P. 118 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 11701-11800   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	gcdeq0 11701	The gcd of two integers is zero iff they are both zero. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀 gcd 𝑁) = 0 ↔ (𝑀 = 0 ∧ 𝑁 = 0)))
Theorem	gcdn0gt0 11702	The gcd of two integers is positive (nonzero) iff they are not both zero. (Contributed by Paul Chapman, 22-Jun-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (¬ (𝑀 = 0 ∧ 𝑁 = 0) ↔ 0 < (𝑀 gcd 𝑁)))
Theorem	gcd0id 11703	The gcd of 0 and an integer is the integer's absolute value. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (𝑁 ∈ ℤ → (0 gcd 𝑁) = (abs‘𝑁))
Theorem	gcdid0 11704	The gcd of an integer and 0 is the integer's absolute value. Theorem 1.4(d)2 in [ApostolNT] p. 16. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ (𝑁 ∈ ℤ → (𝑁 gcd 0) = (abs‘𝑁))
Theorem	nn0gcdid0 11705	The gcd of a nonnegative integer with 0 is itself. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ (𝑁 ∈ ℕ0 → (𝑁 gcd 0) = 𝑁)
Theorem	gcdneg 11706	Negating one operand of the gcd operator does not alter the result. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd -𝑁) = (𝑀 gcd 𝑁))
Theorem	neggcd 11707	Negating one operand of the gcd operator does not alter the result. (Contributed by Paul Chapman, 22-Jun-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (-𝑀 gcd 𝑁) = (𝑀 gcd 𝑁))
Theorem	gcdaddm 11708	Adding a multiple of one operand of the gcd operator to the other does not alter the result. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) = (𝑀 gcd (𝑁 + (𝐾 · 𝑀))))
Theorem	gcdadd 11709	The GCD of two numbers is the same as the GCD of the left and their sum. (Contributed by Scott Fenton, 20-Apr-2014.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) = (𝑀 gcd (𝑁 + 𝑀)))
Theorem	gcdid 11710	The gcd of a number and itself is its absolute value. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ (𝑁 ∈ ℤ → (𝑁 gcd 𝑁) = (abs‘𝑁))
Theorem	gcd1 11711	The gcd of a number with 1 is 1. Theorem 1.4(d)1 in [ApostolNT] p. 16. (Contributed by Mario Carneiro, 19-Feb-2014.)
⊢ (𝑀 ∈ ℤ → (𝑀 gcd 1) = 1)
Theorem	gcdabs 11712	The gcd of two integers is the same as that of their absolute values. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((abs‘𝑀) gcd (abs‘𝑁)) = (𝑀 gcd 𝑁))
Theorem	gcdabs1 11713	gcd of the absolute value of the first operator. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ) → ((abs‘𝑁) gcd 𝑀) = (𝑁 gcd 𝑀))
Theorem	gcdabs2 11714	gcd of the absolute value of the second operator. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ) → (𝑁 gcd (abs‘𝑀)) = (𝑁 gcd 𝑀))
Theorem	modgcd 11715	The gcd remains unchanged if one operand is replaced with its remainder modulo the other. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ) → ((𝑀 mod 𝑁) gcd 𝑁) = (𝑀 gcd 𝑁))
Theorem	1gcd 11716	The GCD of one and an integer is one. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ (𝑀 ∈ ℤ → (1 gcd 𝑀) = 1)
Theorem	gcdmultipled 11717	The greatest common divisor of a nonnegative integer 𝑀 and a multiple of it is 𝑀 itself. (Contributed by Rohan Ridenour, 3-Aug-2023.)
⊢ (𝜑 → 𝑀 ∈ ℕ0)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    ⇒   ⊢ (𝜑 → (𝑀 gcd (𝑁 · 𝑀)) = 𝑀)
Theorem	dvdsgcdidd 11718	The greatest common divisor of a positive integer and another integer it divides is itself. (Contributed by Rohan Ridenour, 3-Aug-2023.)
⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∥ 𝑁)    ⇒   ⊢ (𝜑 → (𝑀 gcd 𝑁) = 𝑀)
Theorem	6gcd4e2 11719	The greatest common divisor of six and four is two. To calculate this gcd, a simple form of Euclid's algorithm is used: (6 gcd 4) = ((4 + 2) gcd 4) = (2 gcd 4) and (2 gcd 4) = (2 gcd (2 + 2)) = (2 gcd 2) = 2. (Contributed by AV, 27-Aug-2020.)
