P. 155 of Theorem List - Intuitionistic Logic Explorer

Type

Label

Description

Statement

Theorem

gausslemma2dlem0d 15401

Auxiliary lemma 4 for gausslemma2d 15418. (Contributed by AV, 9-Jul-2021.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    ⇒   ⊢ (𝜑 → 𝑀 ∈ ℕ₀)

Theorem

gausslemma2dlem0e 15402

Auxiliary lemma 5 for gausslemma2d 15418. (Contributed by AV, 9-Jul-2021.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    ⇒   ⊢ (𝜑 → (𝑀 · 2) < (𝑃 / 2))

Theorem

gausslemma2dlem0f 15403

Auxiliary lemma 6 for gausslemma2d 15418. (Contributed by AV, 9-Jul-2021.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    ⇒   ⊢ (𝜑 → (𝑀 + 1) ≤ 𝐻)

Theorem

gausslemma2dlem0g 15404

Auxiliary lemma 7 for gausslemma2d 15418. (Contributed by AV, 9-Jul-2021.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    ⇒   ⊢ (𝜑 → 𝑀 ≤ 𝐻)

Theorem

gausslemma2dlem0h 15405

Auxiliary lemma 8 for gausslemma2d 15418. (Contributed by AV, 9-Jul-2021.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑁 = (𝐻 − 𝑀)    ⇒   ⊢ (𝜑 → 𝑁 ∈ ℕ₀)

Theorem

gausslemma2dlem0i 15406

Auxiliary lemma 9 for gausslemma2d 15418. (Contributed by AV, 14-Jul-2021.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑁 = (𝐻 − 𝑀)    ⇒   ⊢ (𝜑 → (((2 /_L 𝑃) mod 𝑃) = ((-1↑𝑁) mod 𝑃) → (2 /_L 𝑃) = (-1↑𝑁)))

Theorem

gausslemma2dlem1a 15407*

Lemma for gausslemma2dlem1 15410. (Contributed by AV, 1-Jul-2021.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    ⇒   ⊢ (𝜑 → ran 𝑅 = (1...𝐻))

Theorem

gausslemma2dlem1cl 15408

Lemma for gausslemma2dlem1 15410. Closure of the body of the definition of 𝑅. (Contributed by Jim Kingdon, 10-Aug-2025.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   ⊢ (𝜑 → 𝐴 ∈ (1...𝐻))    ⇒   ⊢ (𝜑 → if((𝐴 · 2) < (𝑃 / 2), (𝐴 · 2), (𝑃 − (𝐴 · 2))) ∈ ℤ)

Theorem

gausslemma2dlem1f1o 15409*

Lemma for gausslemma2dlem1 15410. (Contributed by Jim Kingdon, 9-Aug-2025.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    ⇒   ⊢ (𝜑 → 𝑅:(1...𝐻)–1-1-onto→(1...𝐻))

Theorem

gausslemma2dlem1 15410*

Lemma 1 for gausslemma2d 15418. (Contributed by AV, 5-Jul-2021.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    ⇒   ⊢ (𝜑 → (!‘𝐻) = ∏𝑘 ∈ (1...𝐻)(𝑅‘𝑘))

Theorem

gausslemma2dlem2 15411*

Lemma 2 for gausslemma2d 15418. (Contributed by AV, 4-Jul-2021.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    ⇒   ⊢ (𝜑 → ∀𝑘 ∈ (1...𝑀)(𝑅‘𝑘) = (𝑘 · 2))

Theorem

gausslemma2dlem3 15412*

Lemma 3 for gausslemma2d 15418. (Contributed by AV, 4-Jul-2021.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    ⇒   ⊢ (𝜑 → ∀𝑘 ∈ ((𝑀 + 1)...𝐻)(𝑅‘𝑘) = (𝑃 − (𝑘 · 2)))

Theorem

gausslemma2dlem4 15413*

Lemma 4 for gausslemma2d 15418. (Contributed by AV, 16-Jun-2021.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    ⇒   ⊢ (𝜑 → (!‘𝐻) = (∏𝑘 ∈ (1...𝑀)(𝑅‘𝑘) · ∏𝑘 ∈ ((𝑀 + 1)...𝐻)(𝑅‘𝑘)))

Theorem

gausslemma2dlem5a 15414*

Lemma for gausslemma2dlem5 15415. (Contributed by AV, 8-Jul-2021.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    ⇒   ⊢ (𝜑 → (∏𝑘 ∈ ((𝑀 + 1)...𝐻)(𝑅‘𝑘) mod 𝑃) = (∏𝑘 ∈ ((𝑀 + 1)...𝐻)(-1 · (𝑘 · 2)) mod 𝑃))

Theorem

gausslemma2dlem5 15415*

Lemma 5 for gausslemma2d 15418. (Contributed by AV, 9-Jul-2021.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝑁 = (𝐻 − 𝑀)    ⇒   ⊢ (𝜑 → (∏𝑘 ∈ ((𝑀 + 1)...𝐻)(𝑅‘𝑘) mod 𝑃) = (((-1↑𝑁) · ∏𝑘 ∈ ((𝑀 + 1)...𝐻)(𝑘 · 2)) mod 𝑃))

Theorem

gausslemma2dlem6 15416*

Lemma 6 for gausslemma2d 15418. (Contributed by AV, 16-Jun-2021.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝑁 = (𝐻 − 𝑀)    ⇒   ⊢ (𝜑 → ((!‘𝐻) mod 𝑃) = ((((-1↑𝑁) · (2↑𝐻)) · (!‘𝐻)) mod 𝑃))

Theorem

gausslemma2dlem7 15417*

Lemma 7 for gausslemma2d 15418. (Contributed by AV, 13-Jul-2021.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝑁 = (𝐻 − 𝑀)    ⇒   ⊢ (𝜑 → (((-1↑𝑁) · (2↑𝐻)) mod 𝑃) = 1)

Theorem

gausslemma2d 15418*

Gauss' Lemma (see also theorem 9.6 in [ApostolNT] p. 182) for integer 2: Let p be an odd prime. Let S = {2, 4, 6, ..., p - 1}. Let n denote the number of elements of S whose least positive residue modulo p is greater than p/2. Then ( 2 | p ) = (-1)^n. (Contributed by AV, 14-Jul-2021.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ 𝐻 = ((𝑃 − 1) / 2)    &   ⊢ 𝑅 = (𝑥 ∈ (1...𝐻) ↦ if((𝑥 · 2) < (𝑃 / 2), (𝑥 · 2), (𝑃 − (𝑥 · 2))))    &   ⊢ 𝑀 = (⌊‘(𝑃 / 4))    &   ⊢ 𝑁 = (𝐻 − 𝑀)    ⇒   ⊢ (𝜑 → (2 /_L 𝑃) = (-1↑𝑁))

