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Theorem List for Intuitionistic Logic Explorer - 11801-11900   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremnnoddm1d2 11801 A positive integer is odd iff its successor divided by 2 is a positive integer. (Contributed by AV, 28-Jun-2021.)
(𝑁 ∈ ℕ → (¬ 2 ∥ 𝑁 ↔ ((𝑁 + 1) / 2) ∈ ℕ))
 
Theoremz0even 11802 0 is even. (Contributed by AV, 11-Feb-2020.) (Revised by AV, 23-Jun-2021.)
2 ∥ 0
 
Theoremn2dvds1 11803 2 does not divide 1 (common case). That means 1 is odd. (Contributed by David A. Wheeler, 8-Dec-2018.)
¬ 2 ∥ 1
 
Theoremn2dvdsm1 11804 2 does not divide -1. That means -1 is odd. (Contributed by AV, 15-Aug-2021.)
¬ 2 ∥ -1
 
Theoremz2even 11805 2 is even. (Contributed by AV, 12-Feb-2020.) (Revised by AV, 23-Jun-2021.)
2 ∥ 2
 
Theoremn2dvds3 11806 2 does not divide 3, i.e. 3 is an odd number. (Contributed by AV, 28-Feb-2021.)
¬ 2 ∥ 3
 
Theoremz4even 11807 4 is an even number. (Contributed by AV, 23-Jul-2020.) (Revised by AV, 4-Jul-2021.)
2 ∥ 4
 
Theorem4dvdseven 11808 An integer which is divisible by 4 is an even integer. (Contributed by AV, 4-Jul-2021.)
(4 ∥ 𝑁 → 2 ∥ 𝑁)
 
5.1.3  The division algorithm
 
Theoremdivalglemnn 11809* Lemma for divalg 11815. Existence for a positive denominator. (Contributed by Jim Kingdon, 30-Nov-2021.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ) → ∃𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)))
 
Theoremdivalglemqt 11810 Lemma for divalg 11815. The 𝑄 = 𝑇 case involved in showing uniqueness. (Contributed by Jim Kingdon, 5-Dec-2021.)
(𝜑𝐷 ∈ ℤ)    &   (𝜑𝑅 ∈ ℤ)    &   (𝜑𝑆 ∈ ℤ)    &   (𝜑𝑄 ∈ ℤ)    &   (𝜑𝑇 ∈ ℤ)    &   (𝜑𝑄 = 𝑇)    &   (𝜑 → ((𝑄 · 𝐷) + 𝑅) = ((𝑇 · 𝐷) + 𝑆))       (𝜑𝑅 = 𝑆)
 
Theoremdivalglemnqt 11811 Lemma for divalg 11815. The 𝑄 < 𝑇 case involved in showing uniqueness. (Contributed by Jim Kingdon, 4-Dec-2021.)
(𝜑𝐷 ∈ ℕ)    &   (𝜑𝑅 ∈ ℤ)    &   (𝜑𝑆 ∈ ℤ)    &   (𝜑𝑄 ∈ ℤ)    &   (𝜑𝑇 ∈ ℤ)    &   (𝜑 → 0 ≤ 𝑆)    &   (𝜑𝑅 < 𝐷)    &   (𝜑 → ((𝑄 · 𝐷) + 𝑅) = ((𝑇 · 𝐷) + 𝑆))       (𝜑 → ¬ 𝑄 < 𝑇)
 
Theoremdivalglemeunn 11812* Lemma for divalg 11815. Uniqueness for a positive denominator. (Contributed by Jim Kingdon, 4-Dec-2021.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ) → ∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)))
 
Theoremdivalglemex 11813* Lemma for divalg 11815. The quotient and remainder exist. (Contributed by Jim Kingdon, 30-Nov-2021.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℤ ∧ 𝐷 ≠ 0) → ∃𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)))
 
Theoremdivalglemeuneg 11814* Lemma for divalg 11815. Uniqueness for a negative denominator. (Contributed by Jim Kingdon, 4-Dec-2021.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℤ ∧ 𝐷 < 0) → ∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)))
 
Theoremdivalg 11815* The division algorithm (theorem). Dividing an integer 𝑁 by a nonzero integer 𝐷 produces a (unique) quotient 𝑞 and a unique remainder 0 ≤ 𝑟 < (abs‘𝐷). Theorem 1.14 in [ApostolNT] p. 19. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℤ ∧ 𝐷 ≠ 0) → ∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)))
 
Theoremdivalgb 11816* Express the division algorithm as stated in divalg 11815 in terms of . (Contributed by Paul Chapman, 31-Mar-2011.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℤ ∧ 𝐷 ≠ 0) → (∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)) ↔ ∃!𝑟 ∈ ℕ0 (𝑟 < (abs‘𝐷) ∧ 𝐷 ∥ (𝑁𝑟))))
 
