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Theorem List for Metamath Proof Explorer - 16201-16300   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theorembezoutr 16201 Partial converse to bezout 16179. Existence of a linear combination does not set the GCD, but it does upper bound it. (Contributed by Stefan O'Rear, 23-Sep-2014.)
(((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) ∧ (𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ)) → (𝐴 gcd 𝐵) ∥ ((𝐴 · 𝑋) + (𝐵 · 𝑌)))
 
Theorembezoutr1 16202 Converse of bezout 16179 for when the greater common divisor is one (sufficient condition for relative primality). (Contributed by Stefan O'Rear, 23-Sep-2014.)
(((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) ∧ (𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ)) → (((𝐴 · 𝑋) + (𝐵 · 𝑌)) = 1 → (𝐴 gcd 𝐵) = 1))
 
6.1.9  Algorithms
 
Theoremnn0seqcvgd 16203* A strictly-decreasing nonnegative integer sequence with initial term 𝑁 reaches zero by the 𝑁 th term. Deduction version. (Contributed by Paul Chapman, 31-Mar-2011.)
(𝜑𝐹:ℕ0⟶ℕ0)    &   (𝜑𝑁 = (𝐹‘0))    &   ((𝜑𝑘 ∈ ℕ0) → ((𝐹‘(𝑘 + 1)) ≠ 0 → (𝐹‘(𝑘 + 1)) < (𝐹𝑘)))       (𝜑 → (𝐹𝑁) = 0)
 
Theoremseq1st 16204 A sequence whose iteration function ignores the second argument is only affected by the first point of the initial value function. (Contributed by Mario Carneiro, 11-Feb-2015.)
𝑍 = (ℤ𝑀)    &   𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}))       ((𝑀 ∈ ℤ ∧ 𝐴𝑉) → 𝑅 = seq𝑀((𝐹 ∘ 1st ), {⟨𝑀, 𝐴⟩}))
 
Theoremalgr0 16205 The value of the algorithm iterator 𝑅 at 0 is the initial state 𝐴. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 28-May-2014.)
𝑍 = (ℤ𝑀)    &   𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}))    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐴𝑆)       (𝜑 → (𝑅𝑀) = 𝐴)
 
Theoremalgrf 16206 An algorithm is a step function 𝐹:𝑆𝑆 on a state space 𝑆. An algorithm acts on an initial state 𝐴𝑆 by iteratively applying 𝐹 to give 𝐴, (𝐹𝐴), (𝐹‘(𝐹𝐴)) and so on. An algorithm is said to halt if a fixed point of 𝐹 is reached after a finite number of iterations.

The algorithm iterator 𝑅:ℕ0𝑆 "runs" the algorithm 𝐹 so that (𝑅𝑘) is the state after 𝑘 iterations of 𝐹 on the initial state 𝐴.

Domain and codomain of the algorithm iterator 𝑅. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 28-May-2014.)

𝑍 = (ℤ𝑀)    &   𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}))    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐴𝑆)    &   (𝜑𝐹:𝑆𝑆)       (𝜑𝑅:𝑍𝑆)
 
Theoremalgrp1 16207 The value of the algorithm iterator 𝑅 at (𝐾 + 1). (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 27-Dec-2014.)
𝑍 = (ℤ𝑀)    &   𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}))    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐴𝑆)    &   (𝜑𝐹:𝑆𝑆)       ((𝜑𝐾𝑍) → (𝑅‘(𝐾 + 1)) = (𝐹‘(𝑅𝐾)))
 
Theoremalginv 16208* If 𝐼 is an invariant of 𝐹, then its value is unchanged after any number of iterations of 𝐹. (Contributed by Paul Chapman, 31-Mar-2011.)
𝑅 = seq0((𝐹 ∘ 1st ), (ℕ0 × {𝐴}))    &   𝐹:𝑆𝑆    &   (𝑥𝑆 → (𝐼‘(𝐹𝑥)) = (𝐼𝑥))       ((𝐴𝑆𝐾 ∈ ℕ0) → (𝐼‘(𝑅𝐾)) = (𝐼‘(𝑅‘0)))
 
Theoremalgcvg 16209* One way to prove that an algorithm halts is to construct a countdown function 𝐶:𝑆⟶ℕ0 whose value is guaranteed to decrease for each iteration of 𝐹 until it reaches 0. That is, if 𝑋𝑆 is not a fixed point of 𝐹, then (𝐶‘(𝐹𝑋)) < (𝐶𝑋).

If 𝐶 is a countdown function for algorithm 𝐹, the sequence (𝐶‘(𝑅𝑘)) reaches 0 after at most 𝑁 steps, where 𝑁 is the value of 𝐶 for the initial state 𝐴. (Contributed by Paul Chapman, 22-Jun-2011.)

𝐹:𝑆𝑆    &   𝑅 = seq0((𝐹 ∘ 1st ), (ℕ0 × {𝐴}))    &   𝐶:𝑆⟶ℕ0    &   (𝑧𝑆 → ((𝐶‘(𝐹𝑧)) ≠ 0 → (𝐶‘(𝐹𝑧)) < (𝐶𝑧)))    &   𝑁 = (𝐶𝐴)       (𝐴𝑆 → (𝐶‘(𝑅𝑁)) = 0)
 
Theoremalgcvgblem 16210 Lemma for algcvgb 16211. (Contributed by Paul Chapman, 31-Mar-2011.)
((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → ((𝑁 ≠ 0 → 𝑁 < 𝑀) ↔ ((𝑀 ≠ 0 → 𝑁 < 𝑀) ∧ (𝑀 = 0 → 𝑁 = 0))))
 
