P. 106 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 10501-10600   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	halfleoddlt 10501	An integer is greater than half of an odd number iff it is greater than or equal to the half of the odd number. (Contributed by AV, 1-Jul-2021.)
⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁 ∧ 𝑀 ∈ ℤ) → ((𝑁 / 2) ≤ 𝑀 ↔ (𝑁 / 2) < 𝑀))
Theorem	opoe 10502	The sum of two odds is even. (Contributed by Scott Fenton, 7-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ (((𝐴 ∈ ℤ ∧ ¬ 2 ∥ 𝐴) ∧ (𝐵 ∈ ℤ ∧ ¬ 2 ∥ 𝐵)) → 2 ∥ (𝐴 + 𝐵))
Theorem	omoe 10503	The difference of two odds is even. (Contributed by Scott Fenton, 7-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ (((𝐴 ∈ ℤ ∧ ¬ 2 ∥ 𝐴) ∧ (𝐵 ∈ ℤ ∧ ¬ 2 ∥ 𝐵)) → 2 ∥ (𝐴 − 𝐵))
Theorem	opeo 10504	The sum of an odd and an even is odd. (Contributed by Scott Fenton, 7-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ (((𝐴 ∈ ℤ ∧ ¬ 2 ∥ 𝐴) ∧ (𝐵 ∈ ℤ ∧ 2 ∥ 𝐵)) → ¬ 2 ∥ (𝐴 + 𝐵))
Theorem	omeo 10505	The difference of an odd and an even is odd. (Contributed by Scott Fenton, 7-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ (((𝐴 ∈ ℤ ∧ ¬ 2 ∥ 𝐴) ∧ (𝐵 ∈ ℤ ∧ 2 ∥ 𝐵)) → ¬ 2 ∥ (𝐴 − 𝐵))
Theorem	m1expe 10506	Exponentiation of -1 by an even power. Variant of m1expeven 9672. (Contributed by AV, 25-Jun-2021.)
⊢ (2 ∥ 𝑁 → (-1↑𝑁) = 1)
Theorem	m1expo 10507	Exponentiation of -1 by an odd power. (Contributed by AV, 26-Jun-2021.)
⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → (-1↑𝑁) = -1)
Theorem	m1exp1 10508	Exponentiation of negative one is one iff the exponent is even. (Contributed by AV, 20-Jun-2021.)
⊢ (𝑁 ∈ ℤ → ((-1↑𝑁) = 1 ↔ 2 ∥ 𝑁))
Theorem	nn0enne 10509	A positive integer is an even nonnegative integer iff it is an even positive integer. (Contributed by AV, 30-May-2020.)
⊢ (𝑁 ∈ ℕ → ((𝑁 / 2) ∈ ℕ0 ↔ (𝑁 / 2) ∈ ℕ))
Theorem	nn0ehalf 10510	The half of an even nonnegative integer is a nonnegative integer. (Contributed by AV, 22-Jun-2020.) (Revised by AV, 28-Jun-2021.)
⊢ ((𝑁 ∈ ℕ0 ∧ 2 ∥ 𝑁) → (𝑁 / 2) ∈ ℕ0)
Theorem	nnehalf 10511	The half of an even positive integer is a positive integer. (Contributed by AV, 28-Jun-2021.)
⊢ ((𝑁 ∈ ℕ ∧ 2 ∥ 𝑁) → (𝑁 / 2) ∈ ℕ)
Theorem	nn0o1gt2 10512	An odd nonnegative integer is either 1 or greater than 2. (Contributed by AV, 2-Jun-2020.)
⊢ ((𝑁 ∈ ℕ0 ∧ ((𝑁 + 1) / 2) ∈ ℕ0) → (𝑁 = 1 ∨ 2 < 𝑁))
Theorem	nno 10513	An alternate characterization of an odd integer greater than 1. (Contributed by AV, 2-Jun-2020.)
⊢ ((𝑁 ∈ (ℤ≥‘2) ∧ ((𝑁 + 1) / 2) ∈ ℕ0) → ((𝑁 − 1) / 2) ∈ ℕ)
Theorem	nn0o 10514	An alternate characterization of an odd nonnegative integer. (Contributed by AV, 28-May-2020.) (Proof shortened by AV, 2-Jun-2020.)
