P. 165 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 16401-16500   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	sadasslem 16401	Lemma for sadass 16402. (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐶 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (((𝐴 sadd 𝐵) sadd 𝐶) ∩ (0..^𝑁)) = ((𝐴 sadd (𝐵 sadd 𝐶)) ∩ (0..^𝑁)))
Theorem	sadass 16402	Sequence addition is associative. (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ ((𝐴 ⊆ ℕ0 ∧ 𝐵 ⊆ ℕ0 ∧ 𝐶 ⊆ ℕ0) → ((𝐴 sadd 𝐵) sadd 𝐶) = (𝐴 sadd (𝐵 sadd 𝐶)))
Theorem	sadeq 16403	Any element of a sequence sum only depends on the values of the argument sequences up to and including that point. (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    ⇒   ⊢ (𝜑 → ((𝐴 sadd 𝐵) ∩ (0..^𝑁)) = (((𝐴 ∩ (0..^𝑁)) sadd (𝐵 ∩ (0..^𝑁))) ∩ (0..^𝑁)))
Theorem	bitsres 16404	Restrict the bits of a number to an upper integer set. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℕ0) → ((bits‘𝐴) ∩ (ℤ≥‘𝑁)) = (bits‘((⌊‘(𝐴 / (2↑𝑁))) · (2↑𝑁))))
Theorem	bitsuz 16405	The bits of a number are all at least 𝑁 iff the number is divisible by 2↑𝑁. (Contributed by Mario Carneiro, 21-Sep-2016.)
⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℕ0) → ((2↑𝑁) ∥ 𝐴 ↔ (bits‘𝐴) ⊆ (ℤ≥‘𝑁)))
Theorem	bitsshft 16406*	Shifting a bit sequence to the left (toward the more significant bits) causes the number to be multiplied by a power of two. (Contributed by Mario Carneiro, 22-Sep-2016.)
⊢ ((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℕ0) → {𝑛 ∈ ℕ0 ∣ (𝑛 − 𝑁) ∈ (bits‘𝐴)} = (bits‘(𝐴 · (2↑𝑁))))
Definition	df-smu 16407*	Define the multiplication of two bit sequences, using repeated sequence addition. (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ smul = (𝑥 ∈ 𝒫 ℕ0, 𝑦 ∈ 𝒫 ℕ0 ↦ {𝑘 ∈ ℕ0 ∣ 𝑘 ∈ (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ 𝑥 ∧ (𝑛 − 𝑚) ∈ 𝑦)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘(𝑘 + 1))})
Theorem	smufval 16408*	The multiplication of two bit sequences as repeated sequence addition. (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ 𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ 𝐴 ∧ (𝑛 − 𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    ⇒   ⊢ (𝜑 → (𝐴 smul 𝐵) = {𝑘 ∈ ℕ0 ∣ 𝑘 ∈ (𝑃‘(𝑘 + 1))})
Theorem	smupf 16409*	The sequence of partial sums of the sequence multiplication. (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ 𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ 𝐴 ∧ (𝑛 − 𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    ⇒   ⊢ (𝜑 → 𝑃:ℕ0⟶𝒫 ℕ0)
Theorem	smup0 16410*	The initial element of the partial sum sequence. (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ 𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ 𝐴 ∧ (𝑛 − 𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    ⇒   ⊢ (𝜑 → (𝑃‘0) = ∅)
Theorem	smupp1 16411*	The initial element of the partial sum sequence. (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ 𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ 𝐴 ∧ (𝑛 − 𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (𝑃‘(𝑁 + 1)) = ((𝑃‘𝑁) sadd {𝑛 ∈ ℕ0 ∣ (𝑁 ∈ 𝐴 ∧ (𝑛 − 𝑁) ∈ 𝐵)}))
Theorem	smuval 16412*	Define the addition of two bit sequences, using df-had 1596 and df-cad 1609 bit operations. (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ 𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ 𝐴 ∧ (𝑛 − 𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (𝑁 ∈ (𝐴 smul 𝐵) ↔ 𝑁 ∈ (𝑃‘(𝑁 + 1))))
Theorem	smuval2 16413*	The partial sum sequence stabilizes at 𝑁 after the 𝑁 + 1-th element of the sequence; this stable value is the value of the sequence multiplication. (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ 𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ 𝐴 ∧ (𝑛 − 𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘(𝑁 + 1)))    ⇒   ⊢ (𝜑 → (𝑁 ∈ (𝐴 smul 𝐵) ↔ 𝑁 ∈ (𝑃‘𝑀)))
Theorem	smupvallem 16414*	If 𝐴 only has elements less than 𝑁, then all elements of the partial sum sequence past 𝑁 already equal the final value. (Contributed by Mario Carneiro, 20-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ 𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ 𝐴 ∧ (𝑛 − 𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    &   ⊢ (𝜑 → 𝐴 ⊆ (0..^𝑁))    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘𝑁))    ⇒   ⊢ (𝜑 → (𝑃‘𝑀) = (𝐴 smul 𝐵))
Theorem	smucl 16415	The product of two sequences is a sequence. (Contributed by Mario Carneiro, 19-Sep-2016.)
