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Theorem List for Metamath Proof Explorer - 27601-27700   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theorempntrlog2bndlem4 27601* Lemma for pntrlog2bnd 27605. 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 27602* Lemma for pntrlog2bnd 27605. 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 27603* Lemma for pntrlog2bndlem6 27604. (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 27604* Lemma for pntrlog2bnd 27605. 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 27605* 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 27606* Lemma for pntpbnd 27609. (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 27607* Lemma for pntpbnd 27609. (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 27608* Lemma for pntpbnd 27609. (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 27609* Lemma for pnt 27635. 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 27610 Lemma for pntibnd 27614. (Contributed by Mario Carneiro, 10-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑𝐴 ∈ ℝ+)    &   𝐿 = ((1 / 4) / (𝐴 + 3))       (𝜑𝐿 ∈ (0(,)1))
 
Theorempntibndlem2a 27611* Lemma for pntibndlem2 27612. (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 27612* Lemma for pntibnd 27614. 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 27613* Lemma for pntibnd 27614. Package up pntibndlem2 27612 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 27614* Lemma for pnt 27635. 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 27615 Lemma for pnt 27635. 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 27616* Lemma for pnt 27635. 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 27617* Lemma for pnt 27635. 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 27618* Lemma for pnt 27635. 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 27619* Lemma for pnt 27635. 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 27620* Lemma for pnt 27635. 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 27621* Lemma for pnt 27635. 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 27622* Lemma for pntlemj 27624. (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 27623* Lemma for pntlemj 27624. (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 27624* Lemma for pnt 27635. The induction step. Using pntibnd 27614, 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 27625* Lemma for pnt 27635. Eliminate some assumptions from pntlemj 27624. (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 27626* Lemma for pnt 27635. Add up the pieces in pntlemi 27625 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 27627* Lemma for pnt 27635. 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 27628* Lemma for pnt 27635. 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 27629* Lemma for pnt 27635. Package up pntlemo 27628 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 27630* Lemma for pnt 27635. 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 27631* Lemma for pnt 27635. 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 27632* Lemma for pnt 27635. 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 27633 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 27634 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 27635 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 27636* 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 27637* 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 27638* 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 27639* Lemma for ostth2 27658, although it is just a simple statement about exponentials which does not involve any specifics of ostth2 27658. 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 27640 The base set of the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂflds ℚ)       ℚ = (Base‘𝑄)
 
Theoremqdrng 27641 The rationals form a division ring. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂflds ℚ)       𝑄 ∈ DivRing
 
Theoremqrng0 27642 The zero element of the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂflds ℚ)       0 = (0g𝑄)
 
Theoremqrng1 27643 The unity element of the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂflds ℚ)       1 = (1r𝑄)
 
Theoremqrngneg 27644 The additive inverse in the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂflds ℚ)       (𝑋 ∈ ℚ → ((invg𝑄)‘𝑋) = -𝑋)
 
Theoremqrngdiv 27645 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 27646 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 27647 Induct the product rule abvmul 20749 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 27648* Lemma for ostth 27660. 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 27649* Lemma for ostth 27660. Refine ostthlem1 27648 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 27650 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 27651* 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 27652* 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 27653* 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 27654* - Lemma for ostth 27660: 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 20749 of the absolute value, 𝐹 is equal to 1 on all the integers, and ostthlem1 27648 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 27655* Lemma for ostth2 27658. (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 27656* Lemma for ostth2 27658. (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 27657* Lemma for ostth2 27658. (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 27658* - Lemma for ostth 27660: 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 27659* - Lemma for ostth 27660: 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 27660* 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  SURREAL NUMBERS
 
15.1  Sign sequence representation and Alling's axioms

The surreal numbers can be represented in several equivalent ways. In [Alling], Norman Alling made this notion explicit by giving a set of axioms that all representations admit, then proving that there is an order and birthday preserving bijection between any systems that satisfy these axioms.

In this section, we start with the definition of surreal numbers given in [Gonshor] and derive Alling's axioms. After deriving them we no longer refer to the explicit definition of surreals. In particular, we never take advantage of the fact that the empty set is a surreal number under our definition.

