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	bitsfzolem 16401*	Lemma for bitsfzo 16402. (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 16402	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 16403	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 16404	Every number is associated with a finite set of bits. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℕ0 → (bits‘𝑁) ∈ Fin)
Theorem	bitscmp 16405	The bit complement of 𝑁 is -𝑁 − 1. (Thus, by bitsfi 16404, all negative numbers have cofinite bits representations.) (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℤ → (ℕ0 ∖ (bits‘𝑁)) = (bits‘(-𝑁 − 1)))
Theorem	0bits 16406	The bits of zero. (Contributed by Mario Carneiro, 6-Sep-2016.)
⊢ (bits‘0) = ∅
Theorem	m1bits 16407	The bits of negative one. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (bits‘-1) = ℕ0
Theorem	bitsinv1lem 16408	Lemma for bitsinv1 16409. (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 16409*	There is an explicit inverse to the bits function for nonnegative integers (which can be extended to negative integers using bitscmp 16405), part 1. (Contributed by Mario Carneiro, 7-Sep-2016.)
⊢ (𝑁 ∈ ℕ0 → Σ𝑛 ∈ (bits‘𝑁)(2↑𝑛) = 𝑁)
Theorem	bitsinv2 16410*	There is an explicit inverse to the bits function for nonnegative integers, part 2. (Contributed by Mario Carneiro, 8-Sep-2016.)
⊢ (𝐴 ∈ (𝒫 ℕ0 ∩ Fin) → (bits‘Σ𝑛 ∈ 𝐴 (2↑𝑛)) = 𝐴)
Theorem	bitsf1ocnv 16411*	The bits function restricted to nonnegative integers is a bijection from the integers to the finite sets of integers. It is in fact the inverse of the Ackermann bijection ackbijnn 15791. (Contributed by Mario Carneiro, 8-Sep-2016.)
⊢ ((bits ↾ ℕ0):ℕ0–1-1-onto→(𝒫 ℕ0 ∩ Fin) ∧ ◡(bits ↾ ℕ0) = (𝑥 ∈ (𝒫 ℕ0 ∩ Fin) ↦ Σ𝑛 ∈ 𝑥 (2↑𝑛)))
Theorem	bitsf1o 16412	The bits function restricted to nonnegative integers is a bijection from the integers to the finite sets of integers. It is in fact the inverse of the Ackermann bijection ackbijnn 15791. (Contributed by Mario Carneiro, 8-Sep-2016.)
⊢ (bits ↾ ℕ0):ℕ0–1-1-onto→(𝒫 ℕ0 ∩ Fin)
Theorem	bitsf1 16413	The bits function is an injection from ℤ to 𝒫 ℕ0. It is obviously not a bijection (by Cantor's theorem canth2 9065), and in fact its range is the set of finite and cofinite subsets of ℕ0. (Contributed by Mario Carneiro, 22-Sep-2016.)
⊢ bits:ℤ–1-1→𝒫 ℕ0
Theorem	2ebits 16414	The bits of a power of two. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℕ0 → (bits‘(2↑𝑁)) = {𝑁})
Theorem	bitsinv 16415*	The inverse of the bits function. (Contributed by Mario Carneiro, 8-Sep-2016.)
⊢ 𝐾 = ◡(bits ↾ ℕ0)    ⇒   ⊢ (𝐴 ∈ (𝒫 ℕ0 ∩ Fin) → (𝐾‘𝐴) = Σ𝑘 ∈ 𝐴 (2↑𝑘))
Theorem	bitsinvp1 16416	Recursive definition of the inverse of the bits function. (Contributed by Mario Carneiro, 8-Sep-2016.)
⊢ 𝐾 = ◡(bits ↾ ℕ0)    ⇒   ⊢ ((𝐴 ⊆ ℕ0 ∧ 𝑁 ∈ ℕ0) → (𝐾‘(𝐴 ∩ (0..^(𝑁 + 1)))) = ((𝐾‘(𝐴 ∩ (0..^𝑁))) + if(𝑁 ∈ 𝐴, (2↑𝑁), 0)))
Theorem	sadadd2lem2 16417	The core of the proof of sadadd2 16427. The intuitive justification for this is that cadd is true if at least two arguments are true, and hadd is true if an odd number of arguments are true, so altogether the result is 𝑛 · 𝐴 where 𝑛 is the number of true arguments, which is equivalently obtained by adding together one 𝐴 for each true argument, on the right side. (Contributed by Mario Carneiro, 8-Sep-2016.)
