P. 166 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 16501-16600   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	algrf 16501	An algorithm is a step function 𝐹:𝑆⟶𝑆 on a state space 𝑆. An algorithm acts on an initial state 𝐴 ∈ 𝑆 by iteratively applying 𝐹 to give 𝐴, (𝐹‘𝐴), (𝐹‘(𝐹‘𝐴)) 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:ℕ0⟶𝑆 "runs" the algorithm 𝐹 so that (𝑅‘𝑘) 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 = (ℤ≥‘𝑀)    &   ⊢ 𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}))    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝐴 ∈ 𝑆)    &   ⊢ (𝜑 → 𝐹:𝑆⟶𝑆)    ⇒   ⊢ (𝜑 → 𝑅:𝑍⟶𝑆)
Theorem	algrp1 16502	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 = (ℤ≥‘𝑀)    &   ⊢ 𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}))    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝐴 ∈ 𝑆)    &   ⊢ (𝜑 → 𝐹:𝑆⟶𝑆)    ⇒   ⊢ ((𝜑 ∧ 𝐾 ∈ 𝑍) → (𝑅‘(𝐾 + 1)) = (𝐹‘(𝑅‘𝐾)))
Theorem	alginv 16503*	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:𝑆⟶𝑆    &   ⊢ (𝑥 ∈ 𝑆 → (𝐼‘(𝐹‘𝑥)) = (𝐼‘𝑥))    ⇒   ⊢ ((𝐴 ∈ 𝑆 ∧ 𝐾 ∈ ℕ0) → (𝐼‘(𝑅‘𝐾)) = (𝐼‘(𝑅‘0)))
Theorem	algcvg 16504*	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 𝑋 ∈ 𝑆 is not a fixed point of 𝐹, then (𝐶‘(𝐹‘𝑋)) < (𝐶‘𝑋). If 𝐶 is a countdown function for algorithm 𝐹, the sequence (𝐶‘(𝑅‘𝑘)) 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:𝑆⟶𝑆    &   ⊢ 𝑅 = seq0((𝐹 ∘ 1st ), (ℕ0 × {𝐴}))    &   ⊢ 𝐶:𝑆⟶ℕ0    &   ⊢ (𝑧 ∈ 𝑆 → ((𝐶‘(𝐹‘𝑧)) ≠ 0 → (𝐶‘(𝐹‘𝑧)) < (𝐶‘𝑧)))    &   ⊢ 𝑁 = (𝐶‘𝐴)    ⇒   ⊢ (𝐴 ∈ 𝑆 → (𝐶‘(𝑅‘𝑁)) = 0)
Theorem	algcvgblem 16505	Lemma for algcvgb 16506. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ ((𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0) → ((𝑁 ≠ 0 → 𝑁 < 𝑀) ↔ ((𝑀 ≠ 0 → 𝑁 < 𝑀) ∧ (𝑀 = 0 → 𝑁 = 0))))
Theorem	algcvgb 16506	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:𝑆⟶𝑆    &   ⊢ 𝐶:𝑆⟶ℕ0    ⇒   ⊢ (𝑋 ∈ 𝑆 → (((𝐶‘(𝐹‘𝑋)) ≠ 0 → (𝐶‘(𝐹‘𝑋)) < (𝐶‘𝑋)) ↔ (((𝐶‘𝑋) ≠ 0 → (𝐶‘(𝐹‘𝑋)) < (𝐶‘𝑋)) ∧ ((𝐶‘𝑋) = 0 → (𝐶‘(𝐹‘𝑋)) = 0))))
Theorem	algcvga 16507*	The countdown function 𝐶 remains 0 after 𝑁 steps. (Contributed by Paul Chapman, 22-Jun-2011.)