 
15.1.1  Definitions and initial properties
 
Syntaxcsur 27661 Declare the class of all surreal numbers (see df-no 27664).
class No
 
Syntaxcslt 27662 Declare the less-than relation over surreal numbers (see df-slt 27665).
class <s
 
Syntaxcbday 27663 Declare the birthday function for surreal numbers (see df-bday 27666).
class bday
 
Definitiondf-no 27664* Define the class of surreal numbers. The surreal numbers are a proper class of numbers developed by John H. Conway and introduced by Donald Knuth in 1975. They form a proper class into which all ordered fields can be embedded. The approach we take to defining them was first introduced by Hary Gonshor, and is based on the conception of a "sign expansion" of a surreal number. We define the surreals as ordinal-indexed sequences of 1o and 2o, analogous to Gonshor's ( − ) and ( + ).

After introducing this definition, we will abstract away from it using axioms that Norman Alling developed in "Foundations of Analysis over Surreal Number Fields." This is done in an effort to be agnostic towards the exact implementation of surreals. (Contributed by Scott Fenton, 9-Jun-2011.)

No = {𝑓 ∣ ∃𝑎 ∈ On 𝑓:𝑎⟶{1o, 2o}}
 
Definitiondf-slt 27665* Next, we introduce surreal less-than, a comparison relation over the surreals by lexicographically ordering them. (Contributed by Scott Fenton, 9-Jun-2011.)
<s = {⟨𝑓, 𝑔⟩ ∣ ((𝑓 No 𝑔 No ) ∧ ∃𝑥 ∈ On (∀𝑦𝑥 (𝑓𝑦) = (𝑔𝑦) ∧ (𝑓𝑥){⟨1o, ∅⟩, ⟨1o, 2o⟩, ⟨∅, 2o⟩} (𝑔𝑥)))}
 
Definitiondf-bday 27666 Finally, we introduce the birthday function. This function maps each surreal to an ordinal. In our implementation, this is the domain of the sign function. The important properties of this function are established later. (Contributed by Scott Fenton, 11-Jun-2011.)
bday = (𝑥 No ↦ dom 𝑥)
 
Theoremelno 27667* Membership in the surreals. (Contributed by Scott Fenton, 11-Jun-2011.) (Proof shortened by SF, 14-Apr-2012.) Avoid ax-rep 5289. (Revised by SN, 5-Jun-2025.)
(𝐴 No ↔ ∃𝑥 ∈ On 𝐴:𝑥⟶{1o, 2o})
 
TheoremelnoOLD 27668* Obsolete version of elno 27667 as of 5-Jun-2025. (Contributed by Scott Fenton, 11-Jun-2011.) (Proof modification is discouraged.) (New usage is discouraged.)
(𝐴 No ↔ ∃𝑥 ∈ On 𝐴:𝑥⟶{1o, 2o})
 
Theoremsltval 27669* The value of the surreal less-than relation. (Contributed by Scott Fenton, 14-Jun-2011.)
((𝐴 No 𝐵 No ) → (𝐴 <s 𝐵 ↔ ∃𝑥 ∈ On (∀𝑦𝑥 (𝐴𝑦) = (𝐵𝑦) ∧ (𝐴𝑥){⟨1o, ∅⟩, ⟨1o, 2o⟩, ⟨∅, 2o⟩} (𝐵𝑥))))
 
Theorembdayval 27670 The value of the birthday function within the surreals. (Contributed by Scott Fenton, 14-Jun-2011.)
(𝐴 No → ( bday 𝐴) = dom 𝐴)
 
Theoremnofun 27671 A surreal is a function. (Contributed by Scott Fenton, 16-Jun-2011.)
(𝐴 No → Fun 𝐴)
 
Theoremnodmon 27672 The domain of a surreal is an ordinal. (Contributed by Scott Fenton, 16-Jun-2011.)
(𝐴 No → dom 𝐴 ∈ On)
 
Theoremnorn 27673 The range of a surreal is a subset of the surreal signs. (Contributed by Scott Fenton, 16-Jun-2011.)
(𝐴 No → ran 𝐴 ⊆ {1o, 2o})
 