11.3.6  Quadratic reciprocity

Theorem

lgseisenlem1 15419*

Lemma for lgseisen 15423. If 𝑅(𝑢) = (𝑄 · 𝑢) mod 𝑃 and 𝑀(𝑢) = (-1↑𝑅(𝑢)) · 𝑅(𝑢), then for any even 1 ≤ 𝑢 ≤ 𝑃 − 1, 𝑀(𝑢) is also an even integer 1 ≤ 𝑀(𝑢) ≤ 𝑃 − 1. To simplify these statements, we divide all the even numbers by 2, so that it becomes the statement that 𝑀(𝑥 / 2) = (-1↑𝑅(𝑥 / 2)) · 𝑅(𝑥 / 2) / 2 is an integer between 1 and (𝑃 − 1) / 2. (Contributed by Mario Carneiro, 17-Jun-2015.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑄 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑄)    &   ⊢ 𝑅 = ((𝑄 · (2 · 𝑥)) mod 𝑃)    &   ⊢ 𝑀 = (𝑥 ∈ (1...((𝑃 − 1) / 2)) ↦ ((((-1↑𝑅) · 𝑅) mod 𝑃) / 2))    ⇒   ⊢ (𝜑 → 𝑀:(1...((𝑃 − 1) / 2))⟶(1...((𝑃 − 1) / 2)))

Theorem

lgseisenlem2 15420*

Lemma for lgseisen 15423. The function 𝑀 is an injection (and hence a bijection by the pigeonhole principle). (Contributed by Mario Carneiro, 17-Jun-2015.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑄 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑄)    &   ⊢ 𝑅 = ((𝑄 · (2 · 𝑥)) mod 𝑃)    &   ⊢ 𝑀 = (𝑥 ∈ (1...((𝑃 − 1) / 2)) ↦ ((((-1↑𝑅) · 𝑅) mod 𝑃) / 2))    &   ⊢ 𝑆 = ((𝑄 · (2 · 𝑦)) mod 𝑃)    ⇒   ⊢ (𝜑 → 𝑀:(1...((𝑃 − 1) / 2))–1-1-onto→(1...((𝑃 − 1) / 2)))

Theorem

lgseisenlem3 15421*

Lemma for lgseisen 15423. (Contributed by Mario Carneiro, 17-Jun-2015.) (Proof shortened by AV, 28-Jul-2019.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑄 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑄)    &   ⊢ 𝑅 = ((𝑄 · (2 · 𝑥)) mod 𝑃)    &   ⊢ 𝑀 = (𝑥 ∈ (1...((𝑃 − 1) / 2)) ↦ ((((-1↑𝑅) · 𝑅) mod 𝑃) / 2))    &   ⊢ 𝑆 = ((𝑄 · (2 · 𝑦)) mod 𝑃)    &   ⊢ 𝑌 = (ℤ/nℤ‘𝑃)    &   ⊢ 𝐺 = (mulGrp‘𝑌)    &   ⊢ 𝐿 = (ℤRHom‘𝑌)    ⇒   ⊢ (𝜑 → (𝐺 Σ_g (𝑥 ∈ (1...((𝑃 − 1) / 2)) ↦ (𝐿‘((-1↑𝑅) · 𝑄)))) = (1_r‘𝑌))

Theorem

lgseisenlem4 15422*

Lemma for lgseisen 15423. (Contributed by Mario Carneiro, 18-Jun-2015.) (Proof shortened by AV, 15-Jun-2019.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑄 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑄)    &   ⊢ 𝑅 = ((𝑄 · (2 · 𝑥)) mod 𝑃)    &   ⊢ 𝑀 = (𝑥 ∈ (1...((𝑃 − 1) / 2)) ↦ ((((-1↑𝑅) · 𝑅) mod 𝑃) / 2))    &   ⊢ 𝑆 = ((𝑄 · (2 · 𝑦)) mod 𝑃)    &   ⊢ 𝑌 = (ℤ/nℤ‘𝑃)    &   ⊢ 𝐺 = (mulGrp‘𝑌)    &   ⊢ 𝐿 = (ℤRHom‘𝑌)    ⇒   ⊢ (𝜑 → ((𝑄↑((𝑃 − 1) / 2)) mod 𝑃) = ((-1↑Σ𝑥 ∈ (1...((𝑃 − 1) / 2))(⌊‘((𝑄 / 𝑃) · (2 · 𝑥)))) mod 𝑃))

Theorem

lgseisen 15423*

Eisenstein's lemma, an expression for (𝑃 /_L 𝑄) when 𝑃, 𝑄 are distinct odd primes. (Contributed by Mario Carneiro, 18-Jun-2015.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑄 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑄)    ⇒   ⊢ (𝜑 → (𝑄 /_L 𝑃) = (-1↑Σ𝑥 ∈ (1...((𝑃 − 1) / 2))(⌊‘((𝑄 / 𝑃) · (2 · 𝑥)))))

Theorem

lgsquadlemsfi 15424*

Lemma for lgsquad 15429. 𝑆 is finite. (Contributed by Jim Kingdon, 16-Sep-2025.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑄 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑄)    &   ⊢ 𝑀 = ((𝑃 − 1) / 2)    &   ⊢ 𝑁 = ((𝑄 − 1) / 2)    &   ⊢ 𝑆 = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ (1...𝑀) ∧ 𝑦 ∈ (1...𝑁)) ∧ (𝑦 · 𝑃) < (𝑥 · 𝑄))}    ⇒   ⊢ (𝜑 → 𝑆 ∈ Fin)

Theorem

lgsquadlemofi 15425*

Lemma for lgsquad 15429. There are finitely many members of 𝑆 with odd first part. (Contributed by Jim Kingdon, 16-Sep-2025.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑄 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑄)    &   ⊢ 𝑀 = ((𝑃 − 1) / 2)    &   ⊢ 𝑁 = ((𝑄 − 1) / 2)    &   ⊢ 𝑆 = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ (1...𝑀) ∧ 𝑦 ∈ (1...𝑁)) ∧ (𝑦 · 𝑃) < (𝑥 · 𝑄))}    ⇒   ⊢ (𝜑 → {𝑧 ∈ 𝑆 ∣ ¬ 2 ∥ (1^st ‘𝑧)} ∈ Fin)