⊢ ((𝐴 ⊆ ℕ0 ∧ 𝐵 ⊆ ℕ0) → (𝐴 smul 𝐵) ⊆ ℕ0)
Theorem	smu01lem 16416*	Lemma for smu01 16417 and smu02 16418. (Contributed by Mario Carneiro, 19-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ ((𝜑 ∧ (𝑘 ∈ ℕ0 ∧ 𝑛 ∈ ℕ0)) → ¬ (𝑘 ∈ 𝐴 ∧ (𝑛 − 𝑘) ∈ 𝐵))    ⇒   ⊢ (𝜑 → (𝐴 smul 𝐵) = ∅)
Theorem	smu01 16417	Multiplication of a sequence by 0 on the right. (Contributed by Mario Carneiro, 19-Sep-2016.)
⊢ (𝐴 ⊆ ℕ0 → (𝐴 smul ∅) = ∅)
Theorem	smu02 16418	Multiplication of a sequence by 0 on the left. (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ (𝐴 ⊆ ℕ0 → (∅ smul 𝐴) = ∅)
Theorem	smupval 16419*	Rewrite the elements of the partial sum sequence in terms of sequence multiplication. (Contributed by Mario Carneiro, 20-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ 𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ 𝐴 ∧ (𝑛 − 𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (𝑃‘𝑁) = ((𝐴 ∩ (0..^𝑁)) smul 𝐵))
Theorem	smup1 16420*	Rewrite smupp1 16411 using only smul instead of the internal recursive function 𝑃. (Contributed by Mario Carneiro, 20-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    ⇒   ⊢ (𝜑 → ((𝐴 ∩ (0..^(𝑁 + 1))) smul 𝐵) = (((𝐴 ∩ (0..^𝑁)) smul 𝐵) sadd {𝑛 ∈ ℕ0 ∣ (𝑁 ∈ 𝐴 ∧ (𝑛 − 𝑁) ∈ 𝐵)}))
Theorem	smueqlem 16421*	Any element of a sequence multiplication only depends on the values of the argument sequences up to and including that point. (Contributed by Mario Carneiro, 20-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    &   ⊢ 𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ 𝐴 ∧ (𝑛 − 𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   ⊢ 𝑄 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚 ∈ 𝐴 ∧ (𝑛 − 𝑚) ∈ (𝐵 ∩ (0..^𝑁)))})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    ⇒   ⊢ (𝜑 → ((𝐴 smul 𝐵) ∩ (0..^𝑁)) = (((𝐴 ∩ (0..^𝑁)) smul (𝐵 ∩ (0..^𝑁))) ∩ (0..^𝑁)))
Theorem	smueq 16422	Any element of a sequence multiplication only depends on the values of the argument sequences up to and including that point. (Contributed by Mario Carneiro, 20-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    ⇒   ⊢ (𝜑 → ((𝐴 smul 𝐵) ∩ (0..^𝑁)) = (((𝐴 ∩ (0..^𝑁)) smul (𝐵 ∩ (0..^𝑁))) ∩ (0..^𝑁)))
Theorem	smumullem 16423	Lemma for smumul 16424. (Contributed by Mario Carneiro, 22-Sep-2016.)
⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (((bits‘𝐴) ∩ (0..^𝑁)) smul (bits‘𝐵)) = (bits‘((𝐴 mod (2↑𝑁)) · 𝐵)))
Theorem	smumul 16424	For sequences that correspond to valid integers, the sequence multiplication function produces the sequence for the product. This is effectively a proof of the correctness of the multiplication process, implemented in terms of logic gates for df-sad 16382, whose correctness is verified in sadadd 16398. Outside this range, the sequences cannot be representing integers, but the smul function still "works". This extended function is best interpreted in terms of the ring structure of the 2-adic integers. (Contributed by Mario Carneiro, 22-Sep-2016.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ((bits‘𝐴) smul (bits‘𝐵)) = (bits‘(𝐴 · 𝐵)))
6.1.7  The greatest common divisor operator
Syntax	cgcd 16425	Extend the definition of a class to include the greatest common divisor operator.
class gcd
Definition	df-gcd 16426*	Define the gcd operator. For example, (-6 gcd 9) = 3 (ex-gcd 30515). For an alternate definition, based on the definition in [ApostolNT] p. 15, see dfgcd2 16477. (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	gcdval 16427*	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 16428	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 16429*	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	gcdcllem1 16430*	Lemma for gcdn0cl 16433, gcddvds 16434 and dvdslegcd 16435. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ 𝑆 = {𝑧 ∈ ℤ ∣ ∀𝑛 ∈ 𝐴 𝑧 ∥ 𝑛}    ⇒   ⊢ ((𝐴 ⊆ ℤ ∧ ∃𝑛 ∈ 𝐴 𝑛 ≠ 0) → (𝑆 ≠ ∅ ∧ ∃𝑥 ∈ ℤ ∀𝑦 ∈ 𝑆 𝑦 ≤ 𝑥))
Theorem	gcdcllem2 16431*	Lemma for gcdn0cl 16433, gcddvds 16434 and dvdslegcd 16435. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ 𝑆 = {𝑧 ∈ ℤ ∣ ∀𝑛 ∈ {𝑀, 𝑁}𝑧 ∥ 𝑛}    &   ⊢ 𝑅 = {𝑧 ∈ ℤ ∣ (𝑧 ∥ 𝑀 ∧ 𝑧 ∥ 𝑁)}    ⇒   ⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑅 = 𝑆)
Theorem	gcdcllem3 16432*	Lemma for gcdn0cl 16433, gcddvds 16434 and dvdslegcd 16435. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ 𝑆 = {𝑧 ∈ ℤ ∣ ∀𝑛 ∈ {𝑀, 𝑁}𝑧 ∥ 𝑛}    &   ⊢ 𝑅 = {𝑧 ∈ ℤ ∣ (𝑧 ∥ 𝑀 ∧ 𝑧 ∥ 𝑁)}    ⇒   ⊢ (((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∧ 𝑁 = 0)) → (sup(𝑅, ℝ, < ) ∈ ℕ ∧ (sup(𝑅, ℝ, < ) ∥ 𝑀 ∧ sup(𝑅, ℝ, < ) ∥ 𝑁) ∧ ((𝐾 ∈ ℤ ∧ 𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) → 𝐾 ≤ sup(𝑅, ℝ, < ))))
Theorem	gcdn0cl 16433	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 16434	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 16435	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 16436	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 16437	Closure of the gcd operator. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) ∈ ℕ0)
Theorem	gcdnncl 16438	Closure of the gcd operator. (Contributed by Thierry Arnoux, 2-Feb-2020.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 gcd 𝑁) ∈ ℕ)
Theorem	gcdcld 16439	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 16440	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 16441*	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 16442	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 16443	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 16444	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 16445	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 16446	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 16447	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 16448	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 16449	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 16450	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 16451	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 16452	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 16453	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 16454	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	gcdaddmlem 16455	Lemma for gcdaddm 16456. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ 𝐾 ∈ ℤ    &   ⊢ 𝑀 ∈ ℤ    &   ⊢ 𝑁 ∈ ℤ    ⇒   ⊢ (𝑀 gcd 𝑁) = (𝑀 gcd ((𝐾 · 𝑀) + 𝑁))
