)Λ(𝑛) / 𝑚𝑛 is asymptotic to log↑2(𝑥) / 2 + 𝑂(log𝑥). (Contributed by Mario Carneiro, 30-May-2016.)
(𝑥 ∈ (1(,)+∞) ↦ ((Σ𝑛 ∈ (1...(⌊‘𝑥))(((Λ‘𝑛) / 𝑛) · Σ𝑚 ∈ (1...(⌊‘(𝑥 / 𝑛)))((Λ‘𝑚) / 𝑚)) / (log‘𝑥)) − ((log‘𝑥) / 2))) ∈ 𝑂(1)

Theoremlogsqvma 25212* A formula for log↑2(𝑁) in terms of the primes. Equation 10.4.6 of [Shapiro], p. 418. (Contributed by Mario Carneiro, 13-May-2016.)
(𝑁 ∈ ℕ → Σ𝑑 ∈ {𝑥 ∈ ℕ ∣ 𝑥𝑁} (Σ𝑢 ∈ {𝑥 ∈ ℕ ∣ 𝑥𝑑} ((Λ‘𝑢) · (Λ‘(𝑑 / 𝑢))) + ((Λ‘𝑑) · (log‘𝑑))) = ((log‘𝑁)↑2))

Theoremlogsqvma2 25213* The Möbius inverse of logsqvma 25212. Equation 10.4.8 of [Shapiro], p. 418. (Contributed by Mario Carneiro, 13-May-2016.)
(𝑁 ∈ ℕ → Σ𝑑 ∈ {𝑥 ∈ ℕ ∣ 𝑥𝑁} ((μ‘𝑑) · ((log‘(𝑁 / 𝑑))↑2)) = (Σ𝑑 ∈ {𝑥 ∈ ℕ ∣ 𝑥𝑁} ((Λ‘𝑑) · (Λ‘(𝑁 / 𝑑))) + ((Λ‘𝑁) · (log‘𝑁))))

Theoremlog2sumbnd 25214* Bound on the difference between Σ𝑛𝐴, log↑2(𝑛) and the equivalent integral. (Contributed by Mario Carneiro, 20-May-2016.)
((𝐴 ∈ ℝ+ ∧ 1 ≤ 𝐴) → (abs‘(Σ𝑛 ∈ (1...(⌊‘𝐴))((log‘𝑛)↑2) − (𝐴 · (((log‘𝐴)↑2) + (2 − (2 · (log‘𝐴))))))) ≤ (((log‘𝐴)↑2) + 2))

Theoremselberglem1 25215* Lemma for selberg 25218. Estimation of the asymptotic part of selberglem3 25217. (Contributed by Mario Carneiro, 20-May-2016.)
𝑇 = ((((log‘(𝑥 / 𝑛))↑2) + (2 − (2 · (log‘(𝑥 / 𝑛))))) / 𝑛)       (𝑥 ∈ ℝ+ ↦ (Σ𝑛 ∈ (1...(⌊‘𝑥))((μ‘𝑛) · 𝑇) − (2 · (log‘𝑥)))) ∈ 𝑂(1)

Theoremselberglem2 25216* Lemma for selberg 25218. (Contributed by Mario Carneiro, 23-May-2016.)
𝑇 = ((((log‘(𝑥 / 𝑛))↑2) + (2 − (2 · (log‘(𝑥 / 𝑛))))) / 𝑛)       (𝑥 ∈ ℝ+ ↦ ((Σ𝑛 ∈ (1...(⌊‘𝑥))Σ𝑚 ∈ (1...(⌊‘(𝑥 / 𝑛)))((μ‘𝑛) · ((log‘𝑚)↑2)) / 𝑥) − (2 · (log‘𝑥)))) ∈ 𝑂(1)

Theoremselberglem3 25217* Lemma for selberg 25218. Estimation of the left-hand side of logsqvma2 25213. (Contributed by Mario Carneiro, 23-May-2016.)
(𝑥 ∈ ℝ+ ↦ ((Σ𝑛 ∈ (1...(⌊‘𝑥))Σ𝑑 ∈ {𝑦 ∈ ℕ ∣ 𝑦𝑛} ((μ‘𝑑) · ((log‘(𝑛 / 𝑑))↑2)) / 𝑥) − (2 · (log‘𝑥)))) ∈ 𝑂(1)

Theoremselberg 25218* Selberg's symmetry formula. The statement has many forms, and this one is equivalent to the statement that Σ𝑛𝑥, Λ(𝑛)log𝑛 + Σ𝑚 · 𝑛𝑥, Λ(𝑚)Λ(𝑛) = 2𝑥log𝑥 + 𝑂(𝑥). Equation 10.4.10 of [Shapiro], p. 419. (Contributed by Mario Carneiro, 23-May-2016.)
(𝑥 ∈ ℝ+ ↦ ((Σ𝑛 ∈ (1...(⌊‘𝑥))((Λ‘𝑛) · ((log‘𝑛) + (ψ‘(𝑥 / 𝑛)))) / 𝑥) − (2 · (log‘𝑥)))) ∈ 𝑂(1)

Theoremselbergb 25219* Convert eventual boundedness in selberg 25218 to boundedness on [1, +∞). (We have to bound away from zero because the log terms diverge at zero.) (Contributed by Mario Carneiro, 30-May-2016.)
𝑐 ∈ ℝ+𝑥 ∈ (1[,)+∞)(abs‘((Σ𝑛 ∈ (1...(⌊‘𝑥))((Λ‘𝑛) · ((log‘𝑛) + (ψ‘(𝑥 / 𝑛)))) / 𝑥) − (2 · (log‘𝑥)))) ≤ 𝑐

Theoremselberg2lem 25220* Lemma for selberg2 25221. Equation 10.4.12 of [Shapiro], p. 420. (Contributed by Mario Carneiro, 23-May-2016.)
(𝑥 ∈ ℝ+ ↦ ((Σ𝑛 ∈ (1...(⌊‘𝑥))((Λ‘𝑛) · (log‘𝑛)) − ((ψ‘𝑥) · (log‘𝑥))) / 𝑥)) ∈ 𝑂(1)

Theoremselberg2 25221* Selberg's symmetry formula, using the second Chebyshev function. Equation 10.4.14 of [Shapiro], p. 420. (Contributed by Mario Carneiro, 23-May-2016.)
(𝑥 ∈ ℝ+ ↦ (((((ψ‘𝑥) · (log‘𝑥)) + Σ𝑛 ∈ (1...(⌊‘𝑥))((Λ‘𝑛) · (ψ‘(𝑥 / 𝑛)))) / 𝑥) − (2 · (log‘𝑥)))) ∈ 𝑂(1)

Theoremselberg2b 25222* Convert eventual boundedness in selberg2 25221 to boundedness on any interval [𝐴, +∞). (We have to bound away from zero because the log terms diverge at zero.) (Contributed by Mario Carneiro, 25-May-2016.)
𝑐 ∈ ℝ+𝑥 ∈ (1[,)+∞)(abs‘(((((ψ‘𝑥) · (log‘𝑥)) + Σ𝑛 ∈ (1...(⌊‘𝑥))((Λ‘𝑛) · (ψ‘(𝑥 / 𝑛)))) / 𝑥) − (2 · (log‘𝑥)))) ≤ 𝑐

Theoremchpdifbndlem1 25223* Lemma for chpdifbnd 25225. (Contributed by Mario Carneiro, 25-May-2016.)
(𝜑𝐴 ∈ ℝ+)    &   (𝜑 → 1 ≤ 𝐴)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑 → ∀𝑧 ∈ (1[,)+∞)(abs‘(((((ψ‘𝑧) · (log‘𝑧)) + Σ𝑚 ∈ (1...(⌊‘𝑧))((Λ‘𝑚) · (ψ‘(𝑧 / 𝑚)))) / 𝑧) − (2 · (log‘𝑧)))) ≤ 𝐵)    &   𝐶 = ((𝐵 · (𝐴 + 1)) + ((2 · 𝐴) · (log‘𝐴)))    &   (𝜑𝑋 ∈ (1(,)+∞))    &   (𝜑𝑌 ∈ (𝑋[,](𝐴 · 𝑋)))       (𝜑 → ((ψ‘𝑌) − (ψ‘𝑋)) ≤ ((2 · (𝑌𝑋)) + (𝐶 · (𝑋 / (log‘𝑋)))))

