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Theorem List for Metamath Proof Explorer - 26601-26700   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremselberg 26601* 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 26602* Convert eventual boundedness in selberg 26601 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 26603* Lemma for selberg2 26604. 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 26604* 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 26605* Convert eventual boundedness in selberg2 26604 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 26606* Lemma for chpdifbnd 26608. (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 26607* Lemma for chpdifbnd 26608. (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 26608* 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 26609* A bound on a sum of logs, used in pntlemk 26659. This is not as precise as logdivsum 26586 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 26610* Introduce a log weighting on the summands of Σ𝑚 · 𝑛𝑥, Λ(𝑚)Λ(𝑛), the core of selberg2 26604 (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 26611* Lemma for selberg3 26612. 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 26612* Introduce a log weighting on the summands of Σ𝑚 · 𝑛𝑥, Λ(𝑚)Λ(𝑛), the core of selberg2 26604 (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 26613* Lemma for selberg4 26614. 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 26614* 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 26615* 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 26616 Functionality of the residual. Lemma for pnt 26667. (Contributed by Mario Carneiro, 8-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))       𝑅:ℝ+⟶ℝ
 
Theorempntrmax 26617* 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 26618* 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 26619* 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 26620* 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 26621* 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 26622* 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 26623* 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 26624* The sum of selberg3r 26622 and selberg4r 26623. (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 26625* Define the "Selberg function", whose asymptotic behavior is the content of selberg 26601. (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 26626* 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 26627* 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 26628* 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 26629* 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 26630* The sum of selberg3r 26622 and selberg4r 26623. (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 26631* Lemma for pntrlog2bnd 26637. 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 26632* Lemma for pntrlog2bnd 26637. 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 26633* Lemma for pntrlog2bnd 26637. 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 26634* Lemma for pntrlog2bnd 26637. 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 26635* Lemma for pntrlog2bndlem6 26636. (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 26636* Lemma for pntrlog2bnd 26637. 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 26637* 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 26638* Lemma for pntpbnd 26641. (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 26639* Lemma for pntpbnd 26641. (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 26640* Lemma for pntpbnd 26641. (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 26641* Lemma for pnt 26667. 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 26642 Lemma for pntibnd 26646. (Contributed by Mario Carneiro, 10-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   𝐿 = ((1 / 4) / (𝐴 + 3))       (𝜑𝐿 ∈ (0(,)1))
 
Theorempntibndlem2a 26643* Lemma for pntibndlem2 26644. (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 26644* Lemma for pntibnd 26646. 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 26645* Lemma for pntibnd 26646. Package up pntibndlem2 26644 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 26646* Lemma for pnt 26667. 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 26647 Lemma for pnt 26667. 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 26648* Lemma for pnt 26667. 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 26649* Lemma for pnt 26667. 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 26650* Lemma for pnt 26667. 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 26651* Lemma for pnt 26667. 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 26652* Lemma for pnt 26667. 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 26653* Lemma for pnt 26667. 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 26654* Lemma for pntlemj 26656. (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 26655* Lemma for pntlemj 26656. (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 26656* Lemma for pnt 26667. The induction step. Using pntibnd 26646, 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 26657* Lemma for pnt 26667. Eliminate some assumptions from pntlemj 26656. (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 26658* Lemma for pnt 26667. Add up the pieces in pntlemi 26657 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 26659* Lemma for pnt 26667. 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 26660* Lemma for pnt 26667. 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 26661* Lemma for pnt 26667. Package up pntlemo 26660 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 26662* Lemma for pnt 26667. 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 26663* Lemma for pnt 26667. 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 26664* Lemma for pnt 26667. 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 26665 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 26666 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 26667 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 26668* 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 26669* 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 26670* 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 26671* Lemma for ostth2 26690, although it is just a simple statement about exponentials which does not involve any specifics of ostth2 26690. 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 26672 The base set of the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂflds ℚ)       ℚ = (Base‘𝑄)
 
