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Theorem List for Metamath Proof Explorer - 27901-28000   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theorempntsf 27901* 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 27902* 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+ ↦ (((𝑆‘𝑥) / 𝑥) − (2 · (log‘𝑥)))) ∈ 𝑂(1)
 
Theoremselbergsb 27903* 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‘(((𝑆‘𝑥) / 𝑥) − (2 · (log‘𝑥)))) ≤ 𝑐
 
Theorempntsval2 27904* 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 → (𝑆‘𝐴) = Σ𝑛 ∈ (1...(⌊‘𝐴))(((Λ‘𝑛) · (log‘𝑛)) + Σ𝑚 ∈ {𝑦 ∈ ℕ ∣ 𝑦 ∥ 𝑛} ((Λ‘𝑚) · (Λ‘(𝑛 / 𝑚)))))
 
Theorempntrlog2bndlem1 27905* The sum of selberg3r 27897 and selberg4r 27898. (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‘(𝑅‘𝑥)) · (log‘𝑥)) − (Σ𝑛 ∈ (1...(⌊‘𝑥))((abs‘(𝑅‘(𝑥 / 𝑛))) · ((𝑆‘𝑛) − (𝑆‘(𝑛 − 1)))) / (log‘𝑥))) / 𝑥)) ∈ ≤𝑂(1)
 
Theorempntrlog2bndlem2 27906* Lemma for pntrlog2bnd 27912. 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))    &   (𝜑 → 𝐴 ∈ 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 27907* Lemma for pntrlog2bnd 27912. 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))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   (𝜑 → ∀𝑦 ∈ (1[,)+∞)(abs‘(((𝑆‘𝑦) / 𝑦) − (2 · (log‘𝑦)))) ≤ 𝐴)    ⇒   (𝜑 → (𝑥 ∈ (1(,)+∞) ↦ (Σ𝑛 ∈ (1...(⌊‘𝑥))(((abs‘(𝑅‘(𝑥 / 𝑛))) − (abs‘(𝑅‘(𝑥 / (𝑛 + 1))))) · ((𝑆‘𝑛) − (2 · (𝑛 · (log‘𝑛))))) / (𝑥 · (log‘𝑥)))) ∈ 𝑂(1))
 
Theorempntrlog2bndlem4 27908* Lemma for pntrlog2bnd 27912. 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‘(𝑅‘𝑥)) · (log‘𝑥)) − ((2 / (log‘𝑥)) · Σ𝑛 ∈ (1...(⌊‘𝑥))((abs‘(𝑅‘(𝑥 / 𝑛))) · ((𝑇‘𝑛) − (𝑇‘(𝑛 − 1)))))) / 𝑥)) ∈ ≤𝑂(1)
 
Theorempntrlog2bndlem5 27909* Lemma for pntrlog2bnd 27912. 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))    &   (𝜑 → 𝐵 ∈ ℝ+)    &   (𝜑 → ∀𝑦 ∈ ℝ+ (abs‘((𝑅‘𝑦) / 𝑦)) ≤ 𝐵)    ⇒   (𝜑 → (𝑥 ∈ (1(,)+∞) ↦ ((((abs‘(𝑅‘𝑥)) · (log‘𝑥)) − ((2 / (log‘𝑥)) · Σ𝑛 ∈ (1...(⌊‘𝑥))((abs‘(𝑅‘(𝑥 / 𝑛))) · (log‘𝑛)))) / 𝑥)) ∈ ≤𝑂(1))
 
Theorempntrlog2bndlem6a 27910* Lemma for pntrlog2bndlem6 27911. (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))    &   (𝜑 → 𝐵 ∈ ℝ+)    &   (𝜑 → ∀𝑦 ∈ ℝ+ (abs‘((𝑅‘𝑦) / 𝑦)) ≤ 𝐵)    &   (𝜑 → 𝐴 ∈ ℝ)    &   (𝜑 → 1 ≤ 𝐴)    ⇒   ((𝜑 ∧ 𝑥 ∈ (1(,)+∞)) → (1...(⌊‘𝑥)) = ((1...(⌊‘(𝑥 / 𝐴))) ∪ (((⌊‘(𝑥 / 𝐴)) + 1)...(⌊‘𝑥))))
 
Theorempntrlog2bndlem6 27911* Lemma for pntrlog2bnd 27912. 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))    &   (𝜑 → 𝐵 ∈ ℝ+)    &   (𝜑 → ∀𝑦 ∈ ℝ+ (abs‘((𝑅‘𝑦) / 𝑦)) ≤ 𝐵)    &   (𝜑 → 𝐴 ∈ ℝ)    &   (𝜑 → 1 ≤ 𝐴)    ⇒   (𝜑 → (𝑥 ∈ (1(,)+∞) ↦ ((((abs‘(𝑅‘𝑥)) · (log‘𝑥)) − ((2 / (log‘𝑥)) · Σ𝑛 ∈ (1...(⌊‘(𝑥 / 𝐴)))((abs‘(𝑅‘(𝑥 / 𝑛))) · (log‘𝑛)))) / 𝑥)) ∈ ≤𝑂(1))
 
Theorempntrlog2bnd 27912* 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‘(𝑅‘𝑥)) · (log‘𝑥)) − ((2 / (log‘𝑥)) · Σ𝑛 ∈ (1...(⌊‘(𝑥 / 𝐴)))((abs‘(𝑅‘(𝑥 / 𝑛))) · (log‘𝑛)))) / 𝑥) ≤ 𝑐)
 
Theorempntpbnd1a 27913* Lemma for pntpbnd 27916. (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))    &   (𝜑 → 𝐸 ∈ (0(,)1))    &   𝑋 = (exp‘(2 / 𝐸))    &   (𝜑 → 𝑌 ∈ (𝑋(,)+∞))    &   (𝜑 → 𝑁 ∈ ℕ)    &   (𝜑 → (𝑌 < 𝑁 ∧ 𝑁 ≤ (𝐾 · 𝑌)))    &   (𝜑 → (abs‘(𝑅‘𝑁)) ≤ (abs‘((𝑅‘(𝑁 + 1)) − (𝑅‘𝑁))))    ⇒   (𝜑 → (abs‘((𝑅‘𝑁) / 𝑁)) ≤ 𝐸)
 
Theorempntpbnd1 27914* Lemma for pntpbnd 27916. (Contributed by Mario Carneiro, 11-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑 → 𝐸 ∈ (0(,)1))    &   𝑋 = (exp‘(2 / 𝐸))    &   (𝜑 → 𝑌 ∈ (𝑋(,)+∞))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   (𝜑 → ∀𝑖 ∈ ℕ ∀𝑗 ∈ ℤ (abs‘Σ𝑦 ∈ (𝑖...𝑗)((𝑅‘𝑦) / (𝑦 · (𝑦 + 1)))) ≤ 𝐴)    &   𝐶 = (𝐴 + 2)    &   (𝜑 → 𝐾 ∈ ((exp‘(𝐶 / 𝐸))[,)+∞))    &   (𝜑 → ¬ ∃𝑦 ∈ ℕ ((𝑌 < 𝑦 ∧ 𝑦 ≤ (𝐾 · 𝑌)) ∧ (abs‘((𝑅‘𝑦) / 𝑦)) ≤ 𝐸))    ⇒   (𝜑 → Σ𝑛 ∈ (((⌊‘𝑌) + 1)...(⌊‘(𝐾 · 𝑌)))(abs‘((𝑅‘𝑛) / (𝑛 · (𝑛 + 1)))) ≤ 𝐴)
 
Theorempntpbnd2 27915* Lemma for pntpbnd 27916. (Contributed by Mario Carneiro, 11-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑 → 𝐸 ∈ (0(,)1))    &   𝑋 = (exp‘(2 / 𝐸))    &   (𝜑 → 𝑌 ∈ (𝑋(,)+∞))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   (𝜑 → ∀𝑖 ∈ ℕ ∀𝑗 ∈ ℤ (abs‘Σ𝑦 ∈ (𝑖...𝑗)((𝑅‘𝑦) / (𝑦 · (𝑦 + 1)))) ≤ 𝐴)    &   𝐶 = (𝐴 + 2)    &   (𝜑 → 𝐾 ∈ ((exp‘(𝐶 / 𝐸))[,)+∞))    &   (𝜑 → ¬ ∃𝑦 ∈ ℕ ((𝑌 < 𝑦 ∧ 𝑦 ≤ (𝐾 · 𝑌)) ∧ (abs‘((𝑅‘𝑦) / 𝑦)) ≤ 𝐸))    ⇒    ¬ 𝜑
 
Theorempntpbnd 27916* Lemma for pnt 27942. 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 < 𝑛 ∧ 𝑛 ≤ (𝑘 · 𝑦)) ∧ (abs‘((𝑅‘𝑛) / 𝑛)) ≤ 𝑒)
 
Theorempntibndlem1 27917 Lemma for pntibnd 27921. (Contributed by Mario Carneiro, 10-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   𝐿 = ((1 / 4) / (𝐴 + 3))    ⇒   (𝜑 → 𝐿 ∈ (0(,)1))
 
Theorempntibndlem2a 27918* Lemma for pntibndlem2 27919. (Contributed by Mario Carneiro, 7-Jun-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   𝐿 = ((1 / 4) / (𝐴 + 3))    &   (𝜑 → ∀𝑥 ∈ ℝ+ (abs‘((𝑅‘𝑥) / 𝑥)) ≤ 𝐴)    &   (𝜑 → 𝐵 ∈ ℝ+)    &   𝐾 = (exp‘(𝐵 / (𝐸 / 2)))    &   𝐶 = ((2 · 𝐵) + (log‘2))    &   (𝜑 → 𝐸 ∈ (0(,)1))    &   (𝜑 → 𝑍 ∈ ℝ+)    &   (𝜑 → 𝑁 ∈ ℕ)    ⇒   ((𝜑 ∧ 𝑢 ∈ (𝑁[,]((1 + (𝐿 · 𝐸)) · 𝑁))) → (𝑢 ∈ ℝ ∧ 𝑁 ≤ 𝑢 ∧ 𝑢 ≤ ((1 + (𝐿 · 𝐸)) · 𝑁)))
 
