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Theorem List for Intuitionistic Logic Explorer - 10601-10700   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theorembezoutlembi 10601* Lemma for Bézout's identity. Like bezoutlembz 10600 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 (𝑧𝑑 ↔ (𝑧𝐴𝑧𝐵)) ∧ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑑 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))))

Theorembezoutlemmo 10602* 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.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝐷 ∈ ℕ0)    &   (𝜑 → ∀𝑧 ∈ ℤ (𝑧𝐷 ↔ (𝑧𝐴𝑧𝐵)))    &   (𝜑𝐸 ∈ ℕ0)    &   (𝜑 → ∀𝑧 ∈ ℤ (𝑧𝐸 ↔ (𝑧𝐴𝑧𝐵)))       (𝜑𝐷 = 𝐸)

Theorembezoutlemeu 10603* 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.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝐷 ∈ ℕ0)    &   (𝜑 → ∀𝑧 ∈ ℤ (𝑧𝐷 ↔ (𝑧𝐴𝑧𝐵)))       (𝜑 → ∃!𝑑 ∈ ℕ0𝑧 ∈ ℤ (𝑧𝑑 ↔ (𝑧𝐴𝑧𝐵)))

Theorembezoutlemle 10604* 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.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝐷 ∈ ℕ0)    &   (𝜑 → ∀𝑧 ∈ ℤ (𝑧𝐷 ↔ (𝑧𝐴𝑧𝐵)))    &   (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0))       (𝜑 → ∀𝑧 ∈ ℤ ((𝑧𝐴𝑧𝐵) → 𝑧𝐷))

Theorembezoutlemsup 10605* 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.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝐷 ∈ ℕ0)    &   (𝜑 → ∀𝑧 ∈ ℤ (𝑧𝐷 ↔ (𝑧𝐴𝑧𝐵)))    &   (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0))       (𝜑𝐷 = sup({𝑧 ∈ ℤ ∣ (𝑧𝐴𝑧𝐵)}, ℝ, < ))

Theoremdfgcd3 10606* Alternate definition of the gcd operator. (Contributed by Jim Kingdon, 31-Dec-2021.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) = (𝑑 ∈ ℕ0𝑧 ∈ ℤ (𝑧𝑑 ↔ (𝑧𝑀𝑧𝑁))))

Theorembezout 10607* 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)))

Theoremdvdsgcd 10608 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) → ((𝐾𝑀𝐾𝑁) → 𝐾 ∥ (𝑀 gcd 𝑁)))

Theoremdvdsgcdb 10609 Biconditional form of dvdsgcd 10608. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾𝑀𝐾𝑁) ↔ 𝐾 ∥ (𝑀 gcd 𝑁)))

Theoremdfgcd2 10610* 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 ∧ (𝐷𝑀𝐷𝑁) ∧ ∀𝑒 ∈ ℤ ((𝑒𝑀𝑒𝑁) → 𝑒𝐷))))

Theoremgcdass 10611 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)))

Theoremmulgcd 10612 Distribute multiplication by a nonnegative integer over gcd. (Contributed by Paul Chapman, 22-Jun-2011.) (Proof shortened by Mario Carneiro, 30-May-2014.)
((𝐾 ∈ ℕ0𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 · 𝑀) gcd (𝐾 · 𝑁)) = (𝐾 · (𝑀 gcd 𝑁)))

Theoremabsmulgcd 10613 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))))

Theoremmulgcdr 10614 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))

Theoremgcddiv 10615 Division law for GCD. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℕ) ∧ (𝐶𝐴𝐶𝐵)) → ((𝐴 gcd 𝐵) / 𝐶) = ((𝐴 / 𝐶) gcd (𝐵 / 𝐶)))

Theoremgcdmultiple 10616 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)

Theoremgcdmultiplez 10617 Extend gcdmultiple 10616 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)

Theoremgcdzeq 10618 A positive integer 𝐴 is equal to its gcd with an integer 𝐵 if and only if 𝐴 divides 𝐵. Generalization of gcdeq 10619. (Contributed by AV, 1-Jul-2020.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ) → ((𝐴 gcd 𝐵) = 𝐴𝐴𝐵))

