P. 168 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 16701-16800   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	neggcd 16701	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	gcdaddmlem 16702	Lemma for gcdaddm 16703. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ 𝐾 ∈ ℤ    &   ⊢ 𝑀 ∈ ℤ    &   ⊢ 𝑁 ∈ ℤ    ⇒   ⊢ (𝑀 gcd 𝑁) = (𝑀 gcd ((𝐾 · 𝑀) + 𝑁))
Theorem	gcdaddm 16703	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 16704	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 16705	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	gcdnn0id 16706	The gcd of a nonnegative integer and itself is the integer. (Contributed by SN, 25-Aug-2024.)
⊢ (𝑁 ∈ ℕ0 → (𝑁 gcd 𝑁) = 𝑁)
Theorem	gcd1 16707	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	gcdabs1 16708	The 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 16709	The 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	gcdabs 16710	The gcd of two integers is the same as that of their absolute values. (Contributed by Paul Chapman, 31-Mar-2011.) (Proof shortened by SN, 15-Sep-2024.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((abs‘𝑀) gcd (abs‘𝑁)) = (𝑀 gcd 𝑁))
Theorem	modgcd 16711	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 16712	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 16713	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	gcdmultiplez 16714	The gcd of a multiple of an integer is the integer itself. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) (Proof shortened by AV, 12-Jan-2023.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd (𝑀 · 𝑁)) = 𝑀)
Theorem	gcdmultiple 16715	The gcd of a multiple of a positive integer is the positive integer itself. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) (Proof shortened by AV, 12-Jan-2023.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 gcd (𝑀 · 𝑁)) = 𝑀)
Theorem	dvdsgcdidd 16716	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 16717	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
6.1.8  Bézout's identity
Theorem	bezoutlem1 16718*	Lemma for bezout 16722. (Contributed by Mario Carneiro, 15-Mar-2014.)
⊢ 𝑀 = {𝑧 ∈ ℕ ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑧 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))}    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    ⇒   ⊢ (𝜑 → (𝐴 ≠ 0 → (abs‘𝐴) ∈ 𝑀))
Theorem	bezoutlem2 16719*	Lemma for bezout 16722. (Contributed by Mario Carneiro, 15-Mar-2014.) ( Revised by AV, 30-Sep-2020.)
⊢ 𝑀 = {𝑧 ∈ ℕ ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑧 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))}    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ 𝐺 = inf(𝑀, ℝ, < )    &   ⊢ (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0))    ⇒   ⊢ (𝜑 → 𝐺 ∈ 𝑀)
Theorem	bezoutlem3 16720*	Lemma for bezout 16722. (Contributed by Mario Carneiro, 22-Feb-2014.) ( Revised by AV, 30-Sep-2020.)
⊢ 𝑀 = {𝑧 ∈ ℕ ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑧 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))}    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ 𝐺 = inf(𝑀, ℝ, < )    &   ⊢ (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0))    ⇒   ⊢ (𝜑 → (𝐶 ∈ 𝑀 → 𝐺 ∥ 𝐶))
Theorem	bezoutlem4 16721*	Lemma for bezout 16722. (Contributed by Mario Carneiro, 22-Feb-2014.)
⊢ 𝑀 = {𝑧 ∈ ℕ ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑧 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))}    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ 𝐺 = inf(𝑀, ℝ, < )    &   ⊢ (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0))    ⇒   ⊢ (𝜑 → (𝐴 gcd 𝐵) ∈ 𝑀)
Theorem	bezout 16722*	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. (Contributed by Mario Carneiro, 22-Feb-2014.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ (𝐴 gcd 𝐵) = ((𝐴 · 𝑥) + (𝐵 · 𝑦)))
Theorem	dvdsgcd 16723	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 16724	Biconditional form of dvdsgcd 16723. (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 16725*	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 16726	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 16727	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 16728	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 16729	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 16730	Division law for the gcd operator. (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	gcdzeq 16731	A positive integer 𝐴 is equal to its gcd with an integer 𝐵 if and only if 𝐴 divides 𝐵. Generalization of gcdeq 16732. (Contributed by AV, 1-Jul-2020.)
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ) → ((𝐴 gcd 𝐵) = 𝐴 ↔ 𝐴 ∥ 𝐵))
Theorem	gcdeq 16732	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 16733	Unidirectional form of dvdssq 16748. (Contributed by Scott Fenton, 19-Apr-2014.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 → (𝑀↑2) ∥ (𝑁↑2)))
Theorem	dvdsexpim 16734	If two numbers are divisible, so are their nonnegative exponents. Similar to dvdssqim 16733 for nonnegative exponents. (Contributed by Steven Nguyen, 2-Apr-2023.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝑁 ∈ ℕ0) → (𝐴 ∥ 𝐵 → (𝐴↑𝑁) ∥ (𝐵↑𝑁)))
Theorem	dvdsexpad 16735	Deduction associated with dvdsexpim 16734. (Contributed by SN, 21-Aug-2024.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    &   ⊢ (𝜑 → 𝐴 ∥ 𝐵)    ⇒   ⊢ (𝜑 → (𝐴↑𝑁) ∥ (𝐵↑𝑁))
Theorem	dvdsmulgcd 16736	A divisibility equivalent for odmulg 19776. (Contributed by Stefan O'Rear, 6-Sep-2015.)
⊢ ((𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ) → (𝐴 ∥ (𝐵 · 𝐶) ↔ 𝐴 ∥ (𝐵 · (𝐶 gcd 𝐴))))
Theorem	rpmulgcd 16737	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 16738	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	rprpwr 16739	If 𝐴 and 𝐵 are relatively prime, then so are 𝐴 and 𝐵↑𝑁. Originally a subproof of rppwr 16740. (Contributed by SN, 21-Aug-2024.)