Theoremalgcvgb 16211 Two ways of expressing that 𝐶 is a countdown function for algorithm 𝐹. The first is used in these theorems. The second states the condition more intuitively as a conjunction: if the countdown function's value is currently nonzero, it must decrease at the next step; if it has reached zero, it must remain zero at the next step. (Contributed by Paul Chapman, 31-Mar-2011.)
𝐹:𝑆𝑆    &   𝐶:𝑆⟶ℕ0       (𝑋𝑆 → (((𝐶‘(𝐹𝑋)) ≠ 0 → (𝐶‘(𝐹𝑋)) < (𝐶𝑋)) ↔ (((𝐶𝑋) ≠ 0 → (𝐶‘(𝐹𝑋)) < (𝐶𝑋)) ∧ ((𝐶𝑋) = 0 → (𝐶‘(𝐹𝑋)) = 0))))
 
Theoremalgcvga 16212* The countdown function 𝐶 remains 0 after 𝑁 steps. (Contributed by Paul Chapman, 22-Jun-2011.)
𝐹:𝑆𝑆    &   𝑅 = seq0((𝐹 ∘ 1st ), (ℕ0 × {𝐴}))    &   𝐶:𝑆⟶ℕ0    &   (𝑧𝑆 → ((𝐶‘(𝐹𝑧)) ≠ 0 → (𝐶‘(𝐹𝑧)) < (𝐶𝑧)))    &   𝑁 = (𝐶𝐴)       (𝐴𝑆 → (𝐾 ∈ (ℤ𝑁) → (𝐶‘(𝑅𝐾)) = 0))
 
Theoremalgfx 16213* If 𝐹 reaches a fixed point when the countdown function 𝐶 reaches 0, 𝐹 remains fixed after 𝑁 steps. (Contributed by Paul Chapman, 22-Jun-2011.)
𝐹:𝑆𝑆    &   𝑅 = seq0((𝐹 ∘ 1st ), (ℕ0 × {𝐴}))    &   𝐶:𝑆⟶ℕ0    &   (𝑧𝑆 → ((𝐶‘(𝐹𝑧)) ≠ 0 → (𝐶‘(𝐹𝑧)) < (𝐶𝑧)))    &   𝑁 = (𝐶𝐴)    &   (𝑧𝑆 → ((𝐶𝑧) = 0 → (𝐹𝑧) = 𝑧))       (𝐴𝑆 → (𝐾 ∈ (ℤ𝑁) → (𝑅𝐾) = (𝑅𝑁)))
 
6.1.10  Euclid's Algorithm
 
Theoremeucalgval2 16214* The value of the step function 𝐸 for Euclid's Algorithm on an ordered pair. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 28-May-2014.)
𝐸 = (𝑥 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))       ((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → (𝑀𝐸𝑁) = if(𝑁 = 0, ⟨𝑀, 𝑁⟩, ⟨𝑁, (𝑀 mod 𝑁)⟩))
 
Theoremeucalgval 16215* Euclid's Algorithm eucalg 16220 computes the greatest common divisor of two nonnegative integers by repeatedly replacing the larger of them with its remainder modulo the smaller until the remainder is 0.

The value of the step function 𝐸 for Euclid's Algorithm. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 28-May-2014.)

𝐸 = (𝑥 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))       (𝑋 ∈ (ℕ0 × ℕ0) → (𝐸𝑋) = if((2nd𝑋) = 0, 𝑋, ⟨(2nd𝑋), ( mod ‘𝑋)⟩))
 
Theoremeucalgf 16216* Domain and codomain of the step function 𝐸 for Euclid's Algorithm. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 28-May-2014.)
𝐸 = (𝑥 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))       𝐸:(ℕ0 × ℕ0)⟶(ℕ0 × ℕ0)
 
Theoremeucalginv 16217* The invariant of the step function 𝐸 for Euclid's Algorithm is the gcd operator applied to the state. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 29-May-2014.)
𝐸 = (𝑥 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))       (𝑋 ∈ (ℕ0 × ℕ0) → ( gcd ‘(𝐸𝑋)) = ( gcd ‘𝑋))
 
Theoremeucalglt 16218* The second member of the state decreases with each iteration of the step function 𝐸 for Euclid's Algorithm. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 29-May-2014.)
𝐸 = (𝑥 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))       (𝑋 ∈ (ℕ0 × ℕ0) → ((2nd ‘(𝐸𝑋)) ≠ 0 → (2nd ‘(𝐸𝑋)) < (2nd𝑋)))
 
Theoremeucalgcvga 16219* Once Euclid's Algorithm halts after 𝑁 steps, the second element of the state remains 0 . (Contributed by Paul Chapman, 22-Jun-2011.) (Revised by Mario Carneiro, 29-May-2014.)
𝐸 = (𝑥 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))    &   𝑅 = seq0((𝐸 ∘ 1st ), (ℕ0 × {𝐴}))    &   𝑁 = (2nd𝐴)       (𝐴 ∈ (ℕ0 × ℕ0) → (𝐾 ∈ (ℤ𝑁) → (2nd ‘(𝑅𝐾)) = 0))
 
Theoremeucalg 16220* Euclid's Algorithm computes the greatest common divisor of two nonnegative integers by repeatedly replacing the larger of them with its remainder modulo the smaller until the remainder is 0. Theorem 1.15 in [ApostolNT] p. 20.

Upon halting, the 1st member of the final state (𝑅𝑁) is equal to the gcd of the values comprising the input state 𝑀, 𝑁. This is Metamath 100 proof #69 (greatest common divisor algorithm). (Contributed by Paul Chapman, 31-Mar-2011.) (Proof shortened by Mario Carneiro, 29-May-2014.)