Theoremchpdifbndlem2 25224* Lemma for chpdifbnd 25225. (Contributed by Mario Carneiro, 25-May-2016.)
(𝜑𝐴 ∈ ℝ+)    &   (𝜑 → 1 ≤ 𝐴)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑 → ∀𝑧 ∈ (1[,)+∞)(abs‘(((((ψ‘𝑧) · (log‘𝑧)) + Σ𝑚 ∈ (1...(⌊‘𝑧))((Λ‘𝑚) · (ψ‘(𝑧 / 𝑚)))) / 𝑧) − (2 · (log‘𝑧)))) ≤ 𝐵)    &   𝐶 = ((𝐵 · (𝐴 + 1)) + ((2 · 𝐴) · (log‘𝐴)))       (𝜑 → ∃𝑐 ∈ ℝ+𝑥 ∈ (1(,)+∞)∀𝑦 ∈ (𝑥[,](𝐴 · 𝑥))((ψ‘𝑦) − (ψ‘𝑥)) ≤ ((2 · (𝑦𝑥)) + (𝑐 · (𝑥 / (log‘𝑥)))))

Theoremchpdifbnd 25225* A bound on the difference of nearby ψ values. Theorem 10.5.2 of [Shapiro], p. 427. (Contributed by Mario Carneiro, 25-May-2016.)
((𝐴 ∈ ℝ ∧ 1 ≤ 𝐴) → ∃𝑐 ∈ ℝ+𝑥 ∈ (1(,)+∞)∀𝑦 ∈ (𝑥[,](𝐴 · 𝑥))((ψ‘𝑦) − (ψ‘𝑥)) ≤ ((2 · (𝑦𝑥)) + (𝑐 · (𝑥 / (log‘𝑥)))))

Theoremlogdivbnd 25226* A bound on a sum of logs, used in pntlemk 25276. This is not as precise as logdivsum 25203 in its asymptotic behavior, but it is valid for all 𝑁 and does not require a limit value. (Contributed by Mario Carneiro, 13-Apr-2016.)
(𝑁 ∈ ℕ → Σ𝑛 ∈ (1...𝑁)((log‘𝑛) / 𝑛) ≤ ((((log‘𝑁) + 1)↑2) / 2))

Theoremselberg3lem1 25227* Introduce a log weighting on the summands of Σ𝑚 · 𝑛𝑥, Λ(𝑚)Λ(𝑛), the core of selberg2 25221 (written here as Σ𝑛𝑥, Λ(𝑛)ψ(𝑥 / 𝑛)). Equation 10.4.21 of [Shapiro], p. 422. (Contributed by Mario Carneiro, 30-May-2016.)
(𝜑𝐴 ∈ ℝ+)    &   (𝜑 → ∀𝑦 ∈ (1[,)+∞)(abs‘((Σ𝑘 ∈ (1...(⌊‘𝑦))((Λ‘𝑘) · (log‘𝑘)) − ((ψ‘𝑦) · (log‘𝑦))) / 𝑦)) ≤ 𝐴)       (𝜑 → (𝑥 ∈ (1(,)+∞) ↦ ((((2 / (log‘𝑥)) · Σ𝑛 ∈ (1...(⌊‘𝑥))(((Λ‘𝑛) · (ψ‘(𝑥 / 𝑛))) · (log‘𝑛))) − Σ𝑛 ∈ (1...(⌊‘𝑥))((Λ‘𝑛) · (ψ‘(𝑥 / 𝑛)))) / 𝑥)) ∈ 𝑂(1))

Theoremselberg3lem2 25228* Lemma for selberg3 25229. Equation 10.4.21 of [Shapiro], p. 422. (Contributed by Mario Carneiro, 30-May-2016.)
(𝑥 ∈ (1(,)+∞) ↦ ((((2 / (log‘𝑥)) · Σ𝑛 ∈ (1...(⌊‘𝑥))(((Λ‘𝑛) · (ψ‘(𝑥 / 𝑛))) · (log‘𝑛))) − Σ𝑛 ∈ (1...(⌊‘𝑥))((Λ‘𝑛) · (ψ‘(𝑥 / 𝑛)))) / 𝑥)) ∈ 𝑂(1)

Theoremselberg3 25229* Introduce a log weighting on the summands of Σ𝑚 · 𝑛𝑥, Λ(𝑚)Λ(𝑛), the core of selberg2 25221 (written here as Σ𝑛𝑥, Λ(𝑛)ψ(𝑥 / 𝑛)). Equation 10.6.7 of [Shapiro], p. 422. (Contributed by Mario Carneiro, 30-May-2016.)
(𝑥 ∈ (1(,)+∞) ↦ (((((ψ‘𝑥) · (log‘𝑥)) + ((2 / (log‘𝑥)) · Σ𝑛 ∈ (1...(⌊‘𝑥))(((Λ‘𝑛) · (ψ‘(𝑥 / 𝑛))) · (log‘𝑛)))) / 𝑥) − (2 · (log‘𝑥)))) ∈ 𝑂(1)

Theoremselberg4lem1 25230* Lemma for selberg4 25231. Equation 10.4.20 of [Shapiro], p. 422. (Contributed by Mario Carneiro, 30-May-2016.)
(𝜑𝐴 ∈ ℝ+)    &   (𝜑 → ∀𝑦 ∈ (1[,)+∞)(abs‘((Σ𝑖 ∈ (1...(⌊‘𝑦))((Λ‘𝑖) · ((log‘𝑖) + (ψ‘(𝑦 / 𝑖)))) / 𝑦) − (2 · (log‘𝑦)))) ≤ 𝐴)       (𝜑 → (𝑥 ∈ (1(,)+∞) ↦ ((Σ𝑛 ∈ (1...(⌊‘𝑥))((Λ‘𝑛) · Σ𝑚 ∈ (1...(⌊‘(𝑥 / 𝑛)))((Λ‘𝑚) · ((log‘𝑚) + (ψ‘((𝑥 / 𝑛) / 𝑚))))) / (𝑥 · (log‘𝑥))) − (log‘𝑥))) ∈ 𝑂(1))

Theoremselberg4 25231* The Selberg symmetry formula for products of three primes, instead of two. The sum here can also be written in the symmetric form Σ𝑖𝑗𝑘𝑥, Λ(𝑖)Λ(𝑗)Λ(𝑘); we eliminate one of the nested sums by using the definition of ψ(𝑥) = Σ𝑘𝑥, Λ(𝑘). This statement can thus equivalently be written ψ(𝑥)log↑2(𝑥) = 𝑖𝑗𝑘𝑥, Λ(𝑖)Λ(𝑗)Λ(𝑘) + 𝑂(𝑥log𝑥). Equation 10.4.23 of [Shapiro], p. 422. (Contributed by Mario Carneiro, 30-May-2016.)
(𝑥 ∈ (1(,)+∞) ↦ ((((ψ‘𝑥) · (log‘𝑥)) − ((2 / (log‘𝑥)) · Σ𝑛 ∈ (1...(⌊‘𝑥))((Λ‘𝑛) · Σ𝑚 ∈ (1...(⌊‘(𝑥 / 𝑛)))((Λ‘𝑚) · (ψ‘((𝑥 / 𝑛) / 𝑚)))))) / 𝑥)) ∈ 𝑂(1)

Theorempntrval 25232* Define the residual of the second Chebyshev function. The goal is to have 𝑅(𝑥) ∈ 𝑜(𝑥), or 𝑅(𝑥) / 𝑥𝑟 0. (Contributed by Mario Carneiro, 8-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))       (𝐴 ∈ ℝ+ → (𝑅𝐴) = ((ψ‘𝐴) − 𝐴))

Theorempntrf 25233 Functionality of the residual. Lemma for pnt 25284. (Contributed by Mario Carneiro, 8-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))       𝑅:ℝ+⟶ℝ

Theorempntrmax 25234* There is a bound on the residual valid for all 𝑥. (Contributed by Mario Carneiro, 9-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))       𝑐 ∈ ℝ+𝑥 ∈ ℝ+ (abs‘((𝑅𝑥) / 𝑥)) ≤ 𝑐

Theorempntrsumo1 25235* A bound on a sum over 𝑅. Equation 10.1.16 of [Shapiro], p. 403. (Contributed by Mario Carneiro, 25-May-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))       (𝑥 ∈ ℝ ↦ Σ𝑛 ∈ (1...(⌊‘𝑥))((𝑅𝑛) / (𝑛 · (𝑛 + 1)))) ∈ 𝑂(1)

Theorempntrsumbnd 25236* A bound on a sum over 𝑅. Equation 10.1.16 of [Shapiro], p. 403. (Contributed by Mario Carneiro, 25-May-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))       𝑐 ∈ ℝ+𝑚 ∈ ℤ (abs‘Σ𝑛 ∈ (1...𝑚)((𝑅𝑛) / (𝑛 · (𝑛 + 1)))) ≤ 𝑐

Theorempntrsumbnd2 25237* A bound on a sum over 𝑅. Equation 10.1.16 of [Shapiro], p. 403. (Contributed by Mario Carneiro, 14-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))       𝑐 ∈ ℝ+𝑘 ∈ ℕ ∀𝑚 ∈ ℤ (abs‘Σ𝑛 ∈ (𝑘...𝑚)((𝑅𝑛) / (𝑛 · (𝑛 + 1)))) ≤ 𝑐