Theoremqdrng 26673 The rationals form a division ring. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂflds ℚ)       𝑄 ∈ DivRing
 
Theoremqrng0 26674 The zero element of the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂflds ℚ)       0 = (0g𝑄)
 
Theoremqrng1 26675 The unit element of the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂflds ℚ)       1 = (1r𝑄)
 
Theoremqrngneg 26676 The additive inverse in the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂflds ℚ)       (𝑋 ∈ ℚ → ((invg𝑄)‘𝑋) = -𝑋)
 
Theoremqrngdiv 26677 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 26678 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 26679 Induct the product rule abvmul 20004 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 26680* Lemma for ostth 26692. 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 26681* Lemma for ostth 26692. Refine ostthlem1 26680 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 26682 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 26683* 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)
 
Theorempadicabvf 26684* The p-adic absolute value is an absolute value. (Contributed by Mario Carneiro, 9-Sep-2014.)
𝑄 = (ℂflds ℚ)    &   𝐴 = (AbsVal‘𝑄)    &   𝐽 = (𝑞 ∈ ℙ ↦ (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, (𝑞↑-(𝑞 pCnt 𝑥)))))       𝐽:ℙ⟶𝐴
 
Theorempadicabvcxp 26685* All positive powers of the p-adic absolute value are absolute values. (Contributed by Mario Carneiro, 9-Sep-2014.)
𝑄 = (ℂflds ℚ)    &   𝐴 = (AbsVal‘𝑄)    &   𝐽 = (𝑞 ∈ ℙ ↦ (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, (𝑞↑-(𝑞 pCnt 𝑥)))))       ((𝑃 ∈ ℙ ∧ 𝑅 ∈ ℝ+) → (𝑦 ∈ ℚ ↦ (((𝐽𝑃)‘𝑦)↑𝑐𝑅)) ∈ 𝐴)
 
Theoremostth1 26686* - Lemma for ostth 26692: trivial case. (Not that the proof is trivial, but that we are proving that the function is trivial.) If 𝐹 is equal to 1 on the primes, then by complete induction and the multiplicative property abvmul 20004 of the absolute value, 𝐹 is equal to 1 on all the integers, and ostthlem1 26680 extends this to the other rational numbers. (Contributed by Mario Carneiro, 10-Sep-2014.)
𝑄 = (ℂflds ℚ)    &   𝐴 = (AbsVal‘𝑄)    &   𝐽 = (𝑞 ∈ ℙ ↦ (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, (𝑞↑-(𝑞 pCnt 𝑥)))))    &   𝐾 = (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, 1))    &   (𝜑𝐹𝐴)    &   (𝜑 → ∀𝑛 ∈ ℕ ¬ 1 < (𝐹𝑛))    &   (𝜑 → ∀𝑛 ∈ ℙ ¬ (𝐹𝑛) < 1)       (𝜑𝐹 = 𝐾)
 
Theoremostth2lem2 26687* Lemma for ostth2 26690. (Contributed by Mario Carneiro, 10-Sep-2014.)
𝑄 = (ℂflds ℚ)    &   𝐴 = (AbsVal‘𝑄)    &   𝐽 = (𝑞 ∈ ℙ ↦ (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, (𝑞↑-(𝑞 pCnt 𝑥)))))    &   𝐾 = (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, 1))    &   (𝜑𝐹𝐴)    &   (𝜑𝑁 ∈ (ℤ‘2))    &   (𝜑 → 1 < (𝐹𝑁))    &   𝑅 = ((log‘(𝐹𝑁)) / (log‘𝑁))    &   (𝜑𝑀 ∈ (ℤ‘2))    &   𝑆 = ((log‘(𝐹𝑀)) / (log‘𝑀))    &   𝑇 = if((𝐹𝑀) ≤ 1, 1, (𝐹𝑀))       ((𝜑𝑋 ∈ ℕ0𝑌 ∈ (0...((𝑀𝑋) − 1))) → (𝐹𝑌) ≤ ((𝑀 · 𝑋) · (𝑇𝑋)))
 