Theoremgcdeq 10619 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) = 𝐴𝐴𝐵))

Theoremdvdssqim 10620 Unidirectional form of dvdssq 10627. (Contributed by Scott Fenton, 19-Apr-2014.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀𝑁 → (𝑀↑2) ∥ (𝑁↑2)))

Theoremdvdsmulgcd 10621 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))))

Theoremrpmulgcd 10622 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))

Theoremrplpwr 10623 If 𝐴 and 𝐵 are relatively prime, then so are 𝐴𝑁 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 → ((𝐴𝑁) gcd 𝐵) = 1))

Theoremrppwr 10624 If 𝐴 and 𝐵 are relatively prime, then so are 𝐴𝑁 and 𝐵𝑁. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝐴 gcd 𝐵) = 1 → ((𝐴𝑁) gcd (𝐵𝑁)) = 1))

Theoremsqgcd 10625 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)))

Theoremdvdssqlem 10626 Lemma for dvdssq 10627. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀𝑁 ↔ (𝑀↑2) ∥ (𝑁↑2)))

Theoremdvdssq 10627 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) → (𝑀𝑁 ↔ (𝑀↑2) ∥ (𝑁↑2)))

Theorembezoutr 10628 Partial converse to bezout 10607. Existence of a linear combination does not set the GCD, but it does upper bound it. (Contributed by Stefan O'Rear, 23-Sep-2014.)
(((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) ∧ (𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ)) → (𝐴 gcd 𝐵) ∥ ((𝐴 · 𝑋) + (𝐵 · 𝑌)))

Theorembezoutr1 10629 Converse of bezout 10607 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))

4.1.6  Algorithms

Theoremnn0seqcvgd 10630* A strictly-decreasing nonnegative integer sequence with initial term 𝑁 reaches zero by the 𝑁 th term. Deduction version. (Contributed by Paul Chapman, 31-Mar-2011.)
(𝜑𝐹:ℕ0⟶ℕ0)    &   (𝜑𝑁 = (𝐹‘0))    &   ((𝜑𝑘 ∈ ℕ0) → ((𝐹‘(𝑘 + 1)) ≠ 0 → (𝐹‘(𝑘 + 1)) < (𝐹𝑘)))       (𝜑 → (𝐹𝑁) = 0)

Theoremialgrlem1st 10631 Lemma for ialgr0 10633. Expressing algrflemg 5902 in a form suitable for theorems such as iseq1 9585 or iseqfcl 9587. (Contributed by Jim Kingdon, 22-Jul-2021.)
(𝜑𝐹:𝑆𝑆)       ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥(𝐹 ∘ 1st )𝑦) ∈ 𝑆)

Theoremialgrlemconst 10632 Lemma for ialgr0 10633. Closure of a constant function, in a form suitable for theorems such as iseq1 9585 or iseqfcl 9587. (Contributed by Jim Kingdon, 22-Jul-2021.)
𝑍 = (ℤ𝑀)    &   (𝜑𝐴𝑆)       ((𝜑𝑥 ∈ (ℤ𝑀)) → ((𝑍 × {𝐴})‘𝑥) ∈ 𝑆)

Theoremialgr0 10633 The value of the algorithm iterator 𝑅 at 0 is the initial state 𝐴. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 28-May-2014.)
𝑍 = (ℤ𝑀)    &   𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}), 𝑆)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐴𝑆)    &   (𝜑𝐹:𝑆𝑆)    &   (𝜑𝑆𝑉)       (𝜑 → (𝑅𝑀) = 𝐴)

Theoremialgrf 10634 An algorithm is a step function 𝐹:𝑆𝑆 on a state space 𝑆. An algorithm acts on an initial state 𝐴𝑆 by iteratively applying 𝐹 to give 𝐴, (𝐹𝐴), (𝐹‘(𝐹𝐴)) and so on. An algorithm is said to halt if a fixed point of 𝐹 is reached after a finite number of iterations.