Theoremselbergr 25238* Selberg's symmetry formula, using the residual of the second Chebyshev function. Equation 10.6.2 of [Shapiro], p. 428. (Contributed by Mario Carneiro, 16-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))       (𝑥 ∈ ℝ+ ↦ ((((𝑅𝑥) · (log‘𝑥)) + Σ𝑑 ∈ (1...(⌊‘𝑥))((Λ‘𝑑) · (𝑅‘(𝑥 / 𝑑)))) / 𝑥)) ∈ 𝑂(1)

Theoremselberg3r 25239* Selberg's symmetry formula, using the residual of the second Chebyshev function. Equation 10.6.8 of [Shapiro], p. 429. (Contributed by Mario Carneiro, 30-May-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))       (𝑥 ∈ (1(,)+∞) ↦ ((((𝑅𝑥) · (log‘𝑥)) + ((2 / (log‘𝑥)) · Σ𝑛 ∈ (1...(⌊‘𝑥))(((Λ‘𝑛) · (𝑅‘(𝑥 / 𝑛))) · (log‘𝑛)))) / 𝑥)) ∈ 𝑂(1)

Theoremselberg4r 25240* Selberg's symmetry formula, using the residual of the second Chebyshev function. Equation 10.6.11 of [Shapiro], p. 430. (Contributed by Mario Carneiro, 30-May-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))       (𝑥 ∈ (1(,)+∞) ↦ ((((𝑅𝑥) · (log‘𝑥)) − ((2 / (log‘𝑥)) · Σ𝑛 ∈ (1...(⌊‘𝑥))((Λ‘𝑛) · Σ𝑚 ∈ (1...(⌊‘(𝑥 / 𝑛)))((Λ‘𝑚) · (𝑅‘((𝑥 / 𝑛) / 𝑚)))))) / 𝑥)) ∈ 𝑂(1)

Theoremselberg34r 25241* The sum of selberg3r 25239 and selberg4r 25240. (Contributed by Mario Carneiro, 31-May-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))       (𝑥 ∈ (1(,)+∞) ↦ ((((𝑅𝑥) · (log‘𝑥)) − (Σ𝑛 ∈ (1...(⌊‘𝑥))((𝑅‘(𝑥 / 𝑛)) · (Σ𝑚 ∈ {𝑦 ∈ ℕ ∣ 𝑦𝑛} ((Λ‘𝑚) · (Λ‘(𝑛 / 𝑚))) − ((Λ‘𝑛) · (log‘𝑛)))) / (log‘𝑥))) / 𝑥)) ∈ 𝑂(1)

Theorempntsval 25242* Define the "Selberg function", whose asymptotic behavior is the content of selberg 25218. (Contributed by Mario Carneiro, 31-May-2016.)
𝑆 = (𝑎 ∈ ℝ ↦ Σ𝑖 ∈ (1...(⌊‘𝑎))((Λ‘𝑖) · ((log‘𝑖) + (ψ‘(𝑎 / 𝑖)))))       (𝐴 ∈ ℝ → (𝑆𝐴) = Σ𝑛 ∈ (1...(⌊‘𝐴))((Λ‘𝑛) · ((log‘𝑛) + (ψ‘(𝐴 / 𝑛)))))

Theorempntsf 25243* Functionality of the Selberg function. (Contributed by Mario Carneiro, 31-May-2016.)
𝑆 = (𝑎 ∈ ℝ ↦ Σ𝑖 ∈ (1...(⌊‘𝑎))((Λ‘𝑖) · ((log‘𝑖) + (ψ‘(𝑎 / 𝑖)))))       𝑆:ℝ⟶ℝ

Theoremselbergs 25244* Selberg's symmetry formula, using the definition of the Selberg function. (Contributed by Mario Carneiro, 31-May-2016.)
𝑆 = (𝑎 ∈ ℝ ↦ Σ𝑖 ∈ (1...(⌊‘𝑎))((Λ‘𝑖) · ((log‘𝑖) + (ψ‘(𝑎 / 𝑖)))))       (𝑥 ∈ ℝ+ ↦ (((𝑆𝑥) / 𝑥) − (2 · (log‘𝑥)))) ∈ 𝑂(1)

Theoremselbergsb 25245* Selberg's symmetry formula, using the definition of the Selberg function. (Contributed by Mario Carneiro, 31-May-2016.)
𝑆 = (𝑎 ∈ ℝ ↦ Σ𝑖 ∈ (1...(⌊‘𝑎))((Λ‘𝑖) · ((log‘𝑖) + (ψ‘(𝑎 / 𝑖)))))       𝑐 ∈ ℝ+𝑥 ∈ (1[,)+∞)(abs‘(((𝑆𝑥) / 𝑥) − (2 · (log‘𝑥)))) ≤ 𝑐

Theorempntsval2 25246* The Selberg function can be expressed using the convolution product of the von Mangoldt function with itself. (Contributed by Mario Carneiro, 31-May-2016.)
𝑆 = (𝑎 ∈ ℝ ↦ Σ𝑖 ∈ (1...(⌊‘𝑎))((Λ‘𝑖) · ((log‘𝑖) + (ψ‘(𝑎 / 𝑖)))))       (𝐴 ∈ ℝ → (𝑆𝐴) = Σ𝑛 ∈ (1...(⌊‘𝐴))(((Λ‘𝑛) · (log‘𝑛)) + Σ𝑚 ∈ {𝑦 ∈ ℕ ∣ 𝑦𝑛} ((Λ‘𝑚) · (Λ‘(𝑛 / 𝑚)))))

Theorempntrlog2bndlem1 25247* The sum of selberg3r 25239 and selberg4r 25240. (Contributed by Mario Carneiro, 31-May-2016.)
𝑆 = (𝑎 ∈ ℝ ↦ Σ𝑖 ∈ (1...(⌊‘𝑎))((Λ‘𝑖) · ((log‘𝑖) + (ψ‘(𝑎 / 𝑖)))))    &   𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))       (𝑥 ∈ (1(,)+∞) ↦ ((((abs‘(𝑅𝑥)) · (log‘𝑥)) − (Σ𝑛 ∈ (1...(⌊‘𝑥))((abs‘(𝑅‘(𝑥 / 𝑛))) · ((𝑆𝑛) − (𝑆‘(𝑛 − 1)))) / (log‘𝑥))) / 𝑥)) ∈ ≤𝑂(1)

Theorempntrlog2bndlem2 25248* Lemma for pntrlog2bnd 25254. Bound on the difference between the Selberg function and its approximation, inside a sum. (Contributed by Mario Carneiro, 31-May-2016.)
𝑆 = (𝑎 ∈ ℝ ↦ Σ𝑖 ∈ (1...(⌊‘𝑎))((Λ‘𝑖) · ((log‘𝑖) + (ψ‘(𝑎 / 𝑖)))))    &   𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑 → ∀𝑦 ∈ ℝ+ (ψ‘𝑦) ≤ (𝐴 · 𝑦))       (𝜑 → (𝑥 ∈ (1(,)+∞) ↦ (Σ𝑛 ∈ (1...(⌊‘𝑥))(𝑛 · (abs‘((𝑅‘(𝑥 / (𝑛 + 1))) − (𝑅‘(𝑥 / 𝑛))))) / (𝑥 · (log‘𝑥)))) ∈ 𝑂(1))

Theorempntrlog2bndlem3 25249* Lemma for pntrlog2bnd 25254. Bound on the difference between the Selberg function and its approximation, inside a sum. (Contributed by Mario Carneiro, 31-May-2016.)
𝑆 = (𝑎 ∈ ℝ ↦ Σ𝑖 ∈ (1...(⌊‘𝑎))((Λ‘𝑖) · ((log‘𝑖) + (ψ‘(𝑎 / 𝑖)))))    &   𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑 → ∀𝑦 ∈ (1[,)+∞)(abs‘(((𝑆𝑦) / 𝑦) − (2 · (log‘𝑦)))) ≤ 𝐴)       (𝜑 → (𝑥 ∈ (1(,)+∞) ↦ (Σ𝑛 ∈ (1...(⌊‘𝑥))(((abs‘(𝑅‘(𝑥 / 𝑛))) − (abs‘(𝑅‘(𝑥 / (𝑛 + 1))))) · ((𝑆𝑛) − (2 · (𝑛 · (log‘𝑛))))) / (𝑥 · (log‘𝑥)))) ∈ 𝑂(1))