Theorem

lgsquadlem1 15426*

Lemma for lgsquad 15429. Count the members of 𝑆 with odd coordinates. (Contributed by Mario Carneiro, 19-Jun-2015.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑄 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑄)    &   ⊢ 𝑀 = ((𝑃 − 1) / 2)    &   ⊢ 𝑁 = ((𝑄 − 1) / 2)    &   ⊢ 𝑆 = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ (1...𝑀) ∧ 𝑦 ∈ (1...𝑁)) ∧ (𝑦 · 𝑃) < (𝑥 · 𝑄))}    ⇒   ⊢ (𝜑 → (-1↑Σ𝑢 ∈ (((⌊‘(𝑀 / 2)) + 1)...𝑀)(⌊‘((𝑄 / 𝑃) · (2 · 𝑢)))) = (-1↑(♯‘{𝑧 ∈ 𝑆 ∣ ¬ 2 ∥ (1^st ‘𝑧)})))

Theorem

lgsquadlem2 15427*

Lemma for lgsquad 15429. Count the members of 𝑆 with even coordinates, and combine with lgsquadlem1 15426 to get the total count of lattice points in 𝑆 (up to parity). (Contributed by Mario Carneiro, 18-Jun-2015.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑄 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑄)    &   ⊢ 𝑀 = ((𝑃 − 1) / 2)    &   ⊢ 𝑁 = ((𝑄 − 1) / 2)    &   ⊢ 𝑆 = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ (1...𝑀) ∧ 𝑦 ∈ (1...𝑁)) ∧ (𝑦 · 𝑃) < (𝑥 · 𝑄))}    ⇒   ⊢ (𝜑 → (𝑄 /_L 𝑃) = (-1↑(♯‘𝑆)))

Theorem

lgsquadlem3 15428*

Lemma for lgsquad 15429. (Contributed by Mario Carneiro, 18-Jun-2015.)

⊢ (𝜑 → 𝑃 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑄 ∈ (ℙ ∖ {2}))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑄)    &   ⊢ 𝑀 = ((𝑃 − 1) / 2)    &   ⊢ 𝑁 = ((𝑄 − 1) / 2)    &   ⊢ 𝑆 = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ (1...𝑀) ∧ 𝑦 ∈ (1...𝑁)) ∧ (𝑦 · 𝑃) < (𝑥 · 𝑄))}    ⇒   ⊢ (𝜑 → ((𝑃 /_L 𝑄) · (𝑄 /_L 𝑃)) = (-1↑(𝑀 · 𝑁)))

Theorem

lgsquad 15429

The Law of Quadratic Reciprocity, see also theorem 9.8 in [ApostolNT] p. 185. If 𝑃 and 𝑄 are distinct odd primes, then the product of the Legendre symbols (𝑃 /_L 𝑄) and (𝑄 /_L 𝑃) is the parity of ((𝑃 − 1) / 2) · ((𝑄 − 1) / 2). This uses Eisenstein's proof, which also has a nice geometric interpretation - see https://en.wikipedia.org/wiki/Proofs_of_quadratic_reciprocity. This is Metamath 100 proof #7. (Contributed by Mario Carneiro, 19-Jun-2015.)

⊢ ((𝑃 ∈ (ℙ ∖ {2}) ∧ 𝑄 ∈ (ℙ ∖ {2}) ∧ 𝑃 ≠ 𝑄) → ((𝑃 /_L 𝑄) · (𝑄 /_L 𝑃)) = (-1↑(((𝑃 − 1) / 2) · ((𝑄 − 1) / 2))))

Theorem

lgsquad2lem1 15430

Lemma for lgsquad2 15432. (Contributed by Mario Carneiro, 19-Jun-2015.)

⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝑀)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝑁)    &   ⊢ (𝜑 → (𝑀 gcd 𝑁) = 1)    &   ⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → (𝐴 · 𝐵) = 𝑀)    &   ⊢ (𝜑 → ((𝐴 /_L 𝑁) · (𝑁 /_L 𝐴)) = (-1↑(((𝐴 − 1) / 2) · ((𝑁 − 1) / 2))))    &   ⊢ (𝜑 → ((𝐵 /_L 𝑁) · (𝑁 /_L 𝐵)) = (-1↑(((𝐵 − 1) / 2) · ((𝑁 − 1) / 2))))    ⇒   ⊢ (𝜑 → ((𝑀 /_L 𝑁) · (𝑁 /_L 𝑀)) = (-1↑(((𝑀 − 1) / 2) · ((𝑁 − 1) / 2))))

Theorem

lgsquad2lem2 15431*

Lemma for lgsquad2 15432. (Contributed by Mario Carneiro, 19-Jun-2015.)

⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝑀)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝑁)    &   ⊢ (𝜑 → (𝑀 gcd 𝑁) = 1)    &   ⊢ ((𝜑 ∧ (𝑚 ∈ (ℙ ∖ {2}) ∧ (𝑚 gcd 𝑁) = 1)) → ((𝑚 /_L 𝑁) · (𝑁 /_L 𝑚)) = (-1↑(((𝑚 − 1) / 2) · ((𝑁 − 1) / 2))))    &   ⊢ (𝜓 ↔ ∀𝑥 ∈ (1...𝑘)((𝑥 gcd (2 · 𝑁)) = 1 → ((𝑥 /_L 𝑁) · (𝑁 /_L 𝑥)) = (-1↑(((𝑥 − 1) / 2) · ((𝑁 − 1) / 2)))))    ⇒   ⊢ (𝜑 → ((𝑀 /_L 𝑁) · (𝑁 /_L 𝑀)) = (-1↑(((𝑀 − 1) / 2) · ((𝑁 − 1) / 2))))

Theorem

lgsquad2 15432

Extend lgsquad 15429 to coprime odd integers (the domain of the Jacobi symbol). (Contributed by Mario Carneiro, 19-Jun-2015.)

⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝑀)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝑁)    &   ⊢ (𝜑 → (𝑀 gcd 𝑁) = 1)    ⇒   ⊢ (𝜑 → ((𝑀 /_L 𝑁) · (𝑁 /_L 𝑀)) = (-1↑(((𝑀 − 1) / 2) · ((𝑁 − 1) / 2))))

Theorem

lgsquad3 15433

Extend lgsquad2 15432 to integers which share a factor. (Contributed by Mario Carneiro, 19-Jun-2015.)

⊢ (((𝑀 ∈ ℕ ∧ ¬ 2 ∥ 𝑀) ∧ (𝑁 ∈ ℕ ∧ ¬ 2 ∥ 𝑁)) → (𝑀 /_L 𝑁) = ((-1↑(((𝑀 − 1) / 2) · ((𝑁 − 1) / 2))) · (𝑁 /_L 𝑀)))

Theorem

m1lgs 15434

The first supplement to the law of quadratic reciprocity. Negative one is a square mod an odd prime 𝑃 iff 𝑃≡1 (mod 4). See first case of theorem 9.4 in [ApostolNT] p. 181. (Contributed by Mario Carneiro, 19-Jun-2015.)

