P. 128 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 12701-12800   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	ndvdsp1 12701	Special case of ndvdsadd 12700. 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 12702	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	5ndvds3 12703	5 does not divide 3. (Contributed by AV, 8-Sep-2025.)
⊢ ¬ 5 ∥ 3
Theorem	5ndvds6 12704	5 does not divide 6. (Contributed by AV, 8-Sep-2025.)
⊢ ¬ 5 ∥ 6
Theorem	flodddiv4 12705	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 12706	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 12707	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 12708	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  Bit sequences
Syntax	cbits 12709	Define the binary bits of an integer.
class bits
Definition	df-bits 12710*	Define the binary bits of an integer. The expression 𝑀 ∈ (bits‘𝑁) means that the 𝑀-th bit of 𝑁 is 1 (and its negation means the bit is 0). (Contributed by Mario Carneiro, 4-Sep-2016.)
⊢ bits = (𝑛 ∈ ℤ ↦ {𝑚 ∈ ℕ0 ∣ ¬ 2 ∥ (⌊‘(𝑛 / (2↑𝑚)))})
Theorem	bitsfval 12711*	Expand the definition of the bits of an integer. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℤ → (bits‘𝑁) = {𝑚 ∈ ℕ0 ∣ ¬ 2 ∥ (⌊‘(𝑁 / (2↑𝑚)))})
Theorem	bitsval 12712	Expand the definition of the bits of an integer. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑀 ∈ (bits‘𝑁) ↔ (𝑁 ∈ ℤ ∧ 𝑀 ∈ ℕ0 ∧ ¬ 2 ∥ (⌊‘(𝑁 / (2↑𝑀)))))
Theorem	bitsval2 12713	Expand the definition of the bits of an integer. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℕ0) → (𝑀 ∈ (bits‘𝑁) ↔ ¬ 2 ∥ (⌊‘(𝑁 / (2↑𝑀)))))
Theorem	bitsss 12714	The set of bits of an integer is a subset of ℕ0. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (bits‘𝑁) ⊆ ℕ0
Theorem	bitsf 12715	The bits function is a function from integers to subsets of nonnegative integers. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ bits:ℤ⟶𝒫 ℕ0
Theorem	bitsdc 12716	Whether a bit is set is decidable. (Contributed by Jim Kingdon, 31-Oct-2025.)
⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℕ0) → DECID 𝑀 ∈ (bits‘𝑁))
Theorem	bits0 12717	Value of the zeroth bit. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℤ → (0 ∈ (bits‘𝑁) ↔ ¬ 2 ∥ 𝑁))
Theorem	bits0e 12718	The zeroth bit of an even number is zero. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℤ → ¬ 0 ∈ (bits‘(2 · 𝑁)))
Theorem	bits0o 12719	The zeroth bit of an odd number is one. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℤ → 0 ∈ (bits‘((2 · 𝑁) + 1)))
Theorem	bitsp1 12720	The 𝑀 + 1-th bit of 𝑁 is the 𝑀-th bit of ⌊(𝑁 / 2). (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℕ0) → ((𝑀 + 1) ∈ (bits‘𝑁) ↔ 𝑀 ∈ (bits‘(⌊‘(𝑁 / 2)))))
Theorem	bitsp1e 12721	The 𝑀 + 1-th bit of 2𝑁 is the 𝑀-th bit of 𝑁. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℕ0) → ((𝑀 + 1) ∈ (bits‘(2 · 𝑁)) ↔ 𝑀 ∈ (bits‘𝑁)))
Theorem	bitsp1o 12722	The 𝑀 + 1-th bit of 2𝑁 + 1 is the 𝑀-th bit of 𝑁. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℕ0) → ((𝑀 + 1) ∈ (bits‘((2 · 𝑁) + 1)) ↔ 𝑀 ∈ (bits‘𝑁)))
Theorem	bitsfzolem 12723*	Lemma for bitsfzo 12724. (Contributed by Mario Carneiro, 5-Sep-2016.) (Revised by AV, 1-Oct-2020.)