Theorempntrlog2bndlem4 25250* Lemma for pntrlog2bnd 25254. Bound on the difference between the Selberg function and its approximation, inside a sum. (Contributed by Mario Carneiro, 31-May-2016.)
𝑆 = (𝑎 ∈ ℝ ↦ Σ𝑖 ∈ (1...(⌊‘𝑎))((Λ‘𝑖) · ((log‘𝑖) + (ψ‘(𝑎 / 𝑖)))))    &   𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   𝑇 = (𝑎 ∈ ℝ ↦ if(𝑎 ∈ ℝ+, (𝑎 · (log‘𝑎)), 0))       (𝑥 ∈ (1(,)+∞) ↦ ((((abs‘(𝑅𝑥)) · (log‘𝑥)) − ((2 / (log‘𝑥)) · Σ𝑛 ∈ (1...(⌊‘𝑥))((abs‘(𝑅‘(𝑥 / 𝑛))) · ((𝑇𝑛) − (𝑇‘(𝑛 − 1)))))) / 𝑥)) ∈ ≤𝑂(1)

Theorempntrlog2bndlem5 25251* Lemma for pntrlog2bnd 25254. Bound on the difference between the Selberg function and its approximation, inside a sum. (Contributed by Mario Carneiro, 31-May-2016.)
𝑆 = (𝑎 ∈ ℝ ↦ Σ𝑖 ∈ (1...(⌊‘𝑎))((Λ‘𝑖) · ((log‘𝑖) + (ψ‘(𝑎 / 𝑖)))))    &   𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   𝑇 = (𝑎 ∈ ℝ ↦ if(𝑎 ∈ ℝ+, (𝑎 · (log‘𝑎)), 0))    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑 → ∀𝑦 ∈ ℝ+ (abs‘((𝑅𝑦) / 𝑦)) ≤ 𝐵)       (𝜑 → (𝑥 ∈ (1(,)+∞) ↦ ((((abs‘(𝑅𝑥)) · (log‘𝑥)) − ((2 / (log‘𝑥)) · Σ𝑛 ∈ (1...(⌊‘𝑥))((abs‘(𝑅‘(𝑥 / 𝑛))) · (log‘𝑛)))) / 𝑥)) ∈ ≤𝑂(1))

Theorempntrlog2bndlem6a 25252* Lemma for pntrlog2bndlem6 25253. (Contributed by Mario Carneiro, 7-Jun-2016.)
𝑆 = (𝑎 ∈ ℝ ↦ Σ𝑖 ∈ (1...(⌊‘𝑎))((Λ‘𝑖) · ((log‘𝑖) + (ψ‘(𝑎 / 𝑖)))))    &   𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   𝑇 = (𝑎 ∈ ℝ ↦ if(𝑎 ∈ ℝ+, (𝑎 · (log‘𝑎)), 0))    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑 → ∀𝑦 ∈ ℝ+ (abs‘((𝑅𝑦) / 𝑦)) ≤ 𝐵)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 1 ≤ 𝐴)       ((𝜑𝑥 ∈ (1(,)+∞)) → (1...(⌊‘𝑥)) = ((1...(⌊‘(𝑥 / 𝐴))) ∪ (((⌊‘(𝑥 / 𝐴)) + 1)...(⌊‘𝑥))))

Theorempntrlog2bndlem6 25253* Lemma for pntrlog2bnd 25254. Bound on the difference between the Selberg function and its approximation, inside a sum. (Contributed by Mario Carneiro, 31-May-2016.)
𝑆 = (𝑎 ∈ ℝ ↦ Σ𝑖 ∈ (1...(⌊‘𝑎))((Λ‘𝑖) · ((log‘𝑖) + (ψ‘(𝑎 / 𝑖)))))    &   𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   𝑇 = (𝑎 ∈ ℝ ↦ if(𝑎 ∈ ℝ+, (𝑎 · (log‘𝑎)), 0))    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑 → ∀𝑦 ∈ ℝ+ (abs‘((𝑅𝑦) / 𝑦)) ≤ 𝐵)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 1 ≤ 𝐴)       (𝜑 → (𝑥 ∈ (1(,)+∞) ↦ ((((abs‘(𝑅𝑥)) · (log‘𝑥)) − ((2 / (log‘𝑥)) · Σ𝑛 ∈ (1...(⌊‘(𝑥 / 𝐴)))((abs‘(𝑅‘(𝑥 / 𝑛))) · (log‘𝑛)))) / 𝑥)) ∈ ≤𝑂(1))

Theorempntrlog2bnd 25254* A bound on 𝑅(𝑥)log↑2(𝑥). Equation 10.6.15 of [Shapiro], p. 431. (Contributed by Mario Carneiro, 1-Jun-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))       ((𝐴 ∈ ℝ ∧ 1 ≤ 𝐴) → ∃𝑐 ∈ ℝ+𝑥 ∈ (1(,)+∞)((((abs‘(𝑅𝑥)) · (log‘𝑥)) − ((2 / (log‘𝑥)) · Σ𝑛 ∈ (1...(⌊‘(𝑥 / 𝐴)))((abs‘(𝑅‘(𝑥 / 𝑛))) · (log‘𝑛)))) / 𝑥) ≤ 𝑐)

Theorempntpbnd1a 25255* Lemma for pntpbnd 25258. (Contributed by Mario Carneiro, 11-Apr-2016.) Replace reference to OLD theorem. (Revised by Wolf Lammen, 8-Sep-2020.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐸 ∈ (0(,)1))    &   𝑋 = (exp‘(2 / 𝐸))    &   (𝜑𝑌 ∈ (𝑋(,)+∞))    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑 → (𝑌 < 𝑁𝑁 ≤ (𝐾 · 𝑌)))    &   (𝜑 → (abs‘(𝑅𝑁)) ≤ (abs‘((𝑅‘(𝑁 + 1)) − (𝑅𝑁))))       (𝜑 → (abs‘((𝑅𝑁) / 𝑁)) ≤ 𝐸)

Theorempntpbnd1 25256* Lemma for pntpbnd 25258. (Contributed by Mario Carneiro, 11-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐸 ∈ (0(,)1))    &   𝑋 = (exp‘(2 / 𝐸))    &   (𝜑𝑌 ∈ (𝑋(,)+∞))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑 → ∀𝑖 ∈ ℕ ∀𝑗 ∈ ℤ (abs‘Σ𝑦 ∈ (𝑖...𝑗)((𝑅𝑦) / (𝑦 · (𝑦 + 1)))) ≤ 𝐴)    &   𝐶 = (𝐴 + 2)    &   (𝜑𝐾 ∈ ((exp‘(𝐶 / 𝐸))[,)+∞))    &   (𝜑 → ¬ ∃𝑦 ∈ ℕ ((𝑌 < 𝑦𝑦 ≤ (𝐾 · 𝑌)) ∧ (abs‘((𝑅𝑦) / 𝑦)) ≤ 𝐸))       (𝜑 → Σ𝑛 ∈ (((⌊‘𝑌) + 1)...(⌊‘(𝐾 · 𝑌)))(abs‘((𝑅𝑛) / (𝑛 · (𝑛 + 1)))) ≤ 𝐴)

Theorempntpbnd2 25257* Lemma for pntpbnd 25258. (Contributed by Mario Carneiro, 11-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐸 ∈ (0(,)1))    &   𝑋 = (exp‘(2 / 𝐸))    &   (𝜑𝑌 ∈ (𝑋(,)+∞))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑 → ∀𝑖 ∈ ℕ ∀𝑗 ∈ ℤ (abs‘Σ𝑦 ∈ (𝑖...𝑗)((𝑅𝑦) / (𝑦 · (𝑦 + 1)))) ≤ 𝐴)    &   𝐶 = (𝐴 + 2)    &   (𝜑𝐾 ∈ ((exp‘(𝐶 / 𝐸))[,)+∞))    &   (𝜑 → ¬ ∃𝑦 ∈ ℕ ((𝑌 < 𝑦𝑦 ≤ (𝐾 · 𝑌)) ∧ (abs‘((𝑅𝑦) / 𝑦)) ≤ 𝐸))        ¬ 𝜑

Theorempntpbnd 25258* Lemma for pnt 25284. Establish smallness of 𝑅 at a point. Lemma 10.6.1 in [Shapiro], p. 436. (Contributed by Mario Carneiro, 10-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))       𝑐 ∈ ℝ+𝑒 ∈ (0(,)1)∃𝑥 ∈ ℝ+𝑘 ∈ ((exp‘(𝑐 / 𝑒))[,)+∞)∀𝑦 ∈ (𝑥(,)+∞)∃𝑛 ∈ ℕ ((𝑦 < 𝑛𝑛 ≤ (𝑘 · 𝑦)) ∧ (abs‘((𝑅𝑛) / 𝑛)) ≤ 𝑒)

Theorempntibndlem1 25259 Lemma for pntibnd 25263. (Contributed by Mario Carneiro, 10-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   𝐿 = ((1 / 4) / (𝐴 + 3))       (𝜑𝐿 ∈ (0(,)1))