𝐸 = (𝑥 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))    &   𝑅 = seq0((𝐸 ∘ 1st ), (ℕ0 × {𝐴}))    &   𝐴 = ⟨𝑀, 𝑁       ((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → (1st ‘(𝑅𝑁)) = (𝑀 gcd 𝑁))
 
6.1.11  The least common multiple

According to Wikipedia ("Least common multiple", 27-Aug-2020, https://en.wikipedia.org/wiki/Least_common_multiple): "In arithmetic and number theory, the least common multiple, lowest common multiple, or smallest common multiple of two integers a and b, usually denoted by lcm(a, b), is the smallest positive integer that is divisible by both a and b. Since division of integers by zero is undefined, this definition has meaning only if a and b are both different from zero. However, some authors define lcm(a,0) as 0 for all a, which is the result of taking the lcm to be the least upper bound in the lattice of divisibility. ... The lcm of more than two integers is also well-defined: it is the smallest positive integer hat is divisible by each of them."

In this section, an operation calculating the least common multiple of two integers (df-lcm 16223) as well as a function mapping a set of integers to their least common multiple (df-lcmf 16224) are provided. Both definitions are valid for all integers, including negative integers and 0, obeying the above mentioned convention. It is shown by lcmfpr 16260 that the two definitions are compatible.

 
Syntaxclcm 16221 Extend the definition of a class to include the least common multiple operator.
class lcm
 
Syntaxclcmf 16222 Extend the definition of a class to include the least common multiple function.
class lcm
 
Definitiondf-lcm 16223* Define the lcm operator. For example, (6 lcm 9) = 18 (ex-lcm 28723). (Contributed by Steve Rodriguez, 20-Jan-2020.) (Revised by AV, 16-Sep-2020.)
lcm = (𝑥 ∈ ℤ, 𝑦 ∈ ℤ ↦ if((𝑥 = 0 ∨ 𝑦 = 0), 0, inf({𝑛 ∈ ℕ ∣ (𝑥𝑛𝑦𝑛)}, ℝ, < )))
 
Definitiondf-lcmf 16224* Define the lcm function on a set of integers. (Contributed by AV, 21-Aug-2020.) (Revised by AV, 16-Sep-2020.)
lcm = (𝑧 ∈ 𝒫 ℤ ↦ if(0 ∈ 𝑧, 0, inf({𝑛 ∈ ℕ ∣ ∀𝑚𝑧 𝑚𝑛}, ℝ, < )))
 
Theoremlcmval 16225* Value of the lcm operator. (𝑀 lcm 𝑁) is the least common multiple of 𝑀 and 𝑁. If either 𝑀 or 𝑁 is 0, the result is defined conventionally as 0. Contrast with df-gcd 16130 and gcdval 16131. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Revised by AV, 16-Sep-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 lcm 𝑁) = if((𝑀 = 0 ∨ 𝑁 = 0), 0, inf({𝑛 ∈ ℕ ∣ (𝑀𝑛𝑁𝑛)}, ℝ, < )))
 
Theoremlcmcom 16226 The lcm operator is commutative. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Proof shortened by AV, 16-Sep-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 lcm 𝑁) = (𝑁 lcm 𝑀))
 
Theoremlcm0val 16227 The value, by convention, of the lcm operator when either operand is 0. (Use lcmcom 16226 for a left-hand 0.) (Contributed by Steve Rodriguez, 20-Jan-2020.) (Proof shortened by AV, 16-Sep-2020.)
(𝑀 ∈ ℤ → (𝑀 lcm 0) = 0)
 
Theoremlcmn0val 16228* The value of the lcm operator when both operands are nonzero. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Revised by AV, 16-Sep-2020.)
(((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∨ 𝑁 = 0)) → (𝑀 lcm 𝑁) = inf({𝑛 ∈ ℕ ∣ (𝑀𝑛𝑁𝑛)}, ℝ, < ))
 
Theoremlcmcllem 16229* Lemma for lcmn0cl 16230 and dvdslcm 16231. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Proof shortened by AV, 16-Sep-2020.)
(((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∨ 𝑁 = 0)) → (𝑀 lcm 𝑁) ∈ {𝑛 ∈ ℕ ∣ (𝑀𝑛𝑁𝑛)})
 
Theoremlcmn0cl 16230 Closure of the lcm operator. (Contributed by Steve Rodriguez, 20-Jan-2020.)
(((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∨ 𝑁 = 0)) → (𝑀 lcm 𝑁) ∈ ℕ)
 
Theoremdvdslcm 16231 The lcm of two integers is divisible by each of them. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ (𝑀 lcm 𝑁) ∧ 𝑁 ∥ (𝑀 lcm 𝑁)))
 
Theoremlcmledvds 16232 A positive integer which both operands of the lcm operator divide bounds it. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Proof shortened by AV, 16-Sep-2020.)
(((𝐾 ∈ ℕ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∨ 𝑁 = 0)) → ((𝑀𝐾𝑁𝐾) → (𝑀 lcm 𝑁) ≤ 𝐾))
 
Theoremlcmeq0 16233 The lcm of two integers is zero iff either is zero. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀 lcm 𝑁) = 0 ↔ (𝑀 = 0 ∨ 𝑁 = 0)))
 
Theoremlcmcl 16234 Closure of the lcm operator. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 lcm 𝑁) ∈ ℕ0)
 
Theoremgcddvdslcm 16235 The greatest common divisor of two numbers divides their least common multiple. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) ∥ (𝑀 lcm 𝑁))
 
Theoremlcmneg 16236 Negating one operand of the lcm operator does not alter the result. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 lcm -𝑁) = (𝑀 lcm 𝑁))
 
Theoremneglcm 16237 Negating one operand of the lcm operator does not alter the result. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (-𝑀 lcm 𝑁) = (𝑀 lcm 𝑁))
 
Theoremlcmabs 16238 The lcm of two integers is the same as that of their absolute values. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((abs‘𝑀) lcm (abs‘𝑁)) = (𝑀 lcm 𝑁))
 
Theoremlcmgcdlem 16239 Lemma for lcmgcd 16240 and lcmdvds 16241. Prove them for positive 𝑀, 𝑁, and 𝐾. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Proof shortened by AV, 16-Sep-2020.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (((𝑀 lcm 𝑁) · (𝑀 gcd 𝑁)) = (abs‘(𝑀 · 𝑁)) ∧ ((𝐾 ∈ ℕ ∧ (𝑀𝐾𝑁𝐾)) → (𝑀 lcm 𝑁) ∥ 𝐾)))
 
Theoremlcmgcd 16240 The product of two numbers' least common multiple and greatest common divisor is the absolute value of the product of the two numbers. In particular, that absolute value is the least common multiple of two coprime numbers, for which (𝑀 gcd 𝑁) = 1.