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

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

𝑍 = (ℤ𝑀)    &   𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}), 𝑆)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐴𝑆)    &   (𝜑𝐹:𝑆𝑆)    &   (𝜑𝑆𝑉)       (𝜑𝑅:𝑍𝑆)

Theoremialgrp1 10635 The value of the algorithm iterator 𝑅 at (𝐾 + 1). (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 27-Dec-2014.)
𝑍 = (ℤ𝑀)    &   𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}), 𝑆)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐴𝑆)    &   (𝜑𝐹:𝑆𝑆)    &   (𝜑𝑆𝑉)       ((𝜑𝐾𝑍) → (𝑅‘(𝐾 + 1)) = (𝐹‘(𝑅𝐾)))

Theoremialginv 10636* If 𝐼 is an invariant of 𝐹, its value is unchanged after any number of iterations of 𝐹. (Contributed by Paul Chapman, 31-Mar-2011.)
𝑅 = seq0((𝐹 ∘ 1st ), (ℕ0 × {𝐴}), 𝑆)    &   𝐹:𝑆𝑆    &   𝐼 Fn 𝑆    &   (𝑥𝑆 → (𝐼‘(𝐹𝑥)) = (𝐼𝑥))    &   𝑆𝑉       ((𝐴𝑆𝐾 ∈ ℕ0) → (𝐼‘(𝑅𝐾)) = (𝐼‘(𝑅‘0)))

Theoremialgcvg 10637* One way to prove that an algorithm halts is to construct a countdown function 𝐶:𝑆⟶ℕ0 whose value is guaranteed to decrease for each iteration of 𝐹 until it reaches 0. That is, if 𝑋𝑆 is not a fixed point of 𝐹, then (𝐶‘(𝐹𝑋)) < (𝐶𝑋).

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

𝐹:𝑆𝑆    &   𝑅 = seq0((𝐹 ∘ 1st ), (ℕ0 × {𝐴}), 𝑆)    &   𝐶:𝑆⟶ℕ0    &   (𝑧𝑆 → ((𝐶‘(𝐹𝑧)) ≠ 0 → (𝐶‘(𝐹𝑧)) < (𝐶𝑧)))    &   𝑁 = (𝐶𝐴)    &   𝑆𝑉       (𝐴𝑆 → (𝐶‘(𝑅𝑁)) = 0)

Theoremalgcvgblem 10638 Lemma for algcvgb 10639. (Contributed by Paul Chapman, 31-Mar-2011.)
((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → ((𝑁 ≠ 0 → 𝑁 < 𝑀) ↔ ((𝑀 ≠ 0 → 𝑁 < 𝑀) ∧ (𝑀 = 0 → 𝑁 = 0))))

Theoremalgcvgb 10639 Two ways of expressing that 𝐶 is a countdown function for algorithm 𝐹. The first is used in these theorems. The second states the condition more intuitively as a conjunction: if the countdown function's value is currently nonzero, it must decrease at the next step; if it has reached zero, it must remain zero at the next step. (Contributed by Paul Chapman, 31-Mar-2011.)
𝐹:𝑆𝑆    &   𝐶:𝑆⟶ℕ0       (𝑋𝑆 → (((𝐶‘(𝐹𝑋)) ≠ 0 → (𝐶‘(𝐹𝑋)) < (𝐶𝑋)) ↔ (((𝐶𝑋) ≠ 0 → (𝐶‘(𝐹𝑋)) < (𝐶𝑋)) ∧ ((𝐶𝑋) = 0 → (𝐶‘(𝐹𝑋)) = 0))))

Theoremialgcvga 10640* The countdown function 𝐶 remains 0 after 𝑁 steps. (Contributed by Paul Chapman, 22-Jun-2011.)
𝐹:𝑆𝑆    &   𝑅 = seq0((𝐹 ∘ 1st ), (ℕ0 × {𝐴}), 𝑆)    &   𝐶:𝑆⟶ℕ0    &   (𝑧𝑆 → ((𝐶‘(𝐹𝑧)) ≠ 0 → (𝐶‘(𝐹𝑧)) < (𝐶𝑧)))    &   𝑁 = (𝐶𝐴)    &   𝑆𝑉       (𝐴𝑆 → (𝐾 ∈ (ℤ𝑁) → (𝐶‘(𝑅𝐾)) = 0))