Theoremnofnbday 27674 A surreal is a function over its birthday. (Contributed by Scott Fenton, 16-Jun-2011.)
(𝐴 No 𝐴 Fn ( bday 𝐴))
 
Theoremnodmord 27675 The domain of a surreal has the ordinal property. (Contributed by Scott Fenton, 16-Jun-2011.)
(𝐴 No → Ord dom 𝐴)
 
Theoremelno2 27676 An alternative condition for membership in No . (Contributed by Scott Fenton, 21-Mar-2012.)
(𝐴 No ↔ (Fun 𝐴 ∧ dom 𝐴 ∈ On ∧ ran 𝐴 ⊆ {1o, 2o}))
 
Theoremelno3 27677 Another condition for membership in No . (Contributed by Scott Fenton, 14-Apr-2012.)
(𝐴 No ↔ (𝐴:dom 𝐴⟶{1o, 2o} ∧ dom 𝐴 ∈ On))
 
Theoremsltval2 27678* Alternate expression for surreal less-than. Two surreals obey surreal less-than iff they obey the sign ordering at the first place they differ. (Contributed by Scott Fenton, 17-Jun-2011.)
((𝐴 No 𝐵 No ) → (𝐴 <s 𝐵 ↔ (𝐴 {𝑎 ∈ On ∣ (𝐴𝑎) ≠ (𝐵𝑎)}){⟨1o, ∅⟩, ⟨1o, 2o⟩, ⟨∅, 2o⟩} (𝐵 {𝑎 ∈ On ∣ (𝐴𝑎) ≠ (𝐵𝑎)})))
 
Theoremnofv 27679 The function value of a surreal is either a sign or the empty set. (Contributed by Scott Fenton, 22-Jun-2011.)
(𝐴 No → ((𝐴𝑋) = ∅ ∨ (𝐴𝑋) = 1o ∨ (𝐴𝑋) = 2o))
 
Theoremnosgnn0 27680 is not a surreal sign. (Contributed by Scott Fenton, 16-Jun-2011.)
¬ ∅ ∈ {1o, 2o}
 
Theoremnosgnn0i 27681 If 𝑋 is a surreal sign, then it is not null. (Contributed by Scott Fenton, 3-Aug-2011.)
𝑋 ∈ {1o, 2o}       ∅ ≠ 𝑋
 
Theoremnoreson 27682 The restriction of a surreal to an ordinal is still a surreal. (Contributed by Scott Fenton, 4-Sep-2011.)
((𝐴 No 𝐵 ∈ On) → (𝐴𝐵) ∈ No )
 
Theoremsltintdifex 27683* If 𝐴 <s 𝐵, then the intersection of all the ordinals that have differing signs in 𝐴 and 𝐵 exists. (Contributed by Scott Fenton, 22-Feb-2012.)
((𝐴 No 𝐵 No ) → (𝐴 <s 𝐵 {𝑎 ∈ On ∣ (𝐴𝑎) ≠ (𝐵𝑎)} ∈ V))
 
Theoremsltres 27684 If the restrictions of two surreals to a given ordinal obey surreal less-than, then so do the two surreals themselves. (Contributed by Scott Fenton, 4-Sep-2011.)
((𝐴 No 𝐵 No 𝑋 ∈ On) → ((𝐴𝑋) <s (𝐵𝑋) → 𝐴 <s 𝐵))
 
Theoremnoxp1o 27685 The Cartesian product of an ordinal and {1o} is a surreal. (Contributed by Scott Fenton, 12-Jun-2011.)
(𝐴 ∈ On → (𝐴 × {1o}) ∈ No )
 
Theoremnoseponlem 27686* Lemma for nosepon 27687. Consider a case of proper subset domain. (Contributed by Scott Fenton, 21-Sep-2020.)
((𝐴 No 𝐵 No ∧ dom 𝐴 ∈ dom 𝐵) → ¬ ∀𝑥 ∈ On (𝐴𝑥) = (𝐵𝑥))
 