Multiple methods exist for proving this, and it is often proven either as a consequence of the fundamental theorem of arithmetic 1arith 16556 or of Bézout's identity bezout 16179; see e.g., https://proofwiki.org/wiki/Product_of_GCD_and_LCM 16179 and https://math.stackexchange.com/a/470827 16179. This proof uses the latter to first confirm it for positive integers 𝑀 and 𝑁 (the "Second Proof" in the above Stack Exchange page), then shows that implies it for all nonzero integer inputs, then finally uses lcm0val 16227 to show it applies when either or both inputs are zero. (Contributed by Steve Rodriguez, 20-Jan-2020.)

((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀 lcm 𝑁) · (𝑀 gcd 𝑁)) = (abs‘(𝑀 · 𝑁)))
 
Theoremlcmdvds 16241 The lcm of two integers divides any integer the two divide. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀𝐾𝑁𝐾) → (𝑀 lcm 𝑁) ∥ 𝐾))
 
Theoremlcmid 16242 The lcm of an integer and itself is its absolute value. (Contributed by Steve Rodriguez, 20-Jan-2020.)
(𝑀 ∈ ℤ → (𝑀 lcm 𝑀) = (abs‘𝑀))
 
Theoremlcm1 16243 The lcm of an integer and 1 is the absolute value of the integer. (Contributed by AV, 23-Aug-2020.)
(𝑀 ∈ ℤ → (𝑀 lcm 1) = (abs‘𝑀))
 
Theoremlcmgcdnn 16244 The product of two positive integers' least common multiple and greatest common divisor is the product of the two integers. (Contributed by AV, 27-Aug-2020.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝑀 lcm 𝑁) · (𝑀 gcd 𝑁)) = (𝑀 · 𝑁))
 
Theoremlcmgcdeq 16245 Two integers' absolute values are equal iff their least common multiple and greatest common divisor are equal. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀 lcm 𝑁) = (𝑀 gcd 𝑁) ↔ (abs‘𝑀) = (abs‘𝑁)))
 
Theoremlcmdvdsb 16246 Biconditional form of lcmdvds 16241. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀𝐾𝑁𝐾) ↔ (𝑀 lcm 𝑁) ∥ 𝐾))
 
Theoremlcmass 16247 Associative law for lcm operator. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Proof shortened by AV, 16-Sep-2020.)
((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑃 ∈ ℤ) → ((𝑁 lcm 𝑀) lcm 𝑃) = (𝑁 lcm (𝑀 lcm 𝑃)))
 
Theorem3lcm2e6woprm 16248 The least common multiple of three and two is six. In contrast to 3lcm2e6 16364, this proof does not use the property of 2 and 3 being prime, therefore it is much longer. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Revised by AV, 27-Aug-2020.) (Proof modification is discouraged.) (New usage is discouraged.)
(3 lcm 2) = 6
 
Theorem6lcm4e12 16249 The least common multiple of six and four is twelve. (Contributed by AV, 27-Aug-2020.)
(6 lcm 4) = 12
 
Theoremabsproddvds 16250* The absolute value of the product of the elements of a finite subset of the integers is divisible by each element of this subset. (Contributed by AV, 21-Aug-2020.)
(𝜑𝑍 ⊆ ℤ)    &   (𝜑𝑍 ∈ Fin)    &   𝑃 = (abs‘∏𝑧𝑍 𝑧)       (𝜑 → ∀𝑚𝑍 𝑚𝑃)
 
Theoremabsprodnn 16251* The absolute value of the product of the elements of a finite subset of the integers not containing 0 is a poitive integer. (Contributed by AV, 21-Aug-2020.)
(𝜑𝑍 ⊆ ℤ)    &   (𝜑𝑍 ∈ Fin)    &   𝑃 = (abs‘∏𝑧𝑍 𝑧)    &   (𝜑 → 0 ∉ 𝑍)       (𝜑𝑃 ∈ ℕ)
 
Theoremfissn0dvds 16252* For each finite subset of the integers not containing 0 there is a positive integer which is divisible by each element of this subset. (Contributed by AV, 21-Aug-2020.)
((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍) → ∃𝑛 ∈ ℕ ∀𝑚𝑍 𝑚𝑛)
 
Theoremfissn0dvdsn0 16253* For each finite subset of the integers not containing 0 there is a positive integer which is divisible by each element of this subset. (Contributed by AV, 21-Aug-2020.)
((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍) → {𝑛 ∈ ℕ ∣ ∀𝑚𝑍 𝑚𝑛} ≠ ∅)
 
Theoremlcmfval 16254* Value of the lcm function. (lcm𝑍) is the least common multiple of the integers contained in the finite subset of integers 𝑍. If at least one of the elements of 𝑍 is 0, the result is defined conventionally as 0. (Contributed by AV, 21-Apr-2020.) (Revised by AV, 16-Sep-2020.)
((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin) → (lcm𝑍) = if(0 ∈ 𝑍, 0, inf({𝑛 ∈ ℕ ∣ ∀𝑚𝑍 𝑚𝑛}, ℝ, < )))
 
Theoremlcmf0val 16255 The value, by convention, of the least common multiple for a set containing 0 is 0. (Contributed by AV, 21-Apr-2020.) (Proof shortened by AV, 16-Sep-2020.)
((𝑍 ⊆ ℤ ∧ 0 ∈ 𝑍) → (lcm𝑍) = 0)
 
