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Theorem List for Intuitionistic Logic Explorer - 11901-12000   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremzsupcllemex 11901* Lemma for zsupcl 11902. Existence of the supremum. (Contributed by Jim Kingdon, 7-Dec-2021.)
(𝜑𝑀 ∈ ℤ)    &   (𝑛 = 𝑀 → (𝜓𝜒))    &   (𝜑𝜒)    &   ((𝜑𝑛 ∈ (ℤ𝑀)) → DECID 𝜓)    &   (𝜑 → ∃𝑗 ∈ (ℤ𝑀)∀𝑛 ∈ (ℤ𝑗) ¬ 𝜓)       (𝜑 → ∃𝑥 ∈ ℤ (∀𝑦 ∈ {𝑛 ∈ ℤ ∣ 𝜓} ¬ 𝑥 < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 < 𝑥 → ∃𝑧 ∈ {𝑛 ∈ ℤ ∣ 𝜓}𝑦 < 𝑧)))
 
Theoremzsupcl 11902* Closure of supremum for decidable integer properties. The property which defines the set we are taking the supremum of must (a) be true at 𝑀 (which corresponds to the nonempty condition of classical supremum theorems), (b) decidable at each value after 𝑀, and (c) be false after 𝑗 (which corresponds to the upper bound condition found in classical supremum theorems). (Contributed by Jim Kingdon, 7-Dec-2021.)
(𝜑𝑀 ∈ ℤ)    &   (𝑛 = 𝑀 → (𝜓𝜒))    &   (𝜑𝜒)    &   ((𝜑𝑛 ∈ (ℤ𝑀)) → DECID 𝜓)    &   (𝜑 → ∃𝑗 ∈ (ℤ𝑀)∀𝑛 ∈ (ℤ𝑗) ¬ 𝜓)       (𝜑 → sup({𝑛 ∈ ℤ ∣ 𝜓}, ℝ, < ) ∈ (ℤ𝑀))
 
Theoremzssinfcl 11903* The infimum of a set of integers is an element of the set. (Contributed by Jim Kingdon, 16-Jan-2022.)
(𝜑 → ∃𝑥 ∈ ℝ (∀𝑦𝐵 ¬ 𝑦 < 𝑥 ∧ ∀𝑦 ∈ ℝ (𝑥 < 𝑦 → ∃𝑧𝐵 𝑧 < 𝑦)))    &   (𝜑𝐵 ⊆ ℤ)    &   (𝜑 → inf(𝐵, ℝ, < ) ∈ ℤ)       (𝜑 → inf(𝐵, ℝ, < ) ∈ 𝐵)
 
Theoreminfssuzex 11904* Existence of the infimum of a subset of an upper set of integers. (Contributed by Jim Kingdon, 13-Jan-2022.)
(𝜑𝑀 ∈ ℤ)    &   𝑆 = {𝑛 ∈ (ℤ𝑀) ∣ 𝜓}    &   (𝜑𝐴𝑆)    &   ((𝜑𝑛 ∈ (𝑀...𝐴)) → DECID 𝜓)       (𝜑 → ∃𝑥 ∈ ℝ (∀𝑦𝑆 ¬ 𝑦 < 𝑥 ∧ ∀𝑦 ∈ ℝ (𝑥 < 𝑦 → ∃𝑧𝑆 𝑧 < 𝑦)))
 
Theoreminfssuzledc 11905* The infimum of a subset of an upper set of integers is less than or equal to all members of the subset. (Contributed by Jim Kingdon, 13-Jan-2022.)
(𝜑𝑀 ∈ ℤ)    &   𝑆 = {𝑛 ∈ (ℤ𝑀) ∣ 𝜓}    &   (𝜑𝐴𝑆)    &   ((𝜑𝑛 ∈ (𝑀...𝐴)) → DECID 𝜓)       (𝜑 → inf(𝑆, ℝ, < ) ≤ 𝐴)
 