⊢ (𝜑 → 𝑁 ∈ ℕ0)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ0)    &   ⊢ (𝜑 → (bits‘𝑁) ⊆ (0..^𝑀))    &   ⊢ 𝑆 = inf({𝑛 ∈ ℕ0 ∣ 𝑁 < (2↑𝑛)}, ℝ, < )    ⇒   ⊢ (𝜑 → 𝑁 ∈ (0..^(2↑𝑀)))
Theorem	bitsfzo 12724	The bits of a number are all at positions less than 𝑀 iff the number is nonnegative and less than 2↑𝑀. (Contributed by Mario Carneiro, 5-Sep-2016.) (Proof shortened by AV, 1-Oct-2020.)
⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℕ0) → (𝑁 ∈ (0..^(2↑𝑀)) ↔ (bits‘𝑁) ⊆ (0..^𝑀)))
Theorem	bitsmod 12725	Truncating the bit sequence after some 𝑀 is equivalent to reducing the argument mod 2↑𝑀. (Contributed by Mario Carneiro, 6-Sep-2016.)
⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℕ0) → (bits‘(𝑁 mod (2↑𝑀))) = ((bits‘𝑁) ∩ (0..^𝑀)))
Theorem	bitsfi 12726	Every number is associated with a finite set of bits. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℕ0 → (bits‘𝑁) ∈ Fin)
Theorem	bitscmp 12727	The bit complement of 𝑁 is -𝑁 − 1. (Thus, by bitsfi 12726, all negative numbers have cofinite bits representations.) (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℤ → (ℕ0 ∖ (bits‘𝑁)) = (bits‘(-𝑁 − 1)))
Theorem	0bits 12728	The bits of zero. (Contributed by Mario Carneiro, 6-Sep-2016.)
⊢ (bits‘0) = ∅
Theorem	m1bits 12729	The bits of negative one. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (bits‘-1) = ℕ0
Theorem	bitsinv1lem 12730	Lemma for bitsinv1 12731. (Contributed by Mario Carneiro, 22-Sep-2016.)
⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℕ0) → (𝑁 mod (2↑(𝑀 + 1))) = ((𝑁 mod (2↑𝑀)) + if(𝑀 ∈ (bits‘𝑁), (2↑𝑀), 0)))
Theorem	bitsinv1 12731*	There is an explicit inverse to the bits function for nonnegative integers (which can be extended to negative integers using bitscmp 12727), part 1. (Contributed by Mario Carneiro, 7-Sep-2016.)
⊢ (𝑁 ∈ ℕ0 → Σ𝑛 ∈ (bits‘𝑁)(2↑𝑛) = 𝑁)
5.1.5  The greatest common divisor operator
Syntax	cgcd 12732	Extend the definition of a class to include the greatest common divisor operator.
class gcd
Definition	df-gcd 12733*	Define the gcd operator. For example, (-6 gcd 9) = 3 (ex-gcd 16757). (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 12734	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	dvdsbnd 12735*	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 12736*	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 12737*	Closure of the supremum used in defining gcd. A lemma for gcdval 12738 and gcdn0cl 12741. (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 12738*	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 12739	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 12740*	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 12741	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 12742	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 12743	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 12744	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 12745	Closure of the gcd operator. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) ∈ ℕ0)
Theorem	gcdnncl 12746	Closure of the gcd operator. (Contributed by Thierry Arnoux, 2-Feb-2020.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 gcd 𝑁) ∈ ℕ)
Theorem	gcdcld 12747	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 12748	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 12749*	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 12750	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 12751	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 12752	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 12753	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 12754	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 12755	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 12756	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 12757	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 12758	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 12759	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 12760	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 12761	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 12762	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 12763	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 12764	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 12765	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 12766	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 12767	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 12768	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 12769	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 12770	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 12771	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 12772	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 12773	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 12774	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
