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Theorem List for Metamath Proof Explorer - 16101-16200   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremoddpwp1fsum 16101* An odd power of a number increased by 1 expressed by a product with a finite sum. (Contributed by AV, 15-Aug-2021.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑 → ¬ 2 ∥ 𝑁)       (𝜑 → ((𝐴𝑁) + 1) = ((𝐴 + 1) · Σ𝑘 ∈ (0...(𝑁 − 1))((-1↑𝑘) · (𝐴𝑘))))
 
6.1.5  The division algorithm
 
Theoremdivalglem0 16102 Lemma for divalg 16112. (Contributed by Paul Chapman, 21-Mar-2011.)
𝑁 ∈ ℤ    &   𝐷 ∈ ℤ       ((𝑅 ∈ ℤ ∧ 𝐾 ∈ ℤ) → (𝐷 ∥ (𝑁𝑅) → 𝐷 ∥ (𝑁 − (𝑅 − (𝐾 · (abs‘𝐷))))))
 
Theoremdivalglem1 16103 Lemma for divalg 16112. (Contributed by Paul Chapman, 21-Mar-2011.)
𝑁 ∈ ℤ    &   𝐷 ∈ ℤ    &   𝐷 ≠ 0       0 ≤ (𝑁 + (abs‘(𝑁 · 𝐷)))
 
Theoremdivalglem2 16104* Lemma for divalg 16112. (Contributed by Paul Chapman, 21-Mar-2011.) (Revised by AV, 2-Oct-2020.)
𝑁 ∈ ℤ    &   𝐷 ∈ ℤ    &   𝐷 ≠ 0    &   𝑆 = {𝑟 ∈ ℕ0𝐷 ∥ (𝑁𝑟)}       inf(𝑆, ℝ, < ) ∈ 𝑆
 
Theoremdivalglem4 16105* Lemma for divalg 16112. (Contributed by Paul Chapman, 21-Mar-2011.)
𝑁 ∈ ℤ    &   𝐷 ∈ ℤ    &   𝐷 ≠ 0    &   𝑆 = {𝑟 ∈ ℕ0𝐷 ∥ (𝑁𝑟)}       𝑆 = {𝑟 ∈ ℕ0 ∣ ∃𝑞 ∈ ℤ 𝑁 = ((𝑞 · 𝐷) + 𝑟)}
 
Theoremdivalglem5 16106* Lemma for divalg 16112. (Contributed by Paul Chapman, 21-Mar-2011.) (Revised by AV, 2-Oct-2020.)
𝑁 ∈ ℤ    &   𝐷 ∈ ℤ    &   𝐷 ≠ 0    &   𝑆 = {𝑟 ∈ ℕ0𝐷 ∥ (𝑁𝑟)}    &   𝑅 = inf(𝑆, ℝ, < )       (0 ≤ 𝑅𝑅 < (abs‘𝐷))
 
Theoremdivalglem6 16107 Lemma for divalg 16112. (Contributed by Paul Chapman, 21-Mar-2011.)
𝐴 ∈ ℕ    &   𝑋 ∈ (0...(𝐴 − 1))    &   𝐾 ∈ ℤ       (𝐾 ≠ 0 → ¬ (𝑋 + (𝐾 · 𝐴)) ∈ (0...(𝐴 − 1)))
 
Theoremdivalglem7 16108 Lemma for divalg 16112. (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))))
 
Theoremdivalglem8 16109* Lemma for divalg 16112. (Contributed by Paul Chapman, 21-Mar-2011.)
𝑁 ∈ ℤ    &   𝐷 ∈ ℤ    &   𝐷 ≠ 0    &   𝑆 = {𝑟 ∈ ℕ0𝐷 ∥ (𝑁𝑟)}       (((𝑋𝑆𝑌𝑆) ∧ (𝑋 < (abs‘𝐷) ∧ 𝑌 < (abs‘𝐷))) → (𝐾 ∈ ℤ → ((𝐾 · (abs‘𝐷)) = (𝑌𝑋) → 𝑋 = 𝑌)))
 
Theoremdivalglem9 16110* Lemma for divalg 16112. (Contributed by Paul Chapman, 21-Mar-2011.) (Revised by AV, 2-Oct-2020.)
𝑁 ∈ ℤ    &   𝐷 ∈ ℤ    &   𝐷 ≠ 0    &   𝑆 = {𝑟 ∈ ℕ0𝐷 ∥ (𝑁𝑟)}    &   𝑅 = inf(𝑆, ℝ, < )       ∃!𝑥𝑆 𝑥 < (abs‘𝐷)
 
Theoremdivalglem10 16111* Lemma for divalg 16112. (Contributed by Paul Chapman, 21-Mar-2011.) (Proof shortened by AV, 2-Oct-2020.)
𝑁 ∈ ℤ    &   𝐷 ∈ ℤ    &   𝐷 ≠ 0    &   𝑆 = {𝑟 ∈ ℕ0𝐷 ∥ (𝑁𝑟)}       ∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟))
 
Theoremdivalg 16112* 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 ≤ 𝑟𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)))
 
Theoremdivalgb 16113* Express the division algorithm as stated in divalg 16112 in terms of . (Contributed by Paul Chapman, 31-Mar-2011.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℤ ∧ 𝐷 ≠ 0) → (∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)) ↔ ∃!𝑟 ∈ ℕ0 (𝑟 < (abs‘𝐷) ∧ 𝐷 ∥ (𝑁𝑟))))
 