⊢ ((𝑁 ∈ ℕ0 ∧ ((𝑁 + 1) / 2) ∈ ℕ0) → ((𝑁 − 1) / 2) ∈ ℕ0)
Theorem	nn0ob 10515	Alternate characterizations of an odd nonnegative integer. (Contributed by AV, 4-Jun-2020.)
⊢ (𝑁 ∈ ℕ0 → (((𝑁 + 1) / 2) ∈ ℕ0 ↔ ((𝑁 − 1) / 2) ∈ ℕ0))
Theorem	nn0oddm1d2 10516	A positive integer is odd iff its predecessor divided by 2 is a positive integer. (Contributed by AV, 28-Jun-2021.)
⊢ (𝑁 ∈ ℕ0 → (¬ 2 ∥ 𝑁 ↔ ((𝑁 − 1) / 2) ∈ ℕ0))
Theorem	nnoddm1d2 10517	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))
Theorem	z0even 10518	0 is even. (Contributed by AV, 11-Feb-2020.) (Revised by AV, 23-Jun-2021.)
⊢ 2 ∥ 0
Theorem	n2dvds1 10519	2 does not divide 1 (common case). That means 1 is odd. (Contributed by David A. Wheeler, 8-Dec-2018.)
⊢ ¬ 2 ∥ 1
Theorem	n2dvdsm1 10520	2 does not divide -1. That means -1 is odd. (Contributed by AV, 15-Aug-2021.)
⊢ ¬ 2 ∥ -1
Theorem	z2even 10521	2 is even. (Contributed by AV, 12-Feb-2020.) (Revised by AV, 23-Jun-2021.)
⊢ 2 ∥ 2
Theorem	n2dvds3 10522	2 does not divide 3, i.e. 3 is an odd number. (Contributed by AV, 28-Feb-2021.)
⊢ ¬ 2 ∥ 3
Theorem	z4even 10523	4 is an even number. (Contributed by AV, 23-Jul-2020.) (Revised by AV, 4-Jul-2021.)
⊢ 2 ∥ 4
Theorem	4dvdseven 10524	An integer which is divisible by 4 is an even integer. (Contributed by AV, 4-Jul-2021.)
⊢ (4 ∥ 𝑁 → 2 ∥ 𝑁)
4.1.3  The division algorithm
Theorem	divalglemnn 10525*	Lemma for divalg 10531. Existence for a positive denominator. (Contributed by Jim Kingdon, 30-Nov-2021.)
⊢ ((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ) → ∃𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)))
Theorem	divalglemqt 10526	Lemma for divalg 10531. The 𝑄 = 𝑇 case involved in showing uniqueness. (Contributed by Jim Kingdon, 5-Dec-2021.)
⊢ (𝜑 → 𝐷 ∈ ℤ)    &   ⊢ (𝜑 → 𝑅 ∈ ℤ)    &   ⊢ (𝜑 → 𝑆 ∈ ℤ)    &   ⊢ (𝜑 → 𝑄 ∈ ℤ)    &   ⊢ (𝜑 → 𝑇 ∈ ℤ)    &   ⊢ (𝜑 → 𝑄 = 𝑇)    &   ⊢ (𝜑 → ((𝑄 · 𝐷) + 𝑅) = ((𝑇 · 𝐷) + 𝑆))    ⇒   ⊢ (𝜑 → 𝑅 = 𝑆)
Theorem	divalglemnqt 10527	Lemma for divalg 10531. The 𝑄 < 𝑇 case involved in showing uniqueness. (Contributed by Jim Kingdon, 4-Dec-2021.)