Theoremlcmfn0val 16256* The value of the lcm function for a set without 0. (Contributed by AV, 21-Aug-2020.) (Revised by AV, 16-Sep-2020.)
((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍) → (lcm𝑍) = inf({𝑛 ∈ ℕ ∣ ∀𝑚𝑍 𝑚𝑛}, ℝ, < ))
 
Theoremlcmfnnval 16257* The value of the lcm function for a subset of the positive integers. (Contributed by AV, 21-Aug-2020.) (Revised by AV, 16-Sep-2020.)
((𝑍 ⊆ ℕ ∧ 𝑍 ∈ Fin) → (lcm𝑍) = inf({𝑛 ∈ ℕ ∣ ∀𝑚𝑍 𝑚𝑛}, ℝ, < ))
 
Theoremlcmfcllem 16258* Lemma for lcmfn0cl 16259 and dvdslcmf 16264. (Contributed by AV, 21-Aug-2020.) (Proof shortened by AV, 16-Sep-2020.)
((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍) → (lcm𝑍) ∈ {𝑛 ∈ ℕ ∣ ∀𝑚𝑍 𝑚𝑛})
 
Theoremlcmfn0cl 16259 Closure of the lcm function. (Contributed by AV, 21-Aug-2020.)
((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍) → (lcm𝑍) ∈ ℕ)
 
Theoremlcmfpr 16260 The value of the lcm function for an unordered pair is the value of the lcm operator for both elements. (Contributed by AV, 22-Aug-2020.) (Proof shortened by AV, 16-Sep-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (lcm‘{𝑀, 𝑁}) = (𝑀 lcm 𝑁))
 
Theoremlcmfcl 16261 Closure of the lcm function. (Contributed by AV, 21-Aug-2020.)
((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin) → (lcm𝑍) ∈ ℕ0)
 
Theoremlcmfnncl 16262 Closure of the lcm function. (Contributed by AV, 20-Apr-2020.)
((𝑍 ⊆ ℕ ∧ 𝑍 ∈ Fin) → (lcm𝑍) ∈ ℕ)
 
Theoremlcmfeq0b 16263 The least common multiple of a set of integers is 0 iff at least one of its element is 0. (Contributed by AV, 21-Aug-2020.)
((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin) → ((lcm𝑍) = 0 ↔ 0 ∈ 𝑍))
 
Theoremdvdslcmf 16264* The least common multiple of a set of integers is divisible by each of its elements. (Contributed by AV, 22-Aug-2020.)
((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin) → ∀𝑥𝑍 𝑥 ∥ (lcm𝑍))
 
Theoremlcmfledvds 16265* A positive integer which is divisible by all elements of a set of integers bounds the least common multiple of the set. (Contributed by AV, 22-Aug-2020.) (Proof shortened by AV, 16-Sep-2020.)
((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍) → ((𝐾 ∈ ℕ ∧ ∀𝑚𝑍 𝑚𝐾) → (lcm𝑍) ≤ 𝐾))
 
Theoremlcmf 16266* Characterization of the least common multiple of a set of integers (without 0): A positiven integer is the least common multiple of a set of integers iff it divides each of the elements of the set and every integer which divides each of the elements of the set is greater than or equal to this integer. (Contributed by AV, 22-Aug-2020.)
((𝐾 ∈ ℕ ∧ (𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍)) → (𝐾 = (lcm𝑍) ↔ (∀𝑚𝑍 𝑚𝐾 ∧ ∀𝑘 ∈ ℕ (∀𝑚𝑍 𝑚𝑘𝐾𝑘))))
 
Theoremlcmf0 16267 The least common multiple of the empty set is 1. (Contributed by AV, 22-Aug-2020.) (Proof shortened by AV, 16-Sep-2020.)
(lcm‘∅) = 1
 
Theoremlcmfsn 16268 The least common multiple of a singleton is its absolute value. (Contributed by AV, 22-Aug-2020.)
(𝑀 ∈ ℤ → (lcm‘{𝑀}) = (abs‘𝑀))
 
Theoremlcmftp 16269 The least common multiple of a triple of integers is the least common multiple of the third integer and the least common multiple of the first two integers. Although there would be a shorter proof using lcmfunsn 16277, this explicit proof (not based on induction) should be kept. (Proof modification is discouraged.) (Contributed by AV, 23-Aug-2020.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ) → (lcm‘{𝐴, 𝐵, 𝐶}) = ((𝐴 lcm 𝐵) lcm 𝐶))
 
Theoremlcmfunsnlem1 16270* Lemma for lcmfdvds 16275 and lcmfunsnlem 16274 (Induction step part 1). (Contributed by AV, 25-Aug-2020.)
(((𝑧 ∈ ℤ ∧ 𝑦 ⊆ ℤ ∧ 𝑦 ∈ Fin) ∧ (∀𝑘 ∈ ℤ (∀𝑚𝑦 𝑚𝑘 → (lcm𝑦) ∥ 𝑘) ∧ ∀𝑛 ∈ ℤ (lcm‘(𝑦 ∪ {𝑛})) = ((lcm𝑦) lcm 𝑛))) → ∀𝑘 ∈ ℤ (∀𝑚 ∈ (𝑦 ∪ {𝑧})𝑚𝑘 → (lcm‘(𝑦 ∪ {𝑧})) ∥ 𝑘))
 