Theoremdivalg2 11817* The division algorithm (theorem) for a positive divisor. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ) → ∃!𝑟 ∈ ℕ0 (𝑟 < 𝐷𝐷 ∥ (𝑁𝑟)))
 
Theoremdivalgmod 11818 The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor (compare divalg2 11817 and divalgb 11816). This demonstration theorem justifies the use of mod to yield an explicit remainder from this point forward. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by AV, 21-Aug-2021.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ) → (𝑅 = (𝑁 mod 𝐷) ↔ (𝑅 ∈ ℕ0 ∧ (𝑅 < 𝐷𝐷 ∥ (𝑁𝑅)))))
 
Theoremdivalgmodcl 11819 The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor. Variant of divalgmod 11818. (Contributed by Stefan O'Rear, 17-Oct-2014.) (Proof shortened by AV, 21-Aug-2021.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ ∧ 𝑅 ∈ ℕ0) → (𝑅 = (𝑁 mod 𝐷) ↔ (𝑅 < 𝐷𝐷 ∥ (𝑁𝑅))))
 
Theoremmodremain 11820* The result of the modulo operation is the remainder of the division algorithm. (Contributed by AV, 19-Aug-2021.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ ∧ (𝑅 ∈ ℕ0𝑅 < 𝐷)) → ((𝑁 mod 𝐷) = 𝑅 ↔ ∃𝑧 ∈ ℤ ((𝑧 · 𝐷) + 𝑅) = 𝑁))
 
Theoremndvdssub 11821 Corollary of the division algorithm. If an integer 𝐷 greater than 1 divides 𝑁, then it does not divide any of 𝑁 − 1, 𝑁 − 2... 𝑁 − (𝐷 − 1). (Contributed by Paul Chapman, 31-Mar-2011.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ ∧ (𝐾 ∈ ℕ ∧ 𝐾 < 𝐷)) → (𝐷𝑁 → ¬ 𝐷 ∥ (𝑁𝐾)))
 
Theoremndvdsadd 11822 Corollary of the division algorithm. If an integer 𝐷 greater than 1 divides 𝑁, then it does not divide any of 𝑁 + 1, 𝑁 + 2... 𝑁 + (𝐷 − 1). (Contributed by Paul Chapman, 31-Mar-2011.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ ∧ (𝐾 ∈ ℕ ∧ 𝐾 < 𝐷)) → (𝐷𝑁 → ¬ 𝐷 ∥ (𝑁 + 𝐾)))
 
Theoremndvdsp1 11823 Special case of ndvdsadd 11822. If an integer 𝐷 greater than 1 divides 𝑁, it does not divide 𝑁 + 1. (Contributed by Paul Chapman, 31-Mar-2011.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ ∧ 1 < 𝐷) → (𝐷𝑁 → ¬ 𝐷 ∥ (𝑁 + 1)))
 
Theoremndvdsi 11824 A quick test for non-divisibility. (Contributed by Mario Carneiro, 18-Feb-2014.)
𝐴 ∈ ℕ    &   𝑄 ∈ ℕ0    &   𝑅 ∈ ℕ    &   ((𝐴 · 𝑄) + 𝑅) = 𝐵    &   𝑅 < 𝐴        ¬ 𝐴𝐵
 
Theoremflodddiv4 11825 The floor of an odd integer divided by 4. (Contributed by AV, 17-Jun-2021.)
((𝑀 ∈ ℤ ∧ 𝑁 = ((2 · 𝑀) + 1)) → (⌊‘(𝑁 / 4)) = if(2 ∥ 𝑀, (𝑀 / 2), ((𝑀 − 1) / 2)))
 
Theoremfldivndvdslt 11826 The floor of an integer divided by a nonzero integer not dividing the first integer is less than the integer divided by the positive integer. (Contributed by AV, 4-Jul-2021.)
((𝐾 ∈ ℤ ∧ (𝐿 ∈ ℤ ∧ 𝐿 ≠ 0) ∧ ¬ 𝐿𝐾) → (⌊‘(𝐾 / 𝐿)) < (𝐾 / 𝐿))
 
Theoremflodddiv4lt 11827 The floor of an odd number divided by 4 is less than the odd number divided by 4. (Contributed by AV, 4-Jul-2021.)
((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → (⌊‘(𝑁 / 4)) < (𝑁 / 4))
 
Theoremflodddiv4t2lthalf 11828 The floor of an odd number divided by 4, multiplied by 2 is less than the half of the odd number. (Contributed by AV, 4-Jul-2021.)
((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → ((⌊‘(𝑁 / 4)) · 2) < (𝑁 / 2))
 