⊢ (𝑃 ∈ (ℙ ∖ {2}) → ((-1 /_L 𝑃) = 1 ↔ (𝑃 mod 4) = 1))

Theorem

2lgslem1a1 15435*

Lemma 1 for 2lgslem1a 15437. (Contributed by AV, 16-Jun-2021.)

⊢ ((𝑃 ∈ ℕ ∧ ¬ 2 ∥ 𝑃) → ∀𝑖 ∈ (1...((𝑃 − 1) / 2))(𝑖 · 2) = ((𝑖 · 2) mod 𝑃))

Theorem

2lgslem1a2 15436

Lemma 2 for 2lgslem1a 15437. (Contributed by AV, 18-Jun-2021.)

⊢ ((𝑁 ∈ ℤ ∧ 𝐼 ∈ ℤ) → ((⌊‘(𝑁 / 4)) < 𝐼 ↔ (𝑁 / 2) < (𝐼 · 2)))

Theorem

2lgslem1a 15437*

Lemma 1 for 2lgslem1 15440. (Contributed by AV, 18-Jun-2021.)

⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → {𝑥 ∈ ℤ ∣ ∃𝑖 ∈ (1...((𝑃 − 1) / 2))(𝑥 = (𝑖 · 2) ∧ (𝑃 / 2) < (𝑥 mod 𝑃))} = {𝑥 ∈ ℤ ∣ ∃𝑖 ∈ (((⌊‘(𝑃 / 4)) + 1)...((𝑃 − 1) / 2))𝑥 = (𝑖 · 2)})

Theorem

2lgslem1b 15438*

Lemma 2 for 2lgslem1 15440. (Contributed by AV, 18-Jun-2021.)

⊢ 𝐼 = (𝐴...𝐵)    &   ⊢ 𝐹 = (𝑗 ∈ 𝐼 ↦ (𝑗 · 2))    ⇒   ⊢ 𝐹:𝐼–1-1-onto→{𝑥 ∈ ℤ ∣ ∃𝑖 ∈ 𝐼 𝑥 = (𝑖 · 2)}

Theorem

2lgslem1c 15439

Lemma 3 for 2lgslem1 15440. (Contributed by AV, 19-Jun-2021.)

⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → (⌊‘(𝑃 / 4)) ≤ ((𝑃 − 1) / 2))

Theorem

2lgslem1 15440*

Lemma 1 for 2lgs 15453. (Contributed by AV, 19-Jun-2021.)

⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → (♯‘{𝑥 ∈ ℤ ∣ ∃𝑖 ∈ (1...((𝑃 − 1) / 2))(𝑥 = (𝑖 · 2) ∧ (𝑃 / 2) < (𝑥 mod 𝑃))}) = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4))))

Theorem

2lgslem2 15441

Lemma 2 for 2lgs 15453. (Contributed by AV, 20-Jun-2021.)

⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℙ ∧ ¬ 2 ∥ 𝑃) → 𝑁 ∈ ℤ)

Theorem

2lgslem3a 15442

Lemma for 2lgslem3a1 15446. (Contributed by AV, 14-Jul-2021.)

⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝐾 ∈ ℕ₀ ∧ 𝑃 = ((8 · 𝐾) + 1)) → 𝑁 = (2 · 𝐾))

Theorem

2lgslem3b 15443

Lemma for 2lgslem3b1 15447. (Contributed by AV, 16-Jul-2021.)

⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝐾 ∈ ℕ₀ ∧ 𝑃 = ((8 · 𝐾) + 3)) → 𝑁 = ((2 · 𝐾) + 1))

Theorem

2lgslem3c 15444

Lemma for 2lgslem3c1 15448. (Contributed by AV, 16-Jul-2021.)

⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝐾 ∈ ℕ₀ ∧ 𝑃 = ((8 · 𝐾) + 5)) → 𝑁 = ((2 · 𝐾) + 1))

Theorem

2lgslem3d 15445

Lemma for 2lgslem3d1 15449. (Contributed by AV, 16-Jul-2021.)

⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝐾 ∈ ℕ₀ ∧ 𝑃 = ((8 · 𝐾) + 7)) → 𝑁 = ((2 · 𝐾) + 2))

Theorem

2lgslem3a1 15446

Lemma 1 for 2lgslem3 15450. (Contributed by AV, 15-Jul-2021.)

⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 1) → (𝑁 mod 2) = 0)

Theorem

2lgslem3b1 15447

Lemma 2 for 2lgslem3 15450. (Contributed by AV, 16-Jul-2021.)

⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 3) → (𝑁 mod 2) = 1)

Theorem

2lgslem3c1 15448

Lemma 3 for 2lgslem3 15450. (Contributed by AV, 16-Jul-2021.)

⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 5) → (𝑁 mod 2) = 1)

Theorem

2lgslem3d1 15449

Lemma 4 for 2lgslem3 15450. (Contributed by AV, 15-Jul-2021.)

⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℕ ∧ (𝑃 mod 8) = 7) → (𝑁 mod 2) = 0)

Theorem

2lgslem3 15450

Lemma 3 for 2lgs 15453. (Contributed by AV, 16-Jul-2021.)

⊢ 𝑁 = (((𝑃 − 1) / 2) − (⌊‘(𝑃 / 4)))    ⇒   ⊢ ((𝑃 ∈ ℕ ∧ ¬ 2 ∥ 𝑃) → (𝑁 mod 2) = if((𝑃 mod 8) ∈ {1, 7}, 0, 1))

Theorem

2lgs2 15451

The Legendre symbol for 2 at 2 is 0. (Contributed by AV, 20-Jun-2021.)

⊢ (2 /_L 2) = 0

Theorem

2lgslem4 15452

Lemma 4 for 2lgs 15453: special case of 2lgs 15453 for 𝑃 = 2. (Contributed by AV, 20-Jun-2021.)

⊢ ((2 /_L 2) = 1 ↔ (2 mod 8) ∈ {1, 7})

Theorem

2lgs 15453

The second supplement to the law of quadratic reciprocity (for the Legendre symbol extended to arbitrary primes as second argument). Two is a square modulo a prime 𝑃 iff 𝑃≡±1 (mod 8), see first case of theorem 9.5 in [ApostolNT] p. 181. This theorem justifies our definition of (𝑁 /_L 2) (lgs2 15366) to some degree, by demanding that reciprocity extend to the case 𝑄 = 2. (Proposed by Mario Carneiro, 19-Jun-2015.) (Contributed by AV, 16-Jul-2021.)