⊢ 𝐹:𝑆⟶𝑆    &   ⊢ 𝑅 = seq0((𝐹 ∘ 1st ), (ℕ0 × {𝐴}))    &   ⊢ 𝐶:𝑆⟶ℕ0    &   ⊢ (𝑧 ∈ 𝑆 → ((𝐶‘(𝐹‘𝑧)) ≠ 0 → (𝐶‘(𝐹‘𝑧)) < (𝐶‘𝑧)))    &   ⊢ 𝑁 = (𝐶‘𝐴)    ⇒   ⊢ (𝐴 ∈ 𝑆 → (𝐾 ∈ (ℤ≥‘𝑁) → (𝐶‘(𝑅‘𝐾)) = 0))
Theorem	algfx 16508*	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:𝑆⟶𝑆    &   ⊢ 𝑅 = seq0((𝐹 ∘ 1st ), (ℕ0 × {𝐴}))    &   ⊢ 𝐶:𝑆⟶ℕ0    &   ⊢ (𝑧 ∈ 𝑆 → ((𝐶‘(𝐹‘𝑧)) ≠ 0 → (𝐶‘(𝐹‘𝑧)) < (𝐶‘𝑧)))    &   ⊢ 𝑁 = (𝐶‘𝐴)    &   ⊢ (𝑧 ∈ 𝑆 → ((𝐶‘𝑧) = 0 → (𝐹‘𝑧) = 𝑧))    ⇒   ⊢ (𝐴 ∈ 𝑆 → (𝐾 ∈ (ℤ≥‘𝑁) → (𝑅‘𝐾) = (𝑅‘𝑁)))
6.1.10  Euclid's Algorithm
Theorem	eucalgval2 16509*	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 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0) → (𝑀𝐸𝑁) = if(𝑁 = 0, ⟨𝑀, 𝑁⟩, ⟨𝑁, (𝑀 mod 𝑁)⟩))
Theorem	eucalgval 16510*	Euclid's Algorithm eucalg 16515 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) → (𝐸‘𝑋) = if((2nd ‘𝑋) = 0, 𝑋, ⟨(2nd ‘𝑋), ( mod ‘𝑋)⟩))
Theorem	eucalgf 16511*	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)
Theorem	eucalginv 16512*	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 ‘(𝐸‘𝑋)) = ( gcd ‘𝑋))
Theorem	eucalglt 16513*	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 ‘(𝐸‘𝑋)) ≠ 0 → (2nd ‘(𝐸‘𝑋)) < (2nd ‘𝑋)))
Theorem	eucalgcvga 16514*	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 = (2nd ‘𝐴)    ⇒   ⊢ (𝐴 ∈ (ℕ0 × ℕ0) → (𝐾 ∈ (ℤ≥‘𝑁) → (2nd ‘(𝑅‘𝐾)) = 0))
Theorem	eucalg 16515*	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 first member of the final state (𝑅‘𝑁) 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 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0) → (1st ‘(𝑅‘𝑁)) = (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 16518) as well as a function mapping a set of integers to their least common multiple (df-lcmf 16519) are provided. Both definitions are valid for all integers, including negative integers and 0, obeying the above mentioned convention. It is shown by lcmfpr 16555 that the two definitions are compatible.
Syntax	clcm 16516	Extend the definition of a class to include the least common multiple operator.
class lcm
Syntax	clcmf 16517	Extend the definition of a class to include the least common multiple function.
class lcm
Definition	df-lcm 16518*	Define the lcm operator. For example, (6 lcm 9) = 18 (ex-lcm 30517). (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 ∣ (𝑥 ∥ 𝑛 ∧ 𝑦 ∥ 𝑛)}, ℝ, < )))
Definition	df-lcmf 16519*	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 ∣ ∀𝑚 ∈ 𝑧 𝑚 ∥ 𝑛}, ℝ, < )))
Theorem	lcmval 16520*	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 16423 and gcdval 16424. (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 ∣ (𝑀 ∥ 𝑛 ∧ 𝑁 ∥ 𝑛)}, ℝ, < )))
Theorem	lcmcom 16521	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))
Theorem	lcm0val 16522	The value, by convention, of the lcm operator when either operand is 0. (Use lcmcom 16521 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)
Theorem	lcmn0val 16523*	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 ∣ (𝑀 ∥ 𝑛 ∧ 𝑁 ∥ 𝑛)}, ℝ, < ))
Theorem	lcmcllem 16524*	Lemma for lcmn0cl 16525 and dvdslcm 16526. (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 ∣ (𝑀 ∥ 𝑛 ∧ 𝑁 ∥ 𝑛)})
Theorem	lcmn0cl 16525	Closure of the lcm operator. (Contributed by Steve Rodriguez, 20-Jan-2020.)
