P. 162 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 16101-16200   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	divalglem1 16101	Lemma for divalg 16110. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ 𝑁 ∈ ℤ    &   ⊢ 𝐷 ∈ ℤ    &   ⊢ 𝐷 ≠ 0    ⇒   ⊢ 0 ≤ (𝑁 + (abs‘(𝑁 · 𝐷)))
Theorem	divalglem2 16102*	Lemma for divalg 16110. (Contributed by Paul Chapman, 21-Mar-2011.) (Revised by AV, 2-Oct-2020.)
⊢ 𝑁 ∈ ℤ    &   ⊢ 𝐷 ∈ ℤ    &   ⊢ 𝐷 ≠ 0    &   ⊢ 𝑆 = {𝑟 ∈ ℕ0 ∣ 𝐷 ∥ (𝑁 − 𝑟)}    ⇒   ⊢ inf(𝑆, ℝ, < ) ∈ 𝑆
Theorem	divalglem4 16103*	Lemma for divalg 16110. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ 𝑁 ∈ ℤ    &   ⊢ 𝐷 ∈ ℤ    &   ⊢ 𝐷 ≠ 0    &   ⊢ 𝑆 = {𝑟 ∈ ℕ0 ∣ 𝐷 ∥ (𝑁 − 𝑟)}    ⇒   ⊢ 𝑆 = {𝑟 ∈ ℕ0 ∣ ∃𝑞 ∈ ℤ 𝑁 = ((𝑞 · 𝐷) + 𝑟)}
Theorem	divalglem5 16104*	Lemma for divalg 16110. (Contributed by Paul Chapman, 21-Mar-2011.) (Revised by AV, 2-Oct-2020.)
⊢ 𝑁 ∈ ℤ    &   ⊢ 𝐷 ∈ ℤ    &   ⊢ 𝐷 ≠ 0    &   ⊢ 𝑆 = {𝑟 ∈ ℕ0 ∣ 𝐷 ∥ (𝑁 − 𝑟)}    &   ⊢ 𝑅 = inf(𝑆, ℝ, < )    ⇒   ⊢ (0 ≤ 𝑅 ∧ 𝑅 < (abs‘𝐷))
Theorem	divalglem6 16105	Lemma for divalg 16110. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ 𝐴 ∈ ℕ    &   ⊢ 𝑋 ∈ (0...(𝐴 − 1))    &   ⊢ 𝐾 ∈ ℤ    ⇒   ⊢ (𝐾 ≠ 0 → ¬ (𝑋 + (𝐾 · 𝐴)) ∈ (0...(𝐴 − 1)))
Theorem	divalglem7 16106	Lemma for divalg 16110. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ 𝐷 ∈ ℤ    &   ⊢ 𝐷 ≠ 0    ⇒   ⊢ ((𝑋 ∈ (0...((abs‘𝐷) − 1)) ∧ 𝐾 ∈ ℤ) → (𝐾 ≠ 0 → ¬ (𝑋 + (𝐾 · (abs‘𝐷))) ∈ (0...((abs‘𝐷) − 1))))
Theorem	divalglem8 16107*	Lemma for divalg 16110. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ 𝑁 ∈ ℤ    &   ⊢ 𝐷 ∈ ℤ    &   ⊢ 𝐷 ≠ 0    &   ⊢ 𝑆 = {𝑟 ∈ ℕ0 ∣ 𝐷 ∥ (𝑁 − 𝑟)}    ⇒   ⊢ (((𝑋 ∈ 𝑆 ∧ 𝑌 ∈ 𝑆) ∧ (𝑋 < (abs‘𝐷) ∧ 𝑌 < (abs‘𝐷))) → (𝐾 ∈ ℤ → ((𝐾 · (abs‘𝐷)) = (𝑌 − 𝑋) → 𝑋 = 𝑌)))
Theorem	divalglem9 16108*	Lemma for divalg 16110. (Contributed by Paul Chapman, 21-Mar-2011.) (Revised by AV, 2-Oct-2020.)
⊢ 𝑁 ∈ ℤ    &   ⊢ 𝐷 ∈ ℤ    &   ⊢ 𝐷 ≠ 0    &   ⊢ 𝑆 = {𝑟 ∈ ℕ0 ∣ 𝐷 ∥ (𝑁 − 𝑟)}    &   ⊢ 𝑅 = inf(𝑆, ℝ, < )    ⇒   ⊢ ∃!𝑥 ∈ 𝑆 𝑥 < (abs‘𝐷)
Theorem	divalglem10 16109*	Lemma for divalg 16110. (Contributed by Paul Chapman, 21-Mar-2011.) (Proof shortened by AV, 2-Oct-2020.)
⊢ 𝑁 ∈ ℤ    &   ⊢ 𝐷 ∈ ℤ    &   ⊢ 𝐷 ≠ 0    &   ⊢ 𝑆 = {𝑟 ∈ ℕ0 ∣ 𝐷 ∥ (𝑁 − 𝑟)}    ⇒   ⊢ ∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟))
Theorem	divalg 16110*	The division algorithm (theorem). Dividing an integer 𝑁 by a nonzero integer 𝐷 produces a (unique) quotient 𝑞 and a unique remainder 0 ≤ 𝑟 < (abs‘𝐷). Theorem 1.14 in [ApostolNT] p. 19. The proof does not use / or ⌊ or mod. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ ((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℤ ∧ 𝐷 ≠ 0) → ∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)))
Theorem	divalgb 16111*	Express the division algorithm as stated in divalg 16110 in terms of ∥. (Contributed by Paul Chapman, 31-Mar-2011.)