⊢ (6 gcd 4) = 2
5.1.5  Bézout's identity
Theorem	bezoutlemnewy 11720*	Lemma for Bézout's identity. The is-bezout predicate holds for (𝑦 mod 𝑊). (Contributed by Jim Kingdon, 6-Jan-2022.)
⊢ (𝜑 ↔ ∃𝑠 ∈ ℤ ∃𝑡 ∈ ℤ 𝑟 = ((𝐴 · 𝑠) + (𝐵 · 𝑡)))    &   ⊢ (𝜃 → 𝐴 ∈ ℕ0)    &   ⊢ (𝜃 → 𝐵 ∈ ℕ0)    &   ⊢ (𝜃 → 𝑊 ∈ ℕ)    &   ⊢ (𝜃 → [𝑦 / 𝑟]𝜑)    &   ⊢ (𝜃 → 𝑦 ∈ ℕ0)    &   ⊢ (𝜃 → [𝑊 / 𝑟]𝜑)    ⇒   ⊢ (𝜃 → [(𝑦 mod 𝑊) / 𝑟]𝜑)
Theorem	bezoutlemstep 11721*	Lemma for Bézout's identity. This is the induction step for the proof by induction. (Contributed by Jim Kingdon, 3-Jan-2022.)
⊢ (𝜑 ↔ ∃𝑠 ∈ ℤ ∃𝑡 ∈ ℤ 𝑟 = ((𝐴 · 𝑠) + (𝐵 · 𝑡)))    &   ⊢ (𝜃 → 𝐴 ∈ ℕ0)    &   ⊢ (𝜃 → 𝐵 ∈ ℕ0)    &   ⊢ (𝜃 → 𝑊 ∈ ℕ)    &   ⊢ (𝜃 → [𝑦 / 𝑟]𝜑)    &   ⊢ (𝜃 → 𝑦 ∈ ℕ0)    &   ⊢ (𝜃 → [𝑊 / 𝑟]𝜑)    &   ⊢ (𝜓 ↔ ∀𝑧 ∈ ℕ0 (𝑧 ∥ 𝑟 → (𝑧 ∥ 𝑥 ∧ 𝑧 ∥ 𝑦)))    &   ⊢ ((𝜃 ∧ [(𝑦 mod 𝑊) / 𝑟]𝜑) → ∃𝑟 ∈ ℕ0 ([(𝑦 mod 𝑊) / 𝑥][𝑊 / 𝑦]𝜓 ∧ 𝜑))    &   ⊢ Ⅎ𝑥𝜃    &   ⊢ Ⅎ𝑟𝜃    ⇒   ⊢ (𝜃 → ∃𝑟 ∈ ℕ0 ([𝑊 / 𝑥]𝜓 ∧ 𝜑))
Theorem	bezoutlemmain 11722*	Lemma for Bézout's identity. This is the main result which we prove by induction and which represents the application of the Extended Euclidean algorithm. (Contributed by Jim Kingdon, 30-Dec-2021.)
⊢ (𝜑 ↔ ∃𝑠 ∈ ℤ ∃𝑡 ∈ ℤ 𝑟 = ((𝐴 · 𝑠) + (𝐵 · 𝑡)))    &   ⊢ (𝜓 ↔ ∀𝑧 ∈ ℕ0 (𝑧 ∥ 𝑟 → (𝑧 ∥ 𝑥 ∧ 𝑧 ∥ 𝑦)))    &   ⊢ (𝜃 → 𝐴 ∈ ℕ0)    &   ⊢ (𝜃 → 𝐵 ∈ ℕ0)    ⇒   ⊢ (𝜃 → ∀𝑥 ∈ ℕ0 ([𝑥 / 𝑟]𝜑 → ∀𝑦 ∈ ℕ0 ([𝑦 / 𝑟]𝜑 → ∃𝑟 ∈ ℕ0 (𝜓 ∧ 𝜑))))
Theorem	bezoutlema 11723*	Lemma for Bézout's identity. The is-bezout condition is satisfied by 𝐴. (Contributed by Jim Kingdon, 30-Dec-2021.)