⊢ (𝑃 ∈ ℙ → ((2 /_L 𝑃) = 1 ↔ (𝑃 mod 8) ∈ {1, 7}))

Theorem

2lgsoddprmlem1 15454

Lemma 1 for 2lgsoddprm . (Contributed by AV, 19-Jul-2021.)

⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝑁 = ((8 · 𝐴) + 𝐵)) → (((𝑁↑2) − 1) / 8) = (((8 · (𝐴↑2)) + (2 · (𝐴 · 𝐵))) + (((𝐵↑2) − 1) / 8)))

Theorem

2lgsoddprmlem2 15455

Lemma 2 for 2lgsoddprm . (Contributed by AV, 19-Jul-2021.)

⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁 ∧ 𝑅 = (𝑁 mod 8)) → (2 ∥ (((𝑁↑2) − 1) / 8) ↔ 2 ∥ (((𝑅↑2) − 1) / 8)))

Theorem

2lgsoddprmlem3a 15456

Lemma 1 for 2lgsoddprmlem3 15460. (Contributed by AV, 20-Jul-2021.)

⊢ (((1↑2) − 1) / 8) = 0

Theorem

2lgsoddprmlem3b 15457

Lemma 2 for 2lgsoddprmlem3 15460. (Contributed by AV, 20-Jul-2021.)

⊢ (((3↑2) − 1) / 8) = 1

Theorem

2lgsoddprmlem3c 15458

Lemma 3 for 2lgsoddprmlem3 15460. (Contributed by AV, 20-Jul-2021.)

⊢ (((5↑2) − 1) / 8) = 3

Theorem

2lgsoddprmlem3d 15459

Lemma 4 for 2lgsoddprmlem3 15460. (Contributed by AV, 20-Jul-2021.)

⊢ (((7↑2) − 1) / 8) = (2 · 3)

Theorem

2lgsoddprmlem3 15460

Lemma 3 for 2lgsoddprm . (Contributed by AV, 20-Jul-2021.)

⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁 ∧ 𝑅 = (𝑁 mod 8)) → (2 ∥ (((𝑅↑2) − 1) / 8) ↔ 𝑅 ∈ {1, 7}))

Theorem

2lgsoddprmlem4 15461

Lemma 4 for 2lgsoddprm . (Contributed by AV, 20-Jul-2021.)

⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → (2 ∥ (((𝑁↑2) − 1) / 8) ↔ (𝑁 mod 8) ∈ {1, 7}))

Theorem

2lgsoddprm 15462

The second supplement to the law of quadratic reciprocity for odd primes (common representation, see theorem 9.5 in [ApostolNT] p. 181): The Legendre symbol for 2 at an odd prime is minus one to the power of the square of the odd prime minus one divided by eight ((2 /_L 𝑃) = -1^(((P^2)-1)/8) ). (Contributed by AV, 20-Jul-2021.)

⊢ (𝑃 ∈ (ℙ ∖ {2}) → (2 /_L 𝑃) = (-1↑(((𝑃↑2) − 1) / 8)))

11.3.7  All primes 4n+1 are the sum of two squares

Theorem

2sqlem1 15463*

Lemma for 2sq . (Contributed by Mario Carneiro, 19-Jun-2015.)

⊢ 𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    ⇒   ⊢ (𝐴 ∈ 𝑆 ↔ ∃𝑥 ∈ ℤ[i] 𝐴 = ((abs‘𝑥)↑2))

Theorem

2sqlem2 15464*

Lemma for 2sq . (Contributed by Mario Carneiro, 19-Jun-2015.)

⊢ 𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    ⇒   ⊢ (𝐴 ∈ 𝑆 ↔ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝐴 = ((𝑥↑2) + (𝑦↑2)))

Theorem

mul2sq 15465

Fibonacci's identity (actually due to Diophantus). The product of two sums of two squares is also a sum of two squares. We can take advantage of Gaussian integers here to trivialize the proof. (Contributed by Mario Carneiro, 19-Jun-2015.)

⊢ 𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    ⇒   ⊢ ((𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆) → (𝐴 · 𝐵) ∈ 𝑆)

Theorem

2sqlem3 15466

Lemma for 2sqlem5 15468. (Contributed by Mario Carneiro, 20-Jun-2015.)

⊢ 𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℤ)    &   ⊢ (𝜑 → (𝑁 · 𝑃) = ((𝐴↑2) + (𝐵↑2)))    &   ⊢ (𝜑 → 𝑃 = ((𝐶↑2) + (𝐷↑2)))    &   ⊢ (𝜑 → 𝑃 ∥ ((𝐶 · 𝐵) + (𝐴 · 𝐷)))    ⇒   ⊢ (𝜑 → 𝑁 ∈ 𝑆)

Theorem

2sqlem4 15467

Lemma for 2sqlem5 15468. (Contributed by Mario Carneiro, 20-Jun-2015.)

⊢ 𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐶 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℤ)    &   ⊢ (𝜑 → (𝑁 · 𝑃) = ((𝐴↑2) + (𝐵↑2)))    &   ⊢ (𝜑 → 𝑃 = ((𝐶↑2) + (𝐷↑2)))    ⇒   ⊢ (𝜑 → 𝑁 ∈ 𝑆)

Theorem

2sqlem5 15468

Lemma for 2sq . If a number that is a sum of two squares is divisible by a prime that is a sum of two squares, then the quotient is a sum of two squares. (Contributed by Mario Carneiro, 20-Jun-2015.)

⊢ 𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → (𝑁 · 𝑃) ∈ 𝑆)    &   ⊢ (𝜑 → 𝑃 ∈ 𝑆)    ⇒   ⊢ (𝜑 → 𝑁 ∈ 𝑆)

Theorem

2sqlem6 15469*

Lemma for 2sq . If a number that is a sum of two squares is divisible by a number whose prime divisors are all sums of two squares, then the quotient is a sum of two squares. (Contributed by Mario Carneiro, 20-Jun-2015.)

⊢ 𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    &   ⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → ∀𝑝 ∈ ℙ (𝑝 ∥ 𝐵 → 𝑝 ∈ 𝑆))    &   ⊢ (𝜑 → (𝐴 · 𝐵) ∈ 𝑆)    ⇒   ⊢ (𝜑 → 𝐴 ∈ 𝑆)

Theorem

2sqlem7 15470*

Lemma for 2sq . (Contributed by Mario Carneiro, 19-Jun-2015.)

⊢ 𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    &   ⊢ 𝑌 = {𝑧 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ((𝑥 gcd 𝑦) = 1 ∧ 𝑧 = ((𝑥↑2) + (𝑦↑2)))}    ⇒   ⊢ 𝑌 ⊆ (𝑆 ∩ ℕ)

Theorem

2sqlem8a 15471*

Lemma for 2sqlem8 15472. (Contributed by Mario Carneiro, 4-Jun-2016.)