⊢ (𝜑 → 𝐷 ∈ ℕ)    &   ⊢ (𝜑 → 𝑅 ∈ ℤ)    &   ⊢ (𝜑 → 𝑆 ∈ ℤ)    &   ⊢ (𝜑 → 𝑄 ∈ ℤ)    &   ⊢ (𝜑 → 𝑇 ∈ ℤ)    &   ⊢ (𝜑 → 0 ≤ 𝑆)    &   ⊢ (𝜑 → 𝑅 < 𝐷)    &   ⊢ (𝜑 → ((𝑄 · 𝐷) + 𝑅) = ((𝑇 · 𝐷) + 𝑆))    ⇒   ⊢ (𝜑 → ¬ 𝑄 < 𝑇)
Theorem	divalglemeunn 10528*	Lemma for divalg 10531. Uniqueness for a positive denominator. (Contributed by Jim Kingdon, 4-Dec-2021.)
⊢ ((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ) → ∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)))
Theorem	divalglemex 10529*	Lemma for divalg 10531. The quotient and remainder exist. (Contributed by Jim Kingdon, 30-Nov-2021.)
⊢ ((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℤ ∧ 𝐷 ≠ 0) → ∃𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)))
Theorem	divalglemeuneg 10530*	Lemma for divalg 10531. Uniqueness for a negative denominator. (Contributed by Jim Kingdon, 4-Dec-2021.)
⊢ ((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℤ ∧ 𝐷 < 0) → ∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)))
Theorem	divalg 10531*	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 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)))
Theorem	divalgb 10532*	Express the division algorithm as stated in divalg 10531 in terms of ∥. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ ((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℤ ∧ 𝐷 ≠ 0) → (∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)) ↔ ∃!𝑟 ∈ ℕ0 (𝑟 < (abs‘𝐷) ∧ 𝐷 ∥ (𝑁 − 𝑟))))
Theorem	divalg2 10533*	The division algorithm (theorem) for a positive divisor. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ) → ∃!𝑟 ∈ ℕ0 (𝑟 < 𝐷 ∧ 𝐷 ∥ (𝑁 − 𝑟)))
Theorem	divalgmod 10534	The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor (compare divalg2 10533 and divalgb 10532). 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 < 𝐷 ∧ 𝐷 ∥ (𝑁 − 𝑅)))))
Theorem	divalgmodcl 10535	The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor. Variant of divalgmod 10534. (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 < 𝐷 ∧ 𝐷 ∥ (𝑁 − 𝑅))))
Theorem	modremain 10536*	The result of the modulo operation is the remainder of the division algorithm. (Contributed by AV, 19-Aug-2021.)
⊢ ((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ ∧ (𝑅 ∈ ℕ0 ∧ 𝑅 < 𝐷)) → ((𝑁 mod 𝐷) = 𝑅 ↔ ∃𝑧 ∈ ℤ ((𝑧 · 𝐷) + 𝑅) = 𝑁))
Theorem	ndvdssub 10537	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)) → (𝐷 ∥ 𝑁 → ¬ 𝐷 ∥ (𝑁 − 𝐾)))
Theorem	ndvdsadd 10538	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)) → (𝐷 ∥ 𝑁 → ¬ 𝐷 ∥ (𝑁 + 𝐾)))
Theorem	ndvdsp1 10539	Special case of ndvdsadd 10538. 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) → (𝐷 ∥ 𝑁 → ¬ 𝐷 ∥ (𝑁 + 1)))
Theorem	ndvdsi 10540	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    ⇒   ⊢ ¬ 𝐴 ∥ 𝐵
Theorem	flodddiv4 10541	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)))
Theorem	fldivndvdslt 10542	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) ∧ ¬ 𝐿 ∥ 𝐾) → (⌊‘(𝐾 / 𝐿)) < (𝐾 / 𝐿))
Theorem	flodddiv4lt 10543	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))
Theorem	flodddiv4t2lthalf 10544	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))
4.1.4  The greatest common divisor operator
Syntax	cgcd 10545	Extend the definition of a class to include the greatest common divisor operator.
class gcd
Definition	df-gcd 10546*	Define the gcd operator. For example, (-6 gcd 9) = 3 (ex-gcd 10828). (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ gcd = (𝑥 ∈ ℤ, 𝑦 ∈ ℤ ↦ if((𝑥 = 0 ∧ 𝑦 = 0), 0, sup({𝑛 ∈ ℤ ∣ (𝑛 ∥ 𝑥 ∧ 𝑛 ∥ 𝑦)}, ℝ, < )))
Theorem	gcdmndc 10547	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))
Theorem	zsupcllemstep 10548*	Lemma for zsupcl 10550. Induction step. (Contributed by Jim Kingdon, 7-Dec-2021.)
