P. 163 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 16201-16300   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	ruclem3 16201*	Lemma for ruc 16211. The constructed interval [𝑋, 𝑌] always excludes 𝑀. (Contributed by Mario Carneiro, 28-May-2014.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, ⟨(1st ‘𝑥), 𝑚⟩, ⟨((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)⟩)))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 𝐵 ∈ ℝ)    &   ⊢ (𝜑 → 𝑀 ∈ ℝ)    &   ⊢ 𝑋 = (1st ‘(⟨𝐴, 𝐵⟩𝐷𝑀))    &   ⊢ 𝑌 = (2nd ‘(⟨𝐴, 𝐵⟩𝐷𝑀))    &   ⊢ (𝜑 → 𝐴 < 𝐵)    ⇒   ⊢ (𝜑 → (𝑀 < 𝑋 ∨ 𝑌 < 𝑀))
Theorem	ruclem4 16202*	Lemma for ruc 16211. Initial value of the interval sequence. (Contributed by Mario Carneiro, 28-May-2014.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, ⟨(1st ‘𝑥), 𝑚⟩, ⟨((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)⟩)))    &   ⊢ 𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   ⊢ 𝐺 = seq0(𝐷, 𝐶)    ⇒   ⊢ (𝜑 → (𝐺‘0) = ⟨0, 1⟩)
Theorem	ruclem6 16203*	Lemma for ruc 16211. Domain and codomain of the interval sequence. (Contributed by Mario Carneiro, 28-May-2014.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, ⟨(1st ‘𝑥), 𝑚⟩, ⟨((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)⟩)))    &   ⊢ 𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   ⊢ 𝐺 = seq0(𝐷, 𝐶)    ⇒   ⊢ (𝜑 → 𝐺:ℕ0⟶(ℝ × ℝ))
Theorem	ruclem7 16204*	Lemma for ruc 16211. Successor value for the interval sequence. (Contributed by Mario Carneiro, 28-May-2014.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, ⟨(1st ‘𝑥), 𝑚⟩, ⟨((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)⟩)))    &   ⊢ 𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   ⊢ 𝐺 = seq0(𝐷, 𝐶)    ⇒   ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ0) → (𝐺‘(𝑁 + 1)) = ((𝐺‘𝑁)𝐷(𝐹‘(𝑁 + 1))))
Theorem	ruclem8 16205*	Lemma for ruc 16211. The intervals of the 𝐺 sequence are all nonempty. (Contributed by Mario Carneiro, 28-May-2014.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, ⟨(1st ‘𝑥), 𝑚⟩, ⟨((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)⟩)))    &   ⊢ 𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   ⊢ 𝐺 = seq0(𝐷, 𝐶)    ⇒   ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ0) → (1st ‘(𝐺‘𝑁)) < (2nd ‘(𝐺‘𝑁)))
Theorem	ruclem9 16206*	Lemma for ruc 16211. The first components of the 𝐺 sequence are increasing, and the second components are decreasing. (Contributed by Mario Carneiro, 28-May-2014.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, ⟨(1st ‘𝑥), 𝑚⟩, ⟨((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)⟩)))    &   ⊢ 𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   ⊢ 𝐺 = seq0(𝐷, 𝐶)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ0)    &   ⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀))    ⇒   ⊢ (𝜑 → ((1st ‘(𝐺‘𝑀)) ≤ (1st ‘(𝐺‘𝑁)) ∧ (2nd ‘(𝐺‘𝑁)) ≤ (2nd ‘(𝐺‘𝑀))))
Theorem	ruclem10 16207*	Lemma for ruc 16211. Every first component of the 𝐺 sequence is less than every second component. That is, the sequences form a chain a1 < a2 <... < b2 < b1, where ai are the first components and bi are the second components. (Contributed by Mario Carneiro, 28-May-2014.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, ⟨(1st ‘𝑥), 𝑚⟩, ⟨((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)⟩)))    &   ⊢ 𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   ⊢ 𝐺 = seq0(𝐷, 𝐶)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ0)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (1st ‘(𝐺‘𝑀)) < (2nd ‘(𝐺‘𝑁)))
Theorem	ruclem11 16208*	Lemma for ruc 16211. Closure lemmas for supremum. (Contributed by Mario Carneiro, 28-May-2014.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, ⟨(1st ‘𝑥), 𝑚⟩, ⟨((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)⟩)))    &   ⊢ 𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   ⊢ 𝐺 = seq0(𝐷, 𝐶)    ⇒   ⊢ (𝜑 → (ran (1st ∘ 𝐺) ⊆ ℝ ∧ ran (1st ∘ 𝐺) ≠ ∅ ∧ ∀𝑧 ∈ ran (1st ∘ 𝐺)𝑧 ≤ 1))
Theorem	ruclem12 16209*	Lemma for ruc 16211. The supremum of the increasing sequence 1st ∘ 𝐺 is a real number that is not in the range of 𝐹. (Contributed by Mario Carneiro, 28-May-2014.)
