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	rexpen 16401	The real numbers are equinumerous to their own Cartesian product, even though it is not necessarily true that ℝ is well-orderable (so we cannot use infxpidm2 10099 directly). (Contributed by NM, 30-Jul-2004.) (Revised by Mario Carneiro, 16-Jun-2013.)
⊢ (ℝ × ℝ) ≈ ℝ
Theorem	cpnnen 16402	The complex numbers are equinumerous to the powerset of the positive integers. (Contributed by Mario Carneiro, 16-Jun-2013.)
⊢ ℂ ≈ 𝒫 ℕ
Theorem	rucALT 16403	Alternate proof of ruc 16416. This proof is a simple corollary of rpnnen 16400, which determines the exact cardinality of the reals. For an alternate proof discussed at mmcomplex.html#uncountable 16400, see ruc 16416. (Contributed by NM, 13-Oct-2004.) (Revised by Mario Carneiro, 13-May-2013.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ ℕ ≺ ℝ
Theorem	ruclem1 16404*	Lemma for ruc 16416 (the reals are uncountable). Substitutions for the function 𝐷. (Contributed by Mario Carneiro, 28-May-2014.) (Revised by Fan Zheng, 6-Jun-2016.)
⊢ (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⟩𝐷𝑀) ∈ (ℝ × ℝ) ∧ 𝑋 = if(((𝐴 + 𝐵) / 2) < 𝑀, 𝐴, ((((𝐴 + 𝐵) / 2) + 𝐵) / 2)) ∧ 𝑌 = if(((𝐴 + 𝐵) / 2) < 𝑀, ((𝐴 + 𝐵) / 2), 𝐵)))
Theorem	ruclem2 16405*	Lemma for ruc 16416. Ordering property for the input to 𝐷. (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 → (𝐴 ≤ 𝑋 ∧ 𝑋 < 𝑌 ∧ 𝑌 ≤ 𝐵))
Theorem	ruclem3 16406*	Lemma for ruc 16416. 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 16407*	Lemma for ruc 16416. 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 16408*	Lemma for ruc 16416. 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 16409*	Lemma for ruc 16416. 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 16410*	Lemma for ruc 16416. 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 16411*	Lemma for ruc 16416. 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 16412*	Lemma for ruc 16416. 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 16413*	Lemma for ruc 16416. 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 16414*	Lemma for ruc 16416. 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 16415	Lemma for ruc 16416. There is no function that maps ℕ onto ℝ. (Use nex 1832 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 16416	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 16404 through ruclem13 16415 and this final piece. Our proof is based on the proof of Theorem 5.18 of [Truss] p. 114. See ruclem13 16415 for the function existence version of this theorem. For an informal discussion of this proof, see mmcomplex.html#uncountable 16415. For an alternate proof see rucALT 16403. This is Metamath 100 proof #22. (Contributed by NM, 13-Oct-2004.)
⊢ ℕ ≺ ℝ
Theorem	resdomq 16417	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 16418	There are at least aleph-one real numbers. (Contributed by NM, 2-Feb-2005.)
⊢ (ℵ‘1o) ≼ ℝ
Theorem	aleph1irr 16419	There are at least aleph-one irrationals. (Contributed by NM, 2-Feb-2005.)
⊢ (ℵ‘1o) ≼ (ℝ ∖ ℚ)
Theorem	cnso 16420	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 16421	Lemma for sqrt2irr 16422. 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 16422	The square root of 2 is irrational. See zsqrtelqelz 16939 for a generalization to all non-square integers. The proof's core is proven in sqrt2irrlem 16421, 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/ 16421. (Contributed by NM, 8-Jan-2002.) (Proof shortened by Mario Carneiro, 12-Sep-2015.)
⊢ (√‘2) ∉ ℚ
Theorem	sqrt2re 16423	The square root of 2 exists and is a real number. (Contributed by NM, 3-Dec-2004.)
⊢ (√‘2) ∈ ℝ
Theorem	sqrt2irr0 16424	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 16425	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 16426 for a further refinement. (Contributed by NM, 12-Jan-2002.)
⊢ ((ℕ ⊊ ℤ ∧ ℤ ⊊ ℚ) ∧ (ℚ ⊊ ℝ ∧ ℝ ⊊ ℂ))
Theorem	nthruz 16426	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 16425. (Contributed by NM, 9-May-2004.)