Theoremnosepon 27687* Given two unequal surreals, the minimal ordinal at which they differ is an ordinal. (Contributed by Scott Fenton, 21-Sep-2020.)
((𝐴 No 𝐵 No 𝐴𝐵) → {𝑥 ∈ On ∣ (𝐴𝑥) ≠ (𝐵𝑥)} ∈ On)
 
Theoremnoextend 27688 Extending a surreal by one sign value results in a new surreal. (Contributed by Scott Fenton, 22-Nov-2021.)
𝑋 ∈ {1o, 2o}       (𝐴 No → (𝐴 ∪ {⟨dom 𝐴, 𝑋⟩}) ∈ No )
 
Theoremnoextendseq 27689 Extend a surreal by a sequence of ordinals. (Contributed by Scott Fenton, 30-Nov-2021.)
𝑋 ∈ {1o, 2o}       ((𝐴 No 𝐵 ∈ On) → (𝐴 ∪ ((𝐵 ∖ dom 𝐴) × {𝑋})) ∈ No )
 
Theoremnoextenddif 27690* Calculate the place where a surreal and its extension differ. (Contributed by Scott Fenton, 22-Nov-2021.)
𝑋 ∈ {1o, 2o}       (𝐴 No {𝑥 ∈ On ∣ (𝐴𝑥) ≠ ((𝐴 ∪ {⟨dom 𝐴, 𝑋⟩})‘𝑥)} = dom 𝐴)
 
Theoremnoextendlt 27691 Extending a surreal with a negative sign results in a smaller surreal. (Contributed by Scott Fenton, 22-Nov-2021.)
(𝐴 No → (𝐴 ∪ {⟨dom 𝐴, 1o⟩}) <s 𝐴)
 
Theoremnoextendgt 27692 Extending a surreal with a positive sign results in a bigger surreal. (Contributed by Scott Fenton, 22-Nov-2021.)
(𝐴 No 𝐴 <s (𝐴 ∪ {⟨dom 𝐴, 2o⟩}))
 
Theoremnolesgn2o 27693 Given 𝐴 less-than or equal to 𝐵, equal to 𝐵 up to 𝑋, and 𝐴(𝑋) = 2o, then 𝐵(𝑋) = 2o. (Contributed by Scott Fenton, 6-Dec-2021.)
(((𝐴 No 𝐵 No 𝑋 ∈ On) ∧ ((𝐴𝑋) = (𝐵𝑋) ∧ (𝐴𝑋) = 2o) ∧ ¬ 𝐵 <s 𝐴) → (𝐵𝑋) = 2o)
 
Theoremnolesgn2ores 27694 Given 𝐴 less-than or equal to 𝐵, equal to 𝐵 up to 𝑋, and 𝐴(𝑋) = 2o, then (𝐴 ↾ suc 𝑋) = (𝐵 ↾ suc 𝑋). (Contributed by Scott Fenton, 6-Dec-2021.)
(((𝐴 No 𝐵 No 𝑋 ∈ On) ∧ ((𝐴𝑋) = (𝐵𝑋) ∧ (𝐴𝑋) = 2o) ∧ ¬ 𝐵 <s 𝐴) → (𝐴 ↾ suc 𝑋) = (𝐵 ↾ suc 𝑋))
 
Theoremnogesgn1o 27695 Given 𝐴 greater than or equal to 𝐵, equal to 𝐵 up to 𝑋, and 𝐴(𝑋) = 1o, then 𝐵(𝑋) = 1o. (Contributed by Scott Fenton, 9-Aug-2024.)
(((𝐴 No 𝐵 No 𝑋 ∈ On) ∧ ((𝐴𝑋) = (𝐵𝑋) ∧ (𝐴𝑋) = 1o) ∧ ¬ 𝐴 <s 𝐵) → (𝐵𝑋) = 1o)
 
Theoremnogesgn1ores 27696 Given 𝐴 greater than or equal to 𝐵, equal to 𝐵 up to 𝑋, and 𝐴(𝑋) = 1o, then (𝐴 ↾ suc 𝑋) = (𝐵 ↾ suc 𝑋). (Contributed by Scott Fenton, 6-Dec-2021.)
(((𝐴 No 𝐵 No 𝑋 ∈ On) ∧ ((𝐴𝑋) = (𝐵𝑋) ∧ (𝐴𝑋) = 1o) ∧ ¬ 𝐴 <s 𝐵) → (𝐴 ↾ suc 𝑋) = (𝐵 ↾ suc 𝑋))
 