⊢ (𝜑 → 𝐹:ℕ⟶ℝ)    &   ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, ⟨(1st ‘𝑥), 𝑚⟩, ⟨((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)⟩)))    &   ⊢ 𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   ⊢ 𝐺 = seq0(𝐷, 𝐶)    &   ⊢ 𝑆 = sup(ran (1st ∘ 𝐺), ℝ, < )    ⇒   ⊢ (𝜑 → 𝑆 ∈ (ℝ ∖ ran 𝐹))
Theorem	ruclem13 16210	Lemma for ruc 16211. There is no function that maps ℕ onto ℝ. (Use nex 1795 if you want this in the form ¬ ∃𝑓𝑓:ℕ–onto→ℝ.) (Contributed by NM, 14-Oct-2004.) (Proof shortened by Fan Zheng, 6-Jun-2016.)
⊢ ¬ 𝐹:ℕ–onto→ℝ
Theorem	ruc 16211	The set of positive integers is strictly dominated by the set of real numbers, i.e. the real numbers are uncountable. The proof consists of lemmas ruclem1 16199 through ruclem13 16210 and this final piece. Our proof is based on the proof of Theorem 5.18 of [Truss] p. 114. See ruclem13 16210 for the function existence version of this theorem. For an informal discussion of this proof, see mmcomplex.html#uncountable 16210. For an alternate proof see rucALT 16198. This is Metamath 100 proof #22. (Contributed by NM, 13-Oct-2004.)
⊢ ℕ ≺ ℝ
Theorem	resdomq 16212	The set of rationals is strictly less equinumerous than the set of reals (ℝ strictly dominates ℚ). (Contributed by NM, 18-Dec-2004.)
⊢ ℚ ≺ ℝ
Theorem	aleph1re 16213	There are at least aleph-one real numbers. (Contributed by NM, 2-Feb-2005.)
⊢ (ℵ‘1o) ≼ ℝ
Theorem	aleph1irr 16214	There are at least aleph-one irrationals. (Contributed by NM, 2-Feb-2005.)
⊢ (ℵ‘1o) ≼ (ℝ ∖ ℚ)
Theorem	cnso 16215	The complex numbers can be linearly ordered. (Contributed by Stefan O'Rear, 16-Nov-2014.)
⊢ ∃𝑥 𝑥 Or ℂ
PART 6  ELEMENTARY NUMBER THEORY Here we introduce elementary number theory, in particular the elementary properties of divisibility and elementary prime number theory.
6.1  Elementary properties of divisibility
6.1.1  Irrationality of square root of 2
Theorem	sqrt2irrlem 16216	Lemma for sqrt2irr 16217. This is the core of the proof: if 𝐴 / 𝐵 = √(2), then 𝐴 and 𝐵 are even, so 𝐴 / 2 and 𝐵 / 2 are smaller representatives, which is absurd by the method of infinite descent (here implemented by strong induction). This is Metamath 100 proof #1. (Contributed by NM, 20-Aug-2001.) (Revised by Mario Carneiro, 12-Sep-2015.) (Proof shortened by JV, 4-Jan-2022.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → (√‘2) = (𝐴 / 𝐵))    ⇒   ⊢ (𝜑 → ((𝐴 / 2) ∈ ℤ ∧ (𝐵 / 2) ∈ ℕ))
Theorem	sqrt2irr 16217	The square root of 2 is irrational. See zsqrtelqelz 16721 for a generalization to all non-square integers. The proof's core is proven in sqrt2irrlem 16216, which shows that if 𝐴 / 𝐵 = √(2), then 𝐴 and 𝐵 are even, so 𝐴 / 2 and 𝐵 / 2 are smaller representatives, which is absurd. An older version of this proof was included in The Seventeen Provers of the World compiled by Freek Wiedijk. It is also the first of the "top 100" mathematical theorems whose formalization is tracked by Freek Wiedijk on his Formalizing 100 Theorems page at http://www.cs.ru.nl/~freek/100/ 16216. (Contributed by NM, 8-Jan-2002.) (Proof shortened by Mario Carneiro, 12-Sep-2015.)
⊢ (√‘2) ∉ ℚ
Theorem	sqrt2re 16218	The square root of 2 exists and is a real number. (Contributed by NM, 3-Dec-2004.)