⊢ (𝐴 ∈ ℂ → (if(hadd(𝜑, 𝜓, 𝜒), 𝐴, 0) + if(cadd(𝜑, 𝜓, 𝜒), (2 · 𝐴), 0)) = ((if(𝜑, 𝐴, 0) + if(𝜓, 𝐴, 0)) + if(𝜒, 𝐴, 0)))
Definition	df-sad 16418*	Define the addition of two bit sequences, using df-had 1601 and df-cad 1614 bit operations. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ sadd = (𝑥 ∈ 𝒫 ℕ0, 𝑦 ∈ 𝒫 ℕ0 ↦ {𝑘 ∈ ℕ0 ∣ hadd(𝑘 ∈ 𝑥, 𝑘 ∈ 𝑦, ∅ ∈ (seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚 ∈ 𝑥, 𝑚 ∈ 𝑦, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑘))})
Theorem	sadfval 16419*	Define the addition of two bit sequences, using df-had 1601 and df-cad 1614 bit operations. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ 𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚 ∈ 𝐴, 𝑚 ∈ 𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    ⇒   ⊢ (𝜑 → (𝐴 sadd 𝐵) = {𝑘 ∈ ℕ0 ∣ hadd(𝑘 ∈ 𝐴, 𝑘 ∈ 𝐵, ∅ ∈ (𝐶‘𝑘))})
Theorem	sadcf 16420*	The carry sequence is a sequence of elements of 2o encoding a "sequence of wffs". (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ 𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚 ∈ 𝐴, 𝑚 ∈ 𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    ⇒   ⊢ (𝜑 → 𝐶:ℕ0⟶2o)
Theorem	sadc0 16421*	The initial element of the carry sequence is ⊥. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ 𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚 ∈ 𝐴, 𝑚 ∈ 𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    ⇒   ⊢ (𝜑 → ¬ ∅ ∈ (𝐶‘0))
Theorem	sadcp1 16422*	The carry sequence (which is a sequence of wffs, encoded as 1o and ∅) is defined recursively as the carry operation applied to the previous carry and the two current inputs. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ 𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚 ∈ 𝐴, 𝑚 ∈ 𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (∅ ∈ (𝐶‘(𝑁 + 1)) ↔ cadd(𝑁 ∈ 𝐴, 𝑁 ∈ 𝐵, ∅ ∈ (𝐶‘𝑁))))
Theorem	sadval 16423*	The full adder sequence is the half adder function applied to the inputs and the carry sequence. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ 𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚 ∈ 𝐴, 𝑚 ∈ 𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (𝑁 ∈ (𝐴 sadd 𝐵) ↔ hadd(𝑁 ∈ 𝐴, 𝑁 ∈ 𝐵, ∅ ∈ (𝐶‘𝑁))))
Theorem	sadcaddlem 16424*	Lemma for sadcadd 16425. (Contributed by Mario Carneiro, 8-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ 𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚 ∈ 𝐴, 𝑚 ∈ 𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    &   ⊢ 𝐾 = ◡(bits ↾ ℕ0)    &   ⊢ (𝜑 → (∅ ∈ (𝐶‘𝑁) ↔ (2↑𝑁) ≤ ((𝐾‘(𝐴 ∩ (0..^𝑁))) + (𝐾‘(𝐵 ∩ (0..^𝑁))))))    ⇒   ⊢ (𝜑 → (∅ ∈ (𝐶‘(𝑁 + 1)) ↔ (2↑(𝑁 + 1)) ≤ ((𝐾‘(𝐴 ∩ (0..^(𝑁 + 1)))) + (𝐾‘(𝐵 ∩ (0..^(𝑁 + 1)))))))
Theorem	sadcadd 16425*	Non-recursive definition of the carry sequence. (Contributed by Mario Carneiro, 8-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ 𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚 ∈ 𝐴, 𝑚 ∈ 𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    &   ⊢ 𝐾 = ◡(bits ↾ ℕ0)    ⇒   ⊢ (𝜑 → (∅ ∈ (𝐶‘𝑁) ↔ (2↑𝑁) ≤ ((𝐾‘(𝐴 ∩ (0..^𝑁))) + (𝐾‘(𝐵 ∩ (0..^𝑁))))))
Theorem	sadadd2lem 16426*	Lemma for sadadd2 16427. (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ 𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚 ∈ 𝐴, 𝑚 ∈ 𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    &   ⊢ 𝐾 = ◡(bits ↾ ℕ0)    &   ⊢ (𝜑 → ((𝐾‘((𝐴 sadd 𝐵) ∩ (0..^𝑁))) + if(∅ ∈ (𝐶‘𝑁), (2↑𝑁), 0)) = ((𝐾‘(𝐴 ∩ (0..^𝑁))) + (𝐾‘(𝐵 ∩ (0..^𝑁)))))    ⇒   ⊢ (𝜑 → ((𝐾‘((𝐴 sadd 𝐵) ∩ (0..^(𝑁 + 1)))) + if(∅ ∈ (𝐶‘(𝑁 + 1)), (2↑(𝑁 + 1)), 0)) = ((𝐾‘(𝐴 ∩ (0..^(𝑁 + 1)))) + (𝐾‘(𝐵 ∩ (0..^(𝑁 + 1))))))
Theorem	sadadd2 16427*	Sum of initial segments of the sadd sequence. (Contributed by Mario Carneiro, 8-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ 𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚 ∈ 𝐴, 𝑚 ∈ 𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    &   ⊢ 𝐾 = ◡(bits ↾ ℕ0)    ⇒   ⊢ (𝜑 → ((𝐾‘((𝐴 sadd 𝐵) ∩ (0..^𝑁))) + if(∅ ∈ (𝐶‘𝑁), (2↑𝑁), 0)) = ((𝐾‘(𝐴 ∩ (0..^𝑁))) + (𝐾‘(𝐵 ∩ (0..^𝑁)))))