Theorempntibndlem2 27919* Lemma for pntibnd 27921. The main work, after eliminating all the quantifiers. (Contributed by Mario Carneiro, 10-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   𝐿 = ((1 / 4) / (𝐴 + 3))    &   (𝜑 → ∀𝑥 ∈ ℝ+ (abs‘((𝑅‘𝑥) / 𝑥)) ≤ 𝐴)    &   (𝜑 → 𝐵 ∈ ℝ+)    &   𝐾 = (exp‘(𝐵 / (𝐸 / 2)))    &   𝐶 = ((2 · 𝐵) + (log‘2))    &   (𝜑 → 𝐸 ∈ (0(,)1))    &   (𝜑 → 𝑍 ∈ ℝ+)    &   (𝜑 → 𝑁 ∈ ℕ)    &   (𝜑 → 𝑇 ∈ ℝ+)    &   (𝜑 → ∀𝑥 ∈ (1(,)+∞)∀𝑦 ∈ (𝑥[,](2 · 𝑥))((ψ‘𝑦) − (ψ‘𝑥)) ≤ ((2 · (𝑦 − 𝑥)) + (𝑇 · (𝑥 / (log‘𝑥)))))    &   𝑋 = ((exp‘(𝑇 / (𝐸 / 4))) + 𝑍)    &   (𝜑 → 𝑀 ∈ ((exp‘(𝐶 / 𝐸))[,)+∞))    &   (𝜑 → 𝑌 ∈ (𝑋(,)+∞))    &   (𝜑 → ((𝑌 < 𝑁 ∧ 𝑁 ≤ ((𝑀 / 2) · 𝑌)) ∧ (abs‘((𝑅‘𝑁) / 𝑁)) ≤ (𝐸 / 2)))    ⇒   (𝜑 → ∃𝑧 ∈ ℝ+ ((𝑌 < 𝑧 ∧ ((1 + (𝐿 · 𝐸)) · 𝑧) < (𝑀 · 𝑌)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝐿 · 𝐸)) · 𝑧))(abs‘((𝑅‘𝑢) / 𝑢)) ≤ 𝐸))
 
Theorempntibndlem3 27920* Lemma for pntibnd 27921. Package up pntibndlem2 27919 in quantifiers. (Contributed by Mario Carneiro, 10-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   𝐿 = ((1 / 4) / (𝐴 + 3))    &   (𝜑 → ∀𝑥 ∈ ℝ+ (abs‘((𝑅‘𝑥) / 𝑥)) ≤ 𝐴)    &   (𝜑 → 𝐵 ∈ ℝ+)    &   𝐾 = (exp‘(𝐵 / (𝐸 / 2)))    &   𝐶 = ((2 · 𝐵) + (log‘2))    &   (𝜑 → 𝐸 ∈ (0(,)1))    &   (𝜑 → 𝑍 ∈ ℝ+)    &   (𝜑 → ∀𝑚 ∈ (𝐾[,)+∞)∀𝑣 ∈ (𝑍(,)+∞)∃𝑖 ∈ ℕ ((𝑣 < 𝑖 ∧ 𝑖 ≤ (𝑚 · 𝑣)) ∧ (abs‘((𝑅‘𝑖) / 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‘((𝑅‘𝑢) / 𝑢)) ≤ 𝐸))
 
Theorempntibnd 27921* Lemma for pnt 27942. 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‘((𝑅‘𝑢) / 𝑢)) ≤ 𝑒)
 
Theorempntlemd 27922 Lemma for pnt 27942. 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))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   (𝜑 → 𝐵 ∈ ℝ+)    &   (𝜑 → 𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    ⇒   (𝜑 → (𝐿 ∈ ℝ+ ∧ 𝐷 ∈ ℝ+ ∧ 𝐹 ∈ ℝ+))
 
Theorempntlemc 27923* Lemma for pnt 27942. 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))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   (𝜑 → 𝐵 ∈ ℝ+)    &   (𝜑 → 𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑 → 𝑈 ∈ ℝ+)    &   (𝜑 → 𝑈 ≤ 𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    ⇒   (𝜑 → (𝐸 ∈ ℝ+ ∧ 𝐾 ∈ ℝ+ ∧ (𝐸 ∈ (0(,)1) ∧ 1 < 𝐾 ∧ (𝑈 − 𝐸) ∈ ℝ+)))
 
Theorempntlema 27924* Lemma for pnt 27942. 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))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   (𝜑 → 𝐵 ∈ ℝ+)    &   (𝜑 → 𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑 → 𝑈 ∈ ℝ+)    &   (𝜑 → 𝑈 ≤ 𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+ ∧ 𝑌 < 𝑋))    &   (𝜑 → 𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈 − 𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    ⇒   (𝜑 → 𝑊 ∈ ℝ+)
 
Theorempntlemb 27925* Lemma for pnt 27942. 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))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   (𝜑 → 𝐵 ∈ ℝ+)    &   (𝜑 → 𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑 → 𝑈 ∈ ℝ+)    &   (𝜑 → 𝑈 ≤ 𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+ ∧ 𝑌 < 𝑋))    &   (𝜑 → 𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈 − 𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑 → 𝑍 ∈ (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) ≤ (((𝑈 − 𝐸) · ((𝐿 · (𝐸↑2)) / (32 · 𝐵))) · (log‘𝑍)))))
 
Theorempntlemg 27926* Lemma for pnt 27942. 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))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   (𝜑 → 𝐵 ∈ ℝ+)    &   (𝜑 → 𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑 → 𝑈 ∈ ℝ+)    &   (𝜑 → 𝑈 ≤ 𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+ ∧ 𝑌 < 𝑋))    &   (𝜑 → 𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈 − 𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑 → 𝑍 ∈ (𝑊[,)+∞))    &   𝑀 = ((⌊‘((log‘𝑋) / (log‘𝐾))) + 1)    &   𝑁 = (⌊‘(((log‘𝑍) / (log‘𝐾)) / 2))    ⇒   (𝜑 → (𝑀 ∈ ℕ ∧ 𝑁 ∈ (ℤ≥‘𝑀) ∧ (((log‘𝑍) / (log‘𝐾)) / 4) ≤ (𝑁 − 𝑀)))
 
Theorempntlemh 27927* Lemma for pnt 27942. Bounds on the subintervals in the induction. (Contributed by Mario Carneiro, 13-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   (𝜑 → 𝐵 ∈ ℝ+)    &   (𝜑 → 𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑 → 𝑈 ∈ ℝ+)    &   (𝜑 → 𝑈 ≤ 𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+ ∧ 𝑌 < 𝑋))    &   (𝜑 → 𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈 − 𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑 → 𝑍 ∈ (𝑊[,)+∞))    &   𝑀 = ((⌊‘((log‘𝑋) / (log‘𝐾))) + 1)    &   𝑁 = (⌊‘(((log‘𝑍) / (log‘𝐾)) / 2))    ⇒   ((𝜑 ∧ 𝐽 ∈ (𝑀...𝑁)) → (𝑋 < (𝐾↑𝐽) ∧ (𝐾↑𝐽) ≤ (√‘𝑍)))
 
Theorempntlemn 27928* Lemma for pnt 27942. The "naive" base bound, which we will slightly improve. (Contributed by Mario Carneiro, 13-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   (𝜑 → 𝐵 ∈ ℝ+)    &   (𝜑 → 𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑 → 𝑈 ∈ ℝ+)    &   (𝜑 → 𝑈 ≤ 𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+ ∧ 𝑌 < 𝑋))    &   (𝜑 → 𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈 − 𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑 → 𝑍 ∈ (𝑊[,)+∞))    &   𝑀 = ((⌊‘((log‘𝑋) / (log‘𝐾))) + 1)    &   𝑁 = (⌊‘(((log‘𝑍) / (log‘𝐾)) / 2))    &   (𝜑 → ∀𝑧 ∈ (𝑌[,)+∞)(abs‘((𝑅‘𝑧) / 𝑧)) ≤ 𝑈)    ⇒   ((𝜑 ∧ (𝐽 ∈ ℕ ∧ 𝐽 ≤ (𝑍 / 𝑌))) → 0 ≤ (((𝑈 / 𝐽) − (abs‘((𝑅‘(𝑍 / 𝐽)) / 𝑍))) · (log‘𝐽)))
 
Theorempntlemq 27929* Lemma for pntlemj 27931. (Contributed by Mario Carneiro, 7-Jun-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   (𝜑 → 𝐵 ∈ ℝ+)    &   (𝜑 → 𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑 → 𝑈 ∈ ℝ+)    &   (𝜑 → 𝑈 ≤ 𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+ ∧ 𝑌 < 𝑋))    &   (𝜑 → 𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈 − 𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑 → 𝑍 ∈ (𝑊[,)+∞))    &   𝑀 = ((⌊‘((log‘𝑋) / (log‘𝐾))) + 1)    &   𝑁 = (⌊‘(((log‘𝑍) / (log‘𝐾)) / 2))    &   (𝜑 → ∀𝑧 ∈ (𝑌[,)+∞)(abs‘((𝑅‘𝑧) / 𝑧)) ≤ 𝑈)    &   (𝜑 → ∀𝑦 ∈ (𝑋(,)+∞)∃𝑧 ∈ ℝ+ ((𝑦 < 𝑧 ∧ ((1 + (𝐿 · 𝐸)) · 𝑧) < (𝐾 · 𝑦)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝐿 · 𝐸)) · 𝑧))(abs‘((𝑅‘𝑢) / 𝑢)) ≤ 𝐸))    &   𝑂 = (((⌊‘(𝑍 / (𝐾↑(𝐽 + 1)))) + 1)...(⌊‘(𝑍 / (𝐾↑𝐽))))    &   (𝜑 → 𝑉 ∈ ℝ+)    &   (𝜑 → (((𝐾↑𝐽) < 𝑉 ∧ ((1 + (𝐿 · 𝐸)) · 𝑉) < (𝐾 · (𝐾↑𝐽))) ∧ ∀𝑢 ∈ (𝑉[,]((1 + (𝐿 · 𝐸)) · 𝑉))(abs‘((𝑅‘𝑢) / 𝑢)) ≤ 𝐸))    &   (𝜑 → 𝐽 ∈ (𝑀..^𝑁))    &   𝐼 = (((⌊‘(𝑍 / ((1 + (𝐿 · 𝐸)) · 𝑉))) + 1)...(⌊‘(𝑍 / 𝑉)))    ⇒   (𝜑 → 𝐼 ⊆ 𝑂)
 