Theorempntibndlem2a 25260* Lemma for pntibndlem2 25261. (Contributed by Mario Carneiro, 7-Jun-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   𝐿 = ((1 / 4) / (𝐴 + 3))    &   (𝜑 → ∀𝑥 ∈ ℝ+ (abs‘((𝑅𝑥) / 𝑥)) ≤ 𝐴)    &   (𝜑𝐵 ∈ ℝ+)    &   𝐾 = (exp‘(𝐵 / (𝐸 / 2)))    &   𝐶 = ((2 · 𝐵) + (log‘2))    &   (𝜑𝐸 ∈ (0(,)1))    &   (𝜑𝑍 ∈ ℝ+)    &   (𝜑𝑁 ∈ ℕ)       ((𝜑𝑢 ∈ (𝑁[,]((1 + (𝐿 · 𝐸)) · 𝑁))) → (𝑢 ∈ ℝ ∧ 𝑁𝑢𝑢 ≤ ((1 + (𝐿 · 𝐸)) · 𝑁)))

Theorempntibndlem2 25261* Lemma for pntibnd 25263. The main work, after eliminating all the quantifiers. (Contributed by Mario Carneiro, 10-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   𝐿 = ((1 / 4) / (𝐴 + 3))    &   (𝜑 → ∀𝑥 ∈ ℝ+ (abs‘((𝑅𝑥) / 𝑥)) ≤ 𝐴)    &   (𝜑𝐵 ∈ ℝ+)    &   𝐾 = (exp‘(𝐵 / (𝐸 / 2)))    &   𝐶 = ((2 · 𝐵) + (log‘2))    &   (𝜑𝐸 ∈ (0(,)1))    &   (𝜑𝑍 ∈ ℝ+)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑇 ∈ ℝ+)    &   (𝜑 → ∀𝑥 ∈ (1(,)+∞)∀𝑦 ∈ (𝑥[,](2 · 𝑥))((ψ‘𝑦) − (ψ‘𝑥)) ≤ ((2 · (𝑦𝑥)) + (𝑇 · (𝑥 / (log‘𝑥)))))    &   𝑋 = ((exp‘(𝑇 / (𝐸 / 4))) + 𝑍)    &   (𝜑𝑀 ∈ ((exp‘(𝐶 / 𝐸))[,)+∞))    &   (𝜑𝑌 ∈ (𝑋(,)+∞))    &   (𝜑 → ((𝑌 < 𝑁𝑁 ≤ ((𝑀 / 2) · 𝑌)) ∧ (abs‘((𝑅𝑁) / 𝑁)) ≤ (𝐸 / 2)))       (𝜑 → ∃𝑧 ∈ ℝ+ ((𝑌 < 𝑧 ∧ ((1 + (𝐿 · 𝐸)) · 𝑧) < (𝑀 · 𝑌)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝐿 · 𝐸)) · 𝑧))(abs‘((𝑅𝑢) / 𝑢)) ≤ 𝐸))

Theorempntibndlem3 25262* Lemma for pntibnd 25263. Package up pntibndlem2 25261 in quantifiers. (Contributed by Mario Carneiro, 10-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   𝐿 = ((1 / 4) / (𝐴 + 3))    &   (𝜑 → ∀𝑥 ∈ ℝ+ (abs‘((𝑅𝑥) / 𝑥)) ≤ 𝐴)    &   (𝜑𝐵 ∈ ℝ+)    &   𝐾 = (exp‘(𝐵 / (𝐸 / 2)))    &   𝐶 = ((2 · 𝐵) + (log‘2))    &   (𝜑𝐸 ∈ (0(,)1))    &   (𝜑𝑍 ∈ ℝ+)    &   (𝜑 → ∀𝑚 ∈ (𝐾[,)+∞)∀𝑣 ∈ (𝑍(,)+∞)∃𝑖 ∈ ℕ ((𝑣 < 𝑖𝑖 ≤ (𝑚 · 𝑣)) ∧ (abs‘((𝑅𝑖) / 𝑖)) ≤ (𝐸 / 2)))       (𝜑 → ∃𝑥 ∈ ℝ+𝑘 ∈ ((exp‘(𝐶 / 𝐸))[,)+∞)∀𝑦 ∈ (𝑥(,)+∞)∃𝑧 ∈ ℝ+ ((𝑦 < 𝑧 ∧ ((1 + (𝐿 · 𝐸)) · 𝑧) < (𝑘 · 𝑦)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝐿 · 𝐸)) · 𝑧))(abs‘((𝑅𝑢) / 𝑢)) ≤ 𝐸))

Theorempntibnd 25263* Lemma for pnt 25284. Establish smallness of 𝑅 on an interval. Lemma 10.6.2 in [Shapiro], p. 436. (Contributed by Mario Carneiro, 10-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))       𝑐 ∈ ℝ+𝑙 ∈ (0(,)1)∀𝑒 ∈ (0(,)1)∃𝑥 ∈ ℝ+𝑘 ∈ ((exp‘(𝑐 / 𝑒))[,)+∞)∀𝑦 ∈ (𝑥(,)+∞)∃𝑧 ∈ ℝ+ ((𝑦 < 𝑧 ∧ ((1 + (𝑙 · 𝑒)) · 𝑧) < (𝑘 · 𝑦)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝑙 · 𝑒)) · 𝑧))(abs‘((𝑅𝑢) / 𝑢)) ≤ 𝑒)

Theorempntlemd 25264 Lemma for pnt 25284. Closure for the constants used in the proof. For comparison with Equation 10.6.27 of [Shapiro], p. 434, 𝐴 is C^*, 𝐵 is c1, 𝐿 is λ, 𝐷 is c2, and 𝐹 is c3. (Contributed by Mario Carneiro, 13-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))       (𝜑 → (𝐿 ∈ ℝ+𝐷 ∈ ℝ+𝐹 ∈ ℝ+))

Theorempntlemc 25265* Lemma for pnt 25284. Closure for the constants used in the proof. For comparison with Equation 10.6.27 of [Shapiro], p. 434, 𝑈 is α, 𝐸 is ε, and 𝐾 is K. (Contributed by Mario Carneiro, 13-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑𝑈 ∈ ℝ+)    &   (𝜑𝑈𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))       (𝜑 → (𝐸 ∈ ℝ+𝐾 ∈ ℝ+ ∧ (𝐸 ∈ (0(,)1) ∧ 1 < 𝐾 ∧ (𝑈𝐸) ∈ ℝ+)))

Theorempntlema 25266* Lemma for pnt 25284. Closure for the constants used in the proof. The mammoth expression 𝑊 is a number large enough to satisfy all the lower bounds needed for 𝑍. For comparison with Equation 10.6.27 of [Shapiro], p. 434, 𝑌 is x2, 𝑋 is x1, 𝐶 is the big-O constant in Equation 10.6.29 of [Shapiro], p. 435, and 𝑊 is the unnamed lower bound of "for sufficiently large x" in Equation 10.6.34 of [Shapiro], p. 436. (Contributed by Mario Carneiro, 13-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑𝑈 ∈ ℝ+)    &   (𝜑𝑈𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+𝑌 < 𝑋))    &   (𝜑𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))       (𝜑𝑊 ∈ ℝ+)

Theorempntlemb 25267* Lemma for pnt 25284. Unpack all the lower bounds contained in 𝑊, in the form they will be used. For comparison with Equation 10.6.27 of [Shapiro], p. 434, 𝑍 is x. (Contributed by Mario Carneiro, 13-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑𝑈 ∈ ℝ+)    &   (𝜑𝑈𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+𝑌 < 𝑋))    &   (𝜑𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑𝑍 ∈ (𝑊[,)+∞))       (𝜑 → (𝑍 ∈ ℝ+ ∧ (1 < 𝑍 ∧ e ≤ (√‘𝑍) ∧ (√‘𝑍) ≤ (𝑍 / 𝑌)) ∧ ((4 / (𝐿 · 𝐸)) ≤ (√‘𝑍) ∧ (((log‘𝑋) / (log‘𝐾)) + 2) ≤ (((log‘𝑍) / (log‘𝐾)) / 4) ∧ ((𝑈 · 3) + 𝐶) ≤ (((𝑈𝐸) · ((𝐿 · (𝐸↑2)) / (32 · 𝐵))) · (log‘𝑍)))))

Theorempntlemg 25268* Lemma for pnt 25284. Closure for the constants used in the proof. For comparison with Equation 10.6.27 of [Shapiro], p. 434, 𝑀 is j^* and 𝑁 is ĵ. (Contributed by Mario Carneiro, 13-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑𝑈 ∈ ℝ+)    &   (𝜑𝑈𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+𝑌 < 𝑋))    &   (𝜑𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑𝑍 ∈ (𝑊[,)+∞))    &   𝑀 = ((⌊‘((log‘𝑋) / (log‘𝐾))) + 1)    &   𝑁 = (⌊‘(((log‘𝑍) / (log‘𝐾)) / 2))       (𝜑 → (𝑀 ∈ ℕ ∧ 𝑁 ∈ (ℤ𝑀) ∧ (((log‘𝑍) / (log‘𝐾)) / 4) ≤ (𝑁𝑀)))