Theorem	sadadd3 16428*	Sum of initial segments of the sadd sequence. (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ 𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚 ∈ 𝐴, 𝑚 ∈ 𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    &   ⊢ 𝐾 = ◡(bits ↾ ℕ0)    ⇒   ⊢ (𝜑 → ((𝐾‘((𝐴 sadd 𝐵) ∩ (0..^𝑁))) mod (2↑𝑁)) = (((𝐾‘(𝐴 ∩ (0..^𝑁))) + (𝐾‘(𝐵 ∩ (0..^𝑁)))) mod (2↑𝑁)))
Theorem	sadcl 16429	The sum of two sequences is a sequence. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ ((𝐴 ⊆ ℕ0 ∧ 𝐵 ⊆ ℕ0) → (𝐴 sadd 𝐵) ⊆ ℕ0)
Theorem	sadcom 16430	The adder sequence function is commutative. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ ((𝐴 ⊆ ℕ0 ∧ 𝐵 ⊆ ℕ0) → (𝐴 sadd 𝐵) = (𝐵 sadd 𝐴))
Theorem	saddisjlem 16431*	Lemma for sadadd 16434. (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ (𝜑 → (𝐴 ∩ 𝐵) = ∅)    &   ⊢ 𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚 ∈ 𝐴, 𝑚 ∈ 𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (𝑁 ∈ (𝐴 sadd 𝐵) ↔ 𝑁 ∈ (𝐴 ∪ 𝐵)))
Theorem	saddisj 16432	The sum of disjoint sequences is the union of the sequences. (In this case, there are no carried bits.) (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ (𝜑 → 𝐴 ⊆ ℕ0)    &   ⊢ (𝜑 → 𝐵 ⊆ ℕ0)    &   ⊢ (𝜑 → (𝐴 ∩ 𝐵) = ∅)    ⇒   ⊢ (𝜑 → (𝐴 sadd 𝐵) = (𝐴 ∪ 𝐵))
Theorem	sadaddlem 16433*	Lemma for sadadd 16434. (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ 𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚 ∈ (bits‘𝐴), 𝑚 ∈ (bits‘𝐵), ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   ⊢ 𝐾 = ◡(bits ↾ ℕ0)    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ ℤ)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (((bits‘𝐴) sadd (bits‘𝐵)) ∩ (0..^𝑁)) = (bits‘((𝐴 + 𝐵) mod (2↑𝑁))))
Theorem	sadadd 16434	For sequences that correspond to valid integers, the adder sequence function produces the sequence for the sum. This is effectively a proof of the correctness of the ripple carry adder, implemented with logic gates corresponding to df-had 1601 and df-cad 1614. It is interesting to consider in what sense the sadd function can be said to be "adding" things outside the range of the bits function, that is, when adding sequences that are not eventually constant and so do not denote any integer. The correct interpretation is that the sequences are representations of 2-adic integers, which have a natural ring structure. (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ((bits‘𝐴) sadd (bits‘𝐵)) = (bits‘(𝐴 + 𝐵)))
Theorem	sadid1 16435	The adder sequence function has a left identity, the empty set, which is the representation of the integer zero. (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ (𝐴 ⊆ ℕ0 → (𝐴 sadd ∅) = 𝐴)
Theorem	sadid2 16436	The adder sequence function has a right identity, the empty set, which is the representation of the integer zero. (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ (𝐴 ⊆ ℕ0 → (∅ sadd 𝐴) = 𝐴)
Theorem	sadasslem 16437	Lemma for sadass 16438. (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 16438	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 16439	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 16440	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 16441	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 16442*	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 16443*	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 16444*	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 16445*	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 16446*	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 16447*	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 16448*	Define the addition of two bit sequences, using df-had 1601 and df-cad 1614 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 16449*	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 16450*	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 16451	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 16452*	Lemma for smu01 16453 and smu02 16454. (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 16453	Multiplication of a sequence by 0 on the right. (Contributed by Mario Carneiro, 19-Sep-2016.)
