P. 154 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 15301-15400   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	divalg2 15301*	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 15302	The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor (compare divalg2 15301 and divalgb 15300). 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	divalgmodOLD 15303*	Obsolete proof of divalgmod 15302 as of 21-Aug-2021. (Contributed by Paul Chapman, 31-Mar-2011.) (New usage is discouraged.) (Proof modification is discouraged.)
⊢ ((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ) → (𝑟 = (𝑁 mod 𝐷) ↔ (𝑟 ∈ ℕ0 ∧ (𝑟 < 𝐷 ∧ 𝐷 ∥ (𝑁 − 𝑟)))))
Theorem	divalgmodcl 15304	The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor. Variant of divalgmod 15302. (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 15305*	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 15306	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 15307	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 15308	Special case of ndvdsadd 15307. 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 15309	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 15310	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 15311	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 15312	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 15313	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))
6.1.6  Bit sequences
Syntax	cbits 15314	Define the binary bits of an integer.
class bits
Syntax	csad 15315	Define the sequence addition on bit sequences.
class sadd
Syntax	csmu 15316	Define the sequence multiplication on bit sequences.
class smul
Definition	df-bits 15317*	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 15318*	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 15319	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 15320	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 15321	The set of bits of an integer is a subset of ℕ0. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (bits‘𝑁) ⊆ ℕ0
Theorem	bitsf 15322	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	bits0 15323	Value of the zeroth bit. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℤ → (0 ∈ (bits‘𝑁) ↔ ¬ 2 ∥ 𝑁))
Theorem	bits0e 15324	The zeroth bit of an even number is zero. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℤ → ¬ 0 ∈ (bits‘(2 · 𝑁)))
Theorem	bits0o 15325	The zeroth bit of an odd number is zero. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℤ → 0 ∈ (bits‘((2 · 𝑁) + 1)))
Theorem	bitsp1 15326	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 15327	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 15328	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 15329*	Lemma for bitsfzo 15330. (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 15330	The bits of a number are all 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 15331	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 15332	Every number is associated with a finite set of bits. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℕ0 → (bits‘𝑁) ∈ Fin)
Theorem	bitscmp 15333	The bit complement of 𝑁 is -𝑁 − 1. (Thus, by bitsfi 15332, all negative numbers have cofinite bits representations.) (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℤ → (ℕ0 ∖ (bits‘𝑁)) = (bits‘(-𝑁 − 1)))
Theorem	0bits 15334	The bits of zero. (Contributed by Mario Carneiro, 6-Sep-2016.)
⊢ (bits‘0) = ∅
Theorem	m1bits 15335	The bits of negative one. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (bits‘-1) = ℕ0
Theorem	bitsinv1lem 15336	Lemma for bitsinv1 15337. (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 15337*	There is an explicit inverse to the bits function for nonnegative integers (which can be extended to negative integers using bitscmp 15333), part 1. (Contributed by Mario Carneiro, 7-Sep-2016.)
⊢ (𝑁 ∈ ℕ0 → Σ𝑛 ∈ (bits‘𝑁)(2↑𝑛) = 𝑁)
Theorem	bitsinv2 15338*	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 15339*	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 14730. (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 15340	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 14730. (Contributed by Mario Carneiro, 8-Sep-2016.)
⊢ (bits ↾ ℕ0):ℕ0–1-1-onto→(𝒫 ℕ0 ∩ Fin)
Theorem	bitsf1 15341	The bits function is an injection from ℤ to 𝒫 ℕ0. It is obviously not a bijection (by Cantor's theorem canth2 8266), 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 15342	The bits of a power of two. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℕ0 → (bits‘(2↑𝑁)) = {𝑁})
Theorem	bitsinv 15343*	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 15344	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 15345	The core of the proof of sadadd2 15355. 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 15346*	Define the addition of two bit sequences, using df-had 1670 and df-cad 1683 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 15347*	Define the addition of two bit sequences, using df-had 1670 and df-cad 1683 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 15348*	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 15349*	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 15350*	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 15351*	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 15352*	Lemma for sadcadd 15353. (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 15353*	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 15354*	Lemma for sadadd2 15355. (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 15355*	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 15356*	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 15357	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 15358	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 15359*	Lemma for sadadd 15362. (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 15360	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 15361*	Lemma for sadadd 15362. (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 15362	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 1670 and df-cad 1683. 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 15363	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 15364	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 15365	Lemma for sadass 15366. (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 15366	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 15367	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 15368	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 15369	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 15370*	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 15371*	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 15372*	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 15373*	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 15374*	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 15375*	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 15376*	Define the addition of two bit sequences, using df-had 1670 and df-cad 1683 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 15377*	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 15378*	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 15379	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 15380*	Lemma for smu01 15381 and smu02 15382. (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 15381	Multiplication of a sequence by 0 on the right. (Contributed by Mario Carneiro, 19-Sep-2016.)
⊢ (𝐴 ⊆ ℕ0 → (𝐴 smul ∅) = ∅)
Theorem	smu02 15382	Multiplication of a sequence by 0 on the left. (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ (𝐴 ⊆ ℕ0 → (∅ smul 𝐴) = ∅)
Theorem	smupval 15383*	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 15384*	Rewrite smupp1 15375 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 15385*	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 15386	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 15387	Lemma for smumul 15388. (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 15388	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 15346, whose correctness is verified in sadadd 15362. 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 15389	Extend the definition of a class to include the greatest common divisor operator.
class gcd
Definition	df-gcd 15390*	Define the gcd operator. For example, (-6 gcd 9) = 3 (ex-gcd 27596). For an alternate definition, based on the definition in [ApostolNT] p. 15, see dfgcd2 15436. (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 15391*	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 15392	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 15393*	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 15394*	Lemma for gcdn0cl 15397, gcddvds 15398 and dvdslegcd 15399. (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 15395*	Lemma for gcdn0cl 15397, gcddvds 15398 and dvdslegcd 15399. (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 15396*	Lemma for gcdn0cl 15397, gcddvds 15398 and dvdslegcd 15399. (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 15397	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 15398	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 15399	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 15400	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)))