⊢ (𝜑 ↔ ∃𝑠 ∈ ℤ ∃𝑡 ∈ ℤ 𝑟 = ((𝐴 · 𝑠) + (𝐵 · 𝑡)))    &   ⊢ (𝜃 → 𝐴 ∈ ℕ0)    &   ⊢ (𝜃 → 𝐵 ∈ ℕ0)    ⇒   ⊢ (𝜃 → [𝐴 / 𝑟]𝜑)
Theorem	bezoutlemb 11724*	Lemma for Bézout's identity. The is-bezout condition is satisfied by 𝐵. (Contributed by Jim Kingdon, 30-Dec-2021.)
⊢ (𝜑 ↔ ∃𝑠 ∈ ℤ ∃𝑡 ∈ ℤ 𝑟 = ((𝐴 · 𝑠) + (𝐵 · 𝑡)))    &   ⊢ (𝜃 → 𝐴 ∈ ℕ0)    &   ⊢ (𝜃 → 𝐵 ∈ ℕ0)    ⇒   ⊢ (𝜃 → [𝐵 / 𝑟]𝜑)
Theorem	bezoutlemex 11725*	Lemma for Bézout's identity. Existence of a number which we will later show to be the greater common divisor and its decomposition into cofactors. (Contributed by Mario Carneiro and Jim Kingdon, 3-Jan-2022.)
⊢ ((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ0) → ∃𝑑 ∈ ℕ0 (∀𝑧 ∈ ℕ0 (𝑧 ∥ 𝑑 → (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)) ∧ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑑 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))))
Theorem	bezoutlemzz 11726*	Lemma for Bézout's identity. Like bezoutlemex 11725 but where ' z ' is any integer, not just a nonnegative one. (Contributed by Mario Carneiro and Jim Kingdon, 8-Jan-2022.)
⊢ ((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ0) → ∃𝑑 ∈ ℕ0 (∀𝑧 ∈ ℤ (𝑧 ∥ 𝑑 → (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)) ∧ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑑 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))))
Theorem	bezoutlemaz 11727*	Lemma for Bézout's identity. Like bezoutlemzz 11726 but where ' A ' can be any integer, not just a nonnegative one. (Contributed by Mario Carneiro and Jim Kingdon, 8-Jan-2022.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℕ0) → ∃𝑑 ∈ ℕ0 (∀𝑧 ∈ ℤ (𝑧 ∥ 𝑑 → (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)) ∧ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑑 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))))
Theorem	bezoutlembz 11728*	Lemma for Bézout's identity. Like bezoutlemaz 11727 but where ' B ' can be any integer, not just a nonnegative one. (Contributed by Mario Carneiro and Jim Kingdon, 8-Jan-2022.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ∃𝑑 ∈ ℕ0 (∀𝑧 ∈ ℤ (𝑧 ∥ 𝑑 → (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)) ∧ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑑 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))))
Theorem	bezoutlembi 11729*	Lemma for Bézout's identity. Like bezoutlembz 11728 but the greatest common divisor condition is a biconditional, not just an implication. (Contributed by Mario Carneiro and Jim Kingdon, 8-Jan-2022.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ∃𝑑 ∈ ℕ0 (∀𝑧 ∈ ℤ (𝑧 ∥ 𝑑 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)) ∧ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑑 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))))
Theorem	bezoutlemmo 11730*	Lemma for Bézout's identity. There is at most one nonnegative integer meeting the greatest common divisor condition. (Contributed by Mario Carneiro and Jim Kingdon, 9-Jan-2022.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℕ0)    &   ⊢ (𝜑 → ∀𝑧 ∈ ℤ (𝑧 ∥ 𝐷 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)))    &   ⊢ (𝜑 → 𝐸 ∈ ℕ0)    &   ⊢ (𝜑 → ∀𝑧 ∈ ℤ (𝑧 ∥ 𝐸 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)))    ⇒   ⊢ (𝜑 → 𝐷 = 𝐸)