⊢ 𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    &   ⊢ 𝑌 = {𝑧 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ((𝑥 gcd 𝑦) = 1 ∧ 𝑧 = ((𝑥↑2) + (𝑦↑2)))}    &   ⊢ (𝜑 → ∀𝑏 ∈ (1...(𝑀 − 1))∀𝑎 ∈ 𝑌 (𝑏 ∥ 𝑎 → 𝑏 ∈ 𝑆))    &   ⊢ (𝜑 → 𝑀 ∥ 𝑁)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ_≥‘2))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → (𝐴 gcd 𝐵) = 1)    &   ⊢ (𝜑 → 𝑁 = ((𝐴↑2) + (𝐵↑2)))    &   ⊢ 𝐶 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐷 = (((𝐵 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    ⇒   ⊢ (𝜑 → (𝐶 gcd 𝐷) ∈ ℕ)

Theorem

2sqlem8 15472*

Lemma for 2sq . (Contributed by Mario Carneiro, 20-Jun-2015.)

⊢ 𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    &   ⊢ 𝑌 = {𝑧 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ((𝑥 gcd 𝑦) = 1 ∧ 𝑧 = ((𝑥↑2) + (𝑦↑2)))}    &   ⊢ (𝜑 → ∀𝑏 ∈ (1...(𝑀 − 1))∀𝑎 ∈ 𝑌 (𝑏 ∥ 𝑎 → 𝑏 ∈ 𝑆))    &   ⊢ (𝜑 → 𝑀 ∥ 𝑁)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ_≥‘2))    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → (𝐴 gcd 𝐵) = 1)    &   ⊢ (𝜑 → 𝑁 = ((𝐴↑2) + (𝐵↑2)))    &   ⊢ 𝐶 = (((𝐴 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐷 = (((𝐵 + (𝑀 / 2)) mod 𝑀) − (𝑀 / 2))    &   ⊢ 𝐸 = (𝐶 / (𝐶 gcd 𝐷))    &   ⊢ 𝐹 = (𝐷 / (𝐶 gcd 𝐷))    ⇒   ⊢ (𝜑 → 𝑀 ∈ 𝑆)

Theorem

2sqlem9 15473*

Lemma for 2sq . (Contributed by Mario Carneiro, 19-Jun-2015.)

⊢ 𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    &   ⊢ 𝑌 = {𝑧 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ((𝑥 gcd 𝑦) = 1 ∧ 𝑧 = ((𝑥↑2) + (𝑦↑2)))}    &   ⊢ (𝜑 → ∀𝑏 ∈ (1...(𝑀 − 1))∀𝑎 ∈ 𝑌 (𝑏 ∥ 𝑎 → 𝑏 ∈ 𝑆))    &   ⊢ (𝜑 → 𝑀 ∥ 𝑁)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → 𝑁 ∈ 𝑌)    ⇒   ⊢ (𝜑 → 𝑀 ∈ 𝑆)

Theorem

2sqlem10 15474*

Lemma for 2sq . Every factor of a "proper" sum of two squares (where the summands are coprime) is a sum of two squares. (Contributed by Mario Carneiro, 19-Jun-2015.)

⊢ 𝑆 = ran (𝑤 ∈ ℤ[i] ↦ ((abs‘𝑤)↑2))    &   ⊢ 𝑌 = {𝑧 ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ ((𝑥 gcd 𝑦) = 1 ∧ 𝑧 = ((𝑥↑2) + (𝑦↑2)))}    ⇒   ⊢ ((𝐴 ∈ 𝑌 ∧ 𝐵 ∈ ℕ ∧ 𝐵 ∥ 𝐴) → 𝐵 ∈ 𝑆)

PART 12  GUIDES AND MISCELLANEA

12.1  Guides (conventions, explanations, and examples)

12.1.1  Conventions

This section describes the conventions we use. These conventions often refer to existing mathematical practices, which are discussed in more detail in other references. The following sources lay out how mathematics is developed without the law of the excluded middle. Of course, there are a greater number of sources which assume excluded middle and most of what is in them applies here too (especially in a treatment such as ours which is built on first-order logic and set theory, rather than, say, type theory). Studying how a topic is treated in the Metamath Proof Explorer and the references therein is often a good place to start (and is easy to compare with the Intuitionistic Logic Explorer). The textbooks provide a motivation for what we are doing, whereas Metamath lets you see in detail all hidden and implicit steps. Most standard theorems are accompanied by citations. Some closely followed texts include the following:

• Axioms of propositional calculus - Stanford Encyclopedia of Philosophy or [Heyting].
• Axioms of predicate calculus - our axioms are adapted from the ones in the Metamath Proof Explorer.
• Theorems of propositional calculus - [Heyting].
• Theorems of pure predicate calculus - Metamath Proof Explorer.
• Theorems of equality and substitution - Metamath Proof Explorer.
• Axioms of set theory - [Crosilla].
• Development of set theory - Chapter 10 of [HoTT].
• Construction of real and complex numbers - Chapter 11 of [HoTT]; [BauerTaylor].
• Theorems about real numbers - [Geuvers].

Theorem

conventions 15475

Unless there is a reason to diverge, we follow the conventions of the Metamath Proof Explorer (MPE, set.mm). This list of conventions is intended to be read in conjunction with the corresponding conventions in the Metamath Proof Explorer, and only the differences are described below.

• Minimizing axioms and the axiom of choice. We prefer proofs that depend on fewer and/or weaker axioms, even if the proofs are longer. In particular, our choice of IZF (Intuitionistic Zermelo-Fraenkel) over CZF (Constructive Zermelo-Fraenkel, a weaker system) was just an expedient choice because IZF is easier to formalize in Metamath. You can find some development using CZF in BJ's mathbox starting at wbd 15566 (and the section header just above it). As for the axiom of choice, the full axiom of choice implies excluded middle as seen at acexmid 5924, although some authors will use countable choice or dependent choice. For example, countable choice or excluded middle is needed to show that the Cauchy reals coincide with the Dedekind reals - Corollary 11.4.3 of [HoTT], p. (varies).
• Junk/undefined results. Much of the discussion of this topic in the Metamath Proof Explorer applies except that certain techniques are not available to us. For example, the Metamath Proof Explorer will often say "if a function is evaluated within its domain, a certain result follows; if the function is evaluated outside its domain, the same result follows. Since the function must be evaluated within its domain or outside it, the result follows unconditionally" (the use of excluded middle in this argument is perhaps obvious when stated this way). Often, the easiest fix will be to prove we are evaluating functions within their domains, other times it will be possible to use a theorem like relelfvdm 5593 which says that if a function value produces an inhabited set, then the function is being evaluated within its domain.
• Bibliography references. The bibliography for the Intuitionistic Logic Explorer is separate from the one for the Metamath Proof Explorer but feel free to copy-paste a citation in either direction in order to cite it.