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝐴 gcd 𝐵) = 1 → (𝐴 gcd (𝐵↑𝑁)) = 1))
Theorem	rppwr 16740	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	nn0rppwr 16741	If 𝐴 and 𝐵 are relatively prime, then so are 𝐴↑𝑁 and 𝐵↑𝑁. rppwr 16740 extended to nonnegative integers. Less general than rpexp12i 16906. (Contributed by Steven Nguyen, 4-Apr-2023.)
⊢ ((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0) → ((𝐴 gcd 𝐵) = 1 → ((𝐴↑𝑁) gcd (𝐵↑𝑁)) = 1))
Theorem	sqgcd 16742	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	expgcd 16743	Exponentiation distributes over gcd. sqgcd 16742 extended to nonnegative exponents. (Contributed by Steven Nguyen, 4-Apr-2023.)
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝑁 ∈ ℕ0) → ((𝐴 gcd 𝐵)↑𝑁) = ((𝐴↑𝑁) gcd (𝐵↑𝑁)))
Theorem	nn0expgcd 16744	Exponentiation distributes over gcd. nn0gcdsq 16934 extended to nonnegative exponents. expgcd 16743 extended to nonnegative bases. (Contributed by Steven Nguyen, 5-Apr-2023.)
⊢ ((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0) → ((𝐴 gcd 𝐵)↑𝑁) = ((𝐴↑𝑁) gcd (𝐵↑𝑁)))
Theorem	zexpgcd 16745	Exponentiation distributes over gcd. zgcdsq 16935 extended to nonnegative exponents. nn0expgcd 16744 extended to integer bases by symmetry. (Contributed by Steven Nguyen, 5-Apr-2023.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝑁 ∈ ℕ0) → ((𝐴 gcd 𝐵)↑𝑁) = ((𝐴↑𝑁) gcd (𝐵↑𝑁)))
Theorem	dvdsexpnn 16746	Two positive integers are divisible iff their positive integer powers are. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) Generalized to arbitrary positive integer exponents. (Revised by SN, 20-Aug-2024.)
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝐴 ∥ 𝐵 ↔ (𝐴↑𝑁) ∥ (𝐵↑𝑁)))
Theorem	dvdssqnn 16747	Two positive integers are divisible iff their squares are. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) (Proof shortened by AV, 16-Sep-2026.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 ∥ 𝑁 ↔ (𝑀↑2) ∥ (𝑁↑2)))
Theorem	dvdssq 16748	Two integers 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 16749	Partial converse to bezout 16722. 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 16750	Converse of bezout 16722 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
Theorem	nn0seqcvgd 16751*	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	seq1st 16752	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 = (ℤ≥‘𝑀)    &   ⊢ 𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}))    ⇒   ⊢ ((𝑀 ∈ ℤ ∧ 𝐴 ∈ 𝑉) → 𝑅 = seq𝑀((𝐹 ∘ 1st ), {⟨𝑀, 𝐴⟩}))
Theorem	algr0 16753	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 = (ℤ≥‘𝑀)    &   ⊢ 𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}))    &   ⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝐴 ∈ 𝑆)    ⇒   ⊢ (𝜑 → (𝑅‘𝑀) = 𝐴)
Theorem	algrf 16754	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 16755	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 16756*	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 16757*	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 16758	Lemma for algcvgb 16759. (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 16759	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 16760*	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 16761*	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 16762*	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 16763*	Euclid's Algorithm eucalg 16768 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 16764*	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 16765*	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 16766*	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 16767*	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 16768*	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 16771) as well as a function mapping a set of integers to their least common multiple (df-lcmf 16772) are provided. Both definitions are valid for all integers, including negative integers and 0, obeying the above mentioned convention. It is shown by lcmfpr 16808 that the two definitions are compatible.
Syntax	clcm 16769	Extend the definition of a class to include the least common multiple operator.
class lcm
Syntax	clcmf 16770	Extend the definition of a class to include the least common multiple function.
class lcm
Definition	df-lcm 16771*	Define the lcm operator. For example, (6 lcm 9) = 18 (ex-lcm 31065). (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 16772*	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 16773*	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 16671 and gcdval 16672. (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 16774	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 16775	The value, by convention, of the lcm operator when either operand is 0. (Use lcmcom 16774 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 16776*	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 16777*	Lemma for lcmn0cl 16778 and dvdslcm 16779. (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 16778	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 16779	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 16780	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 16781	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 16782	Closure of the lcm operator. (Contributed by Steve Rodriguez, 20-Jan-2020.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 lcm 𝑁) ∈ ℕ0)
Theorem	gcddvdslcm 16783	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 16784	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 16785	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 16786	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 16787	Lemma for lcmgcd 16788 and lcmdvds 16789. 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 16788	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 17111 or of Bézout's identity bezout 16722; see e.g., https://proofwiki.org/wiki/Product_of_GCD_and_LCM 16722 and https://math.stackexchange.com/a/470827 16722. 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 16775 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 16789	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 16790	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 16791	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 16792	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 16793	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 16794	Biconditional form of lcmdvds 16789. (Contributed by Steve Rodriguez, 20-Jan-2020.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀 ∥ 𝐾 ∧ 𝑁 ∥ 𝐾) ↔ (𝑀 lcm 𝑁) ∥ 𝐾))
Theorem	lcmass 16795	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 16796	The least common multiple of three and two is six. In contrast to 3lcm2e6 16914, 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 16797	The least common multiple of six and four is twelve. (Contributed by AV, 27-Aug-2020.)
⊢ (6 lcm 4) = ;12
Theorem	absproddvds 16798*	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 16799*	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 16800*	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)