Theorempntlemr 27930* Lemma for pntlemj 27931. (Contributed by Mario Carneiro, 7-Jun-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   (𝜑 → 𝐵 ∈ ℝ+)    &   (𝜑 → 𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑 → 𝑈 ∈ ℝ+)    &   (𝜑 → 𝑈 ≤ 𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+ ∧ 𝑌 < 𝑋))    &   (𝜑 → 𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈 − 𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑 → 𝑍 ∈ (𝑊[,)+∞))    &   𝑀 = ((⌊‘((log‘𝑋) / (log‘𝐾))) + 1)    &   𝑁 = (⌊‘(((log‘𝑍) / (log‘𝐾)) / 2))    &   (𝜑 → ∀𝑧 ∈ (𝑌[,)+∞)(abs‘((𝑅‘𝑧) / 𝑧)) ≤ 𝑈)    &   (𝜑 → ∀𝑦 ∈ (𝑋(,)+∞)∃𝑧 ∈ ℝ+ ((𝑦 < 𝑧 ∧ ((1 + (𝐿 · 𝐸)) · 𝑧) < (𝐾 · 𝑦)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝐿 · 𝐸)) · 𝑧))(abs‘((𝑅‘𝑢) / 𝑢)) ≤ 𝐸))    &   𝑂 = (((⌊‘(𝑍 / (𝐾↑(𝐽 + 1)))) + 1)...(⌊‘(𝑍 / (𝐾↑𝐽))))    &   (𝜑 → 𝑉 ∈ ℝ+)    &   (𝜑 → (((𝐾↑𝐽) < 𝑉 ∧ ((1 + (𝐿 · 𝐸)) · 𝑉) < (𝐾 · (𝐾↑𝐽))) ∧ ∀𝑢 ∈ (𝑉[,]((1 + (𝐿 · 𝐸)) · 𝑉))(abs‘((𝑅‘𝑢) / 𝑢)) ≤ 𝐸))    &   (𝜑 → 𝐽 ∈ (𝑀..^𝑁))    &   𝐼 = (((⌊‘(𝑍 / ((1 + (𝐿 · 𝐸)) · 𝑉))) + 1)...(⌊‘(𝑍 / 𝑉)))    ⇒   (𝜑 → ((𝑈 − 𝐸) · (((𝐿 · 𝐸) / 8) · (log‘𝑍))) ≤ ((♯‘𝐼) · ((𝑈 − 𝐸) · ((log‘(𝑍 / 𝑉)) / (𝑍 / 𝑉)))))
 
Theorempntlemj 27931* Lemma for pnt 27942. The induction step. Using pntibnd 27921, we find an interval in 𝐾↑𝐽...𝐾↑(𝐽 + 1) which is sufficiently large and has a much smaller value, 𝑅(𝑧) / 𝑧 ≤ 𝐸 (instead of our original bound 𝑅(𝑧) / 𝑧 ≤ 𝑈). (Contributed by Mario Carneiro, 13-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   (𝜑 → 𝐵 ∈ ℝ+)    &   (𝜑 → 𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑 → 𝑈 ∈ ℝ+)    &   (𝜑 → 𝑈 ≤ 𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+ ∧ 𝑌 < 𝑋))    &   (𝜑 → 𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈 − 𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑 → 𝑍 ∈ (𝑊[,)+∞))    &   𝑀 = ((⌊‘((log‘𝑋) / (log‘𝐾))) + 1)    &   𝑁 = (⌊‘(((log‘𝑍) / (log‘𝐾)) / 2))    &   (𝜑 → ∀𝑧 ∈ (𝑌[,)+∞)(abs‘((𝑅‘𝑧) / 𝑧)) ≤ 𝑈)    &   (𝜑 → ∀𝑦 ∈ (𝑋(,)+∞)∃𝑧 ∈ ℝ+ ((𝑦 < 𝑧 ∧ ((1 + (𝐿 · 𝐸)) · 𝑧) < (𝐾 · 𝑦)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝐿 · 𝐸)) · 𝑧))(abs‘((𝑅‘𝑢) / 𝑢)) ≤ 𝐸))    &   𝑂 = (((⌊‘(𝑍 / (𝐾↑(𝐽 + 1)))) + 1)...(⌊‘(𝑍 / (𝐾↑𝐽))))    &   (𝜑 → 𝑉 ∈ ℝ+)    &   (𝜑 → (((𝐾↑𝐽) < 𝑉 ∧ ((1 + (𝐿 · 𝐸)) · 𝑉) < (𝐾 · (𝐾↑𝐽))) ∧ ∀𝑢 ∈ (𝑉[,]((1 + (𝐿 · 𝐸)) · 𝑉))(abs‘((𝑅‘𝑢) / 𝑢)) ≤ 𝐸))    &   (𝜑 → 𝐽 ∈ (𝑀..^𝑁))    &   𝐼 = (((⌊‘(𝑍 / ((1 + (𝐿 · 𝐸)) · 𝑉))) + 1)...(⌊‘(𝑍 / 𝑉)))    ⇒   (𝜑 → ((𝑈 − 𝐸) · (((𝐿 · 𝐸) / 8) · (log‘𝑍))) ≤ Σ𝑛 ∈ 𝑂 (((𝑈 / 𝑛) − (abs‘((𝑅‘(𝑍 / 𝑛)) / 𝑍))) · (log‘𝑛)))
 
Theorempntlemi 27932* Lemma for pnt 27942. Eliminate some assumptions from pntlemj 27931. (Contributed by Mario Carneiro, 13-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   (𝜑 → 𝐵 ∈ ℝ+)    &   (𝜑 → 𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑 → 𝑈 ∈ ℝ+)    &   (𝜑 → 𝑈 ≤ 𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+ ∧ 𝑌 < 𝑋))    &   (𝜑 → 𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈 − 𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑 → 𝑍 ∈ (𝑊[,)+∞))    &   𝑀 = ((⌊‘((log‘𝑋) / (log‘𝐾))) + 1)    &   𝑁 = (⌊‘(((log‘𝑍) / (log‘𝐾)) / 2))    &   (𝜑 → ∀𝑧 ∈ (𝑌[,)+∞)(abs‘((𝑅‘𝑧) / 𝑧)) ≤ 𝑈)    &   (𝜑 → ∀𝑦 ∈ (𝑋(,)+∞)∃𝑧 ∈ ℝ+ ((𝑦 < 𝑧 ∧ ((1 + (𝐿 · 𝐸)) · 𝑧) < (𝐾 · 𝑦)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝐿 · 𝐸)) · 𝑧))(abs‘((𝑅‘𝑢) / 𝑢)) ≤ 𝐸))    &   𝑂 = (((⌊‘(𝑍 / (𝐾↑(𝐽 + 1)))) + 1)...(⌊‘(𝑍 / (𝐾↑𝐽))))    ⇒   ((𝜑 ∧ 𝐽 ∈ (𝑀..^𝑁)) → ((𝑈 − 𝐸) · (((𝐿 · 𝐸) / 8) · (log‘𝑍))) ≤ Σ𝑛 ∈ 𝑂 (((𝑈 / 𝑛) − (abs‘((𝑅‘(𝑍 / 𝑛)) / 𝑍))) · (log‘𝑛)))
 
Theorempntlemf 27933* Lemma for pnt 27942. Add up the pieces in pntlemi 27932 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))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   (𝜑 → 𝐵 ∈ ℝ+)    &   (𝜑 → 𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑 → 𝑈 ∈ ℝ+)    &   (𝜑 → 𝑈 ≤ 𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+ ∧ 𝑌 < 𝑋))    &   (𝜑 → 𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈 − 𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑 → 𝑍 ∈ (𝑊[,)+∞))    &   𝑀 = ((⌊‘((log‘𝑋) / (log‘𝐾))) + 1)    &   𝑁 = (⌊‘(((log‘𝑍) / (log‘𝐾)) / 2))    &   (𝜑 → ∀𝑧 ∈ (𝑌[,)+∞)(abs‘((𝑅‘𝑧) / 𝑧)) ≤ 𝑈)    &   (𝜑 → ∀𝑦 ∈ (𝑋(,)+∞)∃𝑧 ∈ ℝ+ ((𝑦 < 𝑧 ∧ ((1 + (𝐿 · 𝐸)) · 𝑧) < (𝐾 · 𝑦)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝐿 · 𝐸)) · 𝑧))(abs‘((𝑅‘𝑢) / 𝑢)) ≤ 𝐸))    ⇒   (𝜑 → ((𝑈 − 𝐸) · (((𝐿 · (𝐸↑2)) / (32 · 𝐵)) · ((log‘𝑍)↑2))) ≤ Σ𝑛 ∈ (1...(⌊‘(𝑍 / 𝑌)))(((𝑈 / 𝑛) − (abs‘((𝑅‘(𝑍 / 𝑛)) / 𝑍))) · (log‘𝑛)))
 