Label naming conventions

Here are a few of the label naming conventions:

• Suffixes. We follow the conventions of the Metamath Proof Explorer with a few additions. A biconditional in set.mm which is an implication in iset.mm should have a "r" (for the reverse direction), or "i"/"im" (for the forward direction) appended. A theorem in set.mm which has a decidability condition added should add "dc" to the theorem name. A theorem in set.mm where "nonempty class" is changed to "inhabited class" should add "m" (for member) to the theorem name.
• iset.mm versus set.mm names

  Theorems which are the same as in set.mm should be named the same (that is, where the statement of the theorem is the same; the proof can differ without a new name being called for). Theorems which are different should be named differently (we do have a small number of intentional exceptions to this rule but on the whole it serves us well).

  As for how to choose names so they are different between iset.mm and set.mm, when possible choose a name which reflect the difference in the theorems. For example, if a theorem in set.mm is an equality and the iset.mm analogue is a subset, add "ss" to the iset.mm name. If need be, add "i" to the iset.mm name (usually as a prefix to some portion of the name).

  As with set.mm, we welcome suggestions for better names (such as names which are more consistent with naming conventions).

  We do try to keep set.mm and iset.mm similar where we can. For example, if a theorem exists in both places but the name in set.mm isn't great, we tend to keep that name for iset.mm, or change it in both files together. This is mainly to make it easier to copy theorems, but also to generally help people browse proofs, find theorems, write proofs, etc.

The following table shows some commonly-used abbreviations in labels which are not found in the Metamath Proof Explorer, in alphabetical order. For each abbreviation we provide a mnenomic to help you remember it, the source theorem/assumption defining it, an expression showing what it looks like, whether or not it is a "syntax fragment" (an abbreviation that indicates a particular kind of syntax), and hyperlinks to label examples that use the abbreviation. The abbreviation is bolded if there is a df-NAME definition but the label fragment is not NAME.

For the "g" abbreviation, this is related to the set.mm usage, in which "is a set" conditions are converted from hypotheses to antecedents, but is also used where "is a set" conditions are added relative to similar set.mm theorems.

Abbreviation	Mnenomic/Meaning	Source	Expression	Syntax?	Example(s)
ap	apart	df-pap 7333, df-ap 8628		Yes	apadd1 8654, apne 8669
g	with "is a set" condition			No	1stvalg 6209, brtposg 6321, setsmsbasg 14823
m	inhabited (from "member")		∃𝑥𝑥 ∈ 𝐴	No	r19.2m 3538, negm 9708, ctm 7184, basmex 12764
seq3, sum3	recursive sequence	df-seqfrec 10559		Yes	seq3-1 10573, fsum3 11571
tap	tight apartness	df-tap 7335		Yes	df-tap 7335

• Community. The Metamath mailing list also covers the Intuitionistic Logic Explorer and is at: https://groups.google.com/g/metamath 7335.

(Contributed by Jim Kingdon, 24-Feb-2020.) (New usage is discouraged.)

⊢ 𝜑    ⇒   ⊢ 𝜑

12.1.2  Definitional examples

Theorem

ex-or 15476

Example for ax-io 710. Example by David A. Wheeler. (Contributed by Mario Carneiro, 9-May-2015.)

⊢ (2 = 3 ∨ 4 = 4)

Theorem

ex-an 15477

Example for ax-ia1 106. Example by David A. Wheeler. (Contributed by Mario Carneiro, 9-May-2015.)

⊢ (2 = 2 ∧ 3 = 3)

Theorem

1kp2ke3k 15478

Example for df-dec 9477, 1000 + 2000 = 3000.

This proof disproves (by counterexample) the assertion of Hao Wang, who stated, "There is a theorem in the primitive notation of set theory that corresponds to the arithmetic theorem 1000 + 2000 = 3000. The formula would be forbiddingly long... even if (one) knows the definitions and is asked to simplify the long formula according to them, chances are he will make errors and arrive at some incorrect result." (Hao Wang, "Theory and practice in mathematics" , In Thomas Tymoczko, editor, New Directions in the Philosophy of Mathematics, pp 129-152, Birkauser Boston, Inc., Boston, 1986. (QA8.6.N48). The quote itself is on page 140.)

This is noted in Metamath: A Computer Language for Pure Mathematics by Norman Megill (2007) section 1.1.3. Megill then states, "A number of writers have conveyed the impression that the kind of absolute rigor provided by Metamath is an impossible dream, suggesting that a complete, formal verification of a typical theorem would take millions of steps in untold volumes of books... These writers assume, however, that in order to achieve the kind of complete formal verification they desire one must break down a proof into individual primitive steps that make direct reference to the axioms. This is not necessary. There is no reason not to make use of previously proved theorems rather than proving them over and over... A hierarchy of theorems and definitions permits an exponential growth in the formula sizes and primitive proof steps to be described with only a linear growth in the number of symbols used. Of course, this is how ordinary informal mathematics is normally done anyway, but with Metamath it can be done with absolute rigor and precision."

The proof here starts with (2 + 1) = 3, commutes it, and repeatedly multiplies both sides by ten. This is certainly longer than traditional mathematical proofs, e.g., there are a number of steps explicitly shown here to show that we're allowed to do operations such as multiplication. However, while longer, the proof is clearly a manageable size - even though every step is rigorously derived all the way back to the primitive notions of set theory and logic. And while there's a risk of making errors, the many independent verifiers make it much less likely that an incorrect result will be accepted.

This proof heavily relies on the decimal constructor df-dec 9477 developed by Mario Carneiro in 2015. The underlying Metamath language has an intentionally very small set of primitives; it doesn't even have a built-in construct for numbers. Instead, the digits are defined using these primitives, and the decimal constructor is used to make it easy to express larger numbers as combinations of digits.

(Contributed by David A. Wheeler, 29-Jun-2016.) (Shortened by Mario Carneiro using the arithmetic algorithm in mmj2, 30-Jun-2016.)

⊢ (;;;1000 + ;;;2000) = ;;;3000

Theorem

ex-fl 15479

Example for df-fl 10379. Example by David A. Wheeler. (Contributed by Mario Carneiro, 18-Jun-2015.)

⊢ ((⌊‘(3 / 2)) = 1 ∧ (⌊‘-(3 / 2)) = -2)

Theorem

ex-ceil 15480

Example for df-ceil 10380. (Contributed by AV, 4-Sep-2021.)