Theoreminfssuzcldc 11906* The infimum of a subset of an upper set of integers belongs to the subset. (Contributed by Jim Kingdon, 20-Jan-2022.)
(𝜑𝑀 ∈ ℤ)    &   𝑆 = {𝑛 ∈ (ℤ𝑀) ∣ 𝜓}    &   (𝜑𝐴𝑆)    &   ((𝜑𝑛 ∈ (𝑀...𝐴)) → DECID 𝜓)       (𝜑 → inf(𝑆, ℝ, < ) ∈ 𝑆)
 
Theoremsuprzubdc 11907* The supremum of a bounded-above decidable set of integers is greater than any member of the set. (Contributed by Mario Carneiro, 21-Apr-2015.) (Revised by Jim Kingdon, 5-Oct-2024.)
(𝜑𝐴 ⊆ ℤ)    &   (𝜑 → ∀𝑥 ∈ ℤ DECID 𝑥𝐴)    &   (𝜑 → ∃𝑥 ∈ ℤ ∀𝑦𝐴 𝑦𝑥)    &   (𝜑𝐵𝐴)       (𝜑𝐵 ≤ sup(𝐴, ℝ, < ))
 
Theoremnninfdcex 11908* A decidable set of natural numbers has an infimum. (Contributed by Jim Kingdon, 28-Sep-2024.)
(𝜑𝐴 ⊆ ℕ)    &   (𝜑 → ∀𝑥 ∈ ℕ DECID 𝑥𝐴)    &   (𝜑 → ∃𝑦 𝑦𝐴)       (𝜑 → ∃𝑥 ∈ ℝ (∀𝑦𝐴 ¬ 𝑦 < 𝑥 ∧ ∀𝑦 ∈ ℝ (𝑥 < 𝑦 → ∃𝑧𝐴 𝑧 < 𝑦)))
 
Theoremzsupssdc 11909* An inhabited decidable bounded subset of integers has a supremum in the set. (The proof does not use ax-pre-suploc 7895.) (Contributed by Mario Carneiro, 21-Apr-2015.) (Revised by Jim Kingdon, 5-Oct-2024.)
(𝜑𝐴 ⊆ ℤ)    &   (𝜑 → ∃𝑥 𝑥𝐴)    &   (𝜑 → ∀𝑥 ∈ ℤ DECID 𝑥𝐴)    &   (𝜑 → ∃𝑥 ∈ ℤ ∀𝑦𝐴 𝑦𝑥)       (𝜑 → ∃𝑥𝐴 (∀𝑦𝐴 ¬ 𝑥 < 𝑦 ∧ ∀𝑦𝐵 (𝑦 < 𝑥 → ∃𝑧𝐴 𝑦 < 𝑧)))
 
Theoremsuprzcl2dc 11910* The supremum of a bounded-above decidable set of integers is a member of the set. (This theorem avoids ax-pre-suploc 7895.) (Contributed by Mario Carneiro, 21-Apr-2015.) (Revised by Jim Kingdon, 6-Oct-2024.)
(𝜑𝐴 ⊆ ℤ)    &   (𝜑 → ∀𝑥 ∈ ℤ DECID 𝑥𝐴)    &   (𝜑 → ∃𝑥 ∈ ℤ ∀𝑦𝐴 𝑦𝑥)    &   (𝜑 → ∃𝑥 𝑥𝐴)       (𝜑 → sup(𝐴, ℝ, < ) ∈ 𝐴)
 
Theoremdvdsbnd 11911* There is an upper bound to the divisors of a nonzero integer. (Contributed by Jim Kingdon, 11-Dec-2021.)
((𝐴 ∈ ℤ ∧ 𝐴 ≠ 0) → ∃𝑛 ∈ ℕ ∀𝑚 ∈ (ℤ𝑛) ¬ 𝑚𝐴)
 