⊢ (𝐴 ⊆ ℕ0 → (𝐴 smul ∅) = ∅)
Theorem	smu02 16454	Multiplication of a sequence by 0 on the left. (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ (𝐴 ⊆ ℕ0 → (∅ smul 𝐴) = ∅)
Theorem	smupval 16455*	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 16456*	Rewrite smupp1 16447 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 16457*	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 16458	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 16459	Lemma for smumul 16460. (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 16460	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 16418, whose correctness is verified in sadadd 16434. 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 16461	Extend the definition of a class to include the greatest common divisor operator.
class gcd
Definition	df-gcd 16462*	Define the gcd operator. For example, (-6 gcd 9) = 3 (ex-gcd 30552). For an alternate definition, based on the definition in [ApostolNT] p. 15, see dfgcd2 16513. (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 16463*	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 16464	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 16465*	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 16466*	Lemma for gcdn0cl 16469, gcddvds 16470 and dvdslegcd 16471. (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 16467*	Lemma for gcdn0cl 16469, gcddvds 16470 and dvdslegcd 16471. (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 16468*	Lemma for gcdn0cl 16469, gcddvds 16470 and dvdslegcd 16471. (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 16469	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 16470	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 16471	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 16472	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 16473	Closure of the gcd operator. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) ∈ ℕ0)
Theorem	gcdnncl 16474	Closure of the gcd operator. (Contributed by Thierry Arnoux, 2-Feb-2020.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 gcd 𝑁) ∈ ℕ)
Theorem	gcdcld 16475	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 16476	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 16477*	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 16478	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 16479	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 16480	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 16481	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 16482	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 16483	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 16484	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 16485	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 16486	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 16487	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 16488	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 16489	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 16490	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 16491	Lemma for gcdaddm 16492. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ 𝐾 ∈ ℤ    &   ⊢ 𝑀 ∈ ℤ    &   ⊢ 𝑁 ∈ ℤ    ⇒   ⊢ (𝑀 gcd 𝑁) = (𝑀 gcd ((𝐾 · 𝑀) + 𝑁))
Theorem	gcdaddm 16492	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 16493	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 16494	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 16495	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 16496	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 16497	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 16498	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 16499	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 16500	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)