Theoremialgfx 10641* If 𝐹 reaches a fixed point when the countdown function 𝐶 reaches 0, 𝐹 remains fixed after 𝑁 steps. (Contributed by Paul Chapman, 22-Jun-2011.)
𝐹:𝑆𝑆    &   𝑅 = seq0((𝐹 ∘ 1st ), (ℕ0 × {𝐴}), 𝑆)    &   𝐶:𝑆⟶ℕ0    &   (𝑧𝑆 → ((𝐶‘(𝐹𝑧)) ≠ 0 → (𝐶‘(𝐹𝑧)) < (𝐶𝑧)))    &   𝑁 = (𝐶𝐴)    &   𝑆𝑉    &   (𝑧𝑆 → ((𝐶𝑧) = 0 → (𝐹𝑧) = 𝑧))       (𝐴𝑆 → (𝐾 ∈ (ℤ𝑁) → (𝑅𝐾) = (𝑅𝑁)))

4.1.7  Euclid's Algorithm

Theoremeucalgval2 10642* The value of the step function 𝐸 for Euclid's Algorithm on an ordered pair. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 28-May-2014.)
𝐸 = (𝑥 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))       ((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → (𝑀𝐸𝑁) = if(𝑁 = 0, ⟨𝑀, 𝑁⟩, ⟨𝑁, (𝑀 mod 𝑁)⟩))

Theoremeucalgval 10643* Euclid's Algorithm eucialg 10648 computes the greatest common divisor of two nonnegative integers by repeatedly replacing the larger of them with its remainder modulo the smaller until the remainder is 0.

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

𝐸 = (𝑥 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))       (𝑋 ∈ (ℕ0 × ℕ0) → (𝐸𝑋) = if((2nd𝑋) = 0, 𝑋, ⟨(2nd𝑋), ( mod ‘𝑋)⟩))

Theoremeucalgf 10644* Domain and codomain of the step function 𝐸 for Euclid's Algorithm. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 28-May-2014.)
𝐸 = (𝑥 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))       𝐸:(ℕ0 × ℕ0)⟶(ℕ0 × ℕ0)

Theoremeucalginv 10645* The invariant of the step function 𝐸 for Euclid's Algorithm is the gcd operator applied to the state. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 29-May-2014.)
𝐸 = (𝑥 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))       (𝑋 ∈ (ℕ0 × ℕ0) → ( gcd ‘(𝐸𝑋)) = ( gcd ‘𝑋))

Theoremeucalglt 10646* The second member of the state decreases with each iteration of the step function 𝐸 for Euclid's Algorithm. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 29-May-2014.)
𝐸 = (𝑥 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))       (𝑋 ∈ (ℕ0 × ℕ0) → ((2nd ‘(𝐸𝑋)) ≠ 0 → (2nd ‘(𝐸𝑋)) < (2nd𝑋)))

Theoremeucialgcvga 10647* Once Euclid's Algorithm halts after 𝑁 steps, the second element of the state remains 0 . (Contributed by Jim Kingdon, 11-Jan-2022.)
𝐸 = (𝑥 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))    &   𝑅 = seq0((𝐸 ∘ 1st ), (ℕ0 × {𝐴}), (ℕ0 × ℕ0))    &   𝑁 = (2nd𝐴)       (𝐴 ∈ (ℕ0 × ℕ0) → (𝐾 ∈ (ℤ𝑁) → (2nd ‘(𝑅𝐾)) = 0))

Theoremeucialg 10648* Euclid's Algorithm computes the greatest common divisor of two nonnegative integers by repeatedly replacing the larger of them with its remainder modulo the smaller until the remainder is 0. Theorem 1.15 in [ApostolNT] p. 20.