15.1.2  Ordering
 
Theoremsltsolem1 27697 Lemma for sltso 27698. The "sign expansion" binary relation totally orders the surreal signs. (Contributed by Scott Fenton, 8-Jun-2011.)
{⟨1o, ∅⟩, ⟨1o, 2o⟩, ⟨∅, 2o⟩} Or ({1o, 2o} ∪ {∅})
 
Theoremsltso 27698 Less-than totally orders the surreals. Axiom O of [Alling] p. 184. (Contributed by Scott Fenton, 9-Jun-2011.)
<s Or No
 
15.1.3  Birthday Function
 
Theorembdayfo 27699 The birthday function maps the surreals onto the ordinals. Axiom B of [Alling] p. 184. (Proof shortened on 14-Apr-2012 by SF). (Contributed by Scott Fenton, 11-Jun-2011.)
bday : No onto→On
 
15.1.4  Density
 
Theoremfvnobday 27700 The value of a surreal at its birthday is . (Contributed by Scott Fenton, 14-Jun-2011.) (Proof shortened by SF, 14-Apr-2012.)
(𝐴 No → (𝐴‘( bday 𝐴)) = ∅)
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144 14301-14400 145 14401-14500 146 14501-14600 147 14601-14700 148 14701-14800 149 14801-14900 150 14901-15000 151 15001-15100 152 15101-15200 153 15201-15300 154 15301-15400 155 15401-15500 156 15501-15600 157 15601-15700 158 15701-15800 159 15801-15900 160 15901-16000 161 16001-16100 162 16101-16200 163 16201-16300 164 16301-16400 165 16401-16500 166 16501-16600 167 16601-16700 168 16701-16800 169 16801-16900 170 16901-17000 171 17001-17100 172 17101-17200 173 17201-17300 174 17301-17400 175 17401-17500 176 17501-17600 177 17601-17700 178 17701-17800 179 17801-17900 180 17901-18000 181 18001-18100 182 18101-18200 183 18201-18300 184 18301-18400 185 18401-18500 186 18501-18600 187 18601-18700 188 18701-18800 189 18801-18900 190 18901-19000 191 19001-19100 192 19101-19200 193 19201-19300 194 19301-19400 195 19401-19500 196 19501-19600 197 19601-19700 198 19701-19800 199 19801-19900 200 19901-20000 201 20001-20100 202 20101-20200 203 20201-20300 204 20301-20400 205 20401-20500 206 20501-20600 207 20601-20700 208 20701-20800 209 20801-20900 210 20901-21000 211 21001-21100 212 21101-21200 213 21201-21300 214 21301-21400 215 21401-21500 216 21501-21600 217 21601-21700 218 21701-21800 219 21801-21900 220 21901-22000 221 22001-22100 222 22101-22200 223 22201-22300 224 22301-22400 225 22401-22500 226 22501-22600 227 22601-22700 228 22701-22800 229 22801-22900 230 22901-23000 231 23001-23100 232 23101-23200 233 23201-23300 234 23301-23400 235 23401-23500 236 23501-23600 237 23601-23700 238 23701-23800 239 23801-23900 240 23901-24000 241 24001-24100 242 24101-24200 243 24201-24300 244 24301-24400 245 24401-24500 246 24501-24600 247 24601-24700 248 24701-24800 249 24801-24900 250 24901-25000 251 25001-25100 252 25101-25200 253 25201-25300 254 25301-25400 255 25401-25500 256 25501-25600 257 25601-25700 258 25701-25800 259 25801-25900 260 25901-26000 261 26001-26100 262 26101-26200 263 26201-26300 264 26301-26400 265 26401-26500 266 26501-26600 267 26601-26700 268 26701-26800 269 26801-26900 270 26901-27000 271 27001-27100 272 27101-27200 273 27201-27300 274 27301-27400 275 27401-27500 276 27501-27600 277 27601-27700 278 27701-27800 279 27801-27900 280 27901-28000 281 28001-28100 282 28101-28200 283 28201-28300 284 28301-28400 285 28401-28500 286 28501-28600 287 28601-28700 288 28701-28800 289 28801-28900 