Theorempntlemh 25269* Lemma for pnt 25284. Bounds on the subintervals in the induction. (Contributed by Mario Carneiro, 13-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑𝑈 ∈ ℝ+)    &   (𝜑𝑈𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+𝑌 < 𝑋))    &   (𝜑𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑𝑍 ∈ (𝑊[,)+∞))    &   𝑀 = ((⌊‘((log‘𝑋) / (log‘𝐾))) + 1)    &   𝑁 = (⌊‘(((log‘𝑍) / (log‘𝐾)) / 2))       ((𝜑𝐽 ∈ (𝑀...𝑁)) → (𝑋 < (𝐾𝐽) ∧ (𝐾𝐽) ≤ (√‘𝑍)))

Theorempntlemn 25270* Lemma for pnt 25284. The "naive" base bound, which we will slightly improve. (Contributed by Mario Carneiro, 13-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑𝑈 ∈ ℝ+)    &   (𝜑𝑈𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+𝑌 < 𝑋))    &   (𝜑𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑𝑍 ∈ (𝑊[,)+∞))    &   𝑀 = ((⌊‘((log‘𝑋) / (log‘𝐾))) + 1)    &   𝑁 = (⌊‘(((log‘𝑍) / (log‘𝐾)) / 2))    &   (𝜑 → ∀𝑧 ∈ (𝑌[,)+∞)(abs‘((𝑅𝑧) / 𝑧)) ≤ 𝑈)       ((𝜑 ∧ (𝐽 ∈ ℕ ∧ 𝐽 ≤ (𝑍 / 𝑌))) → 0 ≤ (((𝑈 / 𝐽) − (abs‘((𝑅‘(𝑍 / 𝐽)) / 𝑍))) · (log‘𝐽)))

Theorempntlemq 25271* Lemma for pntlemj 25273. (Contributed by Mario Carneiro, 7-Jun-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑𝑈 ∈ ℝ+)    &   (𝜑𝑈𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+𝑌 < 𝑋))    &   (𝜑𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑𝑍 ∈ (𝑊[,)+∞))    &   𝑀 = ((⌊‘((log‘𝑋) / (log‘𝐾))) + 1)    &   𝑁 = (⌊‘(((log‘𝑍) / (log‘𝐾)) / 2))    &   (𝜑 → ∀𝑧 ∈ (𝑌[,)+∞)(abs‘((𝑅𝑧) / 𝑧)) ≤ 𝑈)    &   (𝜑 → ∀𝑦 ∈ (𝑋(,)+∞)∃𝑧 ∈ ℝ+ ((𝑦 < 𝑧 ∧ ((1 + (𝐿 · 𝐸)) · 𝑧) < (𝐾 · 𝑦)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝐿 · 𝐸)) · 𝑧))(abs‘((𝑅𝑢) / 𝑢)) ≤ 𝐸))    &   𝑂 = (((⌊‘(𝑍 / (𝐾↑(𝐽 + 1)))) + 1)...(⌊‘(𝑍 / (𝐾𝐽))))    &   (𝜑𝑉 ∈ ℝ+)    &   (𝜑 → (((𝐾𝐽) < 𝑉 ∧ ((1 + (𝐿 · 𝐸)) · 𝑉) < (𝐾 · (𝐾𝐽))) ∧ ∀𝑢 ∈ (𝑉[,]((1 + (𝐿 · 𝐸)) · 𝑉))(abs‘((𝑅𝑢) / 𝑢)) ≤ 𝐸))    &   (𝜑𝐽 ∈ (𝑀..^𝑁))    &   𝐼 = (((⌊‘(𝑍 / ((1 + (𝐿 · 𝐸)) · 𝑉))) + 1)...(⌊‘(𝑍 / 𝑉)))       (𝜑𝐼𝑂)

Theorempntlemr 25272* Lemma for pntlemj 25273. (Contributed by Mario Carneiro, 7-Jun-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑𝑈 ∈ ℝ+)    &   (𝜑𝑈𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+𝑌 < 𝑋))    &   (𝜑𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑𝑍 ∈ (𝑊[,)+∞))    &   𝑀 = ((⌊‘((log‘𝑋) / (log‘𝐾))) + 1)    &   𝑁 = (⌊‘(((log‘𝑍) / (log‘𝐾)) / 2))    &   (𝜑 → ∀𝑧 ∈ (𝑌[,)+∞)(abs‘((𝑅𝑧) / 𝑧)) ≤ 𝑈)    &   (𝜑 → ∀𝑦 ∈ (𝑋(,)+∞)∃𝑧 ∈ ℝ+ ((𝑦 < 𝑧 ∧ ((1 + (𝐿 · 𝐸)) · 𝑧) < (𝐾 · 𝑦)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝐿 · 𝐸)) · 𝑧))(abs‘((𝑅𝑢) / 𝑢)) ≤ 𝐸))    &   𝑂 = (((⌊‘(𝑍 / (𝐾↑(𝐽 + 1)))) + 1)...(⌊‘(𝑍 / (𝐾𝐽))))    &   (𝜑𝑉 ∈ ℝ+)    &   (𝜑 → (((𝐾𝐽) < 𝑉 ∧ ((1 + (𝐿 · 𝐸)) · 𝑉) < (𝐾 · (𝐾𝐽))) ∧ ∀𝑢 ∈ (𝑉[,]((1 + (𝐿 · 𝐸)) · 𝑉))(abs‘((𝑅𝑢) / 𝑢)) ≤ 𝐸))    &   (𝜑𝐽 ∈ (𝑀..^𝑁))    &   𝐼 = (((⌊‘(𝑍 / ((1 + (𝐿 · 𝐸)) · 𝑉))) + 1)...(⌊‘(𝑍 / 𝑉)))       (𝜑 → ((𝑈𝐸) · (((𝐿 · 𝐸) / 8) · (log‘𝑍))) ≤ ((#‘𝐼) · ((𝑈𝐸) · ((log‘(𝑍 / 𝑉)) / (𝑍 / 𝑉)))))

Theorempntlemj 25273* Lemma for pnt 25284. The induction step. Using pntibnd 25263, we find an interval in 𝐾𝐽...𝐾↑(𝐽 + 1) which is sufficiently large and has a much smaller value, 𝑅(𝑧) / 𝑧𝐸 (instead of our original bound 𝑅(𝑧) / 𝑧𝑈). (Contributed by Mario Carneiro, 13-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑𝑈 ∈ ℝ+)    &   (𝜑𝑈𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+𝑌 < 𝑋))    &   (𝜑𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑𝑍 ∈ (𝑊[,)+∞))    &   𝑀 = ((⌊‘((log‘𝑋) / (log‘𝐾))) + 1)    &   𝑁 = (⌊‘(((log‘𝑍) / (log‘𝐾)) / 2))    &   (𝜑 → ∀𝑧 ∈ (𝑌[,)+∞)(abs‘((𝑅𝑧) / 𝑧)) ≤ 𝑈)    &   (𝜑 → ∀𝑦 ∈ (𝑋(,)+∞)∃𝑧 ∈ ℝ+ ((𝑦 < 𝑧 ∧ ((1 + (𝐿 · 𝐸)) · 𝑧) < (𝐾 · 𝑦)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝐿 · 𝐸)) · 𝑧))(abs‘((𝑅𝑢) / 𝑢)) ≤ 𝐸))    &   𝑂 = (((⌊‘(𝑍 / (𝐾↑(𝐽 + 1)))) + 1)...(⌊‘(𝑍 / (𝐾𝐽))))    &   (𝜑𝑉 ∈ ℝ+)    &   (𝜑 → (((𝐾𝐽) < 𝑉 ∧ ((1 + (𝐿 · 𝐸)) · 𝑉) < (𝐾 · (𝐾𝐽))) ∧ ∀𝑢 ∈ (𝑉[,]((1 + (𝐿 · 𝐸)) · 𝑉))(abs‘((𝑅𝑢) / 𝑢)) ≤ 𝐸))    &   (𝜑𝐽 ∈ (𝑀..^𝑁))    &   𝐼 = (((⌊‘(𝑍 / ((1 + (𝐿 · 𝐸)) · 𝑉))) + 1)...(⌊‘(𝑍 / 𝑉)))       (𝜑 → ((𝑈𝐸) · (((𝐿 · 𝐸) / 8) · (log‘𝑍))) ≤ Σ𝑛𝑂 (((𝑈 / 𝑛) − (abs‘((𝑅‘(𝑍 / 𝑛)) / 𝑍))) · (log‘𝑛)))

