P. 120 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 11901-12000   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	opeo 11901	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 11902	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 11903	Exponentiation of -1 by an even power. Variant of m1expeven 10566. (Contributed by AV, 25-Jun-2021.)
⊢ (2 ∥ 𝑁 → (-1↑𝑁) = 1)
Theorem	m1expo 11904	Exponentiation of -1 by an odd power. (Contributed by AV, 26-Jun-2021.)
⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → (-1↑𝑁) = -1)
Theorem	m1exp1 11905	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 11906	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 11907	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 11908	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 11909	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 11910	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 11911	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 11912	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 11913	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 11914	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 11915	0 is even. (Contributed by AV, 11-Feb-2020.) (Revised by AV, 23-Jun-2021.)
⊢ 2 ∥ 0
Theorem	n2dvds1 11916	2 does not divide 1 (common case). That means 1 is odd. (Contributed by David A. Wheeler, 8-Dec-2018.)
⊢ ¬ 2 ∥ 1
Theorem	n2dvdsm1 11917	2 does not divide -1. That means -1 is odd. (Contributed by AV, 15-Aug-2021.)
⊢ ¬ 2 ∥ -1
Theorem	z2even 11918	2 is even. (Contributed by AV, 12-Feb-2020.) (Revised by AV, 23-Jun-2021.)
⊢ 2 ∥ 2
Theorem	n2dvds3 11919	2 does not divide 3, i.e. 3 is an odd number. (Contributed by AV, 28-Feb-2021.)
⊢ ¬ 2 ∥ 3
Theorem	z4even 11920	4 is an even number. (Contributed by AV, 23-Jul-2020.) (Revised by AV, 4-Jul-2021.)
⊢ 2 ∥ 4
Theorem	4dvdseven 11921	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
Theorem	divalglemnn 11922*	Lemma for divalg 11928. 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 11923	Lemma for divalg 11928. 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 11924	Lemma for divalg 11928. 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 11925*	Lemma for divalg 11928. 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 11926*	Lemma for divalg 11928. 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 11927*	Lemma for divalg 11928. 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 11928*	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 11929*	Express the division algorithm as stated in divalg 11928 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 11930*	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 11931	The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor (compare divalg2 11930 and divalgb 11929). 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 11932	The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor. Variant of divalgmod 11931. (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 11933*	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 11934	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 11935	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 11936	Special case of ndvdsadd 11935. 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 11937	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 11938	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 11939	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 11940	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 11941	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
Syntax	cgcd 11942	Extend the definition of a class to include the greatest common divisor operator.
class gcd
Definition	df-gcd 11943*	Define the gcd operator. For example, (-6 gcd 9) = 3 (ex-gcd 14453). (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 11944	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 11945*	Lemma for zsupcl 11947. 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 11946*	Lemma for zsupcl 11947. 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 11947*	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.)
⊢ (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 11948*	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 11949*	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 11950*	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 11951*	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	suprzubdc 11952*	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.)
⊢ (𝜑 → 𝐴 ⊆ ℤ)    &   ⊢ (𝜑 → ∀𝑥 ∈ ℤ DECID 𝑥 ∈ 𝐴)    &   ⊢ (𝜑 → ∃𝑥 ∈ ℤ ∀𝑦 ∈ 𝐴 𝑦 ≤ 𝑥)    &   ⊢ (𝜑 → 𝐵 ∈ 𝐴)    ⇒   ⊢ (𝜑 → 𝐵 ≤ sup(𝐴, ℝ, < ))
Theorem	nninfdcex 11953*	A decidable set of natural numbers has an infimum. (Contributed by Jim Kingdon, 28-Sep-2024.)
⊢ (𝜑 → 𝐴 ⊆ ℕ)    &   ⊢ (𝜑 → ∀𝑥 ∈ ℕ DECID 𝑥 ∈ 𝐴)    &   ⊢ (𝜑 → ∃𝑦 𝑦 ∈ 𝐴)    ⇒   ⊢ (𝜑 → ∃𝑥 ∈ ℝ (∀𝑦 ∈ 𝐴 ¬ 𝑦 < 𝑥 ∧ ∀𝑦 ∈ ℝ (𝑥 < 𝑦 → ∃𝑧 ∈ 𝐴 𝑧 < 𝑦)))
Theorem	zsupssdc 11954*	An inhabited decidable bounded subset of integers has a supremum in the set. (The proof does not use ax-pre-suploc 7931.) (Contributed by Mario Carneiro, 21-Apr-2015.) (Revised by Jim Kingdon, 5-Oct-2024.)