Theoremostth2lem3 26688* Lemma for ostth2 26690. (Contributed by Mario Carneiro, 10-Sep-2014.)
𝑄 = (ℂflds ℚ)    &   𝐴 = (AbsVal‘𝑄)    &   𝐽 = (𝑞 ∈ ℙ ↦ (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, (𝑞↑-(𝑞 pCnt 𝑥)))))    &   𝐾 = (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, 1))    &   (𝜑𝐹𝐴)    &   (𝜑𝑁 ∈ (ℤ‘2))    &   (𝜑 → 1 < (𝐹𝑁))    &   𝑅 = ((log‘(𝐹𝑁)) / (log‘𝑁))    &   (𝜑𝑀 ∈ (ℤ‘2))    &   𝑆 = ((log‘(𝐹𝑀)) / (log‘𝑀))    &   𝑇 = if((𝐹𝑀) ≤ 1, 1, (𝐹𝑀))    &   𝑈 = ((log‘𝑁) / (log‘𝑀))       ((𝜑𝑋 ∈ ℕ) → (((𝐹𝑁) / (𝑇𝑐𝑈))↑𝑋) ≤ (𝑋 · ((𝑀 · 𝑇) · (𝑈 + 1))))
 
Theoremostth2lem4 26689* Lemma for ostth2 26690. (Contributed by Mario Carneiro, 10-Sep-2014.)
𝑄 = (ℂflds ℚ)    &   𝐴 = (AbsVal‘𝑄)    &   𝐽 = (𝑞 ∈ ℙ ↦ (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, (𝑞↑-(𝑞 pCnt 𝑥)))))    &   𝐾 = (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, 1))    &   (𝜑𝐹𝐴)    &   (𝜑𝑁 ∈ (ℤ‘2))    &   (𝜑 → 1 < (𝐹𝑁))    &   𝑅 = ((log‘(𝐹𝑁)) / (log‘𝑁))    &   (𝜑𝑀 ∈ (ℤ‘2))    &   𝑆 = ((log‘(𝐹𝑀)) / (log‘𝑀))    &   𝑇 = if((𝐹𝑀) ≤ 1, 1, (𝐹𝑀))    &   𝑈 = ((log‘𝑁) / (log‘𝑀))       (𝜑 → (1 < (𝐹𝑀) ∧ 𝑅𝑆))
 
Theoremostth2 26690* - Lemma for ostth 26692: regular case. (Contributed by Mario Carneiro, 10-Sep-2014.)
𝑄 = (ℂflds ℚ)    &   𝐴 = (AbsVal‘𝑄)    &   𝐽 = (𝑞 ∈ ℙ ↦ (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, (𝑞↑-(𝑞 pCnt 𝑥)))))    &   𝐾 = (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, 1))    &   (𝜑𝐹𝐴)    &   (𝜑𝑁 ∈ (ℤ‘2))    &   (𝜑 → 1 < (𝐹𝑁))    &   𝑅 = ((log‘(𝐹𝑁)) / (log‘𝑁))       (𝜑 → ∃𝑎 ∈ (0(,]1)𝐹 = (𝑦 ∈ ℚ ↦ ((abs‘𝑦)↑𝑐𝑎)))
 
Theoremostth3 26691* - Lemma for ostth 26692: p-adic case. (Contributed by Mario Carneiro, 10-Sep-2014.)
𝑄 = (ℂflds ℚ)    &   𝐴 = (AbsVal‘𝑄)    &   𝐽 = (𝑞 ∈ ℙ ↦ (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, (𝑞↑-(𝑞 pCnt 𝑥)))))    &   𝐾 = (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, 1))    &   (𝜑𝐹𝐴)    &   (𝜑 → ∀𝑛 ∈ ℕ ¬ 1 < (𝐹𝑛))    &   (𝜑𝑃 ∈ ℙ)    &   (𝜑 → (𝐹𝑃) < 1)    &   𝑅 = -((log‘(𝐹𝑃)) / (log‘𝑃))    &   𝑆 = if((𝐹𝑃) ≤ (𝐹𝑝), (𝐹𝑝), (𝐹𝑃))       (𝜑 → ∃𝑎 ∈ ℝ+ 𝐹 = (𝑦 ∈ ℚ ↦ (((𝐽𝑃)‘𝑦)↑𝑐𝑎)))
 