⊢ ((⌈‘(3 / 2)) = 2 ∧ (⌈‘-(3 / 2)) = -1)

Theorem

ex-exp 15481

Example for df-exp 10650. (Contributed by AV, 4-Sep-2021.)

⊢ ((5↑2) = ;25 ∧ (-3↑-2) = (1 / 9))

Theorem

ex-fac 15482

Example for df-fac 10837. (Contributed by AV, 4-Sep-2021.)

⊢ (!‘5) = ;;120

Theorem

ex-bc 15483

Example for df-bc 10859. (Contributed by AV, 4-Sep-2021.)

⊢ (5C3) = ;10

Theorem

ex-dvds 15484

Example for df-dvds 11972: 3 divides into 6. (Contributed by David A. Wheeler, 19-May-2015.)

⊢ 3 ∥ 6

Theorem

ex-gcd 15485

Example for df-gcd 12148. (Contributed by AV, 5-Sep-2021.)

⊢ (-6 gcd 9) = 3

PART 13  SUPPLEMENTARY MATERIAL (USERS' MATHBOXES)

13.1  Mathboxes for user contributions

13.1.1  Mathbox guidelines

Theorem

mathbox 15486

(This theorem is a dummy placeholder for these guidelines. The label of this theorem, "mathbox", is hard-coded into the Metamath program to identify the start of the mathbox section for web page generation.)

A "mathbox" is a user-contributed section that is maintained by its contributor independently from the main part of iset.mm.

For contributors:

By making a contribution, you agree to release it into the public domain, according to the statement at the beginning of iset.mm.

Guidelines:

Mathboxes in iset.mm follow the same practices as in set.mm, so refer to the mathbox guidelines there for more details.

(Contributed by NM, 20-Feb-2007.) (Revised by the Metamath team, 9-Sep-2023.) (New usage is discouraged.)

⊢ 𝜑    ⇒   ⊢ 𝜑

13.2  Mathbox for BJ

13.2.1  Propositional calculus

Theorem

bj-nnsn 15487

As far as implying a negated formula is concerned, a formula is equivalent to its double negation. (Contributed by BJ, 24-Nov-2023.)

⊢ ((𝜑 → ¬ 𝜓) ↔ (¬ ¬ 𝜑 → ¬ 𝜓))

Theorem

bj-nnor 15488

Double negation of a disjunction in terms of implication. (Contributed by BJ, 9-Oct-2019.)

⊢ (¬ ¬ (𝜑 ∨ 𝜓) ↔ (¬ 𝜑 → ¬ ¬ 𝜓))

Theorem

bj-nnim 15489

The double negation of an implication implies the implication with the consequent doubly negated. (Contributed by BJ, 24-Nov-2023.)

⊢ (¬ ¬ (𝜑 → 𝜓) → (𝜑 → ¬ ¬ 𝜓))

Theorem

bj-nnan 15490

The double negation of a conjunction implies the conjunction of the double negations. (Contributed by BJ, 24-Nov-2023.)

⊢ (¬ ¬ (𝜑 ∧ 𝜓) → (¬ ¬ 𝜑 ∧ ¬ ¬ 𝜓))

Theorem

bj-nnclavius 15491

Clavius law with doubly negated consequent. (Contributed by BJ, 4-Dec-2023.)

⊢ ((¬ 𝜑 → 𝜑) → ¬ ¬ 𝜑)

Theorem

bj-imnimnn 15492

If a formula is implied by both a formula and its negation, then it is not refutable. There is another proof using the inference associated with bj-nnclavius 15491 as its last step. (Contributed by BJ, 27-Oct-2024.)

⊢ (𝜑 → 𝜓)    &   ⊢ (¬ 𝜑 → 𝜓)    ⇒   ⊢ ¬ ¬ 𝜓

13.2.1.1  Stable formulas

Some of the following theorems, like bj-sttru 15494 or bj-stfal 15496 could be deduced from their analogues for decidability, but stability is not provable from decidability in minimal calculus, so direct proofs have their interest.

Theorem

bj-trst 15493

A provable formula is stable. (Contributed by BJ, 24-Nov-2023.)

⊢ (𝜑 → STAB 𝜑)

Theorem

bj-sttru 15494

The true truth value is stable. (Contributed by BJ, 5-Aug-2024.)

⊢ STAB ⊤

Theorem

bj-fast 15495

A refutable formula is stable. (Contributed by BJ, 24-Nov-2023.)

⊢ (¬ 𝜑 → STAB 𝜑)

Theorem

bj-stfal 15496

The false truth value is stable. (Contributed by BJ, 5-Aug-2024.)

⊢ STAB ⊥

Theorem

bj-nnst 15497

Double negation of stability of a formula. Intuitionistic logic refutes unstability (but does not prove stability) of any formula. This theorem can also be proved in classical refutability calculus (see https://us.metamath.org/mpeuni/bj-peircestab.html) but not in minimal calculus (see https://us.metamath.org/mpeuni/bj-stabpeirce.html). See nnnotnotr 15744 for the version not using the definition of stability. (Contributed by BJ, 9-Oct-2019.) Prove it in ( → , ¬ ) -intuitionistic calculus with definitions (uses of ax-ia1 106, ax-ia2 107, ax-ia3 108 are via sylibr 134, necessary for definition unpackaging), and in ( → , ↔ , ¬ )-intuitionistic calculus, following a discussion with Jim Kingdon. (Revised by BJ, 27-Oct-2024.)

⊢ ¬ ¬ STAB 𝜑

Theorem

bj-nnbist 15498

If a formula is not refutable, then it is stable if and only if it is provable. By double-negation translation, if 𝜑 is a classical tautology, then ¬ ¬ 𝜑 is an intuitionistic tautology. Therefore, if 𝜑 is a classical tautology, then 𝜑 is intuitionistically equivalent to its stability (and to its decidability, see bj-nnbidc 15511). (Contributed by BJ, 24-Nov-2023.)

⊢ (¬ ¬ 𝜑 → (STAB 𝜑 ↔ 𝜑))

Theorem

bj-stst 15499

Stability of a proposition is stable if and only if that proposition is stable. STAB is idempotent. (Contributed by BJ, 9-Oct-2019.)

⊢ (STAB STAB 𝜑 ↔ STAB 𝜑)

Theorem

bj-stim 15500

A conjunction with a stable consequent is stable. See stabnot 834 for negation , bj-stan 15501 for conjunction , and bj-stal 15503 for universal quantification. (Contributed by BJ, 24-Nov-2023.)

⊢ (STAB 𝜓 → STAB (𝜑 → 𝜓))