290 28901-29000 291 29001-29100 292 29101-29200 293 29201-29300 294 29301-29400 295 29401-29500 296 29501-29600 297 29601-29700 298 29701-29800 299 29801-29900 300 29901-30000 301 30001-30100 302 30101-30200 303 30201-30300 304 30301-30400 305 30401-30500 306 30501-30600 307 30601-30700 308 30701-30800 309 30801-30900 310 30901-31000 311 31001-31100 312 31101-31200 313 31201-31300 314 31301-31400 315 31401-31500 316 31501-31600 317 31601-31700 318 31701-31800 319 31801-31900 320 31901-32000 321 32001-32100 322 32101-32200 323 32201-32300 324 32301-32400 325 32401-32500 326 32501-32600 327 32601-32700 328 32701-32800 329 32801-32900 330 32901-33000 331 33001-33100 332 33101-33200 333 33201-33300 334 33301-33400 335 33401-33500 336 33501-33600 337 33601-33700 338 33701-33800 339 33801-33900 340 33901-34000 341 34001-34100 342 34101-34200 343 34201-34300 344 34301-34400 345 34401-34500 346 34501-34600 347 34601-34700 348 34701-34800 349 34801-34900 350 34901-35000 351 35001-35100 352 35101-35200 353 35201-35300 354 35301-35400 355 35401-35500 356 35501-35600 357 35601-35700 358 35701-35800 359 35801-35900 360 35901-36000 361 36001-36100 362 36101-36200 363 36201-36300 364 36301-36400 365 36401-36500 366 36501-36600 367 36601-36700 368 36701-36800 369 36801-36900 370 36901-37000 371 37001-37100 372 37101-37200 373 37201-37300 374 37301-37400 375 37401-37500 376 37501-37600 377 37601-37700 378 37701-37800 379 37801-37900 380 37901-38000 381 38001-38100 382 38101-38200 383 38201-38300 384 38301-38400 385 38401-38500 386 38501-38600 387 38601-38700 388 38701-38800 389 38801-38900 390 38901-39000 391 39001-39100 392 39101-39200 393 39201-39300 394 39301-39400 395 39401-39500 396 39501-39600 397 39601-39700 398 39701-39800 399 39801-39900 400 39901-40000 401 40001-40100 402 40101-40200 403 40201-40300 404 40301-40400 405 40401-40500 406 40501-40600 407 40601-40700 408 40701-40800 409 40801-40900 410 40901-41000 411 41001-41100 412 41101-41200 413 41201-41300 414 41301-41400 415 41401-41500 416 41501-41600 417 41601-41700 418 41701-41800 419 41801-41900 420 41901-42000 421 42001-42100 422 42101-42200 423 42201-42300 424 42301-42400 425 42401-42500 426 42501-42600 427 42601-42700 428 42701-42800 429 42801-42900 430 42901-43000 431 43001-43100 432 43101-43200 433 43201-43300 434 43301-43400 435 43401-43500 436 43501-43600 437 43601-43700 438 43701-43800 439 43801-43900 440 43901-44000 441 44001-44100 442 44101-44200 443 44201-44300 444 44301-44400 445 44401-44500 446 44501-44600 447 44601-44700 448 44701-44800 449 44801-44900 450 44901-45000 451 45001-45100 452 45101-45200 453 45201-45300 454 45301-45400 455 45401-45500 456 45501-45600 457 45601-45700 458 45701-45800 459 45801-45900 460 45901-46000 461 46001-46100 462 46101-46200 463 46201-46300 464 46301-46400 465 46401-46500 466 46501-46600 467 46601-46700 468 46701-46800 469 46801-46900 470 46901-47000 471 47001-47100 472 47101-47200 473 47201-47300 474 47301-47400 475 47401-47500 476 47501-47600 477 47601-47700 478 47701-47800 479 47801-47900 480 47901-48000 481 48001-48100 482 48101-48200 483 48201-48300 484 48301-48400 485 48401-48490
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