⊢ (((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∨ 𝑁 = 0)) → (𝑀 lcm 𝑁) ∈ ℕ)
Theorem	dvdslcm 16526	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)))
Theorem	lcmledvds 16527	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)) → ((𝑀 ∥ 𝐾 ∧ 𝑁 ∥ 𝐾) → (𝑀 lcm 𝑁) ≤ 𝐾))
Theorem	lcmeq0 16528	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)))
Theorem	lcmcl 16529	Closure of the lcm operator. (Contributed by Steve Rodriguez, 20-Jan-2020.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 lcm 𝑁) ∈ ℕ0)
Theorem	gcddvdslcm 16530	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))
Theorem	lcmneg 16531	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))
Theorem	neglcm 16532	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))
Theorem	lcmabs 16533	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))
Theorem	lcmgcdlem 16534	Lemma for lcmgcd 16535 and lcmdvds 16536. 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 ∧ (𝑀 ∥ 𝐾 ∧ 𝑁 ∥ 𝐾)) → (𝑀 lcm 𝑁) ∥ 𝐾)))
Theorem	lcmgcd 16535	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 16856 or of Bézout's identity bezout 16471; see e.g., https://proofwiki.org/wiki/Product_of_GCD_and_LCM 16471 and https://math.stackexchange.com/a/470827 16471. 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 16522 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)))
Theorem	lcmdvds 16536	The lcm of two integers divides any integer the two divide. (Contributed by Steve Rodriguez, 20-Jan-2020.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀 ∥ 𝐾 ∧ 𝑁 ∥ 𝐾) → (𝑀 lcm 𝑁) ∥ 𝐾))
Theorem	lcmid 16537	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))
Theorem	lcm1 16538	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))
Theorem	lcmgcdnn 16539	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))
Theorem	lcmgcdeq 16540	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)))
Theorem	lcmdvdsb 16541	Biconditional form of lcmdvds 16536. (Contributed by Steve Rodriguez, 20-Jan-2020.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀 ∥ 𝐾 ∧ 𝑁 ∥ 𝐾) ↔ (𝑀 lcm 𝑁) ∥ 𝐾))
Theorem	lcmass 16542	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)))
Theorem	3lcm2e6woprm 16543	The least common multiple of three and two is six. In contrast to 3lcm2e6 16660, 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
Theorem	6lcm4e12 16544	The least common multiple of six and four is twelve. (Contributed by AV, 27-Aug-2020.)
⊢ (6 lcm 4) = ;12
Theorem	absproddvds 16545*	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.)
⊢ (𝜑 → 𝑍 ⊆ ℤ)    &   ⊢ (𝜑 → 𝑍 ∈ Fin)    &   ⊢ 𝑃 = (abs‘∏𝑧 ∈ 𝑍 𝑧)    ⇒   ⊢ (𝜑 → ∀𝑚 ∈ 𝑍 𝑚 ∥ 𝑃)
Theorem	absprodnn 16546*	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.)
⊢ (𝜑 → 𝑍 ⊆ ℤ)    &   ⊢ (𝜑 → 𝑍 ∈ Fin)    &   ⊢ 𝑃 = (abs‘∏𝑧 ∈ 𝑍 𝑧)    &   ⊢ (𝜑 → 0 ∉ 𝑍)    ⇒   ⊢ (𝜑 → 𝑃 ∈ ℕ)
Theorem	fissn0dvds 16547*	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 ∀𝑚 ∈ 𝑍 𝑚 ∥ 𝑛)
Theorem	fissn0dvdsn0 16548*	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 ∣ ∀𝑚 ∈ 𝑍 𝑚 ∥ 𝑛} ≠ ∅)
Theorem	lcmfval 16549*	Value of the lcm function. (lcm‘𝑍) 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) → (lcm‘𝑍) = if(0 ∈ 𝑍, 0, inf({𝑛 ∈ ℕ ∣ ∀𝑚 ∈ 𝑍 𝑚 ∥ 𝑛}, ℝ, < )))
Theorem	lcmf0val 16550	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) → (lcm‘𝑍) = 0)
Theorem	lcmfn0val 16551*	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) → (lcm‘𝑍) = inf({𝑛 ∈ ℕ ∣ ∀𝑚 ∈ 𝑍 𝑚 ∥ 𝑛}, ℝ, < ))
Theorem	lcmfnnval 16552*	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) → (lcm‘𝑍) = inf({𝑛 ∈ ℕ ∣ ∀𝑚 ∈ 𝑍 𝑚 ∥ 𝑛}, ℝ, < ))
Theorem	lcmfcllem 16553*	Lemma for lcmfn0cl 16554 and dvdslcmf 16559. (Contributed by AV, 21-Aug-2020.) (Proof shortened by AV, 16-Sep-2020.)