Theoremostth 26692* Ostrowski's theorem, which classifies all absolute values on . Any such absolute value must either be the trivial absolute value 𝐾, a constant exponent 0 < 𝑎 ≤ 1 times the regular absolute value, or a positive exponent times the p-adic absolute value. (Contributed by Mario Carneiro, 10-Sep-2014.)
𝑄 = (ℂflds ℚ)    &   𝐴 = (AbsVal‘𝑄)    &   𝐽 = (𝑞 ∈ ℙ ↦ (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, (𝑞↑-(𝑞 pCnt 𝑥)))))    &   𝐾 = (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, 1))       (𝐹𝐴 ↔ (𝐹 = 𝐾 ∨ ∃𝑎 ∈ (0(,]1)𝐹 = (𝑦 ∈ ℚ ↦ ((abs‘𝑦)↑𝑐𝑎)) ∨ ∃𝑎 ∈ ℝ+𝑔 ∈ ran 𝐽 𝐹 = (𝑦 ∈ ℚ ↦ ((𝑔𝑦)↑𝑐𝑎))))
 
PART 15  ELEMENTARY GEOMETRY

This part develops elementary geometry based on Tarski's axioms, following [Schwabhauser]. Tarski's geometry is a first-order theory with one sort, the "points". It has two primitive notions, the ternary predicate of "betweenness" and the quaternary predicate of "congruence". To adapt this theory to the framework of set.mm, and to be able to talk of *a* Tarski structure as a space satisfying the given axioms, we use the following definition, stated informally:

A Tarski structure 𝑓 is a set (of points) (Base‘𝑓) together with functions (Itv‘𝑓) and (dist‘𝑓) on ((Base‘𝑓) × (Base‘𝑓)) satisfying certain axioms (given in Definitions df-trkg 26718 et sequentes). This allows us to treat a Tarski structure as a special kind of extensible structure (see df-struct 16776).

The translation to and from Tarski's treatment is as follows (given, again, informally).

Suppose that one is given an extensible structure 𝑓. One defines a betweenness ternary predicate Btw by positing that, for any 𝑥, 𝑦, 𝑧 ∈ (Base‘𝑓), one has "Btw 𝑥𝑦𝑧 " if and only if 𝑦𝑥(Itv‘𝑓)𝑧, and a congruence quaternary predicate Congr by positing that, for any 𝑥, 𝑦, 𝑧, 𝑡 ∈ (Base‘𝑓), one has "Congr 𝑥𝑦𝑧𝑡 " if and only if 𝑥(dist‘𝑓)𝑦 = 𝑧(dist‘𝑓)𝑡. It is easy to check that if 𝑓 satisfies our Tarski axioms, then Btw and Congr satisfy Tarski's Tarski axioms when (Base‘𝑓) is interpreted as the universe of discourse.

Conversely, suppose that one is given a set 𝑎, a ternary predicate Btw, and a quaternary predicate Congr. One defines the extensible structure 𝑓 such that (Base‘𝑓) is 𝑎, and (Itv‘𝑓) is the function which associates with each 𝑥, 𝑦⟩ ∈ (𝑎 × 𝑎) the set of points 𝑧𝑎 such that "Btw 𝑥𝑧𝑦", and (dist‘𝑓) is the function which associates with each 𝑥, 𝑦⟩ ∈ (𝑎 × 𝑎) the set of ordered pairs 𝑧, 𝑡⟩ ∈ (𝑎 × 𝑎) such that "Congr 𝑥𝑦𝑧𝑡". It is easy to check that if Btw and Congr satisfy Tarski's Tarski axioms when 𝑎 is interpreted as the universe of discourse, then 𝑓 satisfies our Tarski axioms.