Theorem	bezoutlemeu 11731*	Lemma for Bézout's identity. There is exactly one nonnegative integer meeting the greatest common divisor condition. (Contributed by Mario Carneiro and Jim Kingdon, 9-Jan-2022.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℕ0)    &   ⊢ (𝜑 → ∀𝑧 ∈ ℤ (𝑧 ∥ 𝐷 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)))    ⇒   ⊢ (𝜑 → ∃!𝑑 ∈ ℕ0 ∀𝑧 ∈ ℤ (𝑧 ∥ 𝑑 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)))
Theorem	bezoutlemle 11732*	Lemma for Bézout's identity. The number satisfying the greatest common divisor condition is the largest number which divides both 𝐴 and 𝐵. (Contributed by Mario Carneiro and Jim Kingdon, 9-Jan-2022.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℕ0)    &   ⊢ (𝜑 → ∀𝑧 ∈ ℤ (𝑧 ∥ 𝐷 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)))    &   ⊢ (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0))    ⇒   ⊢ (𝜑 → ∀𝑧 ∈ ℤ ((𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵) → 𝑧 ≤ 𝐷))
Theorem	bezoutlemsup 11733*	Lemma for Bézout's identity. The number satisfying the greatest common divisor condition is the supremum of divisors of both 𝐴 and 𝐵. (Contributed by Mario Carneiro and Jim Kingdon, 9-Jan-2022.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℕ0)    &   ⊢ (𝜑 → ∀𝑧 ∈ ℤ (𝑧 ∥ 𝐷 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)))    &   ⊢ (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0))    ⇒   ⊢ (𝜑 → 𝐷 = sup({𝑧 ∈ ℤ ∣ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)}, ℝ, < ))
Theorem	dfgcd3 11734*	Alternate definition of the gcd operator. (Contributed by Jim Kingdon, 31-Dec-2021.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) = (℩𝑑 ∈ ℕ0 ∀𝑧 ∈ ℤ (𝑧 ∥ 𝑑 ↔ (𝑧 ∥ 𝑀 ∧ 𝑧 ∥ 𝑁))))
Theorem	bezout 11735*	Bézout's identity: For any integers 𝐴 and 𝐵, there are integers 𝑥, 𝑦 such that (𝐴 gcd 𝐵) = 𝐴 · 𝑥 + 𝐵 · 𝑦. This is Metamath 100 proof #60. The proof is constructive, in the sense that it applies the Extended Euclidian Algorithm to constuct a number which can be shown to be (𝐴 gcd 𝐵) and which satisfies the rest of the theorem. In the presence of excluded middle, it is common to prove Bézout's identity by taking the smallest number which satisfies the Bézout condition, and showing it is the greatest common divisor. But we do not have the ability to show that number exists other than by providing a way to determine it. (Contributed by Mario Carneiro, 22-Feb-2014.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ (𝐴 gcd 𝐵) = ((𝐴 · 𝑥) + (𝐵 · 𝑦)))
Theorem	dvdsgcd 11736	An integer which divides each of two others also divides their gcd. (Contributed by Paul Chapman, 22-Jun-2011.) (Revised by Mario Carneiro, 30-May-2014.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) → 𝐾 ∥ (𝑀 gcd 𝑁)))
Theorem	dvdsgcdb 11737	Biconditional form of dvdsgcd 11736. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) ↔ 𝐾 ∥ (𝑀 gcd 𝑁)))
Theorem	dfgcd2 11738*	Alternate definition of the gcd operator, see definition in [ApostolNT] p. 15. (Contributed by AV, 8-Aug-2021.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝐷 = (𝑀 gcd 𝑁) ↔ (0 ≤ 𝐷 ∧ (𝐷 ∥ 𝑀 ∧ 𝐷 ∥ 𝑁) ∧ ∀𝑒 ∈ ℤ ((𝑒 ∥ 𝑀 ∧ 𝑒 ∥ 𝑁) → 𝑒 ∥ 𝐷))))
Theorem	gcdass 11739	Associative law for gcd operator. Theorem 1.4(b) in [ApostolNT] p. 16. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑃 ∈ ℤ) → ((𝑁 gcd 𝑀) gcd 𝑃) = (𝑁 gcd (𝑀 gcd 𝑃)))
Theorem	mulgcd 11740	Distribute multiplication by a nonnegative integer over gcd. (Contributed by Paul Chapman, 22-Jun-2011.) (Proof shortened by Mario Carneiro, 30-May-2014.)