Theorem	gcdaddm 16456	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 16457	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 16458	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 16459	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	gcdabs1 16460	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 16461	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	gcdabs 16462	The gcd of two integers is the same as that of their absolute values. (Contributed by Paul Chapman, 31-Mar-2011.) (Proof shortened by SN, 15-Sep-2024.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((abs‘𝑀) gcd (abs‘𝑁)) = (𝑀 gcd 𝑁))
Theorem	modgcd 16463	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 16464	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 16465	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	gcdmultiplez 16466	The GCD of a multiple of an integer is the integer itself. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) (Proof shortened by AV, 12-Jan-2023.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd (𝑀 · 𝑁)) = 𝑀)
Theorem	gcdmultiple 16467	The GCD of a multiple of a positive integer is the positive integer itself. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) (Proof shortened by AV, 12-Jan-2023.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 gcd (𝑀 · 𝑁)) = 𝑀)
Theorem	dvdsgcdidd 16468	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 16469	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
6.1.8  Bézout's identity
Theorem	bezoutlem1 16470*	Lemma for bezout 16474. (Contributed by Mario Carneiro, 15-Mar-2014.)
⊢ 𝑀 = {𝑧 ∈ ℕ ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑧 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))}    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    ⇒   ⊢ (𝜑 → (𝐴 ≠ 0 → (abs‘𝐴) ∈ 𝑀))
Theorem	bezoutlem2 16471*	Lemma for bezout 16474. (Contributed by Mario Carneiro, 15-Mar-2014.) ( Revised by AV, 30-Sep-2020.)
⊢ 𝑀 = {𝑧 ∈ ℕ ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑧 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))}    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ 𝐺 = inf(𝑀, ℝ, < )    &   ⊢ (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0))    ⇒   ⊢ (𝜑 → 𝐺 ∈ 𝑀)
Theorem	bezoutlem3 16472*	Lemma for bezout 16474. (Contributed by Mario Carneiro, 22-Feb-2014.) ( Revised by AV, 30-Sep-2020.)
⊢ 𝑀 = {𝑧 ∈ ℕ ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑧 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))}    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ 𝐺 = inf(𝑀, ℝ, < )    &   ⊢ (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0))    ⇒   ⊢ (𝜑 → (𝐶 ∈ 𝑀 → 𝐺 ∥ 𝐶))
Theorem	bezoutlem4 16473*	Lemma for bezout 16474. (Contributed by Mario Carneiro, 22-Feb-2014.)
⊢ 𝑀 = {𝑧 ∈ ℕ ∣ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑧 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))}    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ 𝐺 = inf(𝑀, ℝ, < )    &   ⊢ (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0))    ⇒   ⊢ (𝜑 → (𝐴 gcd 𝐵) ∈ 𝑀)
Theorem	bezout 16474*	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. (Contributed by Mario Carneiro, 22-Feb-2014.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ (𝐴 gcd 𝐵) = ((𝐴 · 𝑥) + (𝐵 · 𝑦)))
Theorem	dvdsgcd 16475	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 16476	Biconditional form of dvdsgcd 16475. (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 16477*	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 16478	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 16479	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 16480	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 16481	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 16482	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	gcdzeq 16483	A positive integer 𝐴 is equal to its gcd with an integer 𝐵 if and only if 𝐴 divides 𝐵. Generalization of gcdeq 16484. (Contributed by AV, 1-Jul-2020.)