Theorempntlemk 27934* Lemma for pnt 27942. Evaluate the naive part of the estimate. (Contributed by Mario Carneiro, 14-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   (𝜑 → 𝐵 ∈ ℝ+)    &   (𝜑 → 𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑 → 𝑈 ∈ ℝ+)    &   (𝜑 → 𝑈 ≤ 𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+ ∧ 𝑌 < 𝑋))    &   (𝜑 → 𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈 − 𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑 → 𝑍 ∈ (𝑊[,)+∞))    &   𝑀 = ((⌊‘((log‘𝑋) / (log‘𝐾))) + 1)    &   𝑁 = (⌊‘(((log‘𝑍) / (log‘𝐾)) / 2))    &   (𝜑 → ∀𝑧 ∈ (𝑌[,)+∞)(abs‘((𝑅‘𝑧) / 𝑧)) ≤ 𝑈)    &   (𝜑 → ∀𝑦 ∈ (𝑋(,)+∞)∃𝑧 ∈ ℝ+ ((𝑦 < 𝑧 ∧ ((1 + (𝐿 · 𝐸)) · 𝑧) < (𝐾 · 𝑦)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝐿 · 𝐸)) · 𝑧))(abs‘((𝑅‘𝑢) / 𝑢)) ≤ 𝐸))    ⇒   (𝜑 → (2 · Σ𝑛 ∈ (1...(⌊‘(𝑍 / 𝑌)))((𝑈 / 𝑛) · (log‘𝑛))) ≤ ((𝑈 · ((log‘𝑍) + 3)) · (log‘𝑍)))
 
Theorempntlemo 27935* Lemma for pnt 27942. 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))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   (𝜑 → 𝐵 ∈ ℝ+)    &   (𝜑 → 𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑 → 𝑈 ∈ ℝ+)    &   (𝜑 → 𝑈 ≤ 𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+ ∧ 𝑌 < 𝑋))    &   (𝜑 → 𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈 − 𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑 → 𝑍 ∈ (𝑊[,)+∞))    &   𝑀 = ((⌊‘((log‘𝑋) / (log‘𝐾))) + 1)    &   𝑁 = (⌊‘(((log‘𝑍) / (log‘𝐾)) / 2))    &   (𝜑 → ∀𝑧 ∈ (𝑌[,)+∞)(abs‘((𝑅‘𝑧) / 𝑧)) ≤ 𝑈)    &   (𝜑 → ∀𝑦 ∈ (𝑋(,)+∞)∃𝑧 ∈ ℝ+ ((𝑦 < 𝑧 ∧ ((1 + (𝐿 · 𝐸)) · 𝑧) < (𝐾 · 𝑦)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝐿 · 𝐸)) · 𝑧))(abs‘((𝑅‘𝑢) / 𝑢)) ≤ 𝐸))    &   (𝜑 → ∀𝑧 ∈ (1(,)+∞)((((abs‘(𝑅‘𝑧)) · (log‘𝑧)) − ((2 / (log‘𝑧)) · Σ𝑖 ∈ (1...(⌊‘(𝑧 / 𝑌)))((abs‘(𝑅‘(𝑧 / 𝑖))) · (log‘𝑖)))) / 𝑧) ≤ 𝐶)    ⇒   (𝜑 → (abs‘((𝑅‘𝑍) / 𝑍)) ≤ (𝑈 − (𝐹 · (𝑈↑3))))
 
Theorempntleme 27936* Lemma for pnt 27942. Package up pntlemo 27935 in quantifiers. (Contributed by Mario Carneiro, 14-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   (𝜑 → 𝐵 ∈ ℝ+)    &   (𝜑 → 𝐿 ∈ (0(,)1))    &   𝐷 = (𝐴 + 1)    &   𝐹 = ((1 − (1 / 𝐷)) · ((𝐿 / (32 · 𝐵)) / (𝐷↑2)))    &   (𝜑 → 𝑈 ∈ ℝ+)    &   (𝜑 → 𝑈 ≤ 𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → (𝑋 ∈ ℝ+ ∧ 𝑌 < 𝑋))    &   (𝜑 → 𝐶 ∈ ℝ+)    &   𝑊 = (((𝑌 + (4 / (𝐿 · 𝐸)))↑2) + (((𝑋 · (𝐾↑2))↑4) + (exp‘(((32 · 𝐵) / ((𝑈 − 𝐸) · (𝐿 · (𝐸↑2)))) · ((𝑈 · 3) + 𝐶)))))    &   (𝜑 → ∀𝑧 ∈ (𝑌[,)+∞)(abs‘((𝑅‘𝑧) / 𝑧)) ≤ 𝑈)    &   (𝜑 → ∀𝑘 ∈ (𝐾[,)+∞)∀𝑦 ∈ (𝑋(,)+∞)∃𝑧 ∈ ℝ+ ((𝑦 < 𝑧 ∧ ((1 + (𝐿 · 𝐸)) · 𝑧) < (𝑘 · 𝑦)) ∧ ∀𝑢 ∈ (𝑧[,]((1 + (𝐿 · 𝐸)) · 𝑧))(abs‘((𝑅‘𝑢) / 𝑢)) ≤ 𝐸))    &   (𝜑 → ∀𝑧 ∈ (1(,)+∞)((((abs‘(𝑅‘𝑧)) · (log‘𝑧)) − ((2 / (log‘𝑧)) · Σ𝑖 ∈ (1...(⌊‘(𝑧 / 𝑌)))((abs‘(𝑅‘(𝑧 / 𝑖))) · (log‘𝑖)))) / 𝑧) ≤ 𝐶)    ⇒   (𝜑 → ∃𝑤 ∈ ℝ+ ∀𝑣 ∈ (𝑤[,)+∞)(abs‘((𝑅‘𝑣) / 𝑣)) ≤ (𝑈 − (𝐹 · (𝑈↑3))))
 
Theorempntlem3 27937* Lemma for pnt 27942. 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))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   (𝜑 → ∀𝑥 ∈ ℝ+ (abs‘((𝑅‘𝑥) / 𝑥)) ≤ 𝐴)    &   𝑇 = {𝑡 ∈ (0[,]𝐴) ∣ ∃𝑦 ∈ ℝ+ ∀𝑧 ∈ (𝑦[,)+∞)(abs‘((𝑅‘𝑧) / 𝑧)) ≤ 𝑡}    &   (𝜑 → 𝐶 ∈ ℝ+)    &   ((𝜑 ∧ 𝑢 ∈ 𝑇) → (𝑢 − (𝐶 · (𝑢↑3))) ∈ 𝑇)    ⇒   (𝜑 → (𝑥 ∈ ℝ+ ↦ ((ψ‘𝑥) / 𝑥)) ⇝𝑟 1)
 
Theorempntlemp 27938* Lemma for pnt 27942. Wrapping up more quantifiers. (Contributed by Mario Carneiro, 14-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   (𝜑 → ∀𝑥 ∈ ℝ+ (abs‘((𝑅‘𝑥) / 𝑥)) ≤ 𝐴)    &   (𝜑 → 𝐵 ∈ ℝ+)    &   (𝜑 → 𝐿 ∈ (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‘((𝑅‘𝑢) / 𝑢)) ≤ 𝑒))    &   (𝜑 → 𝑈 ∈ ℝ+)    &   (𝜑 → 𝑈 ≤ 𝐴)    &   𝐸 = (𝑈 / 𝐷)    &   𝐾 = (exp‘(𝐵 / 𝐸))    &   (𝜑 → (𝑌 ∈ ℝ+ ∧ 1 ≤ 𝑌))    &   (𝜑 → ∀𝑧 ∈ (𝑌[,)+∞)(abs‘((𝑅‘𝑧) / 𝑧)) ≤ 𝑈)    ⇒   (𝜑 → ∃𝑤 ∈ ℝ+ ∀𝑣 ∈ (𝑤[,)+∞)(abs‘((𝑅‘𝑣) / 𝑣)) ≤ (𝑈 − (𝐹 · (𝑈↑3))))
 
Theorempntleml 27939* Lemma for pnt 27942. Equation 10.6.35 in [Shapiro], p. 436. (Contributed by Mario Carneiro, 14-Apr-2016.)
𝑅 = (𝑎 ∈ ℝ+ ↦ ((ψ‘𝑎) − 𝑎))    &   (𝜑 → 𝐴 ∈ ℝ+)    &   (𝜑 → ∀𝑥 ∈ ℝ+ (abs‘((𝑅‘𝑥) / 𝑥)) ≤ 𝐴)    &   (𝜑 → 𝐵 ∈ ℝ+)    &   (𝜑 → 𝐿 ∈ (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‘((𝑅‘𝑢) / 𝑢)) ≤ 𝑒))    ⇒   (𝜑 → (𝑥 ∈ ℝ+ ↦ ((ψ‘𝑥) / 𝑥)) ⇝𝑟 1)
 
Theorempnt3 27940 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 27941 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 27942 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(,)+∞) ↦ ((π‘𝑥) / (𝑥 / (log‘𝑥)))) ⇝𝑟 1
 
14.4.14  Ostrowski's theorem
 
Theoremabvcxp 27943* 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 = (𝑥 ∈ 𝐵 ↦ ((𝐹‘𝑥)↑𝑐𝑆))    ⇒   ((𝐹 ∈ 𝐴 ∧ 𝑆 ∈ (0(,]1)) → 𝐺 ∈ 𝐴)
 
Theorempadicfval 27944* 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 → (𝐽‘𝑃) = (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, (𝑃↑ -(𝑃 pCnt 𝑥)))))
 
Theorempadicval 27945* 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) → ((𝐽‘𝑃)‘𝑋) = if(𝑋 = 0, 0, (𝑃↑ -(𝑃 pCnt 𝑋))))
 