Theoremlcmfunsnlem2lem1 16271* Lemma 1 for lcmfunsnlem2 16273. (Contributed by AV, 26-Aug-2020.)
(((0 ∉ 𝑦𝑧 ≠ 0 ∧ 𝑛 ≠ 0) ∧ (𝑛 ∈ ℤ ∧ ((𝑧 ∈ ℤ ∧ 𝑦 ⊆ ℤ ∧ 𝑦 ∈ Fin) ∧ (∀𝑘 ∈ ℤ (∀𝑚𝑦 𝑚𝑘 → (lcm𝑦) ∥ 𝑘) ∧ ∀𝑛 ∈ ℤ (lcm‘(𝑦 ∪ {𝑛})) = ((lcm𝑦) lcm 𝑛))))) → ∀𝑘 ∈ ℕ (∀𝑖 ∈ ((𝑦 ∪ {𝑧}) ∪ {𝑛})𝑖𝑘 → ((lcm‘(𝑦 ∪ {𝑧})) lcm 𝑛) ≤ 𝑘))
 
Theoremlcmfunsnlem2lem2 16272* Lemma 2 for lcmfunsnlem2 16273. (Contributed by AV, 26-Aug-2020.)
(((0 ∉ 𝑦𝑧 ≠ 0 ∧ 𝑛 ≠ 0) ∧ (𝑛 ∈ ℤ ∧ ((𝑧 ∈ ℤ ∧ 𝑦 ⊆ ℤ ∧ 𝑦 ∈ Fin) ∧ (∀𝑘 ∈ ℤ (∀𝑚𝑦 𝑚𝑘 → (lcm𝑦) ∥ 𝑘) ∧ ∀𝑛 ∈ ℤ (lcm‘(𝑦 ∪ {𝑛})) = ((lcm𝑦) lcm 𝑛))))) → (lcm‘((𝑦 ∪ {𝑧}) ∪ {𝑛})) = ((lcm‘(𝑦 ∪ {𝑧})) lcm 𝑛))
 
Theoremlcmfunsnlem2 16273* Lemma for lcmfunsn 16277 and lcmfunsnlem 16274 (Induction step part 2). (Contributed by AV, 26-Aug-2020.)
(((𝑧 ∈ ℤ ∧ 𝑦 ⊆ ℤ ∧ 𝑦 ∈ Fin) ∧ (∀𝑘 ∈ ℤ (∀𝑚𝑦 𝑚𝑘 → (lcm𝑦) ∥ 𝑘) ∧ ∀𝑛 ∈ ℤ (lcm‘(𝑦 ∪ {𝑛})) = ((lcm𝑦) lcm 𝑛))) → ∀𝑛 ∈ ℤ (lcm‘((𝑦 ∪ {𝑧}) ∪ {𝑛})) = ((lcm‘(𝑦 ∪ {𝑧})) lcm 𝑛))
 
Theoremlcmfunsnlem 16274* Lemma for lcmfdvds 16275 and lcmfunsn 16277. These two theorems must be proven simultaneously by induction on the cardinality of a finite set 𝑌, because they depend on each other. This can be seen by the two parts lcmfunsnlem1 16270 and lcmfunsnlem2 16273 of the induction step, each of them using both induction hypotheses. (Contributed by AV, 26-Aug-2020.)
((𝑌 ⊆ ℤ ∧ 𝑌 ∈ Fin) → (∀𝑘 ∈ ℤ (∀𝑚𝑌 𝑚𝑘 → (lcm𝑌) ∥ 𝑘) ∧ ∀𝑛 ∈ ℤ (lcm‘(𝑌 ∪ {𝑛})) = ((lcm𝑌) lcm 𝑛)))
 
Theoremlcmfdvds 16275* The least common multiple of a set of integers divides any integer which is divisible by all elements of the set. (Contributed by AV, 26-Aug-2020.)
((𝐾 ∈ ℤ ∧ 𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin) → (∀𝑚𝑍 𝑚𝐾 → (lcm𝑍) ∥ 𝐾))
 
Theoremlcmfdvdsb 16276* Biconditional form of lcmfdvds 16275. (Contributed by AV, 26-Aug-2020.)
((𝐾 ∈ ℤ ∧ 𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin) → (∀𝑚𝑍 𝑚𝐾 ↔ (lcm𝑍) ∥ 𝐾))
 
Theoremlcmfunsn 16277 The lcm function for a union of a set of integer and a singleton. (Contributed by AV, 26-Aug-2020.)
((𝑌 ⊆ ℤ ∧ 𝑌 ∈ Fin ∧ 𝑁 ∈ ℤ) → (lcm‘(𝑌 ∪ {𝑁})) = ((lcm𝑌) lcm 𝑁))
 
Theoremlcmfun 16278 The lcm function for a union of sets of integers. (Contributed by AV, 27-Aug-2020.)
(((𝑌 ⊆ ℤ ∧ 𝑌 ∈ Fin) ∧ (𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin)) → (lcm‘(𝑌𝑍)) = ((lcm𝑌) lcm (lcm𝑍)))
 
Theoremlcmfass 16279 Associative law for the lcm function. (Contributed by AV, 27-Aug-2020.)
(((𝑌 ⊆ ℤ ∧ 𝑌 ∈ Fin) ∧ (𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin)) → (lcm‘({(lcm𝑌)} ∪ 𝑍)) = (lcm‘(𝑌 ∪ {(lcm𝑍)})))
 
Theoremlcmf2a3a4e12 16280 The least common multiple of 2 , 3 and 4 is 12. (Contributed by AV, 27-Aug-2020.)
(lcm‘{2, 3, 4}) = 12
 
Theoremlcmflefac 16281 The least common multiple of all positive integers less than or equal to an integer is less than or equal to the factorial of the integer. (Contributed by AV, 16-Aug-2020.) (Revised by AV, 27-Aug-2020.)
(𝑁 ∈ ℕ → (lcm‘(1...𝑁)) ≤ (!‘𝑁))
 
6.1.12  Coprimality and Euclid's lemma

According to Wikipedia "Coprime integers", see https://en.wikipedia.org/wiki/Coprime_integers (16-Aug-2020) "[...] two integers a and b are said to be relatively prime, mutually prime, or coprime [...] if the only positive integer (factor) that divides both of them is 1. Consequently, any prime number that divides one does not divide the other. This is equivalent to their greatest common divisor (gcd) being 1.". In the following, we use this equivalent characterization to say that 𝐴 ∈ ℤ and 𝐵 ∈ ℤ are coprime (or relatively prime) if (𝐴 gcd 𝐵) = 1. The equivalence of the definitions is shown by coprmgcdb 16282. The negation, i.e. two integers are not coprime, can be expressed either by (𝐴 gcd 𝐵) ≠ 1, see ncoprmgcdne1b 16283, or equivalently by 1 < (𝐴 gcd 𝐵), see ncoprmgcdgt1b 16284.