Theorempntlemi 25274* Lemma for pnt 25284. Eliminate some assumptions from pntlemj 25273. (Contributed by Mario Carneiro, 13-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑𝑈 ∈ ℝ+)    &   (𝜑𝑈𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+𝑌 < 𝑋))    &   (𝜑𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑𝑍 ∈ (𝑊[,)+∞))    &   𝑀 = ((⌊‘((log‘𝑋) / (log‘𝐾))) + 1)    &   𝑁 = (⌊‘(((log‘𝑍) / (log‘𝐾)) / 2))    &   (𝜑 → ∀𝑧 ∈ (𝑌[,)+∞)(abs‘((𝑅𝑧) / 𝑧)) ≤ 𝑈)    &   (𝜑 → ∀𝑦 ∈ (𝑋(,)+∞)∃𝑧 ∈ ℝ+ ((𝑦 < 𝑧 ∧ ((1 + (𝐿 · 𝐸)) · 𝑧) < (𝐾 · 𝑦)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝐿 · 𝐸)) · 𝑧))(abs‘((𝑅𝑢) / 𝑢)) ≤ 𝐸))    &   𝑂 = (((⌊‘(𝑍 / (𝐾↑(𝐽 + 1)))) + 1)...(⌊‘(𝑍 / (𝐾𝐽))))       ((𝜑𝐽 ∈ (𝑀..^𝑁)) → ((𝑈𝐸) · (((𝐿 · 𝐸) / 8) · (log‘𝑍))) ≤ Σ𝑛𝑂 (((𝑈 / 𝑛) − (abs‘((𝑅‘(𝑍 / 𝑛)) / 𝑍))) · (log‘𝑛)))

Theorempntlemf 25275* Lemma for pnt 25284. Add up the pieces in pntlemi 25274 to get an estimate slightly better than the naive lower bound 0. (Contributed by Mario Carneiro, 13-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑𝑈 ∈ ℝ+)    &   (𝜑𝑈𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+𝑌 < 𝑋))    &   (𝜑𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑𝑍 ∈ (𝑊[,)+∞))    &   𝑀 = ((⌊‘((log‘𝑋) / (log‘𝐾))) + 1)    &   𝑁 = (⌊‘(((log‘𝑍) / (log‘𝐾)) / 2))    &   (𝜑 → ∀𝑧 ∈ (𝑌[,)+∞)(abs‘((𝑅𝑧) / 𝑧)) ≤ 𝑈)    &   (𝜑 → ∀𝑦 ∈ (𝑋(,)+∞)∃𝑧 ∈ ℝ+ ((𝑦 < 𝑧 ∧ ((1 + (𝐿 · 𝐸)) · 𝑧) < (𝐾 · 𝑦)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝐿 · 𝐸)) · 𝑧))(abs‘((𝑅𝑢) / 𝑢)) ≤ 𝐸))       (𝜑 → ((𝑈𝐸) · (((𝐿 · (𝐸↑2)) / (32 · 𝐵)) · ((log‘𝑍)↑2))) ≤ Σ𝑛 ∈ (1...(⌊‘(𝑍 / 𝑌)))(((𝑈 / 𝑛) − (abs‘((𝑅‘(𝑍 / 𝑛)) / 𝑍))) · (log‘𝑛)))

Theorempntlemk 25276* Lemma for pnt 25284. Evaluate the naive part of the estimate. (Contributed by Mario Carneiro, 14-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑𝑈 ∈ ℝ+)    &   (𝜑𝑈𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+𝑌 < 𝑋))    &   (𝜑𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑𝑍 ∈ (𝑊[,)+∞))    &   𝑀 = ((⌊‘((log‘𝑋) / (log‘𝐾))) + 1)    &   𝑁 = (⌊‘(((log‘𝑍) / (log‘𝐾)) / 2))    &   (𝜑 → ∀𝑧 ∈ (𝑌[,)+∞)(abs‘((𝑅𝑧) / 𝑧)) ≤ 𝑈)    &   (𝜑 → ∀𝑦 ∈ (𝑋(,)+∞)∃𝑧 ∈ ℝ+ ((𝑦 < 𝑧 ∧ ((1 + (𝐿 · 𝐸)) · 𝑧) < (𝐾 · 𝑦)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝐿 · 𝐸)) · 𝑧))(abs‘((𝑅𝑢) / 𝑢)) ≤ 𝐸))       (𝜑 → (2 · Σ𝑛 ∈ (1...(⌊‘(𝑍 / 𝑌)))((𝑈 / 𝑛) · (log‘𝑛))) ≤ ((𝑈 · ((log‘𝑍) + 3)) · (log‘𝑍)))

Theorempntlemo 25277* Lemma for pnt 25284. Combine all the estimates to establish a smaller eventual bound on 𝑅(𝑍) / 𝑍. (Contributed by Mario Carneiro, 14-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑𝑈 ∈ ℝ+)    &   (𝜑𝑈𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+𝑌 < 𝑋))    &   (𝜑𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑𝑍 ∈ (𝑊[,)+∞))    &   𝑀 = ((⌊‘((log‘𝑋) / (log‘𝐾))) + 1)    &   𝑁 = (⌊‘(((log‘𝑍) / (log‘𝐾)) / 2))    &   (𝜑 → ∀𝑧 ∈ (𝑌[,)+∞)(abs‘((𝑅𝑧) / 𝑧)) ≤ 𝑈)    &   (𝜑 → ∀𝑦 ∈ (𝑋(,)+∞)∃𝑧 ∈ ℝ+ ((𝑦 < 𝑧 ∧ ((1 + (𝐿 · 𝐸)) · 𝑧) < (𝐾 · 𝑦)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝐿 · 𝐸)) · 𝑧))(abs‘((𝑅𝑢) / 𝑢)) ≤ 𝐸))    &   (𝜑 → ∀𝑧 ∈ (1(,)+∞)((((abs‘(𝑅𝑧)) · (log‘𝑧)) − ((2 / (log‘𝑧)) · Σ𝑖 ∈ (1...(⌊‘(𝑧 / 𝑌)))((abs‘(𝑅‘(𝑧 / 𝑖))) · (log‘𝑖)))) / 𝑧) ≤ 𝐶)       (𝜑 → (abs‘((𝑅𝑍) / 𝑍)) ≤ (𝑈 − (𝐹 · (𝑈↑3))))

Theorempntleme 25278* Lemma for pnt 25284. Package up pntlemo 25277 in quantifiers. (Contributed by Mario Carneiro, 14-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑𝑈 ∈ ℝ+)    &   (𝜑𝑈𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+𝑌 < 𝑋))    &   (𝜑𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑 → ∀𝑧 ∈ (𝑌[,)+∞)(abs‘((𝑅𝑧) / 𝑧)) ≤ 𝑈)    &   (𝜑 → ∀𝑘 ∈ (𝐾[,)+∞)∀𝑦 ∈ (𝑋(,)+∞)∃𝑧 ∈ ℝ+ ((𝑦 < 𝑧 ∧ ((1 + (𝐿 · 𝐸)) · 𝑧) < (𝑘 · 𝑦)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝐿 · 𝐸)) · 𝑧))(abs‘((𝑅𝑢) / 𝑢)) ≤ 𝐸))    &   (𝜑 → ∀𝑧 ∈ (1(,)+∞)((((abs‘(𝑅𝑧)) · (log‘𝑧)) − ((2 / (log‘𝑧)) · Σ𝑖 ∈ (1...(⌊‘(𝑧 / 𝑌)))((abs‘(𝑅‘(𝑧 / 𝑖))) · (log‘𝑖)))) / 𝑧) ≤ 𝐶)       (𝜑 → ∃𝑤 ∈ ℝ+𝑣 ∈ (𝑤[,)+∞)(abs‘((𝑅𝑣) / 𝑣)) ≤ (𝑈 − (𝐹 · (𝑈↑3))))

Theorempntlem3 25279* Lemma for pnt 25284. Equation 10.6.35 in [Shapiro], p. 436. (Contributed by Mario Carneiro, 8-Apr-2016.) (Proof shortened by AV, 27-Sep-2020.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑 → ∀𝑥 ∈ ℝ+ (abs‘((𝑅𝑥) / 𝑥)) ≤ 𝐴)    &   𝑇 = {𝑡 ∈ (0[,]𝐴) ∣ ∃𝑦 ∈ ℝ+𝑧 ∈ (𝑦[,)+∞)(abs‘((𝑅𝑧) / 𝑧)) ≤ 𝑡}    &   (𝜑𝐶 ∈ ℝ+)    &   ((𝜑𝑢𝑇) → (𝑢 − (𝐶 · (𝑢↑3))) ∈ 𝑇)       (𝜑 → (𝑥 ∈ ℝ+ ↦ ((ψ‘𝑥) / 𝑥)) ⇝𝑟 1)