⊢ ((𝜑 ∧ 𝑛 ∈ (ℤ≥‘𝑀)) → DECID 𝜓)    ⇒   ⊢ (𝐾 ∈ (ℤ≥‘𝑀) → (((𝜑 ∧ ∀𝑛 ∈ (ℤ≥‘𝐾) ¬ 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)))))
Theorem	zsupcllemex 10549*	Lemma for zsupcl 10550. Existence of the supremum. (Contributed by Jim Kingdon, 7-Dec-2021.)
⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝑛 = 𝑀 → (𝜓 ↔ 𝜒))    &   ⊢ (𝜑 → 𝜒)    &   ⊢ ((𝜑 ∧ 𝑛 ∈ (ℤ≥‘𝑀)) → DECID 𝜓)    &   ⊢ (𝜑 → ∃𝑗 ∈ (ℤ≥‘𝑀)∀𝑛 ∈ (ℤ≥‘𝑗) ¬ 𝜓)    ⇒   ⊢ (𝜑 → ∃𝑥 ∈ ℤ (∀𝑦 ∈ {𝑛 ∈ ℤ ∣ 𝜓} ¬ 𝑥 < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 < 𝑥 → ∃𝑧 ∈ {𝑛 ∈ ℤ ∣ 𝜓}𝑦 < 𝑧)))
Theorem	zsupcl 10550*	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 non-empty 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.)
⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝑛 = 𝑀 → (𝜓 ↔ 𝜒))    &   ⊢ (𝜑 → 𝜒)    &   ⊢ ((𝜑 ∧ 𝑛 ∈ (ℤ≥‘𝑀)) → DECID 𝜓)    &   ⊢ (𝜑 → ∃𝑗 ∈ (ℤ≥‘𝑀)∀𝑛 ∈ (ℤ≥‘𝑗) ¬ 𝜓)    ⇒   ⊢ (𝜑 → sup({𝑛 ∈ ℤ ∣ 𝜓}, ℝ, < ) ∈ (ℤ≥‘𝑀))
Theorem	zssinfcl 10551*	The infimum of a set of integers is an element of the set. (Contributed by Jim Kingdon, 16-Jan-2022.)
⊢ (𝜑 → ∃𝑥 ∈ ℝ (∀𝑦 ∈ 𝐵 ¬ 𝑦 < 𝑥 ∧ ∀𝑦 ∈ ℝ (𝑥 < 𝑦 → ∃𝑧 ∈ 𝐵 𝑧 < 𝑦)))    &   ⊢ (𝜑 → 𝐵 ⊆ ℤ)    &   ⊢ (𝜑 → inf(𝐵, ℝ, < ) ∈ ℤ)    ⇒   ⊢ (𝜑 → inf(𝐵, ℝ, < ) ∈ 𝐵)
Theorem	infssuzex 10552*	Existence of the infimum of a subset of an upper set of integers. (Contributed by Jim Kingdon, 13-Jan-2022.)
⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ 𝑆 = {𝑛 ∈ (ℤ≥‘𝑀) ∣ 𝜓}    &   ⊢ (𝜑 → 𝐴 ∈ 𝑆)    &   ⊢ ((𝜑 ∧ 𝑛 ∈ (𝑀...𝐴)) → DECID 𝜓)    ⇒   ⊢ (𝜑 → ∃𝑥 ∈ ℝ (∀𝑦 ∈ 𝑆 ¬ 𝑦 < 𝑥 ∧ ∀𝑦 ∈ ℝ (𝑥 < 𝑦 → ∃𝑧 ∈ 𝑆 𝑧 < 𝑦)))
Theorem	infssuzledc 10553*	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.)
⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ 𝑆 = {𝑛 ∈ (ℤ≥‘𝑀) ∣ 𝜓}    &   ⊢ (𝜑 → 𝐴 ∈ 𝑆)    &   ⊢ ((𝜑 ∧ 𝑛 ∈ (𝑀...𝐴)) → DECID 𝜓)    ⇒   ⊢ (𝜑 → inf(𝑆, ℝ, < ) ≤ 𝐴)
Theorem	infssuzcldc 10554*	The infimum of a subset of an upper set of integers belongs to the subset. (Contributed by Jim Kingdon, 20-Jan-2022.)
⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ 𝑆 = {𝑛 ∈ (ℤ≥‘𝑀) ∣ 𝜓}    &   ⊢ (𝜑 → 𝐴 ∈ 𝑆)    &   ⊢ ((𝜑 ∧ 𝑛 ∈ (𝑀...𝐴)) → DECID 𝜓)    ⇒   ⊢ (𝜑 → inf(𝑆, ℝ, < ) ∈ 𝑆)
Theorem	dvdsbnd 10555*	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 ∈ (ℤ≥‘𝑛) ¬ 𝑚 ∥ 𝐴)
Theorem	gcdsupex 10556*	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 ∣ (𝑛 ∥ 𝑋 ∧ 𝑛 ∥ 𝑌)} ¬ 𝑥 < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 < 𝑥 → ∃𝑧 ∈ {𝑛 ∈ ℤ ∣ (𝑛 ∥ 𝑋 ∧ 𝑛 ∥ 𝑌)}𝑦 < 𝑧)))
Theorem	gcdsupcl 10557*	Closure of the supremum used in defining gcd. A lemma for gcdval 10558 and gcdn0cl 10561. (Contributed by Jim Kingdon, 11-Dec-2021.)
⊢ (((𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ) ∧ ¬ (𝑋 = 0 ∧ 𝑌 = 0)) → sup({𝑛 ∈ ℤ ∣ (𝑛 ∥ 𝑋 ∧ 𝑛 ∥ 𝑌)}, ℝ, < ) ∈ ℕ)
Theorem	gcdval 10558*	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 ∣ (𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑁)}, ℝ, < )))
Theorem	gcd0val 10559	The value, by convention, of the gcd operator when both operands are 0. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (0 gcd 0) = 0
Theorem	gcdn0val 10560*	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 ∣ (𝑛 ∥ 𝑀 ∧ 𝑛 ∥ 𝑁)}, ℝ, < ))
Theorem	gcdn0cl 10561	Closure of the gcd operator. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ (((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∧ 𝑁 = 0)) → (𝑀 gcd 𝑁) ∈ ℕ)
Theorem	gcddvds 10562	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))
Theorem	dvdslegcd 10563	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)) → ((𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) → 𝐾 ≤ (𝑀 gcd 𝑁)))
Theorem	nndvdslegcd 10564	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) → ((𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) → 𝐾 ≤ (𝑀 gcd 𝑁)))
Theorem	gcdcl 10565	Closure of the gcd operator. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) ∈ ℕ0)
Theorem	gcdnncl 10566	Closure of the gcd operator. (Contributed by Thierry Arnoux, 2-Feb-2020.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 gcd 𝑁) ∈ ℕ)
Theorem	gcdcld 10567	Closure of the gcd operator. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    ⇒   ⊢ (𝜑 → (𝑀 gcd 𝑁) ∈ ℕ0)
Theorem	gcd2n0cl 10568	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)
Theorem	zeqzmulgcd 10569*	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)))
Theorem	divgcdz 10570	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)
Theorem	gcdf 10571	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
Theorem	gcdcom 10572	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))
Theorem	divgcdnn 10573	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)
Theorem	divgcdnnr 10574	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)
Theorem	gcdeq0 10575	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)))
Theorem	gcdn0gt0 10576	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)))
Theorem	gcd0id 10577	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))
Theorem	gcdid0 10578	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))
Theorem	nn0gcdid0 10579	The gcd of a nonnegative integer with 0 is itself. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ (𝑁 ∈ ℕ0 → (𝑁 gcd 0) = 𝑁)
Theorem	gcdneg 10580	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))
Theorem	neggcd 10581	Negating one operand of the gcd operator does not alter the result. (Contributed by Paul Chapman, 22-Jun-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (-𝑀 gcd 𝑁) = (𝑀 gcd 𝑁))
Theorem	gcdaddm 10582	Adding a multiple of one operand of the gcd operator to the other does not alter the result. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) = (𝑀 gcd (𝑁 + (𝐾 · 𝑀))))
Theorem	gcdadd 10583	The GCD of two numbers is the same as the GCD of the left and their sum. (Contributed by Scott Fenton, 20-Apr-2014.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) = (𝑀 gcd (𝑁 + 𝑀)))
Theorem	gcdid 10584	The gcd of a number and itself is its absolute value. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ (𝑁 ∈ ℤ → (𝑁 gcd 𝑁) = (abs‘𝑁))
Theorem	gcd1 10585	The gcd of a number with 1 is 1. Theorem 1.4(d)1 in [ApostolNT] p. 16. (Contributed by Mario Carneiro, 19-Feb-2014.)
