P. 165 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 16401-16500   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	ruclem3 16401*	Lemma for ruc 16411. 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 16402*	Lemma for ruc 16411. 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 16403*	Lemma for ruc 16411. 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 16404*	Lemma for ruc 16411. 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 16405*	Lemma for ruc 16411. 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 16406*	Lemma for ruc 16411. 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 16407*	Lemma for ruc 16411. 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 16408*	Lemma for ruc 16411. 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 16409*	Lemma for ruc 16411. 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 16410	Lemma for ruc 16411. There is no function that maps ℕ onto ℝ. (Use nex 1833 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 16411	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 16399 through ruclem13 16410 and this final piece. Our proof is based on the proof of Theorem 5.18 of [Truss] p. 114. See ruclem13 16410 for the function existence version of this theorem. For an informal discussion of this proof, see mmcomplex.html#uncountable 16410. For an alternate proof see rucALT 16398. This is Metamath 100 proof #22. (Contributed by NM, 13-Oct-2004.)
⊢ ℕ ≺ ℝ
Theorem	resdomq 16412	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 16413	There are at least aleph-one real numbers. (Contributed by NM, 2-Feb-2005.)
⊢ (ℵ‘1o) ≼ ℝ
Theorem	aleph1irr 16414	There are at least aleph-one irrationals. (Contributed by NM, 2-Feb-2005.)
⊢ (ℵ‘1o) ≼ (ℝ ∖ ℚ)
Theorem	cnso 16415	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 16416	Lemma for sqrt2irr 16417. 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 16417	The square root of 2 is irrational. See zsqrtelqelz 16934 for a generalization to all non-square integers. The proof's core is proven in sqrt2irrlem 16416, 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/ 16416. (Contributed by NM, 8-Jan-2002.) (Proof shortened by Mario Carneiro, 12-Sep-2015.)
⊢ (√‘2) ∉ ℚ
Theorem	sqrt2re 16418	The square root of 2 exists and is a real number. (Contributed by NM, 3-Dec-2004.)
⊢ (√‘2) ∈ ℝ
Theorem	sqrt2irr0 16419	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 16420	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 16421 for a further refinement. (Contributed by NM, 12-Jan-2002.)
⊢ ((ℕ ⊊ ℤ ∧ ℤ ⊊ ℚ) ∧ (ℚ ⊊ ℝ ∧ ℝ ⊊ ℂ))
Theorem	nthruz 16421	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 16420. (Contributed by NM, 9-May-2004.)
⊢ (ℕ ⊊ ℕ0 ∧ ℕ0 ⊊ ℤ)
6.1.3  The divides relation
Syntax	cdvds 16422	Extend the definition of a class to include the divides relation. See df-dvds 16423.
class ∥
Definition	df-dvds 16423*	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 16424*	Define the divides relation. 𝑀 ∥ 𝑁 means 𝑀 divides into 𝑁 with no remainder. For example, 3 ∥ 6 (ex-dvds 31057). As proven in dvdsval3 16426, 𝑀 ∥ 𝑁 ↔ (𝑁 mod 𝑀) = 0. See divides 16424 and dvdsval2 16425 for other equivalent expressions. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ ∃𝑛 ∈ ℤ (𝑛 · 𝑀) = 𝑁))
Theorem	dvdsval2 16425	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 16426	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 16427	Reverse closure for the divisibility relation. (Contributed by Stefan O'Rear, 5-Sep-2015.)
⊢ (𝑋 ∥ 𝑌 → (𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ))
Theorem	dvdsmod0 16428	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 16429	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 16430	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 16961). (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 16431	Strong form of dvdsval2 16425 for positive integers. (Contributed by Stefan O'Rear, 13-Sep-2014.)
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (𝐵 ∥ 𝐴 ↔ (𝐴 / 𝐵) ∈ ℕ))
Theorem	nndivides 16432*	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 16433	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 16434	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	addmulmodb 16435	An integer plus a product is itself modulo a positive integer iff the product is divisible by the positive integer. (Contributed by AV, 8-Sep-2025.)
⊢ ((𝑁 ∈ ℕ ∧ (𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ)) → (𝑁 ∥ (𝐵 · 𝐶) ↔ ((𝐴 + (𝐵 · 𝐶)) mod 𝑁) = (𝐴 mod 𝑁)))
Theorem	dvds0lem 16436	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 16437*	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 16438*	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 16439	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 16440	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 16441	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 16442	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 16443	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 16444	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 16445	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 16446	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 16447	An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑀 ∥ (𝑀 · 𝑁))
Theorem	dvdsmul2 16448	An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑁 ∥ (𝑀 · 𝑁))
Theorem	iddvdsexp 16449	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 16450	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 16451	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 16452	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 16453	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 16454	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 16455	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 16456	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	difmod0 16457	The difference of two integers modulo a positive integer equals zero iff the two integers are equal modulo the positive integer. (Contributed by AV, 15-Nov-2025.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝑁 ∈ ℕ) → (((𝐴 − 𝐵) mod 𝑁) = 0 ↔ (𝐴 mod 𝑁) = (𝐵 mod 𝑁)))
Theorem	modmulconst 16458	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 16459	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 16460	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 16461	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 16462	Deduction form of dvds2add 16460. (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 16463	Deduction form of dvds2sub 16461. (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 16464	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 16465	The divides relation is transitive, a deduction version of dvdstr 16464. (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 16466	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 16467	Deduction form of dvdsmultr1 16466. (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 16468	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 16469	Deduction form of dvdsmultr2 16468. (Contributed by SN, 23-Aug-2024.)
⊢ (𝜑 → 𝐾 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    &   ⊢ (𝜑 → 𝐾 ∥ 𝑁)    ⇒   ⊢ (𝜑 → 𝐾 ∥ (𝑀 · 𝑁))
Theorem	ordvdsmul 16470	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 16471	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 16472	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 16473	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 16474	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 16475	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 16476	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 16477	Adding a multiple of the base does not affect divisibility. Variant of dvdsadd2b 16476 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 16478*	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 16479	Lemma for dvdsle 16480. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ 𝑀 ∈ ℤ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝐾 ∈ ℤ    ⇒   ⊢ (𝑁 < 𝑀 → (𝐾 · 𝑀) ≠ 𝑁)
Theorem	dvdsle 16480	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 16481	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 16482	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 16483	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 16484	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 16485	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 16486*	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 16487*	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 16488*	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 16489	The only nonnegative integer that divides 1 is 1. (Contributed by Mario Carneiro, 2-Jul-2015.)
⊢ (𝑀 ∈ ℕ0 → (𝑀 ∥ 1 ↔ 𝑀 = 1))
Theorem	alzdvds 16490*	Only 0 is divisible by all integers. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (𝑁 ∈ ℤ → (∀𝑥 ∈ ℤ 𝑥 ∥ 𝑁 ↔ 𝑁 = 0))
Theorem	dvdsext 16491*	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 16492	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 16493	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 16494	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 16495	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 14089 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 16496	A positive integer divides any greater factorial. (Contributed by Paul Chapman, 28-Nov-2012.)
⊢ ((𝐾 ∈ ℕ ∧ 𝑁 ∈ (ℤ≥‘𝐾)) → 𝐾 ∥ (!‘𝑁))
Theorem	dvdsexp2im 16497	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 16498	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 16499	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 16500	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))