5.1.4  The greatest common divisor operator
 
Syntaxcgcd 11829 Extend the definition of a class to include the greatest common divisor operator.
class gcd
 
Definitiondf-gcd 11830* Define the gcd operator. For example, (-6 gcd 9) = 3 (ex-gcd 13318). (Contributed by Paul Chapman, 21-Mar-2011.)
gcd = (𝑥 ∈ ℤ, 𝑦 ∈ ℤ ↦ if((𝑥 = 0 ∧ 𝑦 = 0), 0, sup({𝑛 ∈ ℤ ∣ (𝑛𝑥𝑛𝑦)}, ℝ, < )))
 
Theoremgcdmndc 11831 Decidablity lemma used in various proofs related to gcd. (Contributed by Jim Kingdon, 12-Dec-2021.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → DECID (𝑀 = 0 ∧ 𝑁 = 0))
 
Theoremzsupcllemstep 11832* Lemma for zsupcl 11834. Induction step. (Contributed by Jim Kingdon, 7-Dec-2021.)
((𝜑𝑛 ∈ (ℤ𝑀)) → DECID 𝜓)       (𝐾 ∈ (ℤ𝑀) → (((𝜑 ∧ ∀𝑛 ∈ (ℤ𝐾) ¬ 𝜓) → ∃𝑥 ∈ ℤ (∀𝑦 ∈ {𝑛 ∈ ℤ ∣ 𝜓} ¬ 𝑥 < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 < 𝑥 → ∃𝑧 ∈ {𝑛 ∈ ℤ ∣ 𝜓}𝑦 < 𝑧))) → ((𝜑 ∧ ∀𝑛 ∈ (ℤ‘(𝐾 + 1)) ¬ 𝜓) → ∃𝑥 ∈ ℤ (∀𝑦 ∈ {𝑛 ∈ ℤ ∣ 𝜓} ¬ 𝑥 < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 < 𝑥 → ∃𝑧 ∈ {𝑛 ∈ ℤ ∣ 𝜓}𝑦 < 𝑧)))))
 
Theoremzsupcllemex 11833* Lemma for zsupcl 11834. 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 11834* 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 11835* 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 11836* 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 11837* 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 11838* 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)
 
Theoremnninfdcex 11839* 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)))
 
Theoremdvdsbnd 11840* 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 11841* 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 11842* Closure of the supremum used in defining gcd. A lemma for gcdval 11843 and gcdn0cl 11846. (Contributed by Jim Kingdon, 11-Dec-2021.)
(((𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ) ∧ ¬ (𝑋 = 0 ∧ 𝑌 = 0)) → sup({𝑛 ∈ ℤ ∣ (𝑛𝑋𝑛𝑌)}, ℝ, < ) ∈ ℕ)
 
Theoremgcdval 11843* 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 11844 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 11845* 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 11846 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 11847 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 11848 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 11849 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 11850 Closure of the gcd operator. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) ∈ ℕ0)
 
Theoremgcdnncl 11851 Closure of the gcd operator. (Contributed by Thierry Arnoux, 2-Feb-2020.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 gcd 𝑁) ∈ ℕ)
 
Theoremgcdcld 11852 Closure of the gcd operator. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)       (𝜑 → (𝑀 gcd 𝑁) ∈ ℕ0)
 
Theoremgcd2n0cl 11853 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 11854* 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 11855 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 11856 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 11857 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 11858 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 11859 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 11860 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 11861 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 11862 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 11863 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 11864 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 11865 The gcd of a nonnegative integer with 0 is itself. (Contributed by Paul Chapman, 31-Mar-2011.)
(𝑁 ∈ ℕ0 → (𝑁 gcd 0) = 𝑁)
 
Theoremgcdneg 11866 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 11867 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 11868 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 11869 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 11870 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 11871 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 11872 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 11873 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 11874 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 11875 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 11876 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 11877 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 11878 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 11879 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 11880* 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 11881* 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 11882* 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 11883* 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 11884* 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 11885* 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 11886* Lemma for Bézout's identity. Like bezoutlemex 11885 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 11887* Lemma for Bézout's identity. Like bezoutlemzz 11886 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 11888* Lemma for Bézout's identity. Like bezoutlemaz 11887 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 11889* Lemma for Bézout's identity. Like bezoutlembz 11888 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 11890* 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 11891* 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 11892* 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 11893* 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 11894* 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 11895* 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 11896 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 11897 Biconditional form of dvdsgcd 11896. (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 11898* 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 11899 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 11900 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)))
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