⊢ ((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℤ ∧ 𝐷 ≠ 0) → (∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)) ↔ ∃!𝑟 ∈ ℕ0 (𝑟 < (abs‘𝐷) ∧ 𝐷 ∥ (𝑁 − 𝑟))))
Theorem	divalg2 16112*	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 16113	The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor (compare divalg2 16112 and divalgb 16111). This demonstration theorem justifies the use of mod to yield an explicit remainder from this point forward. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by AV, 21-Aug-2021.)
⊢ ((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ) → (𝑅 = (𝑁 mod 𝐷) ↔ (𝑅 ∈ ℕ0 ∧ (𝑅 < 𝐷 ∧ 𝐷 ∥ (𝑁 − 𝑅)))))
Theorem	divalgmodcl 16114	The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor. Variant of divalgmod 16113. (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 16115*	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 16116	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 16117	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 16118	Special case of ndvdsadd 16117. 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 16119	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 16120	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 16121	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 16122	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 16123	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.) (Proof shortened by AV, 10-Jul-2022.)
⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → ((⌊‘(𝑁 / 4)) · 2) < (𝑁 / 2))
6.1.6  Bit sequences
Syntax	cbits 16124	Define the binary bits of an integer.
class bits
Syntax	csad 16125	Define the sequence addition on bit sequences.
class sadd
Syntax	csmu 16126	Define the sequence multiplication on bit sequences.
class smul
Definition	df-bits 16127*	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 16128*	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 16129	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 16130	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 16131	The set of bits of an integer is a subset of ℕ0. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (bits‘𝑁) ⊆ ℕ0
Theorem	bitsf 16132	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 16133	Value of the zeroth bit. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℤ → (0 ∈ (bits‘𝑁) ↔ ¬ 2 ∥ 𝑁))
Theorem	bits0e 16134	The zeroth bit of an even number is zero. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℤ → ¬ 0 ∈ (bits‘(2 · 𝑁)))
Theorem	bits0o 16135	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 16136	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 16137	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 16138	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 16139*	Lemma for bitsfzo 16140. (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 16140	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 16141	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 16142	Every number is associated with a finite set of bits. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℕ0 → (bits‘𝑁) ∈ Fin)
Theorem	bitscmp 16143	The bit complement of 𝑁 is -𝑁 − 1. (Thus, by bitsfi 16142, all negative numbers have cofinite bits representations.) (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℤ → (ℕ0 ∖ (bits‘𝑁)) = (bits‘(-𝑁 − 1)))
Theorem	0bits 16144	The bits of zero. (Contributed by Mario Carneiro, 6-Sep-2016.)
⊢ (bits‘0) = ∅
Theorem	m1bits 16145	The bits of negative one. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (bits‘-1) = ℕ0
Theorem	bitsinv1lem 16146	Lemma for bitsinv1 16147. (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 16147*	There is an explicit inverse to the bits function for nonnegative integers (which can be extended to negative integers using bitscmp 16143), part 1. (Contributed by Mario Carneiro, 7-Sep-2016.)
⊢ (𝑁 ∈ ℕ0 → Σ𝑛 ∈ (bits‘𝑁)(2↑𝑛) = 𝑁)
Theorem	bitsinv2 16148*	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 16149*	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 15538. (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 16150	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 15538. (Contributed by Mario Carneiro, 8-Sep-2016.)
⊢ (bits ↾ ℕ0):ℕ0–1-1-onto→(𝒫 ℕ0 ∩ Fin)
Theorem	bitsf1 16151	The bits function is an injection from ℤ to 𝒫 ℕ0. It is obviously not a bijection (by Cantor's theorem canth2 8899), 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 16152	The bits of a power of two. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℕ0 → (bits‘(2↑𝑁)) = {𝑁})
Theorem	bitsinv 16153*	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 16154	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 16155	The core of the proof of sadadd2 16165. 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 16156*	Define the addition of two bit sequences, using df-had 1599 and df-cad 1613 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 16157*	Define the addition of two bit sequences, using df-had 1599 and df-cad 1613 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 16158*	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 16159*	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 16160*	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 16161*	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 16162*	Lemma for sadcadd 16163. (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 16163*	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 16164*	Lemma for sadadd2 16165. (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 16165*	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 16166*	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 16167	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 16168	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 16169*	Lemma for sadadd 16172. (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 16170	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 16171*	Lemma for sadadd 16172. (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 16172	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 1599 and df-cad 1613. 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 16173	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 16174	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 16175	Lemma for sadass 16176. (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 16176	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 16177	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 16178	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 16179	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 16180*	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 16181*	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 16182*	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 16183*	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 16184*	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 16185*	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 16186*	Define the addition of two bit sequences, using df-had 1599 and df-cad 1613 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 16187*	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 16188*	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 16189	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 16190*	Lemma for smu01 16191 and smu02 16192. (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 16191	Multiplication of a sequence by 0 on the right. (Contributed by Mario Carneiro, 19-Sep-2016.)
⊢ (𝐴 ⊆ ℕ0 → (𝐴 smul ∅) = ∅)
Theorem	smu02 16192	Multiplication of a sequence by 0 on the left. (Contributed by Mario Carneiro, 9-Sep-2016.)
⊢ (𝐴 ⊆ ℕ0 → (∅ smul 𝐴) = ∅)
Theorem	smupval 16193*	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 16194*	Rewrite smupp1 16185 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 16195*	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 16196	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 16197	Lemma for smumul 16198. (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 16198	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 16156, whose correctness is verified in sadadd 16172. 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 16199	Extend the definition of a class to include the greatest common divisor operator.
class gcd
Definition	df-gcd 16200*	Define the gcd operator. For example, (-6 gcd 9) = 3 (ex-gcd 28817). For an alternate definition, based on the definition in [ApostolNT] p. 15, see dfgcd2 16252. (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, < )))