Theoremostth2lem1 27946* Lemma for ostth2 27965, although it is just a simple statement about exponentials which does not involve any specifics of ostth2 27965. If a power is upper bounded by a linear term, the exponent must be less than one. Or in big-O notation, 𝑛 ∈ 𝑜(𝐴↑𝑛) for any 1 < 𝐴. (Contributed by Mario Carneiro, 10-Sep-2014.)
(𝜑 → 𝐴 ∈ ℝ)    &   (𝜑 → 𝐵 ∈ ℝ)    &   ((𝜑 ∧ 𝑛 ∈ ℕ) → (𝐴↑𝑛) ≤ (𝑛 · 𝐵))    ⇒   (𝜑 → 𝐴 ≤ 1)
 
Theoremqrngbas 27947 The base set of the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂfld ↾s ℚ)    ⇒   ℚ = (Base‘𝑄)
 
Theoremqdrng 27948 The rationals form a division ring. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂfld ↾s ℚ)    ⇒   𝑄 ∈ DivRing
 
Theoremqrng0 27949 The zero element of the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂfld ↾s ℚ)    ⇒   0 = (0g‘𝑄)
 
Theoremqrng1 27950 The unity element of the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂfld ↾s ℚ)    ⇒   1 = (1r‘𝑄)
 
Theoremqrngneg 27951 The additive inverse in the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂfld ↾s ℚ)    ⇒   (𝑋 ∈ ℚ → ((invg‘𝑄)‘𝑋) = -𝑋)
 
Theoremqrngdiv 27952 The division operation in the field of rationals. (Contributed by Mario Carneiro, 8-Sep-2014.)
𝑄 = (ℂfld ↾s ℚ)    ⇒   ((𝑋 ∈ ℚ ∧ 𝑌 ∈ ℚ ∧ 𝑌 ≠ 0) → (𝑋(/r‘𝑄)𝑌) = (𝑋 / 𝑌))
 
Theoremqabvle 27953 By using induction on 𝑁, we show a long-range inequality coming from the triangle inequality. (Contributed by Mario Carneiro, 10-Sep-2014.)
𝑄 = (ℂfld ↾s ℚ)    &   𝐴 = (AbsVal‘𝑄)    ⇒   ((𝐹 ∈ 𝐴 ∧ 𝑁 ∈ ℕ0) → (𝐹‘𝑁) ≤ 𝑁)
 
Theoremqabvexp 27954 Induct the product rule abvmul 21079 to find the absolute value of a power. (Contributed by Mario Carneiro, 10-Sep-2014.)
𝑄 = (ℂfld ↾s ℚ)    &   𝐴 = (AbsVal‘𝑄)    ⇒   ((𝐹 ∈ 𝐴 ∧ 𝑀 ∈ ℚ ∧ 𝑁 ∈ ℕ0) → (𝐹‘(𝑀↑𝑁)) = ((𝐹‘𝑀)↑𝑁))
 
Theoremostthlem1 27955* Lemma for ostth 27967. 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 = (ℂfld ↾s ℚ)    &   𝐴 = (AbsVal‘𝑄)    &   (𝜑 → 𝐹 ∈ 𝐴)    &   (𝜑 → 𝐺 ∈ 𝐴)    &   ((𝜑 ∧ 𝑛 ∈ (ℤ≥‘2)) → (𝐹‘𝑛) = (𝐺‘𝑛))    ⇒   (𝜑 → 𝐹 = 𝐺)
 
Theoremostthlem2 27956* Lemma for ostth 27967. Refine ostthlem1 27955 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 = (ℂfld ↾s ℚ)    &   𝐴 = (AbsVal‘𝑄)    &   (𝜑 → 𝐹 ∈ 𝐴)    &   (𝜑 → 𝐺 ∈ 𝐴)    &   ((𝜑 ∧ 𝑝 ∈ ℙ) → (𝐹‘𝑝) = (𝐺‘𝑝))    ⇒   (𝜑 → 𝐹 = 𝐺)
 
Theoremqabsabv 27957 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 = (ℂfld ↾s ℚ)    &   𝐴 = (AbsVal‘𝑄)    ⇒   (abs ↾ ℚ) ∈ 𝐴
 
Theorempadicabv 27958* The p-adic absolute value (with arbitrary base) is an absolute value. (Contributed by Mario Carneiro, 9-Sep-2014.)
𝑄 = (ℂfld ↾s ℚ)    &   𝐴 = (AbsVal‘𝑄)    &   𝐹 = (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, (𝑁↑(𝑃 pCnt 𝑥))))    ⇒   ((𝑃 ∈ ℙ ∧ 𝑁 ∈ (0(,)1)) → 𝐹 ∈ 𝐴)
 
Theorempadicabvf 27959* The p-adic absolute value is an absolute value. (Contributed by Mario Carneiro, 9-Sep-2014.)
𝑄 = (ℂfld ↾s ℚ)    &   𝐴 = (AbsVal‘𝑄)    &   𝐽 = (𝑞 ∈ ℙ ↦ (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, (𝑞↑ -(𝑞 pCnt 𝑥)))))    ⇒   𝐽:ℙ⟶𝐴
 
Theorempadicabvcxp 27960* All positive powers of the p-adic absolute value are absolute values. (Contributed by Mario Carneiro, 9-Sep-2014.)
𝑄 = (ℂfld ↾s ℚ)    &   𝐴 = (AbsVal‘𝑄)    &   𝐽 = (𝑞 ∈ ℙ ↦ (𝑥 ∈ ℚ ↦ if(𝑥 = 0, 0, (𝑞↑ -(𝑞 pCnt 𝑥)))))    ⇒   ((𝑃 ∈ ℙ ∧ 𝑅 ∈ ℝ+) → (𝑦 ∈ ℚ ↦ (((𝐽‘𝑃)‘𝑦)↑𝑐𝑅)) ∈ 𝐴)
 
Theoremostth1 27961* - Lemma for ostth 27967: 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 21079 of the absolute value, 𝐹 is equal to 1 on all the integers, and ostthlem1 27955 extends this to the other rational numbers. (Contributed by Mario Carneiro, 10-Sep-2014.)
𝑄 = (ℂfld ↾s 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))    &   (𝜑 → 𝐹 ∈ 𝐴)    &   (𝜑 → ∀𝑛 ∈ ℕ ¬ 1 < (𝐹‘𝑛))    &   (𝜑 → ∀𝑛 ∈ ℙ ¬ (𝐹‘𝑛) < 1)    ⇒   (𝜑 → 𝐹 = 𝐾)
 
Theoremostth2lem2 27962* Lemma for ostth2 27965. (Contributed by Mario Carneiro, 10-Sep-2014.)
𝑄 = (ℂfld ↾s 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))    &   (𝜑 → 𝐹 ∈ 𝐴)    &   (𝜑 → 𝑁 ∈ (ℤ≥‘2))    &   (𝜑 → 1 < (𝐹‘𝑁))    &   𝑅 = ((log‘(𝐹‘𝑁)) / (log‘𝑁))    &   (𝜑 → 𝑀 ∈ (ℤ≥‘2))    &   𝑆 = ((log‘(𝐹‘𝑀)) / (log‘𝑀))    &   𝑇 = if((𝐹‘𝑀) ≤ 1, 1, (𝐹‘𝑀))    ⇒   ((𝜑 ∧ 𝑋 ∈ ℕ0 ∧ 𝑌 ∈ (0...((𝑀↑𝑋) − 1))) → (𝐹‘𝑌) ≤ ((𝑀 · 𝑋) · (𝑇↑𝑋)))
 
Theoremostth2lem3 27963* Lemma for ostth2 27965. (Contributed by Mario Carneiro, 10-Sep-2014.)
𝑄 = (ℂfld ↾s 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))    &   (𝜑 → 𝐹 ∈ 𝐴)    &   (𝜑 → 𝑁 ∈ (ℤ≥‘2))    &   (𝜑 → 1 < (𝐹‘𝑁))    &   𝑅 = ((log‘(𝐹‘𝑁)) / (log‘𝑁))    &   (𝜑 → 𝑀 ∈ (ℤ≥‘2))    &   𝑆 = ((log‘(𝐹‘𝑀)) / (log‘𝑀))    &   𝑇 = if((𝐹‘𝑀) ≤ 1, 1, (𝐹‘𝑀))    &   𝑈 = ((log‘𝑁) / (log‘𝑀))    ⇒   ((𝜑 ∧ 𝑋 ∈ ℕ) → (((𝐹‘𝑁) / (𝑇↑𝑐𝑈))↑𝑋) ≤ (𝑋 · ((𝑀 · 𝑇) · (𝑈 + 1))))
 
Theoremostth2lem4 27964* Lemma for ostth2 27965. (Contributed by Mario Carneiro, 10-Sep-2014.)
𝑄 = (ℂfld ↾s 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))    &   (𝜑 → 𝐹 ∈ 𝐴)    &   (𝜑 → 𝑁 ∈ (ℤ≥‘2))    &   (𝜑 → 1 < (𝐹‘𝑁))    &   𝑅 = ((log‘(𝐹‘𝑁)) / (log‘𝑁))    &   (𝜑 → 𝑀 ∈ (ℤ≥‘2))    &   𝑆 = ((log‘(𝐹‘𝑀)) / (log‘𝑀))    &   𝑇 = if((𝐹‘𝑀) ≤ 1, 1, (𝐹‘𝑀))    &   𝑈 = ((log‘𝑁) / (log‘𝑀))    ⇒   (𝜑 → (1 < (𝐹‘𝑀) ∧ 𝑅 ≤ 𝑆))
 