⊢ (𝜑 → 𝐴 ⊆ ℤ)    &   ⊢ (𝜑 → ∃𝑥 𝑥 ∈ 𝐴)    &   ⊢ (𝜑 → ∀𝑥 ∈ ℤ DECID 𝑥 ∈ 𝐴)    &   ⊢ (𝜑 → ∃𝑥 ∈ ℤ ∀𝑦 ∈ 𝐴 𝑦 ≤ 𝑥)    ⇒   ⊢ (𝜑 → ∃𝑥 ∈ 𝐴 (∀𝑦 ∈ 𝐴 ¬ 𝑥 < 𝑦 ∧ ∀𝑦 ∈ 𝐵 (𝑦 < 𝑥 → ∃𝑧 ∈ 𝐴 𝑦 < 𝑧)))
Theorem	suprzcl2dc 11955*	The supremum of a bounded-above decidable set of integers is a member of the set. (This theorem avoids ax-pre-suploc 7931.) (Contributed by Mario Carneiro, 21-Apr-2015.) (Revised by Jim Kingdon, 6-Oct-2024.)
⊢ (𝜑 → 𝐴 ⊆ ℤ)    &   ⊢ (𝜑 → ∀𝑥 ∈ ℤ DECID 𝑥 ∈ 𝐴)    &   ⊢ (𝜑 → ∃𝑥 ∈ ℤ ∀𝑦 ∈ 𝐴 𝑦 ≤ 𝑥)    &   ⊢ (𝜑 → ∃𝑥 𝑥 ∈ 𝐴)    ⇒   ⊢ (𝜑 → sup(𝐴, ℝ, < ) ∈ 𝐴)
Theorem	dvdsbnd 11956*	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 11957*	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 11958*	Closure of the supremum used in defining gcd. A lemma for gcdval 11959 and gcdn0cl 11962. (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 11959*	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 11960	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 11961*	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 11962	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 11963	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 11964	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 11965	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 11966	Closure of the gcd operator. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) ∈ ℕ0)
Theorem	gcdnncl 11967	Closure of the gcd operator. (Contributed by Thierry Arnoux, 2-Feb-2020.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 gcd 𝑁) ∈ ℕ)
Theorem	gcdcld 11968	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 11969	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 11970*	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 11971	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 11972	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 11973	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	gcdcomd 11974	The gcd operator is commutative, deduction version. (Contributed by SN, 24-Aug-2024.)
⊢ (𝜑 → 𝑀 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    ⇒   ⊢ (𝜑 → (𝑀 gcd 𝑁) = (𝑁 gcd 𝑀))
Theorem	divgcdnn 11975	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 11976	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 11977	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 11978	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 11979	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 11980	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 11981	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 11982	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 11983	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 11984	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 11985	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 11986	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 11987	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 11988	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 11989	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 11990	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 11991	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 11992	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	gcdmultipled 11993	The greatest common divisor of a nonnegative integer 𝑀 and a multiple of it is 𝑀 itself. (Contributed by Rohan Ridenour, 3-Aug-2023.)
⊢ (𝜑 → 𝑀 ∈ ℕ0)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    ⇒   ⊢ (𝜑 → (𝑀 gcd (𝑁 · 𝑀)) = 𝑀)
Theorem	dvdsgcdidd 11994	The greatest common divisor of a positive integer and another integer it divides is itself. (Contributed by Rohan Ridenour, 3-Aug-2023.)
⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → 𝑁 ∈ ℤ)    &   ⊢ (𝜑 → 𝑀 ∥ 𝑁)    ⇒   ⊢ (𝜑 → (𝑀 gcd 𝑁) = 𝑀)
Theorem	6gcd4e2 11995	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
Theorem	bezoutlemnewy 11996*	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 11997*	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 11998*	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 11999*	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 12000*	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)