⊢ (√‘2) ∈ ℝ
Theorem	sqrt2irr0 16219	The square root of 2 is an irrational number. (Contributed by AV, 23-Dec-2022.)
⊢ (√‘2) ∈ (ℝ ∖ ℚ)
6.1.2  Some Number sets are chains of proper subsets
Theorem	nthruc 16220	The sequence ℕ, ℤ, ℚ, ℝ, and ℂ forms a chain of proper subsets. In each case the proper subset relationship is shown by demonstrating a number that belongs to one set but not the other. We show that zero belongs to ℤ but not ℕ, one-half belongs to ℚ but not ℤ, the square root of 2 belongs to ℝ but not ℚ, and finally that the imaginary number i belongs to ℂ but not ℝ. See nthruz 16221 for a further refinement. (Contributed by NM, 12-Jan-2002.)
⊢ ((ℕ ⊊ ℤ ∧ ℤ ⊊ ℚ) ∧ (ℚ ⊊ ℝ ∧ ℝ ⊊ ℂ))
Theorem	nthruz 16221	The sequence ℕ, ℕ0, and ℤ forms a chain of proper subsets. In each case the proper subset relationship is shown by demonstrating a number that belongs to one set but not the other. We show that zero belongs to ℕ0 but not ℕ and minus one belongs to ℤ but not ℕ0. This theorem refines the chain of proper subsets nthruc 16220. (Contributed by NM, 9-May-2004.)
⊢ (ℕ ⊊ ℕ0 ∧ ℕ0 ⊊ ℤ)
6.1.3  The divides relation
Syntax	cdvds 16222	Extend the definition of a class to include the divides relation. See df-dvds 16223.
class ∥
Definition	df-dvds 16223*	Define the divides relation, see definition in [ApostolNT] p. 14. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ∥ = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ ℤ ∧ 𝑦 ∈ ℤ) ∧ ∃𝑛 ∈ ℤ (𝑛 · 𝑥) = 𝑦)}
Theorem	divides 16224*	Define the divides relation. 𝑀 ∥ 𝑁 means 𝑀 divides into 𝑁 with no remainder. For example, 3 ∥ 6 (ex-dvds 30253). As proven in dvdsval3 16226, 𝑀 ∥ 𝑁 ↔ (𝑁 mod 𝑀) = 0. See divides 16224 and dvdsval2 16225 for other equivalent expressions. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ ∃𝑛 ∈ ℤ (𝑛 · 𝑀) = 𝑁))
Theorem	dvdsval2 16225	One nonzero integer divides another integer if and only if their quotient is an integer. (Contributed by Jeff Hankins, 29-Sep-2013.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑀 ≠ 0 ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ (𝑁 / 𝑀) ∈ ℤ))
Theorem	dvdsval3 16226	One nonzero integer divides another integer if and only if the remainder upon division is zero, see remark in [ApostolNT] p. 106. (Contributed by Mario Carneiro, 22-Feb-2014.) (Revised by Mario Carneiro, 15-Jul-2014.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ (𝑁 mod 𝑀) = 0))
Theorem	dvdszrcl 16227	Reverse closure for the divisibility relation. (Contributed by Stefan O'Rear, 5-Sep-2015.)
⊢ (𝑋 ∥ 𝑌 → (𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ))
Theorem	dvdsmod0 16228	If a positive integer divides another integer, then the remainder upon division is zero. (Contributed by AV, 3-Mar-2022.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑀 ∥ 𝑁) → (𝑁 mod 𝑀) = 0)
Theorem	p1modz1 16229	If a number greater than 1 divides another number, the second number increased by 1 is 1 modulo the first number. (Contributed by AV, 19-Mar-2022.)
⊢ ((𝑀 ∥ 𝐴 ∧ 1 < 𝑀) → ((𝐴 + 1) mod 𝑀) = 1)
Theorem	dvdsmodexp 16230	If a positive integer divides another integer, this other integer is equal to its positive powers modulo the positive integer. (Formerly part of the proof for fermltl 16744). (Contributed by Mario Carneiro, 28-Feb-2014.) (Revised by AV, 19-Mar-2022.)
⊢ ((𝑁 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝑁 ∥ 𝐴) → ((𝐴↑𝐵) mod 𝑁) = (𝐴 mod 𝑁))
Theorem	nndivdvds 16231	Strong form of dvdsval2 16225 for positive integers. (Contributed by Stefan O'Rear, 13-Sep-2014.)
