P. 121 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 12001-12100   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	bezoutlemmain 12001*	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 12002*	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 12003*	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 12004*	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 12005*	Lemma for Bézout's identity. Like bezoutlemex 12004 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 12006*	Lemma for Bézout's identity. Like bezoutlemzz 12005 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 12007*	Lemma for Bézout's identity. Like bezoutlemaz 12006 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 12008*	Lemma for Bézout's identity. Like bezoutlembz 12007 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 12009*	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 12010*	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 12011*	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 12012*	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 12013*	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 12014*	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 12015	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 12016	Biconditional form of dvdsgcd 12015. (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 12017*	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 12018	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 12019	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 12020	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 12021	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 12022	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 12023	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 12024	Extend gcdmultiple 12023 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 12025	A positive integer 𝐴 is equal to its gcd with an integer 𝐵 if and only if 𝐴 divides 𝐵. Generalization of gcdeq 12026. (Contributed by AV, 1-Jul-2020.)
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ) → ((𝐴 gcd 𝐵) = 𝐴 ↔ 𝐴 ∥ 𝐵))
Theorem	gcdeq 12026	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 12027	Unidirectional form of dvdssq 12034. (Contributed by Scott Fenton, 19-Apr-2014.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 → (𝑀↑2) ∥ (𝑁↑2)))
Theorem	dvdsmulgcd 12028	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 12029	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 12030	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 12031	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 12032	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 12033	Lemma for dvdssq 12034. (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 12034	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 12035	Partial converse to bezout 12014. 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 12036	Converse of bezout 12014 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  Decidable sets of integers
Theorem	nnmindc 12037*	An inhabited decidable subset of the natural numbers has a minimum. (Contributed by Jim Kingdon, 23-Sep-2024.)
⊢ ((𝐴 ⊆ ℕ ∧ ∀𝑥 ∈ ℕ DECID 𝑥 ∈ 𝐴 ∧ ∃𝑦 𝑦 ∈ 𝐴) → inf(𝐴, ℝ, < ) ∈ 𝐴)
Theorem	nnminle 12038*	The infimum of a decidable subset of the natural numbers is less than an element of the set. The infimum is also a minimum as shown at nnmindc 12037. (Contributed by Jim Kingdon, 26-Sep-2024.)
⊢ ((𝐴 ⊆ ℕ ∧ ∀𝑥 ∈ ℕ DECID 𝑥 ∈ 𝐴 ∧ 𝐵 ∈ 𝐴) → inf(𝐴, ℝ, < ) ≤ 𝐵)
Theorem	nnwodc 12039*	Well-ordering principle: any inhabited decidable set of positive integers has a least element. Theorem I.37 (well-ordering principle) of [Apostol] p. 34. (Contributed by NM, 17-Aug-2001.) (Revised by Jim Kingdon, 23-Oct-2024.)
⊢ ((𝐴 ⊆ ℕ ∧ ∃𝑤 𝑤 ∈ 𝐴 ∧ ∀𝑗 ∈ ℕ DECID 𝑗 ∈ 𝐴) → ∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 𝑥 ≤ 𝑦)
Theorem	uzwodc 12040*	Well-ordering principle: any inhabited decidable subset of an upper set of integers has a least element. (Contributed by NM, 8-Oct-2005.) (Revised by Jim Kingdon, 22-Oct-2024.)
⊢ ((𝑆 ⊆ (ℤ≥‘𝑀) ∧ ∃𝑥 𝑥 ∈ 𝑆 ∧ ∀𝑥 ∈ (ℤ≥‘𝑀)DECID 𝑥 ∈ 𝑆) → ∃𝑗 ∈ 𝑆 ∀𝑘 ∈ 𝑆 𝑗 ≤ 𝑘)
Theorem	nnwofdc 12041*	Well-ordering principle: any inhabited decidable set of positive integers has a least element. This version allows 𝑥 and 𝑦 to be present in 𝐴 as long as they are effectively not free. (Contributed by NM, 17-Aug-2001.) (Revised by Mario Carneiro, 15-Oct-2016.)
⊢ Ⅎ𝑥𝐴    &   ⊢ Ⅎ𝑦𝐴    ⇒   ⊢ ((𝐴 ⊆ ℕ ∧ ∃𝑧 𝑧 ∈ 𝐴 ∧ ∀𝑗 ∈ ℕ DECID 𝑗 ∈ 𝐴) → ∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 𝑥 ≤ 𝑦)
Theorem	nnwosdc 12042*	Well-ordering principle: any inhabited decidable set of positive integers has a least element (schema form). (Contributed by NM, 17-Aug-2001.) (Revised by Jim Kingdon, 25-Oct-2024.)
⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))    ⇒   ⊢ ((∃𝑥 ∈ ℕ 𝜑 ∧ ∀𝑥 ∈ ℕ DECID 𝜑) → ∃𝑥 ∈ ℕ (𝜑 ∧ ∀𝑦 ∈ ℕ (𝜓 → 𝑥 ≤ 𝑦)))
5.1.7  Algorithms
Theorem	nn0seqcvgd 12043*	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 12044	Lemma for ialgr0 12046. Expressing algrflemg 6233 in a form suitable for theorems such as seq3-1 10462 or seqf 10463. (Contributed by Jim Kingdon, 22-Jul-2021.)
⊢ (𝜑 → 𝐹:𝑆⟶𝑆)    ⇒   ⊢ ((𝜑 ∧ (𝑥 ∈ 𝑆 ∧ 𝑦 ∈ 𝑆)) → (𝑥(𝐹 ∘ 1st )𝑦) ∈ 𝑆)
Theorem	ialgrlemconst 12045	Lemma for ialgr0 12046. Closure of a constant function, in a form suitable for theorems such as seq3-1 10462 or seqf 10463. (Contributed by Jim Kingdon, 22-Jul-2021.)
⊢ 𝑍 = (ℤ≥‘𝑀)    &   ⊢ (𝜑 → 𝐴 ∈ 𝑆)    ⇒   ⊢ ((𝜑 ∧ 𝑥 ∈ (ℤ≥‘𝑀)) → ((𝑍 × {𝐴})‘𝑥) ∈ 𝑆)
Theorem	ialgr0 12046	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 12047	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 12048	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 12049*	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 12050*	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 12051	Lemma for algcvgb 12052. (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 12052	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 12053*	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 12054*	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)))
5.1.8  Euclid's Algorithm
Theorem	eucalgval2 12055*	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 12056*	Euclid's Algorithm eucalg 12061 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 12057*	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 12058*	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 12059*	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 12060*	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 12061*	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))
5.1.9  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 12063). The definition is valid for all integers, including negative integers and 0, obeying the above mentioned convention.
Syntax	clcm 12062	Extend the definition of a class to include the least common multiple operator.
class lcm
Definition	df-lcm 12063*	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 12064	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 12065*	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 11946 and gcdval 11962. (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 12066	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 12067	The value, by convention, of the lcm operator when either operand is 0. (Use lcmcom 12066 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 12068*	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 12069*	Lemma for lcmn0cl 12070 and dvdslcm 12071. (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 12070	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 12071	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 12072	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 12073	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 12074	Closure of the lcm operator. (Contributed by Steve Rodriguez, 20-Jan-2020.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 lcm 𝑁) ∈ ℕ0)
Theorem	gcddvdslcm 12075	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 12076	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 12077	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 12078	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 12079	Lemma for lcmgcd 12080 and lcmdvds 12081. 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 12080	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 12014; see, e.g., https://proofwiki.org/wiki/Product_of_GCD_and_LCM 12014 and https://math.stackexchange.com/a/470827 12014. 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 12067 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 12081	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 12082	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 12083	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 12084	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 12085	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 12086	Biconditional form of lcmdvds 12081. (Contributed by Steve Rodriguez, 20-Jan-2020.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀 ∥ 𝐾 ∧ 𝑁 ∥ 𝐾) ↔ (𝑀 lcm 𝑁) ∥ 𝐾))
Theorem	lcmass 12087	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 12088	The least common multiple of three and two is six. This proof does not use the property of 2 and 3 being prime. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Revised by AV, 27-Aug-2020.)
⊢ (3 lcm 2) = 6
Theorem	6lcm4e12 12089	The least common multiple of six and four is twelve. (Contributed by AV, 27-Aug-2020.)
⊢ (6 lcm 4) = ;12
5.1.10  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 12090. The negation, i.e. two integers are not coprime, can be expressed either by (𝐴 gcd 𝐵) ≠ 1, see ncoprmgcdne1b 12091, or equivalently by 1 < (𝐴 gcd 𝐵), see ncoprmgcdgt1b 12092. A proof of Euclid's lemma based on coprimality is provided in coprmdvds 12094 (as opposed to Euclid's lemma for primes).
Theorem	coprmgcdb 12090*	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 12091*	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. (Contributed by AV, 9-Aug-2020.)
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (∃𝑖 ∈ (ℤ≥‘2)(𝑖 ∥ 𝐴 ∧ 𝑖 ∥ 𝐵) ↔ (𝐴 gcd 𝐵) ≠ 1))
Theorem	ncoprmgcdgt1b 12092*	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 12093	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 12094	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. (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 12095	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 12096	One half of rpmulgcd2 12097, 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 12097	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 12098	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 12099*	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 12100	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))