⊢ (𝑀 ∈ ℤ → (𝑀 gcd 1) = 1)
Theorem	gcdabs 10586	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))
Theorem	gcdabs1 10587	gcd of the absolute value of the first operator. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ) → ((abs‘𝑁) gcd 𝑀) = (𝑁 gcd 𝑀))
Theorem	gcdabs2 10588	gcd of the absolute value of the second operator. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ) → (𝑁 gcd (abs‘𝑀)) = (𝑁 gcd 𝑀))
Theorem	modgcd 10589	The gcd remains unchanged if one operand is replaced with its remainder modulo the other. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ) → ((𝑀 mod 𝑁) gcd 𝑁) = (𝑀 gcd 𝑁))
Theorem	1gcd 10590	The GCD of one and an integer is one. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ (𝑀 ∈ ℤ → (1 gcd 𝑀) = 1)
Theorem	6gcd4e2 10591	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
4.1.5  Bézout's identity
Theorem	bezoutlemnewy 10592*	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)))    &   ⊢ (𝜃 → 𝐴 ∈ ℕ0)    &   ⊢ (𝜃 → 𝐵 ∈ ℕ0)    &   ⊢ (𝜃 → 𝑊 ∈ ℕ)    &   ⊢ (𝜃 → [𝑦 / 𝑟]𝜑)    &   ⊢ (𝜃 → 𝑦 ∈ ℕ0)    &   ⊢ (𝜃 → [𝑊 / 𝑟]𝜑)    ⇒   ⊢ (𝜃 → [(𝑦 mod 𝑊) / 𝑟]𝜑)
Theorem	bezoutlemstep 10593*	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)))    &   ⊢ (𝜃 → 𝐴 ∈ ℕ0)    &   ⊢ (𝜃 → 𝐵 ∈ ℕ0)    &   ⊢ (𝜃 → 𝑊 ∈ ℕ)    &   ⊢ (𝜃 → [𝑦 / 𝑟]𝜑)    &   ⊢ (𝜃 → 𝑦 ∈ ℕ0)    &   ⊢ (𝜃 → [𝑊 / 𝑟]𝜑)    &   ⊢ (𝜓 ↔ ∀𝑧 ∈ ℕ0 (𝑧 ∥ 𝑟 → (𝑧 ∥ 𝑥 ∧ 𝑧 ∥ 𝑦)))    &   ⊢ ((𝜃 ∧ [(𝑦 mod 𝑊) / 𝑟]𝜑) → ∃𝑟 ∈ ℕ0 ([(𝑦 mod 𝑊) / 𝑥][𝑊 / 𝑦]𝜓 ∧ 𝜑))    &   ⊢ Ⅎ𝑥𝜃    &   ⊢ Ⅎ𝑟𝜃    ⇒   ⊢ (𝜃 → ∃𝑟 ∈ ℕ0 ([𝑊 / 𝑥]𝜓 ∧ 𝜑))
Theorem	bezoutlemmain 10594*	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 10595*	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 10596*	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 10597*	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 10598*	Lemma for Bézout's identity. Like bezoutlemex 10597 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 10599*	Lemma for Bézout's identity. Like bezoutlemzz 10598 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 10600*	Lemma for Bézout's identity. Like bezoutlemaz 10599 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))))