⊢ ((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍) → (lcm‘𝑍) ∈ {𝑛 ∈ ℕ ∣ ∀𝑚 ∈ 𝑍 𝑚 ∥ 𝑛})
Theorem	lcmfn0cl 16554	Closure of the lcm function. (Contributed by AV, 21-Aug-2020.)
⊢ ((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍) → (lcm‘𝑍) ∈ ℕ)
Theorem	lcmfpr 16555	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))
Theorem	lcmfcl 16556	Closure of the lcm function. (Contributed by AV, 21-Aug-2020.)
⊢ ((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin) → (lcm‘𝑍) ∈ ℕ0)
Theorem	lcmfnncl 16557	Closure of the lcm function. (Contributed by AV, 20-Apr-2020.)
⊢ ((𝑍 ⊆ ℕ ∧ 𝑍 ∈ Fin) → (lcm‘𝑍) ∈ ℕ)
Theorem	lcmfeq0b 16558	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) → ((lcm‘𝑍) = 0 ↔ 0 ∈ 𝑍))
Theorem	dvdslcmf 16559*	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) → ∀𝑥 ∈ 𝑍 𝑥 ∥ (lcm‘𝑍))
Theorem	lcmfledvds 16560*	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 ∧ ∀𝑚 ∈ 𝑍 𝑚 ∥ 𝐾) → (lcm‘𝑍) ≤ 𝐾))
Theorem	lcmf 16561*	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 = (lcm‘𝑍) ↔ (∀𝑚 ∈ 𝑍 𝑚 ∥ 𝐾 ∧ ∀𝑘 ∈ ℕ (∀𝑚 ∈ 𝑍 𝑚 ∥ 𝑘 → 𝐾 ≤ 𝑘))))
Theorem	lcmf0 16562	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
Theorem	lcmfsn 16563	The least common multiple of a singleton is its absolute value. (Contributed by AV, 22-Aug-2020.)
⊢ (𝑀 ∈ ℤ → (lcm‘{𝑀}) = (abs‘𝑀))
Theorem	lcmftp 16564	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 16572, 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))
Theorem	lcmfunsnlem1 16565*	Lemma for lcmfdvds 16570 and lcmfunsnlem 16569 (Induction step part 1). (Contributed by AV, 25-Aug-2020.)
⊢ (((𝑧 ∈ ℤ ∧ 𝑦 ⊆ ℤ ∧ 𝑦 ∈ Fin) ∧ (∀𝑘 ∈ ℤ (∀𝑚 ∈ 𝑦 𝑚 ∥ 𝑘 → (lcm‘𝑦) ∥ 𝑘) ∧ ∀𝑛 ∈ ℤ (lcm‘(𝑦 ∪ {𝑛})) = ((lcm‘𝑦) lcm 𝑛))) → ∀𝑘 ∈ ℤ (∀𝑚 ∈ (𝑦 ∪ {𝑧})𝑚 ∥ 𝑘 → (lcm‘(𝑦 ∪ {𝑧})) ∥ 𝑘))
Theorem	lcmfunsnlem2lem1 16566*	Lemma 1 for lcmfunsnlem2 16568. (Contributed by AV, 26-Aug-2020.)