Theoremgcdsupex 11912* Existence of the supremum used in defining gcd. (Contributed by Jim Kingdon, 12-Dec-2021.)
(((𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ) ∧ ¬ (𝑋 = 0 ∧ 𝑌 = 0)) → ∃𝑥 ∈ ℤ (∀𝑦 ∈ {𝑛 ∈ ℤ ∣ (𝑛𝑋𝑛𝑌)} ¬ 𝑥 < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 < 𝑥 → ∃𝑧 ∈ {𝑛 ∈ ℤ ∣ (𝑛𝑋𝑛𝑌)}𝑦 < 𝑧)))
 
Theoremgcdsupcl 11913* Closure of the supremum used in defining gcd. A lemma for gcdval 11914 and gcdn0cl 11917. (Contributed by Jim Kingdon, 11-Dec-2021.)
(((𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ) ∧ ¬ (𝑋 = 0 ∧ 𝑌 = 0)) → sup({𝑛 ∈ ℤ ∣ (𝑛𝑋𝑛𝑌)}, ℝ, < ) ∈ ℕ)
 
Theoremgcdval 11914* The value of the gcd operator. (𝑀 gcd 𝑁) is the greatest common divisor of 𝑀 and 𝑁. If 𝑀 and 𝑁 are both 0, the result is defined conventionally as 0. (Contributed by Paul Chapman, 21-Mar-2011.) (Revised by Mario Carneiro, 10-Nov-2013.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) = if((𝑀 = 0 ∧ 𝑁 = 0), 0, sup({𝑛 ∈ ℤ ∣ (𝑛𝑀𝑛𝑁)}, ℝ, < )))
 
Theoremgcd0val 11915 The value, by convention, of the gcd operator when both operands are 0. (Contributed by Paul Chapman, 21-Mar-2011.)
(0 gcd 0) = 0
 
Theoremgcdn0val 11916* The value of the gcd operator when at least one operand is nonzero. (Contributed by Paul Chapman, 21-Mar-2011.)
(((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∧ 𝑁 = 0)) → (𝑀 gcd 𝑁) = sup({𝑛 ∈ ℤ ∣ (𝑛𝑀𝑛𝑁)}, ℝ, < ))
 
Theoremgcdn0cl 11917 Closure of the gcd operator. (Contributed by Paul Chapman, 21-Mar-2011.)
(((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∧ 𝑁 = 0)) → (𝑀 gcd 𝑁) ∈ ℕ)
 
Theoremgcddvds 11918 The gcd of two integers divides each of them. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀 gcd 𝑁) ∥ 𝑀 ∧ (𝑀 gcd 𝑁) ∥ 𝑁))
 
Theoremdvdslegcd 11919 An integer which divides both operands of the gcd operator is bounded by it. (Contributed by Paul Chapman, 21-Mar-2011.)
(((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∧ 𝑁 = 0)) → ((𝐾𝑀𝐾𝑁) → 𝐾 ≤ (𝑀 gcd 𝑁)))
 
Theoremnndvdslegcd 11920 A positive integer which divides both positive operands of the gcd operator is bounded by it. (Contributed by AV, 9-Aug-2020.)
((𝐾 ∈ ℕ ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝐾𝑀𝐾𝑁) → 𝐾 ≤ (𝑀 gcd 𝑁)))
 
Theoremgcdcl 11921 Closure of the gcd operator. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) ∈ ℕ0)
 
Theoremgcdnncl 11922 Closure of the gcd operator. (Contributed by Thierry Arnoux, 2-Feb-2020.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 gcd 𝑁) ∈ ℕ)
 
Theoremgcdcld 11923 Closure of the gcd operator. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)       (𝜑 → (𝑀 gcd 𝑁) ∈ ℕ0)
 
Theoremgcd2n0cl 11924 Closure of the gcd operator if the second operand is not 0. (Contributed by AV, 10-Jul-2021.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑁 ≠ 0) → (𝑀 gcd 𝑁) ∈ ℕ)
 
Theoremzeqzmulgcd 11925* An integer is the product of an integer and the gcd of it and another integer. (Contributed by AV, 11-Jul-2021.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ∃𝑛 ∈ ℤ 𝐴 = (𝑛 · (𝐴 gcd 𝐵)))
 