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (𝐵 ∥ 𝐴 ↔ (𝐴 / 𝐵) ∈ ℕ))
Theorem	nndivides 16232*	Definition of the divides relation for positive integers. (Contributed by AV, 26-Jul-2021.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 ∥ 𝑁 ↔ ∃𝑛 ∈ ℕ (𝑛 · 𝑀) = 𝑁))
Theorem	moddvds 16233	Two ways to say 𝐴≡𝐵 (mod 𝑁), see also definition in [ApostolNT] p. 106. (Contributed by Mario Carneiro, 18-Feb-2014.)
⊢ ((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ((𝐴 mod 𝑁) = (𝐵 mod 𝑁) ↔ 𝑁 ∥ (𝐴 − 𝐵)))
Theorem	modm1div 16234	An integer greater than one divides another integer minus one iff the second integer modulo the first integer is one. (Contributed by AV, 30-May-2023.)
⊢ ((𝑁 ∈ (ℤ≥‘2) ∧ 𝐴 ∈ ℤ) → ((𝐴 mod 𝑁) = 1 ↔ 𝑁 ∥ (𝐴 − 1)))
Theorem	dvds0lem 16235	A lemma to assist theorems of ∥ with no antecedents. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ (𝐾 · 𝑀) = 𝑁) → 𝑀 ∥ 𝑁)
Theorem	dvds1lem 16236*	A lemma to assist theorems of ∥ with one antecedent. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (𝜑 → (𝐽 ∈ ℤ ∧ 𝐾 ∈ ℤ))    &   ⊢ (𝜑 → (𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ))    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ℤ) → 𝑍 ∈ ℤ)    &   ⊢ ((𝜑 ∧ 𝑥 ∈ ℤ) → ((𝑥 · 𝐽) = 𝐾 → (𝑍 · 𝑀) = 𝑁))    ⇒   ⊢ (𝜑 → (𝐽 ∥ 𝐾 → 𝑀 ∥ 𝑁))
Theorem	dvds2lem 16237*	A lemma to assist theorems of ∥ with two antecedents. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (𝜑 → (𝐼 ∈ ℤ ∧ 𝐽 ∈ ℤ))    &   ⊢ (𝜑 → (𝐾 ∈ ℤ ∧ 𝐿 ∈ ℤ))    &   ⊢ (𝜑 → (𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ))    &   ⊢ ((𝜑 ∧ (𝑥 ∈ ℤ ∧ 𝑦 ∈ ℤ)) → 𝑍 ∈ ℤ)    &   ⊢ ((𝜑 ∧ (𝑥 ∈ ℤ ∧ 𝑦 ∈ ℤ)) → (((𝑥 · 𝐼) = 𝐽 ∧ (𝑦 · 𝐾) = 𝐿) → (𝑍 · 𝑀) = 𝑁))    ⇒   ⊢ (𝜑 → ((𝐼 ∥ 𝐽 ∧ 𝐾 ∥ 𝐿) → 𝑀 ∥ 𝑁))
Theorem	iddvds 16238	An integer divides itself. Theorem 1.1(a) in [ApostolNT] p. 14 (reflexive property of the divides relation). (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (𝑁 ∈ ℤ → 𝑁 ∥ 𝑁)
Theorem	1dvds 16239	1 divides any integer. Theorem 1.1(f) in [ApostolNT] p. 14. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (𝑁 ∈ ℤ → 1 ∥ 𝑁)
Theorem	dvds0 16240	Any integer divides 0. Theorem 1.1(g) in [ApostolNT] p. 14. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (𝑁 ∈ ℤ → 𝑁 ∥ 0)
Theorem	negdvdsb 16241	An integer divides another iff its negation does. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ -𝑀 ∥ 𝑁))
Theorem	dvdsnegb 16242	An integer divides another iff it divides its negation. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ 𝑀 ∥ -𝑁))
Theorem	absdvdsb 16243	An integer divides another iff its absolute value does. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ (abs‘𝑀) ∥ 𝑁))
Theorem	dvdsabsb 16244	An integer divides another iff it divides its absolute value. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ 𝑀 ∥ (abs‘𝑁)))
Theorem	0dvds 16245	Only 0 is divisible by 0. Theorem 1.1(h) in [ApostolNT] p. 14. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (𝑁 ∈ ℤ → (0 ∥ 𝑁 ↔ 𝑁 = 0))
Theorem	dvdsmul1 16246	An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑀 ∥ (𝑀 · 𝑁))
Theorem	dvdsmul2 16247	An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑁 ∥ (𝑀 · 𝑁))
Theorem	iddvdsexp 16248	An integer divides a positive integer power of itself. (Contributed by Paul Chapman, 26-Oct-2012.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ) → 𝑀 ∥ (𝑀↑𝑁))