⊢ (((0 ∉ 𝑦 ∧ 𝑧 ≠ 0 ∧ 𝑛 ≠ 0) ∧ (𝑛 ∈ ℤ ∧ ((𝑧 ∈ ℤ ∧ 𝑦 ⊆ ℤ ∧ 𝑦 ∈ Fin) ∧ (∀𝑘 ∈ ℤ (∀𝑚 ∈ 𝑦 𝑚 ∥ 𝑘 → (lcm‘𝑦) ∥ 𝑘) ∧ ∀𝑛 ∈ ℤ (lcm‘(𝑦 ∪ {𝑛})) = ((lcm‘𝑦) lcm 𝑛))))) → ∀𝑘 ∈ ℕ (∀𝑖 ∈ ((𝑦 ∪ {𝑧}) ∪ {𝑛})𝑖 ∥ 𝑘 → ((lcm‘(𝑦 ∪ {𝑧})) lcm 𝑛) ≤ 𝑘))
Theorem	lcmfunsnlem2lem2 16567*	Lemma 2 for lcmfunsnlem2 16568. (Contributed by AV, 26-Aug-2020.)
⊢ (((0 ∉ 𝑦 ∧ 𝑧 ≠ 0 ∧ 𝑛 ≠ 0) ∧ (𝑛 ∈ ℤ ∧ ((𝑧 ∈ ℤ ∧ 𝑦 ⊆ ℤ ∧ 𝑦 ∈ Fin) ∧ (∀𝑘 ∈ ℤ (∀𝑚 ∈ 𝑦 𝑚 ∥ 𝑘 → (lcm‘𝑦) ∥ 𝑘) ∧ ∀𝑛 ∈ ℤ (lcm‘(𝑦 ∪ {𝑛})) = ((lcm‘𝑦) lcm 𝑛))))) → (lcm‘((𝑦 ∪ {𝑧}) ∪ {𝑛})) = ((lcm‘(𝑦 ∪ {𝑧})) lcm 𝑛))
Theorem	lcmfunsnlem2 16568*	Lemma for lcmfunsn 16572 and lcmfunsnlem 16569 (Induction step part 2). (Contributed by AV, 26-Aug-2020.)
⊢ (((𝑧 ∈ ℤ ∧ 𝑦 ⊆ ℤ ∧ 𝑦 ∈ Fin) ∧ (∀𝑘 ∈ ℤ (∀𝑚 ∈ 𝑦 𝑚 ∥ 𝑘 → (lcm‘𝑦) ∥ 𝑘) ∧ ∀𝑛 ∈ ℤ (lcm‘(𝑦 ∪ {𝑛})) = ((lcm‘𝑦) lcm 𝑛))) → ∀𝑛 ∈ ℤ (lcm‘((𝑦 ∪ {𝑧}) ∪ {𝑛})) = ((lcm‘(𝑦 ∪ {𝑧})) lcm 𝑛))
Theorem	lcmfunsnlem 16569*	Lemma for lcmfdvds 16570 and lcmfunsn 16572. 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 16565 and lcmfunsnlem2 16568 of the induction step, each of them using both induction hypotheses. (Contributed by AV, 26-Aug-2020.)
⊢ ((𝑌 ⊆ ℤ ∧ 𝑌 ∈ Fin) → (∀𝑘 ∈ ℤ (∀𝑚 ∈ 𝑌 𝑚 ∥ 𝑘 → (lcm‘𝑌) ∥ 𝑘) ∧ ∀𝑛 ∈ ℤ (lcm‘(𝑌 ∪ {𝑛})) = ((lcm‘𝑌) lcm 𝑛)))
Theorem	lcmfdvds 16570*	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) → (∀𝑚 ∈ 𝑍 𝑚 ∥ 𝐾 → (lcm‘𝑍) ∥ 𝐾))
Theorem	lcmfdvdsb 16571*	Biconditional form of lcmfdvds 16570. (Contributed by AV, 26-Aug-2020.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin) → (∀𝑚 ∈ 𝑍 𝑚 ∥ 𝐾 ↔ (lcm‘𝑍) ∥ 𝐾))
Theorem	lcmfunsn 16572	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})) = ((lcm‘𝑌) lcm 𝑁))
Theorem	lcmfun 16573	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‘(𝑌 ∪ 𝑍)) = ((lcm‘𝑌) lcm (lcm‘𝑍)))
Theorem	lcmfass 16574	Associative law for the lcm function. (Contributed by AV, 27-Aug-2020.)