Theorempntlemp 25280* Lemma for pnt 25284. Wrapping up more quantifiers. (Contributed by Mario Carneiro, 14-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑 → ∀𝑥 ∈ ℝ+ (abs‘((𝑅𝑥) / 𝑥)) ≤ 𝐴)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑 → ∀𝑒 ∈ (0(,)1)∃𝑥 ∈ ℝ+𝑘 ∈ ((exp‘(𝐵 / 𝑒))[,)+∞)∀𝑦 ∈ (𝑥(,)+∞)∃𝑧 ∈ ℝ+ ((𝑦 < 𝑧 ∧ ((1 + (𝐿 · 𝑒)) · 𝑧) < (𝑘 · 𝑦)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝐿 · 𝑒)) · 𝑧))(abs‘((𝑅𝑢) / 𝑢)) ≤ 𝑒))    &   (𝜑𝑈 ∈ ℝ+)    &   (𝜑𝑈𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → ∀𝑧 ∈ (𝑌[,)+∞)(abs‘((𝑅𝑧) / 𝑧)) ≤ 𝑈)       (𝜑 → ∃𝑤 ∈ ℝ+𝑣 ∈ (𝑤[,)+∞)(abs‘((𝑅𝑣) / 𝑣)) ≤ (𝑈 − (𝐹 · (𝑈↑3))))

Theorempntleml 25281* Lemma for pnt 25284. Equation 10.6.35 in [Shapiro], p. 436. (Contributed by Mario Carneiro, 14-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   (𝜑 → ∀𝑥 ∈ ℝ+ (abs‘((𝑅𝑥) / 𝑥)) ≤ 𝐴)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑 → ∀𝑒 ∈ (0(,)1)∃𝑥 ∈ ℝ+𝑘 ∈ ((exp‘(𝐵 / 𝑒))[,)+∞)∀𝑦 ∈ (𝑥(,)+∞)∃𝑧 ∈ ℝ+ ((𝑦 < 𝑧 ∧ ((1 + (𝐿 · 𝑒)) · 𝑧) < (𝑘 · 𝑦)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝐿 · 𝑒)) · 𝑧))(abs‘((𝑅𝑢) / 𝑢)) ≤ 𝑒))       (𝜑 → (𝑥 ∈ ℝ+ ↦ ((ψ‘𝑥) / 𝑥)) ⇝𝑟 1)

Theorempnt3 25282 The Prime Number Theorem, version 3: the second Chebyshev function tends asymptotically to 𝑥. (Contributed by Mario Carneiro, 1-Jun-2016.)
(𝑥 ∈ ℝ+ ↦ ((ψ‘𝑥) / 𝑥)) ⇝𝑟 1

Theorempnt2 25283 The Prime Number Theorem, version 2: the first Chebyshev function tends asymptotically to 𝑥. (Contributed by Mario Carneiro, 1-Jun-2016.)
(𝑥 ∈ ℝ+ ↦ ((θ‘𝑥) / 𝑥)) ⇝𝑟 1

Theorempnt 25284 The Prime Number Theorem: the number of prime numbers less than 𝑥 tends asymptotically to 𝑥 / log(𝑥) as 𝑥 goes to infinity. This is Metamath 100 proof #5. (Contributed by Mario Carneiro, 1-Jun-2016.)
(𝑥 ∈ (1(,)+∞) ↦ ((π𝑥) / (𝑥 / (log‘𝑥)))) ⇝𝑟 1

14.4.14  Ostrowski's theorem

Theoremabvcxp 25285* Raising an absolute value to a power less than one yields another absolute value. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝐴 = (AbsVal‘𝑅)    &   𝐵 = (Base‘𝑅)    &   𝐺 = (𝑥𝐵 ↦ ((𝐹𝑥)↑𝑐𝑆))       ((𝐹𝐴𝑆 ∈ (0(,]1)) → 𝐺𝐴)

Theorempadicfval 25286* Value of the p-adic absolute value. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝐽 = (𝑞 ∈ ℙ ↦ (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, (𝑞↑-(𝑞 pCnt 𝑥)))))       (𝑃 ∈ ℙ → (𝐽𝑃) = (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, (𝑃↑-(𝑃 pCnt 𝑥)))))

Theorempadicval 25287* Value of the p-adic absolute value. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝐽 = (𝑞 ∈ ℙ ↦ (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, (𝑞↑-(𝑞 pCnt 𝑥)))))       ((𝑃 ∈ ℙ ∧ 𝑋 ∈ ℚ) → ((𝐽𝑃)‘𝑋) = if(𝑋 = 0, 0, (𝑃↑-(𝑃 pCnt 𝑋))))

Theoremostth2lem1 25288* Lemma for ostth2 25307, although it is just a simple statement about exponentials which does not involve any specifics of ostth2 25307. If a power is upper bounded by a linear term, the exponent must be less than one. Or in big-O notation, 𝑛𝑜(𝐴𝑛) for any 1 < 𝐴. (Contributed by Mario Carneiro, 10-Sep-2014.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐵 ∈ ℝ)    &   ((𝜑𝑛 ∈ ℕ) → (𝐴𝑛) ≤ (𝑛 · 𝐵))       (𝜑𝐴 ≤ 1)

Theoremqrngbas 25289 The base set of the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂflds ℚ)       ℚ = (Base‘𝑄)

Theoremqdrng 25290 The rationals form a division ring. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂflds ℚ)       𝑄 ∈ DivRing

Theoremqrng0 25291 The zero element of the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂflds ℚ)       0 = (0g𝑄)

Theoremqrng1 25292 The unit element of the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂflds ℚ)       1 = (1r𝑄)

Theoremqrngneg 25293 The additive inverse in the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂflds ℚ)       (𝑋 ∈ ℚ → ((invg𝑄)‘𝑋) = -𝑋)

Theoremqrngdiv 25294 The division operation in the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂflds ℚ)       ((𝑋 ∈ ℚ ∧ 𝑌 ∈ ℚ ∧ 𝑌 ≠ 0) → (𝑋(/r𝑄)𝑌) = (𝑋 / 𝑌))

Theoremqabvle 25295 By using induction on 𝑁, we show a long-range inequality coming from the triangle inequality. (Contributed by Mario Carneiro, 10-Sep-2014.)
𝑄 = (ℂflds ℚ)    &   𝐴 = (AbsVal‘𝑄)       ((𝐹𝐴𝑁 ∈ ℕ0) → (𝐹𝑁) ≤ 𝑁)

Theoremqabvexp 25296 Induct the product rule abvmul 18810 to find the absolute value of a power. (Contributed by Mario Carneiro, 10-Sep-2014.)
𝑄 = (ℂflds ℚ)    &   𝐴 = (AbsVal‘𝑄)       ((𝐹𝐴𝑀 ∈ ℚ ∧ 𝑁 ∈ ℕ0) → (𝐹‘(𝑀𝑁)) = ((𝐹𝑀)↑𝑁))

Theoremostthlem1 25297* Lemma for ostth 25309. If two absolute values agree on the positive integers greater than one, then they agree for all rational numbers and thus are equal as functions. (Contributed by Mario Carneiro, 9-Sep-2014.)
𝑄 = (ℂflds ℚ)    &   𝐴 = (AbsVal‘𝑄)    &   (𝜑𝐹𝐴)    &   (𝜑𝐺𝐴)    &   ((𝜑𝑛 ∈ (ℤ‘2)) → (𝐹𝑛) = (𝐺𝑛))       (𝜑𝐹 = 𝐺)

Theoremostthlem2 25298* Lemma for ostth 25309. Refine ostthlem1 25297 so that it is sufficient to only show equality on the primes. (Contributed by Mario Carneiro, 9-Sep-2014.) (Revised by Mario Carneiro, 20-Jun-2015.)
𝑄 = (ℂflds ℚ)    &   𝐴 = (AbsVal‘𝑄)    &   (𝜑𝐹𝐴)    &   (𝜑𝐺𝐴)    &   ((𝜑𝑝 ∈ ℙ) → (𝐹𝑝) = (𝐺𝑝))       (𝜑𝐹 = 𝐺)

Theoremqabsabv 25299 The regular absolute value function on the rationals is in fact an absolute value under our definition. (Contributed by Mario Carneiro, 9-Sep-2014.)
𝑄 = (ℂflds ℚ)    &   𝐴 = (AbsVal‘𝑄)       (abs ↾ ℚ) ∈ 𝐴

Theorempadicabv 25300* The p-adic absolute value (with arbitrary base) is an absolute value. (Contributed by Mario Carneiro, 9-Sep-2014.)
𝑄 = (ℂflds ℚ)    &   𝐴 = (AbsVal‘𝑄)    &   𝐹 = (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, (𝑁↑(𝑃 pCnt 𝑥))))       ((𝑃 ∈ ℙ ∧ 𝑁 ∈ (0(,)1)) → 𝐹𝐴)

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