Theoremdivgcdz 11926 An integer divided by the gcd of it and a nonzero integer is an integer. (Contributed by AV, 11-Jul-2021.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐵 ≠ 0) → (𝐴 / (𝐴 gcd 𝐵)) ∈ ℤ)
 
Theoremgcdf 11927 Domain and codomain of the gcd operator. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 16-Nov-2013.)
gcd :(ℤ × ℤ)⟶ℕ0
 
Theoremgcdcom 11928 The gcd operator is commutative. Theorem 1.4(a) in [ApostolNT] p. 16. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) = (𝑁 gcd 𝑀))
 
Theoremgcdcomd 11929 The gcd operator is commutative, deduction version. (Contributed by SN, 24-Aug-2024.)
(𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)       (𝜑 → (𝑀 gcd 𝑁) = (𝑁 gcd 𝑀))
 
Theoremdivgcdnn 11930 A positive integer divided by the gcd of it and another integer is a positive integer. (Contributed by AV, 10-Jul-2021.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ) → (𝐴 / (𝐴 gcd 𝐵)) ∈ ℕ)
 
Theoremdivgcdnnr 11931 A positive integer divided by the gcd of it and another integer is a positive integer. (Contributed by AV, 10-Jul-2021.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ) → (𝐴 / (𝐵 gcd 𝐴)) ∈ ℕ)
 
Theoremgcdeq0 11932 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)))
 
Theoremgcdn0gt0 11933 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)))
 
Theoremgcd0id 11934 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))
 
Theoremgcdid0 11935 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))
 
Theoremnn0gcdid0 11936 The gcd of a nonnegative integer with 0 is itself. (Contributed by Paul Chapman, 31-Mar-2011.)
(𝑁 ∈ ℕ0 → (𝑁 gcd 0) = 𝑁)
 
Theoremgcdneg 11937 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))
 
Theoremneggcd 11938 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))
 
Theoremgcdaddm 11939 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))))
 
Theoremgcdadd 11940 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)))
 
Theoremgcdid 11941 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))
 
Theoremgcd1 11942 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)
 
Theoremgcdabs 11943 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))
 
Theoremgcdabs1 11944 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))
 
Theoremgcdabs2 11945 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))
 
Theoremmodgcd 11946 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))
 
Theorem1gcd 11947 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)
 
Theoremgcdmultipled 11948 The greatest common divisor of a nonnegative integer 𝑀 and a multiple of it is 𝑀 itself. (Contributed by Rohan Ridenour, 3-Aug-2023.)
(𝜑𝑀 ∈ ℕ0)    &   (𝜑𝑁 ∈ ℤ)       (𝜑 → (𝑀 gcd (𝑁 · 𝑀)) = 𝑀)
 
Theoremdvdsgcdidd 11949 The greatest common divisor of a positive integer and another integer it divides is itself. (Contributed by Rohan Ridenour, 3-Aug-2023.)
(𝜑𝑀 ∈ ℕ)    &   (𝜑𝑁 ∈ ℤ)    &   (𝜑𝑀𝑁)       (𝜑 → (𝑀 gcd 𝑁) = 𝑀)
 
Theorem6gcd4e2 11950 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
 
Theorembezoutlemnewy 11951* 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)))    &   (𝜃𝐴 ∈ ℕ0)    &   (𝜃𝐵 ∈ ℕ0)    &   (𝜃𝑊 ∈ ℕ)    &   (𝜃 → [𝑦 / 𝑟]𝜑)    &   (𝜃𝑦 ∈ ℕ0)    &   (𝜃[𝑊 / 𝑟]𝜑)       (𝜃[(𝑦 mod 𝑊) / 𝑟]𝜑)
 