⊢ ((𝐾 ∈ ℕ0 ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 · 𝑀) gcd (𝐾 · 𝑁)) = (𝐾 · (𝑀 gcd 𝑁)))
Theorem	absmulgcd 11741	Distribute absolute value of multiplication over gcd. Theorem 1.4(c) in [ApostolNT] p. 16. (Contributed by Paul Chapman, 22-Jun-2011.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 · 𝑀) gcd (𝐾 · 𝑁)) = (abs‘(𝐾 · (𝑀 gcd 𝑁))))
Theorem	mulgcdr 11742	Reverse distribution law for the gcd operator. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℕ0) → ((𝐴 · 𝐶) gcd (𝐵 · 𝐶)) = ((𝐴 gcd 𝐵) · 𝐶))
Theorem	gcddiv 11743	Division law for GCD. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℕ) ∧ (𝐶 ∥ 𝐴 ∧ 𝐶 ∥ 𝐵)) → ((𝐴 gcd 𝐵) / 𝐶) = ((𝐴 / 𝐶) gcd (𝐵 / 𝐶)))
Theorem	gcdmultiple 11744	The GCD of a multiple of a number is the number itself. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 gcd (𝑀 · 𝑁)) = 𝑀)
Theorem	gcdmultiplez 11745	Extend gcdmultiple 11744 so 𝑁 can be an integer. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd (𝑀 · 𝑁)) = 𝑀)
Theorem	gcdzeq 11746	A positive integer 𝐴 is equal to its gcd with an integer 𝐵 if and only if 𝐴 divides 𝐵. Generalization of gcdeq 11747. (Contributed by AV, 1-Jul-2020.)
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ) → ((𝐴 gcd 𝐵) = 𝐴 ↔ 𝐴 ∥ 𝐵))
Theorem	gcdeq 11747	𝐴 is equal to its gcd with 𝐵 if and only if 𝐴 divides 𝐵. (Contributed by Mario Carneiro, 23-Feb-2014.) (Proof shortened by AV, 8-Aug-2021.)
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → ((𝐴 gcd 𝐵) = 𝐴 ↔ 𝐴 ∥ 𝐵))
Theorem	dvdssqim 11748	Unidirectional form of dvdssq 11755. (Contributed by Scott Fenton, 19-Apr-2014.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 → (𝑀↑2) ∥ (𝑁↑2)))
Theorem	dvdsmulgcd 11749	Relationship between the order of an element and that of a multiple. (a divisibility equivalent). (Contributed by Stefan O'Rear, 6-Sep-2015.)
⊢ ((𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ) → (𝐴 ∥ (𝐵 · 𝐶) ↔ 𝐴 ∥ (𝐵 · (𝐶 gcd 𝐴))))
Theorem	rpmulgcd 11750	If 𝐾 and 𝑀 are relatively prime, then the GCD of 𝐾 and 𝑀 · 𝑁 is the GCD of 𝐾 and 𝑁. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ (((𝐾 ∈ ℕ ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ (𝐾 gcd 𝑀) = 1) → (𝐾 gcd (𝑀 · 𝑁)) = (𝐾 gcd 𝑁))
Theorem	rplpwr 11751	If 𝐴 and 𝐵 are relatively prime, then so are 𝐴↑𝑁 and 𝐵. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝐴 gcd 𝐵) = 1 → ((𝐴↑𝑁) gcd 𝐵) = 1))
Theorem	rppwr 11752	If 𝐴 and 𝐵 are relatively prime, then so are 𝐴↑𝑁 and 𝐵↑𝑁. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝐴 gcd 𝐵) = 1 → ((𝐴↑𝑁) gcd (𝐵↑𝑁)) = 1))
Theorem	sqgcd 11753	Square distributes over GCD. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝑀 gcd 𝑁)↑2) = ((𝑀↑2) gcd (𝑁↑2)))
Theorem	dvdssqlem 11754	Lemma for dvdssq 11755. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 ∥ 𝑁 ↔ (𝑀↑2) ∥ (𝑁↑2)))
Theorem	dvdssq 11755	Two numbers are divisible iff their squares are. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ (𝑀↑2) ∥ (𝑁↑2)))
Theorem	bezoutr 11756	Partial converse to bezout 11735. 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)))
Theorem	bezoutr1 11757	Converse of bezout 11735 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))
5.1.6  Algorithms
Theorem	nn0seqcvgd 11758*	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.)