We intentionally choose to represent congruence (without loss of generality) as 𝑥(dist‘𝑓)𝑦 = 𝑧(dist‘𝑓)𝑡 instead of "Congr 𝑥𝑦𝑧𝑡", as it is more convenient. It is always possible to define dist for any particular geometry to produce equal results when conguence is desired, and in many cases there is an obvious interpretation of "distance" between two points that can be useful in other situations. Encoding congruence as an equality of distances makes it easier to use these theorems in cases where there is a preferred distance function. We prove that representing a congruence relationship using a distance in the form 𝑥(dist‘𝑓)𝑦 = 𝑧(dist‘𝑓)𝑡 causes no loss of generality in tgjustc1 26740 and tgjustc2 26741, which in turn are supported by tgjustf 26738 and tgjustr 26739.

A similar representation of congruence (using a "distance" function) is used in Axiom A1 of [Beeson2016] p. 5, which discusses how a large number of formalized proofs were found in Tarskian Geometry using OTTER. Their detailed proofs in Tarski Geometry, along with other information, are available at https://www.michaelbeeson.com/research/FormalTarski/ 26739.

Most theorems are in deduction form, as this is a very general, simple, and convenient format to use in Metamath. An assertion in deduction form can be easily converted into an assertion in inference form (removing the antecedents 𝜑) by insert a ⊤ → in each hypothesis, using a1i 11, then using mptru 1546 to remove the final ⊤ → prefix. In some cases we represent, without loss of generality, an implication antecedent in [Schwabhauser] as a hypothesis. The implication can be retrieved from the by using simpr 484, the theorem as stated, and ex 412.

For descriptions of individual axioms, we refer to the specific definitions below. A particular feature of Tarski's axioms is modularity, so by using various subsets of the set of axioms, we can define the classes of "absolute dimensionless Tarski structures" (df-trkg 26718), of "Euclidean dimensionless Tarski structures" (df-trkge 26716) and of "Tarski structures of dimension no less than N" (df-trkgld 26717).

In this system, angles are not a primitive notion, but instead a derived notion (see df-cgra 27073 and iscgra 27074). To maintain its simplicity, in this system congruence between shapes (a finite sequence of points) is the case where corresponding segments between all corresponding points are congruent. This includes triangles (a shape of 3 distinct points). Note that this definition has no direct regard for angles. For more details and rationale, see df-cgrg 26776.

The first section is devoted to the definitions of these various structures. The second section ("Tarskian geometry") develops the synthetic treatment of geometry. The remaining sections prove that real Euclidean spaces and complex Hilbert spaces, with intended interpretations, are Euclidean Tarski structures.

Most of the work in this part is due to Thierry Arnoux, with earlier work by Mario Carneiro and Scott Fenton. See also the credits in the comment of each statement.

 
15.1  Definition and Tarski's Axioms of Geometry
 
Syntaxcstrkg 26693 Extends class notation with the class of Tarski geometries.
class TarskiG
 
Syntaxcstrkgc 26694 Extends class notation with the class of geometries fulfilling the congruence axioms.
class TarskiGC
 
Syntaxcstrkgb 26695 Extends class notation with the class of geometries fulfilling the betweenness axioms.
class TarskiGB
 
Syntaxcstrkgcb 26696 Extends class notation with the class of geometries fulfilling the congruence and betweenness axioms.
class TarskiGCB
 
Syntaxcstrkgld 26697 Extends class notation with the relation for geometries fulfilling the lower dimension axioms.
class DimTarskiG
 
Syntaxcstrkge 26698 Extends class notation with the class of geometries fulfilling Euclid's axiom.
class TarskiGE
 
Syntaxcitv 26699 Declare the syntax for the Interval (segment) index extractor.
class Itv
 
Syntaxclng 26700 Declare the syntax for the Line function.
class LineG
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