A proof of Euclid's lemma based on coprimality is provided in coprmdvds 16286 (see euclemma 16346 for a version of Euclid's lemma for primes).

 
Theoremcoprmgcdb 16282* Two positive integers are coprime, i.e. the only positive integer that divides both of them is 1, iff their greatest common divisor is 1. (Contributed by AV, 9-Aug-2020.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (∀𝑖 ∈ ℕ ((𝑖𝐴𝑖𝐵) → 𝑖 = 1) ↔ (𝐴 gcd 𝐵) = 1))
 
Theoremncoprmgcdne1b 16283* Two positive integers are not coprime, i.e. there is an integer greater than 1 which divides both integers, iff their greatest common divisor is not 1. See prmdvdsncoprmbd 16359 for a version where the existential quantifier is restricted to primes. (Contributed by AV, 9-Aug-2020.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (∃𝑖 ∈ (ℤ‘2)(𝑖𝐴𝑖𝐵) ↔ (𝐴 gcd 𝐵) ≠ 1))
 
Theoremncoprmgcdgt1b 16284* Two positive integers are not coprime, i.e. there is an integer greater than 1 which divides both integers, iff their greatest common divisor is greater than 1. (Contributed by AV, 9-Aug-2020.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (∃𝑖 ∈ (ℤ‘2)(𝑖𝐴𝑖𝐵) ↔ 1 < (𝐴 gcd 𝐵)))
 
Theoremcoprmdvds1 16285 If two positive integers are coprime, i.e. their greatest common divisor is 1, the only positive integer that divides both of them is 1. (Contributed by AV, 4-Aug-2021.)
((𝐹 ∈ ℕ ∧ 𝐺 ∈ ℕ ∧ (𝐹 gcd 𝐺) = 1) → ((𝐼 ∈ ℕ ∧ 𝐼𝐹𝐼𝐺) → 𝐼 = 1))
 
Theoremcoprmdvds 16286 Euclid's Lemma (see ProofWiki "Euclid's Lemma", 10-Jul-2021, https://proofwiki.org/wiki/Euclid's_Lemma): If an integer divides the product of two integers and is coprime to one of them, then it divides the other. See also theorem 1.5 in [ApostolNT] p. 16. Generalization of euclemma 16346. (Contributed by Paul Chapman, 22-Jun-2011.) (Proof shortened by AV, 10-Jul-2021.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 ∥ (𝑀 · 𝑁) ∧ (𝐾 gcd 𝑀) = 1) → 𝐾𝑁))
 
Theoremcoprmdvds2 16287 If an integer is divisible by two coprime integers, then it is divisible by their product. (Contributed by Mario Carneiro, 24-Feb-2014.)
(((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝐾 ∈ ℤ) ∧ (𝑀 gcd 𝑁) = 1) → ((𝑀𝐾𝑁𝐾) → (𝑀 · 𝑁) ∥ 𝐾))
 
Theoremmulgcddvds 16288 One half of rpmulgcd2 16289, which does not need the coprimality assumption. (Contributed by Mario Carneiro, 2-Jul-2015.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝐾 gcd (𝑀 · 𝑁)) ∥ ((𝐾 gcd 𝑀) · (𝐾 gcd 𝑁)))
 
Theoremrpmulgcd2 16289 If 𝑀 is relatively prime to 𝑁, then the GCD of 𝐾 with 𝑀 · 𝑁 is the product of the GCDs with 𝑀 and 𝑁 respectively. (Contributed by Mario Carneiro, 2-Jul-2015.)
(((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ (𝑀 gcd 𝑁) = 1) → (𝐾 gcd (𝑀 · 𝑁)) = ((𝐾 gcd 𝑀) · (𝐾 gcd 𝑁)))
 
Theoremqredeq 16290 Two equal reduced fractions have the same numerator and denominator. (Contributed by Jeff Hankins, 29-Sep-2013.)
(((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ ∧ (𝑀 gcd 𝑁) = 1) ∧ (𝑃 ∈ ℤ ∧ 𝑄 ∈ ℕ ∧ (𝑃 gcd 𝑄) = 1) ∧ (𝑀 / 𝑁) = (𝑃 / 𝑄)) → (𝑀 = 𝑃𝑁 = 𝑄))
 
Theoremqredeu 16291* Every rational number has a unique reduced form. (Contributed by Jeff Hankins, 29-Sep-2013.)
(𝐴 ∈ ℚ → ∃!𝑥 ∈ (ℤ × ℕ)(((1st𝑥) gcd (2nd𝑥)) = 1 ∧ 𝐴 = ((1st𝑥) / (2nd𝑥))))
 
Theoremrpmul 16292 If 𝐾 is relatively prime to 𝑀 and to 𝑁, it is also relatively prime to their product. (Contributed by Mario Carneiro, 24-Feb-2014.) (Proof shortened by Mario Carneiro, 2-Jul-2015.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (((𝐾 gcd 𝑀) = 1 ∧ (𝐾 gcd 𝑁) = 1) → (𝐾 gcd (𝑀 · 𝑁)) = 1))
 
Theoremrpdvds 16293 If 𝐾 is relatively prime to 𝑁 then it is also relatively prime to any divisor 𝑀 of 𝑁. (Contributed by Mario Carneiro, 19-Jun-2015.)
(((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ((𝐾 gcd 𝑁) = 1 ∧ 𝑀𝑁)) → (𝐾 gcd 𝑀) = 1)
 