⊢ (((𝑌 ⊆ ℤ ∧ 𝑌 ∈ Fin) ∧ (𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin)) → (lcm‘({(lcm‘𝑌)} ∪ 𝑍)) = (lcm‘(𝑌 ∪ {(lcm‘𝑍)})))
Theorem	lcmf2a3a4e12 16575	The least common multiple of 2 , 3 and 4 is 12. (Contributed by AV, 27-Aug-2020.)
⊢ (lcm‘{2, 3, 4}) = ;12
Theorem	lcmflefac 16576	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 16577. The negation, i.e. two integers are not coprime, can be expressed either by (𝐴 gcd 𝐵) ≠ 1, see ncoprmgcdne1b 16578, or equivalently by 1 < (𝐴 gcd 𝐵), see ncoprmgcdgt1b 16579. A proof of Euclid's lemma based on coprimality is provided in coprmdvds 16581 (see euclemma 16641 for a version of Euclid's lemma for primes).
Theorem	coprmgcdb 16577*	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 ((𝑖 ∥ 𝐴 ∧ 𝑖 ∥ 𝐵) → 𝑖 = 1) ↔ (𝐴 gcd 𝐵) = 1))
Theorem	ncoprmgcdne1b 16578*	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 16655 for a version where the existential quantifier is restricted to primes. (Contributed by AV, 9-Aug-2020.)
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (∃𝑖 ∈ (ℤ≥‘2)(𝑖 ∥ 𝐴 ∧ 𝑖 ∥ 𝐵) ↔ (𝐴 gcd 𝐵) ≠ 1))
Theorem	ncoprmgcdgt1b 16579*	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)(𝑖 ∥ 𝐴 ∧ 𝑖 ∥ 𝐵) ↔ 1 < (𝐴 gcd 𝐵)))
Theorem	coprmdvds1 16580	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 ∧ 𝐼 ∥ 𝐹 ∧ 𝐼 ∥ 𝐺) → 𝐼 = 1))
Theorem	coprmdvds 16581	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 16641. (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) → 𝐾 ∥ 𝑁))
Theorem	coprmdvds2 16582	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) → ((𝑀 ∥ 𝐾 ∧ 𝑁 ∥ 𝐾) → (𝑀 · 𝑁) ∥ 𝐾))
Theorem	mulgcddvds 16583	One half of rpmulgcd2 16584, 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)))
Theorem	rpmulgcd2 16584	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)))
Theorem	qredeq 16585	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 = 𝑃 ∧ 𝑁 = 𝑄))
Theorem	qredeu 16586*	Every rational number has a unique reduced form. (Contributed by Jeff Hankins, 29-Sep-2013.)
⊢ (𝐴 ∈ ℚ → ∃!𝑥 ∈ (ℤ × ℕ)(((1st ‘𝑥) gcd (2nd ‘𝑥)) = 1 ∧ 𝐴 = ((1st ‘𝑥) / (2nd ‘𝑥))))
Theorem	rpmul 16587	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))
Theorem	rpdvds 16588	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 ∧ 𝑀 ∥ 𝑁)) → (𝐾 gcd 𝑀) = 1)
Theorem	coprmprod 16589*	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 ∧ ∀𝑚 ∈ 𝑀 ((𝐹‘𝑚) gcd 𝑁) = 1) → (∀𝑚 ∈ 𝑀 ∀𝑛 ∈ (𝑀 ∖ {𝑚})((𝐹‘𝑚) gcd (𝐹‘𝑛)) = 1 → (∏𝑚 ∈ 𝑀 (𝐹‘𝑚) gcd 𝑁) = 1))
Theorem	coprmproddvdslem 16590*	Lemma for coprmproddvds 16591: Induction step. (Contributed by AV, 19-Aug-2020.)