Theoremostth2 27965* - Lemma for ostth 27967: regular case. (Contributed by Mario Carneiro, 10-Sep-2014.)
𝑄 = (ℂfld ↾s 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))    &   (𝜑 → 𝐹 ∈ 𝐴)    &   (𝜑 → 𝑁 ∈ (ℤ≥‘2))    &   (𝜑 → 1 < (𝐹‘𝑁))    &   𝑅 = ((log‘(𝐹‘𝑁)) / (log‘𝑁))    ⇒   (𝜑 → ∃𝑎 ∈ (0(,]1)𝐹 = (𝑦 ∈ ℚ ↦ ((abs‘𝑦)↑𝑐𝑎)))
 
Theoremostth3 27966* - Lemma for ostth 27967: p-adic case. (Contributed by Mario Carneiro, 10-Sep-2014.)
𝑄 = (ℂfld ↾s 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))    &   (𝜑 → 𝐹 ∈ 𝐴)    &   (𝜑 → ∀𝑛 ∈ ℕ ¬ 1 < (𝐹‘𝑛))    &   (𝜑 → 𝑃 ∈ ℙ)    &   (𝜑 → (𝐹‘𝑃) < 1)    &   𝑅 = -((log‘(𝐹‘𝑃)) / (log‘𝑃))    &   𝑆 = if((𝐹‘𝑃) ≤ (𝐹‘𝑝), (𝐹‘𝑝), (𝐹‘𝑃))    ⇒   (𝜑 → ∃𝑎 ∈ ℝ+ 𝐹 = (𝑦 ∈ ℚ ↦ (((𝐽‘𝑃)‘𝑦)↑𝑐𝑎)))
 
Theoremostth 27967* Ostrowski's theorem, which classifies all absolute values on ℚ. 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 = (ℂfld ↾s 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))    ⇒   (𝐹 ∈ 𝐴 ↔ (𝐹 = 𝐾 ∨ ∃𝑎 ∈ (0(,]1)𝐹 = (𝑦 ∈ ℚ ↦ ((abs‘𝑦)↑𝑐𝑎)) ∨ ∃𝑎 ∈ ℝ+ ∃𝑔 ∈ ran 𝐽 𝐹 = (𝑦 ∈ ℚ ↦ ((𝑔‘𝑦)↑𝑐𝑎))))
 
14.4.15  Fermat's Last Theorem
 
14.4.15.1  Basic reductions for Fermat's Last Theorem
 
Theoreminfdesc 27968* Infinite descent. The hypotheses say that 𝑆 is lower bounded, and that if 𝜓 holds for an integer in 𝑆, it holds for a smaller integer in 𝑆. By infinite descent, eventually we cannot go any smaller, therefore 𝜓 holds for no integer in 𝑆. (Contributed by SN, 20-Aug-2024.)
(𝑦 = 𝑥 → (𝜓 ↔ 𝜒))    &   (𝑦 = 𝑧 → (𝜓 ↔ 𝜃))    &   (𝜑 → 𝑆 ⊆ (ℤ≥‘𝑀))    &   ((𝜑 ∧ (𝑥 ∈ 𝑆 ∧ 𝜒)) → ∃𝑧 ∈ 𝑆 (𝜃 ∧ 𝑧 < 𝑥))    ⇒   (𝜑 → {𝑦 ∈ 𝑆 ∣ 𝜓} = ∅)
 
Theoremfltdiv 27969 A counterexample to FLT stays valid when scaled. The hypotheses are more general than they need to be for convenience. (Contributed by SN, 20-Aug-2024.)
(𝜑 → 𝑆 ∈ ℂ)    &   (𝜑 → 𝑆 ≠ 0)    &   (𝜑 → 𝐴 ∈ ℂ)    &   (𝜑 → 𝐵 ∈ ℂ)    &   (𝜑 → 𝐶 ∈ ℂ)    &   (𝜑 → 𝑁 ∈ ℕ0)    &   (𝜑 → ((𝐴↑𝑁) + (𝐵↑𝑁)) = (𝐶↑𝑁))    ⇒   (𝜑 → (((𝐴 / 𝑆)↑𝑁) + ((𝐵 / 𝑆)↑𝑁)) = ((𝐶 / 𝑆)↑𝑁))
 
Theoremflt0 27970 A counterexample for FLT does not exist for 𝑁 = 0. (Contributed by SN, 20-Aug-2024.)
(𝜑 → 𝐴 ∈ ℂ)    &   (𝜑 → 𝐵 ∈ ℂ)    &   (𝜑 → 𝐶 ∈ ℂ)    &   (𝜑 → 𝑁 ∈ ℕ0)    &   (𝜑 → ((𝐴↑𝑁) + (𝐵↑𝑁)) = (𝐶↑𝑁))    ⇒   (𝜑 → 𝑁 ∈ ℕ)
 
Theoremfltdvdsabdvdsc 27971 Any factor of both 𝐴 and 𝐵 also divides 𝐶. This establishes the validity of fltabcoprmex 27972. (Contributed by SN, 21-Aug-2024.)
(𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    &   (𝜑 → 𝑁 ∈ ℕ)    &   (𝜑 → ((𝐴↑𝑁) + (𝐵↑𝑁)) = (𝐶↑𝑁))    ⇒   (𝜑 → (𝐴 gcd 𝐵) ∥ 𝐶)
 
Theoremfltabcoprmex 27972 A counterexample to FLT implies a counterexample to FLT with 𝐴, 𝐵 (assigned to 𝐴 / (𝐴 gcd 𝐵) and 𝐵 / (𝐴 gcd 𝐵)) coprime (by divgcdcoprm0 16841). (Contributed by SN, 20-Aug-2024.)
(𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    &   (𝜑 → 𝑁 ∈ ℕ0)    &   (𝜑 → ((𝐴↑𝑁) + (𝐵↑𝑁)) = (𝐶↑𝑁))    ⇒   (𝜑 → (((𝐴 / (𝐴 gcd 𝐵))↑𝑁) + ((𝐵 / (𝐴 gcd 𝐵))↑𝑁)) = ((𝐶 / (𝐴 gcd 𝐵))↑𝑁))
 
Theoremfltaccoprm 27973 A counterexample to FLT with 𝐴, 𝐵 coprime also has 𝐴, 𝐶 coprime. (Contributed by SN, 20-Aug-2024.)
(𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    &   (𝜑 → 𝑁 ∈ ℕ0)    &   (𝜑 → ((𝐴↑𝑁) + (𝐵↑𝑁)) = (𝐶↑𝑁))    &   (𝜑 → (𝐴 gcd 𝐵) = 1)    ⇒   (𝜑 → (𝐴 gcd 𝐶) = 1)
 
Theoremfltbccoprm 27974 A counterexample to FLT with 𝐴, 𝐵 coprime also has 𝐵, 𝐶 coprime. Proven from fltaccoprm 27973 using commutativity of addition. (Contributed by SN, 20-Aug-2024.)
(𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    &   (𝜑 → 𝑁 ∈ ℕ0)    &   (𝜑 → ((𝐴↑𝑁) + (𝐵↑𝑁)) = (𝐶↑𝑁))    &   (𝜑 → (𝐴 gcd 𝐵) = 1)    ⇒   (𝜑 → (𝐵 gcd 𝐶) = 1)
 
Theoremfltabcoprm 27975 A counterexample to FLT with 𝐴, 𝐶 coprime also has 𝐴, 𝐵 coprime. Converse of fltaccoprm 27973. (Contributed by SN, 22-Aug-2024.)
(𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    &   (𝜑 → (𝐴 gcd 𝐶) = 1)    &   (𝜑 → ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2))    ⇒   (𝜑 → (𝐴 gcd 𝐵) = 1)
 
Theoremfltne 27976 If a counterexample to FLT exists, its addends are not equal. (Contributed by SN, 1-Jun-2023.)
(𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    &   (𝜑 → 𝑁 ∈ (ℤ≥‘2))    &   (𝜑 → ((𝐴↑𝑁) + (𝐵↑𝑁)) = (𝐶↑𝑁))    ⇒   (𝜑 → 𝐴 ≠ 𝐵)
 
Theoremflt4lem1 27977 Satisfy the antecedent used in several pythagtrip 17013 lemmas. (Contributed by SN, 21-Aug-2024.)
(𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    &   (𝜑 → ¬ 2 ∥ 𝐴)    &   (𝜑 → (𝐴 gcd 𝐵) = 1)    &   (𝜑 → ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2))    ⇒   (𝜑 → ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)))
 
Theoremflt4lem2 27978 If 𝐴 is even, 𝐵 is odd. (Contributed by SN, 22-Aug-2024.)
(𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 2 ∥ 𝐴)    &   (𝜑 → (𝐴 gcd 𝐵) = 1)    ⇒   (𝜑 → ¬ 2 ∥ 𝐵)
 
Theoremflt4lem3 27979 Equivalent to pythagtriplem4 16998. Show that 𝐶 + 𝐴 and 𝐶 − 𝐴 are coprime. (Contributed by SN, 22-Aug-2024.)
(𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    &   (𝜑 → 2 ∥ 𝐴)    &   (𝜑 → (𝐴 gcd 𝐵) = 1)    &   (𝜑 → ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2))    ⇒   (𝜑 → ((𝐶 + 𝐴) gcd (𝐶 − 𝐴)) = 1)
 
Theoremflt4lem4 27980 If the product of two coprime factors is a perfect square, the factors are perfect squares. (Contributed by SN, 22-Aug-2024.)
(𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    &   (𝜑 → (𝐴 gcd 𝐵) = 1)    &   (𝜑 → (𝐴 · 𝐵) = (𝐶↑2))    ⇒   (𝜑 → (𝐴 = ((𝐴 gcd 𝐶)↑2) ∧ 𝐵 = ((𝐵 gcd 𝐶)↑2)))
 
Theoremflt4lem5 27981 In the context of the lemmas of pythagtrip 17013, 𝑀 and 𝑁 are coprime. (Contributed by SN, 23-Aug-2024.)
𝑀 = (((√‘(𝐶 + 𝐵)) + (√‘(𝐶 − 𝐵))) / 2)    &   𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶 − 𝐵))) / 2)    ⇒   (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (𝑀 gcd 𝑁) = 1)
 