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ) → ((𝐴 gcd 𝐵) = 𝐴 ↔ 𝐴 ∥ 𝐵))
Theorem	gcdeq 16484	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) = 𝐴 ↔ 𝐴 ∥ 𝐵))
Theorem	dvdssqim 16485	Unidirectional form of dvdssq 16498. (Contributed by Scott Fenton, 19-Apr-2014.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 → (𝑀↑2) ∥ (𝑁↑2)))
Theorem	dvdsexpim 16486	If two numbers are divisible, so are their nonnegative exponents. Similar to dvdssqim 16485 for nonnegative exponents. (Contributed by Steven Nguyen, 2-Apr-2023.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝑁 ∈ ℕ0) → (𝐴 ∥ 𝐵 → (𝐴↑𝑁) ∥ (𝐵↑𝑁)))
Theorem	dvdsmulgcd 16487	A divisibility equivalent for odmulg 19489. (Contributed by Stefan O'Rear, 6-Sep-2015.)
⊢ ((𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ) → (𝐴 ∥ (𝐵 · 𝐶) ↔ 𝐴 ∥ (𝐵 · (𝐶 gcd 𝐴))))
Theorem	rpmulgcd 16488	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))
Theorem	rplpwr 16489	If 𝐴 and 𝐵 are relatively prime, then so are 𝐴↑𝑁 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 → ((𝐴↑𝑁) gcd 𝐵) = 1))
Theorem	rprpwr 16490	If 𝐴 and 𝐵 are relatively prime, then so are 𝐴 and 𝐵↑𝑁. Originally a subproof of rppwr 16491. (Contributed by SN, 21-Aug-2024.)
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝐴 gcd 𝐵) = 1 → (𝐴 gcd (𝐵↑𝑁)) = 1))
Theorem	rppwr 16491	If 𝐴 and 𝐵 are relatively prime, then so are 𝐴↑𝑁 and 𝐵↑𝑁. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝐴 gcd 𝐵) = 1 → ((𝐴↑𝑁) gcd (𝐵↑𝑁)) = 1))
Theorem	nn0rppwr 16492	If 𝐴 and 𝐵 are relatively prime, then so are 𝐴↑𝑁 and 𝐵↑𝑁. rppwr 16491 extended to nonnegative integers. Less general than rpexp12i 16655. (Contributed by Steven Nguyen, 4-Apr-2023.)
⊢ ((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0) → ((𝐴 gcd 𝐵) = 1 → ((𝐴↑𝑁) gcd (𝐵↑𝑁)) = 1))
Theorem	sqgcd 16493	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)))
Theorem	expgcd 16494	Exponentiation distributes over GCD. sqgcd 16493 extended to nonnegative exponents. (Contributed by Steven Nguyen, 4-Apr-2023.)
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝑁 ∈ ℕ0) → ((𝐴 gcd 𝐵)↑𝑁) = ((𝐴↑𝑁) gcd (𝐵↑𝑁)))
Theorem	nn0expgcd 16495	Exponentiation distributes over GCD. nn0gcdsq 16683 extended to nonnegative exponents. expgcd 16494 extended to nonnegative bases. (Contributed by Steven Nguyen, 5-Apr-2023.)
⊢ ((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0) → ((𝐴 gcd 𝐵)↑𝑁) = ((𝐴↑𝑁) gcd (𝐵↑𝑁)))
Theorem	zexpgcd 16496	Exponentiation distributes over GCD. zgcdsq 16684 extended to nonnegative exponents. nn0expgcd 16495 extended to integer bases by symmetry. (Contributed by Steven Nguyen, 5-Apr-2023.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝑁 ∈ ℕ0) → ((𝐴 gcd 𝐵)↑𝑁) = ((𝐴↑𝑁) gcd (𝐵↑𝑁)))
Theorem	dvdssqlem 16497	Lemma for dvdssq 16498. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 ∥ 𝑁 ↔ (𝑀↑2) ∥ (𝑁↑2)))
Theorem	dvdssq 16498	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) → (𝑀 ∥ 𝑁 ↔ (𝑀↑2) ∥ (𝑁↑2)))
Theorem	bezoutr 16499	Partial converse to bezout 16474. 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)))
Theorem	bezoutr1 16500	Converse of bezout 16474 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))