Theorem	muldvds1 16249	If a product divides an integer, so does one of its factors. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 · 𝑀) ∥ 𝑁 → 𝐾 ∥ 𝑁))
Theorem	muldvds2 16250	If a product divides an integer, so does one of its factors. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 · 𝑀) ∥ 𝑁 → 𝑀 ∥ 𝑁))
Theorem	dvdscmul 16251	Multiplication by a constant maintains the divides relation. Theorem 1.1(d) in [ApostolNT] p. 14 (multiplication property of the divides relation). (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝐾 ∈ ℤ) → (𝑀 ∥ 𝑁 → (𝐾 · 𝑀) ∥ (𝐾 · 𝑁)))
Theorem	dvdsmulc 16252	Multiplication by a constant maintains the divides relation. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝐾 ∈ ℤ) → (𝑀 ∥ 𝑁 → (𝑀 · 𝐾) ∥ (𝑁 · 𝐾)))
Theorem	dvdscmulr 16253	Cancellation law for the divides relation. Theorem 1.1(e) in [ApostolNT] p. 14. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ (𝐾 ∈ ℤ ∧ 𝐾 ≠ 0)) → ((𝐾 · 𝑀) ∥ (𝐾 · 𝑁) ↔ 𝑀 ∥ 𝑁))
Theorem	dvdsmulcr 16254	Cancellation law for the divides relation. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ (𝐾 ∈ ℤ ∧ 𝐾 ≠ 0)) → ((𝑀 · 𝐾) ∥ (𝑁 · 𝐾) ↔ 𝑀 ∥ 𝑁))
Theorem	summodnegmod 16255	The sum of two integers modulo a positive integer equals zero iff the first of the two integers equals the negative of the other integer modulo the positive integer. (Contributed by AV, 25-Jul-2021.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝑁 ∈ ℕ) → (((𝐴 + 𝐵) mod 𝑁) = 0 ↔ (𝐴 mod 𝑁) = (-𝐵 mod 𝑁)))
Theorem	modmulconst 16256	Constant multiplication in a modulo operation, see theorem 5.3 in [ApostolNT] p. 108. (Contributed by AV, 21-Jul-2021.)
⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℕ) ∧ 𝑀 ∈ ℕ) → ((𝐴 mod 𝑀) = (𝐵 mod 𝑀) ↔ ((𝐶 · 𝐴) mod (𝐶 · 𝑀)) = ((𝐶 · 𝐵) mod (𝐶 · 𝑀))))
Theorem	dvds2ln 16257	If an integer divides each of two other integers, it divides any linear combination of them. Theorem 1.1(c) in [ApostolNT] p. 14 (linearity property of the divides relation). (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (((𝐼 ∈ ℤ ∧ 𝐽 ∈ ℤ) ∧ (𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ)) → ((𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) → 𝐾 ∥ ((𝐼 · 𝑀) + (𝐽 · 𝑁))))
Theorem	dvds2add 16258	If an integer divides each of two other integers, it divides their sum. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) → 𝐾 ∥ (𝑀 + 𝑁)))
Theorem	dvds2sub 16259	If an integer divides each of two other integers, it divides their difference. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) → 𝐾 ∥ (𝑀 − 𝑁)))
Theorem	dvds2addd 16260	Deduction form of dvds2add 16258. (Contributed by SN, 21-Aug-2024.)
⊢ (𝜑 → 𝐾 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    &   ⊢ (𝜑 → 𝐾 ∥ 𝑀)    &   ⊢ (𝜑 → 𝐾 ∥ 𝑁)    ⇒   ⊢ (𝜑 → 𝐾 ∥ (𝑀 + 𝑁))
Theorem	dvds2subd 16261	Deduction form of dvds2sub 16259. (Contributed by Stanislas Polu, 9-Mar-2020.)
⊢ (𝜑 → 𝐾 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    &   ⊢ (𝜑 → 𝐾 ∥ 𝑀)    &   ⊢ (𝜑 → 𝐾 ∥ 𝑁)    ⇒   ⊢ (𝜑 → 𝐾 ∥ (𝑀 − 𝑁))
Theorem	dvdstr 16262	The divides relation is transitive. Theorem 1.1(b) in [ApostolNT] p. 14 (transitive property of the divides relation). (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 ∥ 𝑀 ∧ 𝑀 ∥ 𝑁) → 𝐾 ∥ 𝑁))
Theorem	dvdstrd 16263	The divides relation is transitive, a deduction version of dvdstr 16262. (Contributed by metakunt, 12-May-2024.)