Theoremcoprmprod 16294* The product of the elements of a sequence of pairwise coprime positive integers is coprime to a positive integer which is coprime to all integers of the sequence. (Contributed by AV, 18-Aug-2020.)
(((𝑀 ∈ Fin ∧ 𝑀 ⊆ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝐹:ℕ⟶ℕ ∧ ∀𝑚𝑀 ((𝐹𝑚) gcd 𝑁) = 1) → (∀𝑚𝑀𝑛 ∈ (𝑀 ∖ {𝑚})((𝐹𝑚) gcd (𝐹𝑛)) = 1 → (∏𝑚𝑀 (𝐹𝑚) gcd 𝑁) = 1))
 
Theoremcoprmproddvdslem 16295* Lemma for coprmproddvds 16296: Induction step. (Contributed by AV, 19-Aug-2020.)
((𝑦 ∈ Fin ∧ ¬ 𝑧𝑦) → ((((𝑦 ⊆ ℕ ∧ (𝐾 ∈ ℕ ∧ 𝐹:ℕ⟶ℕ)) ∧ (∀𝑚𝑦𝑛 ∈ (𝑦 ∖ {𝑚})((𝐹𝑚) gcd (𝐹𝑛)) = 1 ∧ ∀𝑚𝑦 (𝐹𝑚) ∥ 𝐾)) → ∏𝑚𝑦 (𝐹𝑚) ∥ 𝐾) → ((((𝑦 ∪ {𝑧}) ⊆ ℕ ∧ (𝐾 ∈ ℕ ∧ 𝐹:ℕ⟶ℕ)) ∧ (∀𝑚 ∈ (𝑦 ∪ {𝑧})∀𝑛 ∈ ((𝑦 ∪ {𝑧}) ∖ {𝑚})((𝐹𝑚) gcd (𝐹𝑛)) = 1 ∧ ∀𝑚 ∈ (𝑦 ∪ {𝑧})(𝐹𝑚) ∥ 𝐾)) → ∏𝑚 ∈ (𝑦 ∪ {𝑧})(𝐹𝑚) ∥ 𝐾)))
 
Theoremcoprmproddvds 16296* If a positive integer is divisible by each element of a set of pairwise coprime positive integers, then it is divisible by their product. (Contributed by AV, 19-Aug-2020.)
(((𝑀 ⊆ ℕ ∧ 𝑀 ∈ Fin) ∧ (𝐾 ∈ ℕ ∧ 𝐹:ℕ⟶ℕ) ∧ (∀𝑚𝑀𝑛 ∈ (𝑀 ∖ {𝑚})((𝐹𝑚) gcd (𝐹𝑛)) = 1 ∧ ∀𝑚𝑀 (𝐹𝑚) ∥ 𝐾)) → ∏𝑚𝑀 (𝐹𝑚) ∥ 𝐾)
 
6.1.13  Cancellability of congruences
 
Theoremcongr 16297* Definition of congruence by integer multiple (see ProofWiki "Congruence (Number Theory)", 11-Jul-2021, https://proofwiki.org/wiki/Definition:Congruence_(Number_Theory)): An integer 𝐴 is congruent to an integer 𝐵 modulo 𝑀 if their difference is a multiple of 𝑀. See also the definition in [ApostolNT] p. 104: "... 𝑎 is congruent to 𝑏 modulo 𝑚, and we write 𝑎𝑏 (mod 𝑚) if 𝑚 divides the difference 𝑎𝑏", or Wikipedia "Modular arithmetic - Congruence", https://en.wikipedia.org/wiki/Modular_arithmetic#Congruence, 11-Jul-2021,: "Given an integer n > 1, called a modulus, two integers are said to be congruent modulo n, if n is a divisor of their difference (i.e., if there is an integer k such that a-b = kn)". (Contributed by AV, 11-Jul-2021.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝑀 ∈ ℕ) → ((𝐴 mod 𝑀) = (𝐵 mod 𝑀) ↔ ∃𝑛 ∈ ℤ (𝑛 · 𝑀) = (𝐴𝐵)))
 
Theoremdivgcdcoprm0 16298 Integers divided by gcd are coprime. (Contributed by AV, 12-Jul-2021.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐵 ≠ 0) → ((𝐴 / (𝐴 gcd 𝐵)) gcd (𝐵 / (𝐴 gcd 𝐵))) = 1)
 
Theoremdivgcdcoprmex 16299* Integers divided by gcd are coprime (see ProofWiki "Integers Divided by GCD are Coprime", 11-Jul-2021, https://proofwiki.org/wiki/Integers_Divided_by_GCD_are_Coprime): Any pair of integers, not both zero, can be reduced to a pair of coprime ones by dividing them by their gcd. (Contributed by AV, 12-Jul-2021.)
((𝐴 ∈ ℤ ∧ (𝐵 ∈ ℤ ∧ 𝐵 ≠ 0) ∧ 𝑀 = (𝐴 gcd 𝐵)) → ∃𝑎 ∈ ℤ ∃𝑏 ∈ ℤ (𝐴 = (𝑀 · 𝑎) ∧ 𝐵 = (𝑀 · 𝑏) ∧ (𝑎 gcd 𝑏) = 1))
 
Theoremcncongr1 16300 One direction of the bicondition in cncongr 16302. Theorem 5.4 in [ApostolNT] p. 109. (Contributed by AV, 13-Jul-2021.)
(((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ) ∧ (𝑁 ∈ ℕ ∧ 𝑀 = (𝑁 / (𝐶 gcd 𝑁)))) → (((𝐴 · 𝐶) mod 𝑁) = ((𝐵 · 𝐶) mod 𝑁) → (𝐴 mod 𝑀) = (𝐵 mod 𝑀)))
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