⊢ (𝜑 → 𝐹:ℕ0⟶ℕ0)    &   ⊢ (𝜑 → 𝑁 = (𝐹‘0))    &   ⊢ ((𝜑 ∧ 𝑘 ∈ ℕ0) → ((𝐹‘(𝑘 + 1)) ≠ 0 → (𝐹‘(𝑘 + 1)) < (𝐹‘𝑘)))    ⇒   ⊢ (𝜑 → (𝐹‘𝑁) = 0)
Theorem	ialgrlem1st 11759	Lemma for ialgr0 11761. Expressing algrflemg 6135 in a form suitable for theorems such as seq3-1 10264 or seqf 10265. (Contributed by Jim Kingdon, 22-Jul-2021.)
⊢ (𝜑 → 𝐹:𝑆⟶𝑆)    ⇒   ⊢ ((𝜑 ∧ (𝑥 ∈ 𝑆 ∧ 𝑦 ∈ 𝑆)) → (𝑥(𝐹 ∘ 1st )𝑦) ∈ 𝑆)
Theorem	ialgrlemconst 11760	Lemma for ialgr0 11761. Closure of a constant function, in a form suitable for theorems such as seq3-1 10264 or seqf 10265. (Contributed by Jim Kingdon, 22-Jul-2021.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝐴 ∈ 𝑆)    ⇒   ⊢ ((𝜑 ∧ 𝑥 ∈ (ℤ≥‘𝑀)) → ((𝑍 × {𝐴})‘𝑥) ∈ 𝑆)
Theorem	ialgr0 11761	The value of the algorithm iterator 𝑅 at 0 is the initial state 𝐴. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Jim Kingdon, 12-Mar-2023.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ 𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}))    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝐴 ∈ 𝑆)    &   ⊢ (𝜑 → 𝐹:𝑆⟶𝑆)    ⇒   ⊢ (𝜑 → (𝑅‘𝑀) = 𝐴)
Theorem	algrf 11762	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 11763	The value of the algorithm iterator 𝑅 at (𝐾 + 1). (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Jim Kingdon, 12-Mar-2023.)
⊢ 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 11764*	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 11765*	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 11766	Lemma for algcvgb 11767. (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 11767	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 11768*	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 11769*	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)))
5.1.7  Euclid's Algorithm
Theorem	eucalgval2 11770*	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 11771*	Euclid's Algorithm eucalg 11776 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 11772*	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 11773*	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 11774*	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 11775*	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 11776*	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 (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))
5.1.8  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." In this section, an operation calculating the least common multiple of two integers (df-lcm 11778). The definition is valid for all integers, including negative integers and 0, obeying the above mentioned convention.
Syntax	clcm 11777	Extend the definition of a class to include the least common multiple operator.
class lcm
Definition	df-lcm 11778*	Define the lcm operator. For example, (6 lcm 9) = 18. (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, < )))
Theorem	lcmmndc 11779	Decidablity lemma used in various proofs related to lcm. (Contributed by Jim Kingdon, 21-Jan-2022.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → DECID (𝑀 = 0 ∨ 𝑁 = 0))
Theorem	lcmval 11780*	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 11672 and gcdval 11684. (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 11781	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 11782	The value, by convention, of the lcm operator when either operand is 0. (Use lcmcom 11781 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 11783*	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 11784*	Lemma for lcmn0cl 11785 and dvdslcm 11786. (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 11785	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 11786	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 11787	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 11788	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 11789	Closure of the lcm operator. (Contributed by Steve Rodriguez, 20-Jan-2020.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 lcm 𝑁) ∈ ℕ0)
Theorem	gcddvdslcm 11790	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 11791	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 11792	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 11793	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 11794	Lemma for lcmgcd 11795 and lcmdvds 11796. 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 11795	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 or of Bézout's identity bezout 11735; see e.g. https://proofwiki.org/wiki/Product_of_GCD_and_LCM 11735 and https://math.stackexchange.com/a/470827 11735. 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 11782 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 11796	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 11797	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 11798	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 11799	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 11800	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)))