Theorembezoutlemstep 11952* 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)))    &   (𝜃𝐴 ∈ ℕ0)    &   (𝜃𝐵 ∈ ℕ0)    &   (𝜃𝑊 ∈ ℕ)    &   (𝜃 → [𝑦 / 𝑟]𝜑)    &   (𝜃𝑦 ∈ ℕ0)    &   (𝜃[𝑊 / 𝑟]𝜑)    &   (𝜓 ↔ ∀𝑧 ∈ ℕ0 (𝑧𝑟 → (𝑧𝑥𝑧𝑦)))    &   ((𝜃[(𝑦 mod 𝑊) / 𝑟]𝜑) → ∃𝑟 ∈ ℕ0 ([(𝑦 mod 𝑊) / 𝑥][𝑊 / 𝑦]𝜓𝜑))    &   𝑥𝜃    &   𝑟𝜃       (𝜃 → ∃𝑟 ∈ ℕ0 ([𝑊 / 𝑥]𝜓𝜑))
 
Theorembezoutlemmain 11953* 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 (𝑧𝑟 → (𝑧𝑥𝑧𝑦)))    &   (𝜃𝐴 ∈ ℕ0)    &   (𝜃𝐵 ∈ ℕ0)       (𝜃 → ∀𝑥 ∈ ℕ0 ([𝑥 / 𝑟]𝜑 → ∀𝑦 ∈ ℕ0 ([𝑦 / 𝑟]𝜑 → ∃𝑟 ∈ ℕ0 (𝜓𝜑))))
 
Theorembezoutlema 11954* 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)))    &   (𝜃𝐴 ∈ ℕ0)    &   (𝜃𝐵 ∈ ℕ0)       (𝜃[𝐴 / 𝑟]𝜑)
 
Theorembezoutlemb 11955* 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)))    &   (𝜃𝐴 ∈ ℕ0)    &   (𝜃𝐵 ∈ ℕ0)       (𝜃[𝐵 / 𝑟]𝜑)
 
Theorembezoutlemex 11956* 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 ∈ ℕ0𝐵 ∈ ℕ0) → ∃𝑑 ∈ ℕ0 (∀𝑧 ∈ ℕ0 (𝑧𝑑 → (𝑧𝐴𝑧𝐵)) ∧ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑑 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))))
 
Theorembezoutlemzz 11957* Lemma for Bézout's identity. Like bezoutlemex 11956 but where ' z ' is any integer, not just a nonnegative one. (Contributed by Mario Carneiro and Jim Kingdon, 8-Jan-2022.)
((𝐴 ∈ ℕ0𝐵 ∈ ℕ0) → ∃𝑑 ∈ ℕ0 (∀𝑧 ∈ ℤ (𝑧𝑑 → (𝑧𝐴𝑧𝐵)) ∧ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑑 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))))
 
Theorembezoutlemaz 11958* Lemma for Bézout's identity. Like bezoutlemzz 11957 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 (𝑧𝑑 → (𝑧𝐴𝑧𝐵)) ∧ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑑 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))))
 
Theorembezoutlembz 11959* Lemma for Bézout's identity. Like bezoutlemaz 11958 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 (𝑧𝑑 → (𝑧𝐴𝑧𝐵)) ∧ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑑 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))))
 
Theorembezoutlembi 11960* Lemma for Bézout's identity. Like bezoutlembz 11959 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 11961* 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 11962* 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 11963* 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 11964* 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 11965* 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 11966* 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 11967 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 11968 Biconditional form of dvdsgcd 11967. (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 11969* 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 11970 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 11971 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 11972 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 11973 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 11974 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 11975 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 11976 Extend gcdmultiple 11975 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 11977 A positive integer 𝐴 is equal to its gcd with an integer 𝐵 if and only if 𝐴 divides 𝐵. Generalization of gcdeq 11978. (Contributed by AV, 1-Jul-2020.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ) → ((𝐴 gcd 𝐵) = 𝐴𝐴𝐵))
 
Theoremgcdeq 11978 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 11979 Unidirectional form of dvdssq 11986. (Contributed by Scott Fenton, 19-Apr-2014.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀𝑁 → (𝑀↑2) ∥ (𝑁↑2)))
 