Upon halting, the 1st member of the final state (𝑅𝑁) is equal to the gcd of the values comprising the input state 𝑀, 𝑁. This is Metamath 100 proof #69 (greatest common divisor algorithm). (Contributed by Jim Kingdon, 11-Jan-2022.)

𝐸 = (𝑥 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))    &   𝑅 = seq0((𝐸 ∘ 1st ), (ℕ0 × {𝐴}), (ℕ0 × ℕ0))    &   𝑁 = (2nd𝐴)    &   𝐴 = ⟨𝑀, 𝑁       ((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → (1st ‘(𝑅𝑁)) = (𝑀 gcd 𝑁))

4.1.8  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 10650). The definition is valid for all integers, including negative integers and 0, obeying the above mentioned convention.

Syntaxclcm 10649 Extend the definition of a class to include the least common multiple operator.
class lcm

Definitiondf-lcm 10650* 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 ∣ (𝑥𝑛𝑦𝑛)}, ℝ, < )))

Theoremlcmmndc 10651 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))

Theoremlcmval 10652* 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 10546 and gcdval 10558. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Revised by AV, 16-Sep-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 lcm 𝑁) = if((𝑀 = 0 ∨ 𝑁 = 0), 0, inf({𝑛 ∈ ℕ ∣ (𝑀𝑛𝑁𝑛)}, ℝ, < )))

Theoremlcmcom 10653 The lcm operator is commutative. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Proof shortened by AV, 16-Sep-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 lcm 𝑁) = (𝑁 lcm 𝑀))

Theoremlcm0val 10654 The value, by convention, of the lcm operator when either operand is 0. (Use lcmcom 10653 for a left-hand 0.) (Contributed by Steve Rodriguez, 20-Jan-2020.) (Proof shortened by AV, 16-Sep-2020.)
(𝑀 ∈ ℤ → (𝑀 lcm 0) = 0)

Theoremlcmn0val 10655* The value of the lcm operator when both operands are nonzero. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Revised by AV, 16-Sep-2020.)
(((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∨ 𝑁 = 0)) → (𝑀 lcm 𝑁) = inf({𝑛 ∈ ℕ ∣ (𝑀𝑛𝑁𝑛)}, ℝ, < ))

Theoremlcmcllem 10656* Lemma for lcmn0cl 10657 and dvdslcm 10658. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Proof shortened by AV, 16-Sep-2020.)
(((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∨ 𝑁 = 0)) → (𝑀 lcm 𝑁) ∈ {𝑛 ∈ ℕ ∣ (𝑀𝑛𝑁𝑛)})

Theoremlcmn0cl 10657 Closure of the lcm operator. (Contributed by Steve Rodriguez, 20-Jan-2020.)
(((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∨ 𝑁 = 0)) → (𝑀 lcm 𝑁) ∈ ℕ)

Theoremdvdslcm 10658 The lcm of two integers is divisible by each of them. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ (𝑀 lcm 𝑁) ∧ 𝑁 ∥ (𝑀 lcm 𝑁)))

Theoremlcmledvds 10659 A positive integer which both operands of the lcm operator divide bounds it. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Proof shortened by AV, 16-Sep-2020.)
(((𝐾 ∈ ℕ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∨ 𝑁 = 0)) → ((𝑀𝐾𝑁𝐾) → (𝑀 lcm 𝑁) ≤ 𝐾))

Theoremlcmeq0 10660 The lcm of two integers is zero iff either is zero. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀 lcm 𝑁) = 0 ↔ (𝑀 = 0 ∨ 𝑁 = 0)))

Theoremlcmcl 10661 Closure of the lcm operator. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 lcm 𝑁) ∈ ℕ0)

Theoremgcddvdslcm 10662 The greatest common divisor of two numbers divides their least common multiple. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) ∥ (𝑀 lcm 𝑁))

Theoremlcmneg 10663 Negating one operand of the lcm operator does not alter the result. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 lcm -𝑁) = (𝑀 lcm 𝑁))