⊢ (ℕ ⊊ ℕ0 ∧ ℕ0 ⊊ ℤ)
6.1.3  The divides relation
Syntax	cdvds 16427	Extend the definition of a class to include the divides relation. See df-dvds 16428.
class ∥
Definition	df-dvds 16428*	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 16429*	Define the divides relation. 𝑀 ∥ 𝑁 means 𝑀 divides into 𝑁 with no remainder. For example, 3 ∥ 6 (ex-dvds 31062). As proven in dvdsval3 16431, 𝑀 ∥ 𝑁 ↔ (𝑁 mod 𝑀) = 0. See divides 16429 and dvdsval2 16430 for other equivalent expressions. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ ∃𝑛 ∈ ℤ (𝑛 · 𝑀) = 𝑁))
Theorem	dvdsval2 16430	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 16431	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 16432	Reverse closure for the divisibility relation. (Contributed by Stefan O'Rear, 5-Sep-2015.)
⊢ (𝑋 ∥ 𝑌 → (𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ))
Theorem	dvdsmod0 16433	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 16434	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 16435	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 16966). (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 16436	Strong form of dvdsval2 16430 for positive integers. (Contributed by Stefan O'Rear, 13-Sep-2014.)
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (𝐵 ∥ 𝐴 ↔ (𝐴 / 𝐵) ∈ ℕ))
Theorem	nndivides 16437*	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 16438	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 16439	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 16440	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 16441	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 16442*	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 16443*	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 16444	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 16445	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 16446	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 16447	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 16448	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 16449	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 16450	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 16451	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 16452	An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑀 ∥ (𝑀 · 𝑁))
Theorem	dvdsmul2 16453	An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑁 ∥ (𝑀 · 𝑁))
Theorem	iddvdsexp 16454	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 16455	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 16456	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 16457	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 16458	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 16459	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 16460	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 16461	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 16462	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 16463	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 16464	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 16465	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 16466	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 16467	Deduction form of dvds2add 16465. (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 16468	Deduction form of dvds2sub 16466. (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 16469	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 16470	The divides relation is transitive, a deduction version of dvdstr 16469. (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 16471	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 16472	Deduction form of dvdsmultr1 16471. (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 16473	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 16474	Deduction form of dvdsmultr2 16473. (Contributed by SN, 23-Aug-2024.)
⊢ (𝜑 → 𝐾 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    &   ⊢ (𝜑 → 𝐾 ∥ 𝑁)    ⇒   ⊢ (𝜑 → 𝐾 ∥ (𝑀 · 𝑁))
Theorem	ordvdsmul 16475	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 16476	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 16477	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 16478	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 16479	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 16480	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 16481	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 16482	Adding a multiple of the base does not affect divisibility. Variant of dvdsadd2b 16481 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 16483*	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 16484	Lemma for dvdsle 16485. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ 𝑀 ∈ ℤ    &   ⊢ 𝑁 ∈ ℕ    &   ⊢ 𝐾 ∈ ℤ    ⇒   ⊢ (𝑁 < 𝑀 → (𝐾 · 𝑀) ≠ 𝑁)
Theorem	dvdsle 16485	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 16486	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 16487	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 16488	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 16489	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 16490	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 16491*	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 16492*	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 16493*	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 16494	The only nonnegative integer that divides 1 is 1. (Contributed by Mario Carneiro, 2-Jul-2015.)
⊢ (𝑀 ∈ ℕ0 → (𝑀 ∥ 1 ↔ 𝑀 = 1))
Theorem	alzdvds 16495*	Only 0 is divisible by all integers. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (𝑁 ∈ ℤ → (∀𝑥 ∈ ℤ 𝑥 ∥ 𝑁 ↔ 𝑁 = 0))
Theorem	dvdsext 16496*	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 16497	No number from 1 to 𝑀 − 1 is a multiple of 𝑀. (Contributed by Mario Carneiro, 24-Jan-2015.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ (1...(𝑀 − 1))) → ¬ 𝑀 ∥ 𝑁)
Theorem	fzo0dvdseq 16498	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 16499	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 16500	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 14092 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))