⊢ ((𝑦 ∈ Fin ∧ ¬ 𝑧 ∈ 𝑦) → ((((𝑦 ⊆ ℕ ∧ (𝐾 ∈ ℕ ∧ 𝐹:ℕ⟶ℕ)) ∧ (∀𝑚 ∈ 𝑦 ∀𝑛 ∈ (𝑦 ∖ {𝑚})((𝐹‘𝑚) gcd (𝐹‘𝑛)) = 1 ∧ ∀𝑚 ∈ 𝑦 (𝐹‘𝑚) ∥ 𝐾)) → ∏𝑚 ∈ 𝑦 (𝐹‘𝑚) ∥ 𝐾) → ((((𝑦 ∪ {𝑧}) ⊆ ℕ ∧ (𝐾 ∈ ℕ ∧ 𝐹:ℕ⟶ℕ)) ∧ (∀𝑚 ∈ (𝑦 ∪ {𝑧})∀𝑛 ∈ ((𝑦 ∪ {𝑧}) ∖ {𝑚})((𝐹‘𝑚) gcd (𝐹‘𝑛)) = 1 ∧ ∀𝑚 ∈ (𝑦 ∪ {𝑧})(𝐹‘𝑚) ∥ 𝐾)) → ∏𝑚 ∈ (𝑦 ∪ {𝑧})(𝐹‘𝑚) ∥ 𝐾)))
Theorem	coprmproddvds 16591*	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) ∧ (∀𝑚 ∈ 𝑀 ∀𝑛 ∈ (𝑀 ∖ {𝑚})((𝐹‘𝑚) gcd (𝐹‘𝑛)) = 1 ∧ ∀𝑚 ∈ 𝑀 (𝐹‘𝑚) ∥ 𝐾)) → ∏𝑚 ∈ 𝑀 (𝐹‘𝑚) ∥ 𝐾)
6.1.13  Cancellability of congruences
Theorem	congr 16592*	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 𝑎≡𝑏 (mod 𝑚) if 𝑚 divides the difference 𝑎 − 𝑏", 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) = (𝐴 − 𝐵)))
Theorem	divgcdcoprm0 16593	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)
Theorem	divgcdcoprmex 16594*	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))
Theorem	cncongr1 16595	One direction of the bicondition in cncongr 16597. 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)))
Theorem	cncongr2 16596	The other direction of the bicondition in cncongr 16597. (Contributed by AV, 11-Jul-2021.)
⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ) ∧ (𝑁 ∈ ℕ ∧ 𝑀 = (𝑁 / (𝐶 gcd 𝑁)))) → ((𝐴 mod 𝑀) = (𝐵 mod 𝑀) → ((𝐴 · 𝐶) mod 𝑁) = ((𝐵 · 𝐶) mod 𝑁)))
Theorem	cncongr 16597	Cancellability of Congruences (see ProofWiki "Cancellability of Congruences, https://proofwiki.org/wiki/Cancellability_of_Congruences, 10-Jul-2021): Two products with a common factor are congruent modulo a positive integer iff the other factors are congruent modulo the integer divided by the greatest common divisor of the integer and the common factor. See also Theorem 5.4 "Cancellation law" 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)))
Theorem	cncongrcoprm 16598	Corollary 1 of Cancellability of Congruences: Two products with a common factor are congruent modulo an integer being coprime to the common factor iff the other factors are congruent modulo the integer. (Contributed by AV, 13-Jul-2021.)
⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ) ∧ (𝑁 ∈ ℕ ∧ (𝐶 gcd 𝑁) = 1)) → (((𝐴 · 𝐶) mod 𝑁) = ((𝐵 · 𝐶) mod 𝑁) ↔ (𝐴 mod 𝑁) = (𝐵 mod 𝑁)))
6.2  Elementary prime number theory
6.2.1  Elementary properties Remark: to represent odd prime numbers, i.e., all prime numbers except 2, the idiom 𝑃 ∈ (ℙ ∖ {2}) is used. It is a little bit shorter than (𝑃 ∈ ℙ ∧ 𝑃 ≠ 2). Both representations can be converted into each other by eldifsn 4730.
Syntax	cprime 16599	Extend the definition of a class to include the set of prime numbers.
class ℙ
Definition	df-prm 16600*	Define the set of prime numbers. (Contributed by Paul Chapman, 22-Jun-2011.)
⊢ ℙ = {𝑝 ∈ ℕ ∣ {𝑛 ∈ ℕ ∣ 𝑛 ∥ 𝑝} ≈ 2o}