Theoremneglcm 10664 Negating one operand of the lcm operator does not alter the result. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (-𝑀 lcm 𝑁) = (𝑀 lcm 𝑁))

Theoremlcmabs 10665 The lcm of two integers is the same as that of their absolute values. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((abs‘𝑀) lcm (abs‘𝑁)) = (𝑀 lcm 𝑁))

Theoremlcmgcdlem 10666 Lemma for lcmgcd 10667 and lcmdvds 10668. Prove them for positive 𝑀, 𝑁, and 𝐾. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Proof shortened by AV, 16-Sep-2020.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (((𝑀 lcm 𝑁) · (𝑀 gcd 𝑁)) = (abs‘(𝑀 · 𝑁)) ∧ ((𝐾 ∈ ℕ ∧ (𝑀𝐾𝑁𝐾)) → (𝑀 lcm 𝑁) ∥ 𝐾)))

Theoremlcmgcd 10667 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 10607; see e.g. https://proofwiki.org/wiki/Product_of_GCD_and_LCM and https://math.stackexchange.com/a/470827. 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 10654 to show it applies when either or both inputs are zero. (Contributed by Steve Rodriguez, 20-Jan-2020.)

((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀 lcm 𝑁) · (𝑀 gcd 𝑁)) = (abs‘(𝑀 · 𝑁)))

Theoremlcmdvds 10668 The lcm of two integers divides any integer the two divide. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀𝐾𝑁𝐾) → (𝑀 lcm 𝑁) ∥ 𝐾))

Theoremlcmid 10669 The lcm of an integer and itself is its absolute value. (Contributed by Steve Rodriguez, 20-Jan-2020.)
(𝑀 ∈ ℤ → (𝑀 lcm 𝑀) = (abs‘𝑀))

Theoremlcm1 10670 The lcm of an integer and 1 is the absolute value of the integer. (Contributed by AV, 23-Aug-2020.)
(𝑀 ∈ ℤ → (𝑀 lcm 1) = (abs‘𝑀))

Theoremlcmgcdnn 10671 The product of two positive integers' least common multiple and greatest common divisor is the product of the two integers. (Contributed by AV, 27-Aug-2020.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝑀 lcm 𝑁) · (𝑀 gcd 𝑁)) = (𝑀 · 𝑁))

Theoremlcmgcdeq 10672 Two integers' absolute values are equal iff their least common multiple and greatest common divisor are equal. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀 lcm 𝑁) = (𝑀 gcd 𝑁) ↔ (abs‘𝑀) = (abs‘𝑁)))

Theoremlcmdvdsb 10673 Biconditional form of lcmdvds 10668. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀𝐾𝑁𝐾) ↔ (𝑀 lcm 𝑁) ∥ 𝐾))

Theoremlcmass 10674 Associative law for lcm operator. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Proof shortened by AV, 16-Sep-2020.)
((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑃 ∈ ℤ) → ((𝑁 lcm 𝑀) lcm 𝑃) = (𝑁 lcm (𝑀 lcm 𝑃)))

Theorem3lcm2e6woprm 10675 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

Theorem6lcm4e12 10676 The least common multiple of six and four is twelve. (Contributed by AV, 27-Aug-2020.)
(6 lcm 4) = 12

4.1.9  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 10677. The negation, i.e. two integers are not coprime, can be expressed either by (𝐴 gcd 𝐵) ≠ 1, see ncoprmgcdne1b 10678, or equivalently by 1 < (𝐴 gcd 𝐵), see ncoprmgcdgt1b 10679.

A proof of Euclid's lemma based on coprimality is provided in coprmdvds 10681 (as opposed to Euclid's lemma for primes).