5.1.6  Bézout's identity
Theorem	bezoutlemnewy 12775*	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 12776*	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 12777*	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 12778*	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 12779*	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 12780*	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 12781*	Lemma for Bézout's identity. Like bezoutlemex 12780 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 12782*	Lemma for Bézout's identity. Like bezoutlemzz 12781 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 12783*	Lemma for Bézout's identity. Like bezoutlemaz 12782 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))))
Theorem	bezoutlembi 12784*	Lemma for Bézout's identity. Like bezoutlembz 12783 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 (𝑧 ∥ 𝑑 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)) ∧ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑑 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))))
Theorem	bezoutlemmo 12785*	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.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℕ0)    &   ⊢ (𝜑 → ∀𝑧 ∈ ℤ (𝑧 ∥ 𝐷 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)))    &   ⊢ (𝜑 → 𝐸 ∈ ℕ0)    &   ⊢ (𝜑 → ∀𝑧 ∈ ℤ (𝑧 ∥ 𝐸 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)))    ⇒   ⊢ (𝜑 → 𝐷 = 𝐸)
Theorem	bezoutlemeu 12786*	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.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℕ0)    &   ⊢ (𝜑 → ∀𝑧 ∈ ℤ (𝑧 ∥ 𝐷 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)))    ⇒   ⊢ (𝜑 → ∃!𝑑 ∈ ℕ0 ∀𝑧 ∈ ℤ (𝑧 ∥ 𝑑 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)))
Theorem	bezoutlemle 12787*	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.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℕ0)    &   ⊢ (𝜑 → ∀𝑧 ∈ ℤ (𝑧 ∥ 𝐷 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)))    &   ⊢ (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0))    ⇒   ⊢ (𝜑 → ∀𝑧 ∈ ℤ ((𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵) → 𝑧 ≤ 𝐷))
Theorem	bezoutlemsup 12788*	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.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝐷 ∈ ℕ0)    &   ⊢ (𝜑 → ∀𝑧 ∈ ℤ (𝑧 ∥ 𝐷 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)))    &   ⊢ (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0))    ⇒   ⊢ (𝜑 → 𝐷 = sup({𝑧 ∈ ℤ ∣ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)}, ℝ, < ))
Theorem	dfgcd3 12789*	Alternate definition of the gcd operator. (Contributed by Jim Kingdon, 31-Dec-2021.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) = (℩𝑑 ∈ ℕ0 ∀𝑧 ∈ ℤ (𝑧 ∥ 𝑑 ↔ (𝑧 ∥ 𝑀 ∧ 𝑧 ∥ 𝑁))))
Theorem	bezout 12790*	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)))
Theorem	dvdsgcd 12791	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) → ((𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) → 𝐾 ∥ (𝑀 gcd 𝑁)))
Theorem	dvdsgcdb 12792	Biconditional form of dvdsgcd 12791. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) ↔ 𝐾 ∥ (𝑀 gcd 𝑁)))
Theorem	dfgcd2 12793*	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 ∧ (𝐷 ∥ 𝑀 ∧ 𝐷 ∥ 𝑁) ∧ ∀𝑒 ∈ ℤ ((𝑒 ∥ 𝑀 ∧ 𝑒 ∥ 𝑁) → 𝑒 ∥ 𝐷))))
Theorem	gcdass 12794	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)))
Theorem	mulgcd 12795	Distribute multiplication by a nonnegative integer over gcd. (Contributed by Paul Chapman, 22-Jun-2011.) (Proof shortened by Mario Carneiro, 30-May-2014.)
⊢ ((𝐾 ∈ ℕ0 ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 · 𝑀) gcd (𝐾 · 𝑁)) = (𝐾 · (𝑀 gcd 𝑁)))
Theorem	absmulgcd 12796	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))))
Theorem	mulgcdr 12797	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))
Theorem	gcddiv 12798	Division law for GCD. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℕ) ∧ (𝐶 ∥ 𝐴 ∧ 𝐶 ∥ 𝐵)) → ((𝐴 gcd 𝐵) / 𝐶) = ((𝐴 / 𝐶) gcd (𝐵 / 𝐶)))
Theorem	gcdmultiple 12799	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)
Theorem	gcdmultiplez 12800	Extend gcdmultiple 12799 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)