Theoremflt4lem5elem 27982 Version of fltaccoprm 27973 and fltbccoprm 27974 where 𝑀 is not squared. This can be proved in general for any polynomial in three variables: using prmdvdsncoprmbd 16904, dvds2addd 16463, and prmdvdsexp 16892, we can show that if two variables are coprime, the third is also coprime to the two. (Contributed by SN, 24-Aug-2024.)
(𝜑 → 𝑀 ∈ ℕ)    &   (𝜑 → 𝑅 ∈ ℕ)    &   (𝜑 → 𝑆 ∈ ℕ)    &   (𝜑 → 𝑀 = ((𝑅↑2) + (𝑆↑2)))    &   (𝜑 → (𝑅 gcd 𝑆) = 1)    ⇒   (𝜑 → ((𝑅 gcd 𝑀) = 1 ∧ (𝑆 gcd 𝑀) = 1))
 
Theoremflt4lem5a 27983 Part 1 of Equation 1 of https://crypto.stanford.edu/pbc/notes/numberfield/fermatn4.html. (Contributed by SN, 22-Aug-2024.)
𝑀 = (((√‘(𝐶 + (𝐵↑2))) + (√‘(𝐶 − (𝐵↑2)))) / 2)    &   𝑁 = (((√‘(𝐶 + (𝐵↑2))) − (√‘(𝐶 − (𝐵↑2)))) / 2)    &   𝑅 = (((√‘(𝑀 + 𝑁)) + (√‘(𝑀 − 𝑁))) / 2)    &   𝑆 = (((√‘(𝑀 + 𝑁)) − (√‘(𝑀 − 𝑁))) / 2)    &   (𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    &   (𝜑 → ¬ 2 ∥ 𝐴)    &   (𝜑 → (𝐴 gcd 𝐵) = 1)    &   (𝜑 → ((𝐴↑4) + (𝐵↑4)) = (𝐶↑2))    ⇒   (𝜑 → ((𝐴↑2) + (𝑁↑2)) = (𝑀↑2))
 
Theoremflt4lem5b 27984 Part 2 of Equation 1 of https://crypto.stanford.edu/pbc/notes/numberfield/fermatn4.html. (Contributed by SN, 22-Aug-2024.)
𝑀 = (((√‘(𝐶 + (𝐵↑2))) + (√‘(𝐶 − (𝐵↑2)))) / 2)    &   𝑁 = (((√‘(𝐶 + (𝐵↑2))) − (√‘(𝐶 − (𝐵↑2)))) / 2)    &   𝑅 = (((√‘(𝑀 + 𝑁)) + (√‘(𝑀 − 𝑁))) / 2)    &   𝑆 = (((√‘(𝑀 + 𝑁)) − (√‘(𝑀 − 𝑁))) / 2)    &   (𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    &   (𝜑 → ¬ 2 ∥ 𝐴)    &   (𝜑 → (𝐴 gcd 𝐵) = 1)    &   (𝜑 → ((𝐴↑4) + (𝐵↑4)) = (𝐶↑2))    ⇒   (𝜑 → (2 · (𝑀 · 𝑁)) = (𝐵↑2))
 
Theoremflt4lem5c 27985 Part 2 of Equation 2 of https://crypto.stanford.edu/pbc/notes/numberfield/fermatn4.html. (Contributed by SN, 22-Aug-2024.)
𝑀 = (((√‘(𝐶 + (𝐵↑2))) + (√‘(𝐶 − (𝐵↑2)))) / 2)    &   𝑁 = (((√‘(𝐶 + (𝐵↑2))) − (√‘(𝐶 − (𝐵↑2)))) / 2)    &   𝑅 = (((√‘(𝑀 + 𝑁)) + (√‘(𝑀 − 𝑁))) / 2)    &   𝑆 = (((√‘(𝑀 + 𝑁)) − (√‘(𝑀 − 𝑁))) / 2)    &   (𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    &   (𝜑 → ¬ 2 ∥ 𝐴)    &   (𝜑 → (𝐴 gcd 𝐵) = 1)    &   (𝜑 → ((𝐴↑4) + (𝐵↑4)) = (𝐶↑2))    ⇒   (𝜑 → 𝑁 = (2 · (𝑅 · 𝑆)))
 
Theoremflt4lem5d 27986 Part 3 of Equation 2 of https://crypto.stanford.edu/pbc/notes/numberfield/fermatn4.html. (Contributed by SN, 23-Aug-2024.)
𝑀 = (((√‘(𝐶 + (𝐵↑2))) + (√‘(𝐶 − (𝐵↑2)))) / 2)    &   𝑁 = (((√‘(𝐶 + (𝐵↑2))) − (√‘(𝐶 − (𝐵↑2)))) / 2)    &   𝑅 = (((√‘(𝑀 + 𝑁)) + (√‘(𝑀 − 𝑁))) / 2)    &   𝑆 = (((√‘(𝑀 + 𝑁)) − (√‘(𝑀 − 𝑁))) / 2)    &   (𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    &   (𝜑 → ¬ 2 ∥ 𝐴)    &   (𝜑 → (𝐴 gcd 𝐵) = 1)    &   (𝜑 → ((𝐴↑4) + (𝐵↑4)) = (𝐶↑2))    ⇒   (𝜑 → 𝑀 = ((𝑅↑2) + (𝑆↑2)))
 
Theoremflt4lem5e 27987 Satisfy the hypotheses of flt4lem4 27980. (Contributed by SN, 23-Aug-2024.)
𝑀 = (((√‘(𝐶 + (𝐵↑2))) + (√‘(𝐶 − (𝐵↑2)))) / 2)    &   𝑁 = (((√‘(𝐶 + (𝐵↑2))) − (√‘(𝐶 − (𝐵↑2)))) / 2)    &   𝑅 = (((√‘(𝑀 + 𝑁)) + (√‘(𝑀 − 𝑁))) / 2)    &   𝑆 = (((√‘(𝑀 + 𝑁)) − (√‘(𝑀 − 𝑁))) / 2)    &   (𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    &   (𝜑 → ¬ 2 ∥ 𝐴)    &   (𝜑 → (𝐴 gcd 𝐵) = 1)    &   (𝜑 → ((𝐴↑4) + (𝐵↑4)) = (𝐶↑2))    ⇒   (𝜑 → (((𝑅 gcd 𝑆) = 1 ∧ (𝑅 gcd 𝑀) = 1 ∧ (𝑆 gcd 𝑀) = 1) ∧ (𝑅 ∈ ℕ ∧ 𝑆 ∈ ℕ ∧ 𝑀 ∈ ℕ) ∧ ((𝑀 · (𝑅 · 𝑆)) = ((𝐵 / 2)↑2) ∧ (𝐵 / 2) ∈ ℕ)))
 
Theoremflt4lem5f 27988 Final equation of https://crypto.stanford.edu/pbc/notes/numberfield/fermatn4.html. Given 𝐴↑4 + 𝐵↑4 = 𝐶↑2, provide a smaller solution. This satisfies the infinite descent condition. (Contributed by SN, 24-Aug-2024.)
𝑀 = (((√‘(𝐶 + (𝐵↑2))) + (√‘(𝐶 − (𝐵↑2)))) / 2)    &   𝑁 = (((√‘(𝐶 + (𝐵↑2))) − (√‘(𝐶 − (𝐵↑2)))) / 2)    &   𝑅 = (((√‘(𝑀 + 𝑁)) + (√‘(𝑀 − 𝑁))) / 2)    &   𝑆 = (((√‘(𝑀 + 𝑁)) − (√‘(𝑀 − 𝑁))) / 2)    &   (𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    &   (𝜑 → ¬ 2 ∥ 𝐴)    &   (𝜑 → (𝐴 gcd 𝐵) = 1)    &   (𝜑 → ((𝐴↑4) + (𝐵↑4)) = (𝐶↑2))    ⇒   (𝜑 → ((𝑀 gcd (𝐵 / 2))↑2) = (((𝑅 gcd (𝐵 / 2))↑4) + ((𝑆 gcd (𝐵 / 2))↑4)))
 
Theoremflt4lem6 27989 Remove shared factors in a solution to 𝐴↑4 + 𝐵↑4 = 𝐶↑2. (Contributed by SN, 24-Jul-2024.)
(𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    &   (𝜑 → ((𝐴↑4) + (𝐵↑4)) = (𝐶↑2))    ⇒   (𝜑 → (((𝐴 / (𝐴 gcd 𝐵)) ∈ ℕ ∧ (𝐵 / (𝐴 gcd 𝐵)) ∈ ℕ ∧ (𝐶 / ((𝐴 gcd 𝐵)↑2)) ∈ ℕ) ∧ (((𝐴 / (𝐴 gcd 𝐵))↑4) + ((𝐵 / (𝐴 gcd 𝐵))↑4)) = ((𝐶 / ((𝐴 gcd 𝐵)↑2))↑2)))
 
Theoremflt4lem7 27990* Convert flt4lem5f 27988 into a convenient form for nna4b4nsq 27991. (Contributed by SN, 25-Aug-2024.)
(𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    &   (𝜑 → ¬ 2 ∥ 𝐴)    &   (𝜑 → (𝐴 gcd 𝐵) = 1)    &   (𝜑 → ((𝐴↑4) + (𝐵↑4)) = (𝐶↑2))    ⇒   (𝜑 → ∃𝑙 ∈ ℕ (∃𝑔 ∈ ℕ ∃ℎ ∈ ℕ (¬ 2 ∥ 𝑔 ∧ ((𝑔 gcd ℎ) = 1 ∧ ((𝑔↑4) + (ℎ↑4)) = (𝑙↑2))) ∧ 𝑙 < 𝐶))
 