Theoremcoprmgcdb 10677* Two positive integers are coprime, i.e. the only positive integer that divides both of them is 1, iff their greatest common divisor is 1. (Contributed by AV, 9-Aug-2020.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (∀𝑖 ∈ ℕ ((𝑖𝐴𝑖𝐵) → 𝑖 = 1) ↔ (𝐴 gcd 𝐵) = 1))

Theoremncoprmgcdne1b 10678* 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)(𝑖𝐴𝑖𝐵) ↔ (𝐴 gcd 𝐵) ≠ 1))

Theoremncoprmgcdgt1b 10679* Two positive integers are not coprime, i.e. there is an integer greater than 1 which divides both integers, iff their greatest common divisor is greater than 1. (Contributed by AV, 9-Aug-2020.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (∃𝑖 ∈ (ℤ‘2)(𝑖𝐴𝑖𝐵) ↔ 1 < (𝐴 gcd 𝐵)))

Theoremcoprmdvds1 10680 If two positive integers are coprime, i.e. their greatest common divisor is 1, the only positive integer that divides both of them is 1. (Contributed by AV, 4-Aug-2021.)
((𝐹 ∈ ℕ ∧ 𝐺 ∈ ℕ ∧ (𝐹 gcd 𝐺) = 1) → ((𝐼 ∈ ℕ ∧ 𝐼𝐹𝐼𝐺) → 𝐼 = 1))

Theoremcoprmdvds 10681 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) → 𝐾𝑁))

Theoremcoprmdvds2 10682 If an integer is divisible by two coprime integers, then it is divisible by their product. (Contributed by Mario Carneiro, 24-Feb-2014.)
(((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝐾 ∈ ℤ) ∧ (𝑀 gcd 𝑁) = 1) → ((𝑀𝐾𝑁𝐾) → (𝑀 · 𝑁) ∥ 𝐾))

Theoremmulgcddvds 10683 One half of rpmulgcd2 10684, which does not need the coprimality assumption. (Contributed by Mario Carneiro, 2-Jul-2015.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝐾 gcd (𝑀 · 𝑁)) ∥ ((𝐾 gcd 𝑀) · (𝐾 gcd 𝑁)))

Theoremrpmulgcd2 10684 If 𝑀 is relatively prime to 𝑁, then the GCD of 𝐾 with 𝑀 · 𝑁 is the product of the GCDs with 𝑀 and 𝑁 respectively. (Contributed by Mario Carneiro, 2-Jul-2015.)
(((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ (𝑀 gcd 𝑁) = 1) → (𝐾 gcd (𝑀 · 𝑁)) = ((𝐾 gcd 𝑀) · (𝐾 gcd 𝑁)))

Theoremqredeq 10685 Two equal reduced fractions have the same numerator and denominator. (Contributed by Jeff Hankins, 29-Sep-2013.)
(((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ ∧ (𝑀 gcd 𝑁) = 1) ∧ (𝑃 ∈ ℤ ∧ 𝑄 ∈ ℕ ∧ (𝑃 gcd 𝑄) = 1) ∧ (𝑀 / 𝑁) = (𝑃 / 𝑄)) → (𝑀 = 𝑃𝑁 = 𝑄))

Theoremqredeu 10686* Every rational number has a unique reduced form. (Contributed by Jeff Hankins, 29-Sep-2013.)
(𝐴 ∈ ℚ → ∃!𝑥 ∈ (ℤ × ℕ)(((1st𝑥) gcd (2nd𝑥)) = 1 ∧ 𝐴 = ((1st𝑥) / (2nd𝑥))))

Theoremrpmul 10687 If 𝐾 is relatively prime to 𝑀 and to 𝑁, it is also relatively prime to their product. (Contributed by Mario Carneiro, 24-Feb-2014.) (Proof shortened by Mario Carneiro, 2-Jul-2015.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (((𝐾 gcd 𝑀) = 1 ∧ (𝐾 gcd 𝑁) = 1) → (𝐾 gcd (𝑀 · 𝑁)) = 1))

Theoremrpdvds 10688 If 𝐾 is relatively prime to 𝑁 then it is also relatively prime to any divisor 𝑀 of 𝑁. (Contributed by Mario Carneiro, 19-Jun-2015.)
(((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ((𝐾 gcd 𝑁) = 1 ∧ 𝑀𝑁)) → (𝐾 gcd 𝑀) = 1)