Theoremdvdsmulgcd 11980 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 11981 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 11982 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 11983 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 11984 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 11985 Lemma for dvdssq 11986. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀𝑁 ↔ (𝑀↑2) ∥ (𝑁↑2)))
 
Theoremdvdssq 11986 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 11987 Partial converse to bezout 11966. 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 11988 Converse of bezout 11966 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
 
Theoremnnmindc 11989* An inhabited decidable subset of the natural numbers has a minimum. (Contributed by Jim Kingdon, 23-Sep-2024.)
((𝐴 ⊆ ℕ ∧ ∀𝑥 ∈ ℕ DECID 𝑥𝐴 ∧ ∃𝑦 𝑦𝐴) → inf(𝐴, ℝ, < ) ∈ 𝐴)
 
Theoremnnminle 11990* 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 11989. (Contributed by Jim Kingdon, 26-Sep-2024.)
((𝐴 ⊆ ℕ ∧ ∀𝑥 ∈ ℕ DECID 𝑥𝐴𝐵𝐴) → inf(𝐴, ℝ, < ) ≤ 𝐵)
 
Theoremnnwodc 11991* 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 𝑤𝐴 ∧ ∀𝑗 ∈ ℕ DECID 𝑗𝐴) → ∃𝑥𝐴𝑦𝐴 𝑥𝑦)
 
Theoremuzwodc 11992* 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 ⊆ (ℤ𝑀) ∧ ∃𝑥 𝑥𝑆 ∧ ∀𝑥 ∈ (ℤ𝑀)DECID 𝑥𝑆) → ∃𝑗𝑆𝑘𝑆 𝑗𝑘)
 
Theoremnnwofdc 11993* 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.)
𝑥𝐴    &   𝑦𝐴       ((𝐴 ⊆ ℕ ∧ ∃𝑧 𝑧𝐴 ∧ ∀𝑗 ∈ ℕ DECID 𝑗𝐴) → ∃𝑥𝐴𝑦𝐴 𝑥𝑦)
 
Theoremnnwosdc 11994* 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 → (𝜑𝜓))       ((∃𝑥 ∈ ℕ 𝜑 ∧ ∀𝑥 ∈ ℕ DECID 𝜑) → ∃𝑥 ∈ ℕ (𝜑 ∧ ∀𝑦 ∈ ℕ (𝜓𝑥𝑦)))
 
5.1.7  Algorithms
 
Theoremnn0seqcvgd 11995* 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 11996 Lemma for ialgr0 11998. Expressing algrflemg 6209 in a form suitable for theorems such as seq3-1 10416 or seqf 10417. (Contributed by Jim Kingdon, 22-Jul-2021.)
(𝜑𝐹:𝑆𝑆)       ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥(𝐹 ∘ 1st )𝑦) ∈ 𝑆)
 
Theoremialgrlemconst 11997 Lemma for ialgr0 11998. Closure of a constant function, in a form suitable for theorems such as seq3-1 10416 or seqf 10417. (Contributed by Jim Kingdon, 22-Jul-2021.)
𝑍 = (ℤ𝑀)    &   (𝜑𝐴𝑆)       ((𝜑𝑥 ∈ (ℤ𝑀)) → ((𝑍 × {𝐴})‘𝑥) ∈ 𝑆)
 
Theoremialgr0 11998 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 = (ℤ𝑀)    &   𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}))    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐴𝑆)    &   (𝜑𝐹:𝑆𝑆)       (𝜑 → (𝑅𝑀) = 𝐴)
 
Theoremalgrf 11999 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}))    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐴𝑆)    &   (𝜑𝐹:𝑆𝑆)       (𝜑𝑅:𝑍𝑆)
 
Theoremalgrp1 12000 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 = (ℤ𝑀)    &   𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}))    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐴𝑆)    &   (𝜑𝐹:𝑆𝑆)       ((𝜑𝐾𝑍) → (𝑅‘(𝐾 + 1)) = (𝐹‘(𝑅𝐾)))
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