⊢ (𝜑 → 𝐾 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    &   ⊢ (𝜑 → 𝐾 ∥ 𝑀)    &   ⊢ (𝜑 → 𝑀 ∥ 𝑁)    ⇒   ⊢ (𝜑 → 𝐾 ∥ 𝑁)
Theorem	dvdsmultr1 16264	If an integer divides another, it divides a multiple of it. (Contributed by Paul Chapman, 17-Nov-2012.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝐾 ∥ 𝑀 → 𝐾 ∥ (𝑀 · 𝑁)))
Theorem	dvdsmultr1d 16265	Deduction form of dvdsmultr1 16264. (Contributed by Stanislas Polu, 9-Mar-2020.)
⊢ (𝜑 → 𝐾 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    &   ⊢ (𝜑 → 𝐾 ∥ 𝑀)    ⇒   ⊢ (𝜑 → 𝐾 ∥ (𝑀 · 𝑁))
Theorem	dvdsmultr2 16266	If an integer divides another, it divides a multiple of it. (Contributed by Paul Chapman, 17-Nov-2012.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝐾 ∥ 𝑁 → 𝐾 ∥ (𝑀 · 𝑁)))
Theorem	dvdsmultr2d 16267	Deduction form of dvdsmultr2 16266. (Contributed by SN, 23-Aug-2024.)
⊢ (𝜑 → 𝐾 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    &   ⊢ (𝜑 → 𝐾 ∥ 𝑁)    ⇒   ⊢ (𝜑 → 𝐾 ∥ (𝑀 · 𝑁))
Theorem	ordvdsmul 16268	If an integer divides either of two others, it divides their product. (Contributed by Paul Chapman, 17-Nov-2012.) (Proof shortened by Mario Carneiro, 17-Jul-2014.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 ∥ 𝑀 ∨ 𝐾 ∥ 𝑁) → 𝐾 ∥ (𝑀 · 𝑁)))
Theorem	dvdssub2 16269	If an integer divides a difference, then it divides one term iff it divides the other. (Contributed by Mario Carneiro, 13-Jul-2014.)
⊢ (((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ 𝐾 ∥ (𝑀 − 𝑁)) → (𝐾 ∥ 𝑀 ↔ 𝐾 ∥ 𝑁))
Theorem	dvdsadd 16270	An integer divides another iff it divides their sum. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 13-Jul-2014.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ 𝑀 ∥ (𝑀 + 𝑁)))
Theorem	dvdsaddr 16271	An integer divides another iff it divides their sum. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ 𝑀 ∥ (𝑁 + 𝑀)))
Theorem	dvdssub 16272	An integer divides another iff it divides their difference. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ 𝑀 ∥ (𝑀 − 𝑁)))
Theorem	dvdssubr 16273	An integer divides another iff it divides their difference. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ 𝑀 ∥ (𝑁 − 𝑀)))
Theorem	dvdsadd2b 16274	Adding a multiple of the base does not affect divisibility. (Contributed by Stefan O'Rear, 23-Sep-2014.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ (𝐶 ∈ ℤ ∧ 𝐴 ∥ 𝐶)) → (𝐴 ∥ 𝐵 ↔ 𝐴 ∥ (𝐶 + 𝐵)))
Theorem	dvdsaddre2b 16275	Adding a multiple of the base does not affect divisibility. Variant of dvdsadd2b 16274 only requiring 𝐵 to be a real number (not necessarily an integer). (Contributed by AV, 19-Jul-2021.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℝ ∧ (𝐶 ∈ ℤ ∧ 𝐴 ∥ 𝐶)) → (𝐴 ∥ 𝐵 ↔ 𝐴 ∥ (𝐶 + 𝐵)))
Theorem	fsumdvds 16276*	If every term in a sum is divisible by 𝑁, then so is the sum. (Contributed by Mario Carneiro, 17-Jan-2015.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℤ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝑁 ∥ 𝐵)    ⇒   ⊢ (𝜑 → 𝑁 ∥ Σ𝑘 ∈ 𝐴 𝐵)
Theorem	dvdslelem 16277	Lemma for dvdsle 16278. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ 𝑀 ∈ ℤ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝐾 ∈ ℤ    ⇒   ⊢ (𝑁 < 𝑀 → (𝐾 · 𝑀) ≠ 𝑁)
Theorem	dvdsle 16278	The divisors of a positive integer are bounded by it. The proof does not use /. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ) → (𝑀 ∥ 𝑁 → 𝑀 ≤ 𝑁))
Theorem	dvdsleabs 16279	The divisors of a nonzero integer are bounded by its absolute value. Theorem 1.1(i) in [ApostolNT] p. 14 (comparison property of the divides relation). (Contributed by Paul Chapman, 21-Mar-2011.) (Proof shortened by Fan Zheng, 3-Jul-2016.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑁 ≠ 0) → (𝑀 ∥ 𝑁 → 𝑀 ≤ (abs‘𝑁)))
Theorem	dvdsleabs2 16280	Transfer divisibility to an order constraint on absolute values. (Contributed by Stefan O'Rear, 24-Sep-2014.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑁 ≠ 0) → (𝑀 ∥ 𝑁 → (abs‘𝑀) ≤ (abs‘𝑁)))
Theorem	dvdsabseq 16281	If two integers divide each other, they must be equal, up to a difference in sign. Theorem 1.1(j) in [ApostolNT] p. 14. (Contributed by Mario Carneiro, 30-May-2014.) (Revised by AV, 7-Aug-2021.)