4.1.10  Cancellability of congruences

Theoremcongr 10689* Definition of congruence by integer multiple (see ProofWiki "Congruence (Number Theory)", 11-Jul-2021, https://proofwiki.org/wiki/Definition:Congruence_(Number_Theory)): An integer 𝐴 is congruent to an integer 𝐵 modulo 𝑀 if their difference is a multiple of 𝑀. See also the definition in [ApostolNT] p. 104: "... 𝑎 is congruent to 𝑏 modulo 𝑚, and we write 𝑎𝑏 (mod 𝑚) if 𝑚 divides the difference 𝑎𝑏", or Wikipedia "Modular arithmetic - Congruence", https://en.wikipedia.org/wiki/Modular_arithmetic#Congruence, 11-Jul-2021,: "Given an integer n > 1, called a modulus, two integers are said to be congruent modulo n, if n is a divisor of their difference (i.e., if there is an integer k such that a-b = kn)". (Contributed by AV, 11-Jul-2021.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝑀 ∈ ℕ) → ((𝐴 mod 𝑀) = (𝐵 mod 𝑀) ↔ ∃𝑛 ∈ ℤ (𝑛 · 𝑀) = (𝐴𝐵)))

Theoremdivgcdcoprm0 10690 Integers divided by gcd are coprime. (Contributed by AV, 12-Jul-2021.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐵 ≠ 0) → ((𝐴 / (𝐴 gcd 𝐵)) gcd (𝐵 / (𝐴 gcd 𝐵))) = 1)

Theoremdivgcdcoprmex 10691* Integers divided by gcd are coprime (see ProofWiki "Integers Divided by GCD are Coprime", 11-Jul-2021, https://proofwiki.org/wiki/Integers_Divided_by_GCD_are_Coprime): Any pair of integers, not both zero, can be reduced to a pair of coprime ones by dividing them by their gcd. (Contributed by AV, 12-Jul-2021.)
((𝐴 ∈ ℤ ∧ (𝐵 ∈ ℤ ∧ 𝐵 ≠ 0) ∧ 𝑀 = (𝐴 gcd 𝐵)) → ∃𝑎 ∈ ℤ ∃𝑏 ∈ ℤ (𝐴 = (𝑀 · 𝑎) ∧ 𝐵 = (𝑀 · 𝑏) ∧ (𝑎 gcd 𝑏) = 1))

Theoremcncongr1 10692 One direction of the bicondition in cncongr 10694. 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)))

Theoremcncongr2 10693 The other direction of the bicondition in cncongr 10694. (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)))

Theoremcncongr 10694 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 greates 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)))

Theoremcncongrcoprm 10695 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)))

4.2  Elementary prime number theory

4.2.1  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 3535.

Syntaxcprime 10696 Extend the definition of a class to include the set of prime numbers.
class

Definitiondf-prm 10697* Define the set of prime numbers. (Contributed by Paul Chapman, 22-Jun-2011.)
ℙ = {𝑝 ∈ ℕ ∣ {𝑛 ∈ ℕ ∣ 𝑛𝑝} ≈ 2𝑜}

Theoremisprm 10698* The predicate "is a prime number". A prime number is a positive integer with exactly two positive divisors. (Contributed by Paul Chapman, 22-Jun-2011.)
(𝑃 ∈ ℙ ↔ (𝑃 ∈ ℕ ∧ {𝑛 ∈ ℕ ∣ 𝑛𝑃} ≈ 2𝑜))

Theoremprmnn 10699 A prime number is a positive integer. (Contributed by Paul Chapman, 22-Jun-2011.)
(𝑃 ∈ ℙ → 𝑃 ∈ ℕ)

Theoremprmz 10700 A prime number is an integer. (Contributed by Paul Chapman, 22-Jun-2011.) (Proof shortened by Jonathan Yan, 16-Jul-2017.)
(𝑃 ∈ ℙ → 𝑃 ∈ ℤ)

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