Theoremnna4b4nsq 27991 Strengthening of Fermat's last theorem for exponent 4, where the sum is only assumed to be a square. (Contributed by SN, 23-Aug-2024.)
(𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    ⇒   (𝜑 → ((𝐴↑4) + (𝐵↑4)) ≠ (𝐶↑2))
 
Theoremflt4 27992 Fermat's last theorem for the exponent four. (Contributed by AV, 15-Sep-2026.)
(𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    ⇒   (𝜑 → ((𝐴↑4) + (𝐵↑4)) ≠ (𝐶↑4))
 
Theoremflt4ALT 27993* Fermat's last theorem for the exponent four, derived from Fermat's right triangle theorem (which is not proved yet, see fermrtt: after a proof is available, the hypothesis flt4ALT.r can be removed - TODO-AV). (Contributed by AV, 15-Sep-2026.) (Proof modification is discouraged.) (New usage is discouraged.)
(𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    &   (𝜑 → ∀𝑎 ∈ ℕ ∀𝑏 ∈ ℕ ∀𝑐 ∈ ℕ ((𝑎↑4) − (𝑏↑4)) ≠ (𝑐↑2))    ⇒   (𝜑 → ((𝐴↑4) + (𝐵↑4)) ≠ (𝐶↑4))
 
14.4.15.2  Proof of Fermat's Last Theorem (work in progress)

On September 4th, 2026 Anthropic published that their AI "Claude" created a complete and machine-verified formal proof for Fermat's Last Theorem: it consists of about 13 million lines of code, and about 30.000 verified theorems, created within about 11 days, built on Lean's library "mathlib" as well as preliminary work of the Buzzard community (in 2024, Kevin Buzzard started a project at Imperial College to formalize a complete proof in Lean, planned for many years and founded up to 2029). For more details see https://www.anthropic.com/research/formalizing-fermats-last-theorem.

Asking Claude (Opus 5) itself if it is possible to prove Fermat's last theorem also with Metamath, it answered: "Short version: in principle yes, in practice no. Foundations aren't the problem. set.mm is built on ZFC, and in the Grothendieck extension (ax-groth 10909, Tarski's axiom) universes are available too - exactly what Wiles's proof appeared to need beyond ZFC. (Colin McLarty has shown in any case that this can be pushed down into far weaker systems.) So there is no logical obstacle, only a labor one. What has been proved so far is the case n = 4 (see flt4 27992). ... Why it's nevertheless a size too big: Metamath has no tactics; every step stands there fully expanded. More serious is that set.mm simply lacks the necessary infrastructure: no schemes, no modular forms, no Galois representations, no class field theory. That is the actual proof - the number theory of FLT is only the last line. Mathlib had a large part of it after years of community work, and even building on that, the Lean formalization comprised some 30,000 additional theorems; set.mm in total holds only an order of magnitude more than that. ... So it would be feasible - but it would be a project on the same scale as everything assembled in set.mm since 1993."

Nevertheless, this and the previous subsections provide some starting points for a formal proof by Metamath: According to the blueprint provided by Kevin Buzzard, see https://imperialcollegelondon.github.io/FLT/blueprint/ 27992, Lemma 2.1 is already proven (see fltoprm 27996). The next step would be to prove Lemma 2.2 (FLT for n = 3, see flt3 above). Corollary 2.3 follows immediately from Lemmas 2.1 and 2.2 (see fltoprmgt3 27997).

Finally, asking Claude if it makes sense to translate the Lean proof into the Metamath language, it says: "Possible in principle - sensible, not really. Foundationally there's no obstacle. Mario Carneiro showed that Lean's type theory is consistent relative to ZFC with n inaccessible cardinals for all n < ω. set.mm, via Tarski's axiom (ax-groth 10909), supplies arbitrarily large inaccessibles - so the required model is within reach. And there's precedent: in 2014/16 Carneiro translated proofs from the OpenTheory format (HOL Light, HOL4, Isabelle) into Metamath, first into a HOL formalization hol.mm, then embedded into set.mm. The exercise has been done once already. But Lean is a different league from HOL. Three problems: * The translation isn't a rewrite. ... * The transfer problem. ... * The scale. Lean's elaborator performs an enormous amount of invisible work: instance resolution, implicit arguments, universe unification, definitional unfolding. Metamath has none of it. 13 million lines of Lean on top of over a million lines of mathlib plausibly expand into the billions of steps - against the few tens of thousands of statements that make up all of set.mm. The real objection, though, isn't feasibility but payoff. ... Nothing for set.mm. The database is curated and maintained for readability; a generated dump would contradict its purpose and wouldn't be admitted into the main database anyway. What does transfer is the blueprint. The lemma decomposition, the definitional choices, the proof strategy - that's language-independent, and it's precisely the part that took years of work to produce. Buzzard's blueprint for the FLT project is literally that document. A native Metamath development following it would be the sensible path, and it would reasonably start at n = 3.".

The last sentence is actually the way we want to continue (as described above)...

 
Theoremfltoprmlem1 27994* Lemma 1 for fltoprm 27996: Every positive integer is either a power of 2 or has an odd prime factor. (Contributed by AV, 15-Sep-2026.)
(𝑁 ∈ ℕ → (∃𝑝 ∈ ℙ (2 < 𝑝 ∧ 𝑝 ∥ 𝑁) ∨ ∃𝑛 ∈ ℕ0 𝑁 = (2↑𝑛)))
 
Theoremfltoprmlem2 27995 Lemma 2 for fltoprm 27996: If an integer greater than or equal to 3 is a power of 2, it is a multiple of 4. (Contributed by AV, 15-Sep-2026.)
((𝑁 ∈ (ℤ≥‘3) ∧ 𝐾 ∈ ℕ0 ∧ 𝑁 = (2↑𝐾)) → 4 ∥ 𝑁)
 
Theoremfltoprm 27996* Fermat's last theorem holds for any exponent greater than 2 if it holds for odd prime exponents. (Contributed by AV, 15-Sep-2026.)
(𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    &   (𝜑 → 𝑁 ∈ (ℤ≥‘3))    &   (𝜑 → ∀𝑎 ∈ ℕ ∀𝑏 ∈ ℕ ∀𝑐 ∈ ℕ ∀𝑝 ∈ ℙ (2 < 𝑝 → ((𝑎↑𝑝) + (𝑏↑𝑝)) ≠ (𝑐↑𝑝)))    ⇒   (𝜑 → ((𝐴↑𝑁) + (𝐵↑𝑁)) ≠ (𝐶↑𝑁))
 
Theoremfltoprmgt3 27997* Fermat's last theorem holds for any exponent greater than 2 if it holds for all odd prime exponents greater than 3. (TODO-AV: after a proof is available for 𝑁 = 3, see flt3, the hypothesis fltoprmgt3.3 can be removed.) (Contributed by AV, 15-Sep-2026.)
(𝜑 → 𝐴 ∈ ℕ)    &   (𝜑 → 𝐵 ∈ ℕ)    &   (𝜑 → 𝐶 ∈ ℕ)    &   (𝜑 → 𝑁 ∈ (ℤ≥‘3))    &   (𝜑 → ∀𝑎 ∈ ℕ ∀𝑏 ∈ ℕ ∀𝑐 ∈ ℕ ∀𝑝 ∈ ℙ (3 < 𝑝 → ((𝑎↑𝑝) + (𝑏↑𝑝)) ≠ (𝑐↑𝑝)))    &   (𝜑 → ((𝑎↑3) + (𝑏↑3)) ≠ (𝑐↑3))    ⇒   (𝜑 → ((𝐴↑𝑁) + (𝐵↑𝑁)) ≠ (𝐶↑𝑁))
 
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 27998 Declare the class of all surreal numbers (see df-no 28001).
class No
 
Syntaxclts 27999 Declare the less-than relation over surreal numbers (see df-lts 28002).
class <s
 
Syntaxcbday 28000 Declare the birthday function for surreal numbers (see df-bday 28003).
class bday
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78 7701-7800 79 7801-7900 80 7901-8000 81 8001-8100 82 8101-8200 83 8201-8300 84 8301-8400 85 8401-8500 86 8501-8600 87 8601-8700 88 8701-8800 89 8801-8900 90 8901-9000 91 9001-9100 92 9101-9200 93 9201-9300 94 9301-9400 95 9401-9500 96 9501-9600 97 9601-9700 98 9701-9800 99 9801-9900 100 9901-10000 101 10001-10100 102 10101-10200 103 10201-10300 104 10301-10400 105 10401-10500 106 10501-10600 107 10601-10700 108 10701-10800 109 10801-10900 110 10901-11000 111 11001-11100 112 11101-11200 113 11201-11300 114 11301-11400 115 11401-11500 116 11501-11600 117 11601-11700 118 11701-11800 119 11801-11900 120 11901-12000 121 12001-12100 122 12101-12200 123 12201-12300 124 12301-12400 125 12401-12500 126 12501-12600 127 12601-12700 128 12701-12800 129 12801-12900 130 12901-13000 131 13001-13100 132 13101-13200 133 13201-13300 134 13301-13400 135 13401-13500 136 13501-13600 137 13601-13700 138 13701-13800 139 13801-13900 140 13901-14000 141 14001-14100 142 14101-14200 143 14201-14300 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 48401-48500 486 48501-48600 487 48601-48700 488 48701-48800 489 48801-48900 490 48901-49000 491 49001-49100 492 49101-49200 493 49201-49300 494 49301-49400 495 49401-49500 496 49501-49600 497 49601-49700 498 49701-49800 499 49801-49900 500 49901-50000 501 50001-50100 502 50101-50200 503 50201-50300 504 50301-50400 505 50401-50500 506 50501-50600 507 50601-50700 508 50701-50800 509 50801-50900 510 50901-51000 511 51001-51066
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