⊢ ((𝑀 ∥ 𝑁 ∧ 𝑁 ∥ 𝑀) → (abs‘𝑀) = (abs‘𝑁))
Theorem	dvdseq 16282	If two nonnegative integers divide each other, they must be equal. (Contributed by Mario Carneiro, 30-May-2014.) (Proof shortened by AV, 7-Aug-2021.)
⊢ (((𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0) ∧ (𝑀 ∥ 𝑁 ∧ 𝑁 ∥ 𝑀)) → 𝑀 = 𝑁)
Theorem	divconjdvds 16283	If a nonzero integer 𝑀 divides another integer 𝑁, the other integer 𝑁 divided by the nonzero integer 𝑀 (i.e. the divisor conjugate of 𝑁 to 𝑀) divides the other integer 𝑁. Theorem 1.1(k) in [ApostolNT] p. 14. (Contributed by AV, 7-Aug-2021.)
⊢ ((𝑀 ∥ 𝑁 ∧ 𝑀 ≠ 0) → (𝑁 / 𝑀) ∥ 𝑁)
Theorem	dvdsdivcl 16284*	The complement of a divisor of 𝑁 is also a divisor of 𝑁. (Contributed by Mario Carneiro, 2-Jul-2015.) (Proof shortened by AV, 9-Aug-2021.)
⊢ ((𝑁 ∈ ℕ ∧ 𝐴 ∈ {𝑥 ∈ ℕ ∣ 𝑥 ∥ 𝑁}) → (𝑁 / 𝐴) ∈ {𝑥 ∈ ℕ ∣ 𝑥 ∥ 𝑁})
Theorem	dvdsflip 16285*	An involution of the divisors of a number. (Contributed by Stefan O'Rear, 12-Sep-2015.) (Proof shortened by Mario Carneiro, 13-May-2016.)
⊢ 𝐴 = {𝑥 ∈ ℕ ∣ 𝑥 ∥ 𝑁}    &   ⊢ 𝐹 = (𝑦 ∈ 𝐴 ↦ (𝑁 / 𝑦))    ⇒   ⊢ (𝑁 ∈ ℕ → 𝐹:𝐴–1-1-onto→𝐴)
Theorem	dvdsssfz1 16286*	The set of divisors of a number is a subset of a finite set. (Contributed by Mario Carneiro, 22-Sep-2014.)
⊢ (𝐴 ∈ ℕ → {𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐴} ⊆ (1...𝐴))
Theorem	dvds1 16287	The only nonnegative integer that divides 1 is 1. (Contributed by Mario Carneiro, 2-Jul-2015.)
⊢ (𝑀 ∈ ℕ0 → (𝑀 ∥ 1 ↔ 𝑀 = 1))
Theorem	alzdvds 16288*	Only 0 is divisible by all integers. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (𝑁 ∈ ℤ → (∀𝑥 ∈ ℤ 𝑥 ∥ 𝑁 ↔ 𝑁 = 0))
Theorem	dvdsext 16289*	Poset extensionality for division. (Contributed by Stefan O'Rear, 6-Sep-2015.)
⊢ ((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ0) → (𝐴 = 𝐵 ↔ ∀𝑥 ∈ ℕ0 (𝐴 ∥ 𝑥 ↔ 𝐵 ∥ 𝑥)))
Theorem	fzm1ndvds 16290	No number between 1 and 𝑀 − 1 divides 𝑀. (Contributed by Mario Carneiro, 24-Jan-2015.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ (1...(𝑀 − 1))) → ¬ 𝑀 ∥ 𝑁)
Theorem	fzo0dvdseq 16291	Zero is the only one of the first 𝐴 nonnegative integers that is divisible by 𝐴. (Contributed by Stefan O'Rear, 6-Sep-2015.)
⊢ (𝐵 ∈ (0..^𝐴) → (𝐴 ∥ 𝐵 ↔ 𝐵 = 0))
Theorem	fzocongeq 16292	Two different elements of a half-open range are not congruent mod its length. (Contributed by Stefan O'Rear, 6-Sep-2015.)
⊢ ((𝐴 ∈ (𝐶..^𝐷) ∧ 𝐵 ∈ (𝐶..^𝐷)) → ((𝐷 − 𝐶) ∥ (𝐴 − 𝐵) ↔ 𝐴 = 𝐵))
Theorem	addmodlteqALT 16293	Two nonnegative integers less than the modulus are equal iff the sums of these integer with another integer are equal modulo the modulus. Shorter proof of addmodlteq 13935 based on the "divides" relation. (Contributed by AV, 14-Mar-2021.) (New usage is discouraged.) (Proof modification is discouraged.)
⊢ ((𝐼 ∈ (0..^𝑁) ∧ 𝐽 ∈ (0..^𝑁) ∧ 𝑆 ∈ ℤ) → (((𝐼 + 𝑆) mod 𝑁) = ((𝐽 + 𝑆) mod 𝑁) ↔ 𝐼 = 𝐽))
Theorem	dvdsfac 16294	A positive integer divides any greater factorial. (Contributed by Paul Chapman, 28-Nov-2012.)
⊢ ((𝐾 ∈ ℕ ∧ 𝑁 ∈ (ℤ≥‘𝐾)) → 𝐾 ∥ (!‘𝑁))
Theorem	dvdsexp2im 16295	If an integer divides another integer, then it also divides any of its powers. (Contributed by Scott Fenton, 7-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ) → (𝐾 ∥ 𝑀 → 𝐾 ∥ (𝑀↑𝑁)))
Theorem	dvdsexp 16296	A power divides a power with a greater exponent. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ ((𝐴 ∈ ℤ ∧ 𝑀 ∈ ℕ0 ∧ 𝑁 ∈ (ℤ≥‘𝑀)) → (𝐴↑𝑀) ∥ (𝐴↑𝑁))
Theorem	dvdsmod 16297	Any number 𝐾 whose mod base 𝑁 is divisible by a divisor 𝑃 of the base is also divisible by 𝑃. This means that primes will also be relatively prime to the base when reduced mod 𝑁 for any base. (Contributed by Mario Carneiro, 13-Mar-2014.)
⊢ (((𝑃 ∈ ℕ ∧ 𝑁 ∈ ℕ ∧ 𝐾 ∈ ℤ) ∧ 𝑃 ∥ 𝑁) → (𝑃 ∥ (𝐾 mod 𝑁) ↔ 𝑃 ∥ 𝐾))
Theorem	mulmoddvds 16298	If an integer is divisible by a positive integer, the product of this integer with another integer modulo the positive integer is 0. (Contributed by Alexander van der Vekens, 30-Aug-2018.) (Proof shortened by AV, 18-Mar-2022.)
⊢ ((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → (𝑁 ∥ 𝐴 → ((𝐴 · 𝐵) mod 𝑁) = 0))
Theorem	3dvds 16299*	A rule for divisibility by 3 of a number written in base 10. This is Metamath 100 proof #85. (Contributed by Mario Carneiro, 14-Jul-2014.) (Revised by Mario Carneiro, 17-Jan-2015.) (Revised by AV, 8-Sep-2021.)
⊢ ((𝑁 ∈ ℕ0 ∧ 𝐹:(0...𝑁)⟶ℤ) → (3 ∥ Σ𝑘 ∈ (0...𝑁)((𝐹‘𝑘) · (;10↑𝑘)) ↔ 3 ∥ Σ𝑘 ∈ (0...𝑁)(𝐹‘𝑘)))
Theorem	3dvdsdec 16300	A decimal number is divisible by three iff the sum of its two "digits" is divisible by three. The term "digits" in its narrow sense is only correct if 𝐴 and 𝐵 actually are digits (i.e. nonnegative integers less than 10). However, this theorem holds for arbitrary nonnegative integers 𝐴 and 𝐵, especially if 𝐴 is itself a decimal number, e.g., 𝐴 = ;𝐶𝐷. (Contributed by AV, 14-Jun-2021.) (Revised by AV, 8-Sep-2021.)
⊢ 𝐴 ∈ ℕ0    &   ⊢ 𝐵 ∈ ℕ0    ⇒   ⊢ (3 ∥ ;𝐴𝐵 ↔ 3 ∥ (𝐴 + 𝐵))