Theoremdivalg2 16114* The division algorithm (theorem) for a positive divisor. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ) → ∃!𝑟 ∈ ℕ0 (𝑟 < 𝐷𝐷 ∥ (𝑁𝑟)))
 
Theoremdivalgmod 16115 The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor (compare divalg2 16114 and divalgb 16113). 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 < 𝐷𝐷 ∥ (𝑁𝑅)))))
 
Theoremdivalgmodcl 16116 The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor. Variant of divalgmod 16115. (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 < 𝐷𝐷 ∥ (𝑁𝑅))))
 
Theoremmodremain 16117* The result of the modulo operation is the remainder of the division algorithm. (Contributed by AV, 19-Aug-2021.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ ∧ (𝑅 ∈ ℕ0𝑅 < 𝐷)) → ((𝑁 mod 𝐷) = 𝑅 ↔ ∃𝑧 ∈ ℤ ((𝑧 · 𝐷) + 𝑅) = 𝑁))
 
Theoremndvdssub 16118 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)) → (𝐷𝑁 → ¬ 𝐷 ∥ (𝑁𝐾)))
 
Theoremndvdsadd 16119 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)) → (𝐷𝑁 → ¬ 𝐷 ∥ (𝑁 + 𝐾)))
 
Theoremndvdsp1 16120 Special case of ndvdsadd 16119. 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) → (𝐷𝑁 → ¬ 𝐷 ∥ (𝑁 + 1)))
 
Theoremndvdsi 16121 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        ¬ 𝐴𝐵
 
Theoremflodddiv4 16122 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)))
 
Theoremfldivndvdslt 16123 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) ∧ ¬ 𝐿𝐾) → (⌊‘(𝐾 / 𝐿)) < (𝐾 / 𝐿))
 
Theoremflodddiv4lt 16124 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))
 
Theoremflodddiv4t2lthalf 16125 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
 
Syntaxcbits 16126 Define the binary bits of an integer.
class bits
 
Syntaxcsad 16127 Define the sequence addition on bit sequences.
class sadd
 
Syntaxcsmu 16128 Define the sequence multiplication on bit sequences.
class smul
 
Definitiondf-bits 16129* 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)))})
 
Theorembitsfval 16130* 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)))})
 
Theorembitsval 16131 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)))))
 
Theorembitsval2 16132 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)))))
 
Theorembitsss 16133 The set of bits of an integer is a subset of 0. (Contributed by Mario Carneiro, 5-Sep-2016.)
(bits‘𝑁) ⊆ ℕ0
 
Theorembitsf 16134 The bits function is a function from integers to subsets of nonnegative integers. (Contributed by Mario Carneiro, 5-Sep-2016.)
bits:ℤ⟶𝒫 ℕ0
 
Theorembits0 16135 Value of the zeroth bit. (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝑁 ∈ ℤ → (0 ∈ (bits‘𝑁) ↔ ¬ 2 ∥ 𝑁))
 
Theorembits0e 16136 The zeroth bit of an even number is zero. (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝑁 ∈ ℤ → ¬ 0 ∈ (bits‘(2 · 𝑁)))
 
Theorembits0o 16137 The zeroth bit of an odd number is zero. (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝑁 ∈ ℤ → 0 ∈ (bits‘((2 · 𝑁) + 1)))
 
Theorembitsp1 16138 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)))))
 
Theorembitsp1e 16139 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)))
 
Theorembitsp1o 16140 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)))
 
Theorembitsfzolem 16141* Lemma for bitsfzo 16142. (Contributed by Mario Carneiro, 5-Sep-2016.) (Revised by AV, 1-Oct-2020.)
(𝜑𝑁 ∈ ℕ0)    &   (𝜑𝑀 ∈ ℕ0)    &   (𝜑 → (bits‘𝑁) ⊆ (0..^𝑀))    &   𝑆 = inf({𝑛 ∈ ℕ0𝑁 < (2↑𝑛)}, ℝ, < )       (𝜑𝑁 ∈ (0..^(2↑𝑀)))
 
Theorembitsfzo 16142 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)))
 
Theorembitsmod 16143 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)))
 
Theorembitsfi 16144 Every number is associated with a finite set of bits. (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝑁 ∈ ℕ0 → (bits‘𝑁) ∈ Fin)
 
Theorembitscmp 16145 The bit complement of 𝑁 is -𝑁 − 1. (Thus, by bitsfi 16144, all negative numbers have cofinite bits representations.) (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝑁 ∈ ℤ → (ℕ0 ∖ (bits‘𝑁)) = (bits‘(-𝑁 − 1)))
 
Theorem0bits 16146 The bits of zero. (Contributed by Mario Carneiro, 6-Sep-2016.)
(bits‘0) = ∅
 
Theoremm1bits 16147 The bits of negative one. (Contributed by Mario Carneiro, 5-Sep-2016.)
(bits‘-1) = ℕ0
 
Theorembitsinv1lem 16148 Lemma for bitsinv1 16149. (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)))
 
Theorembitsinv1 16149* There is an explicit inverse to the bits function for nonnegative integers (which can be extended to negative integers using bitscmp 16145), part 1. (Contributed by Mario Carneiro, 7-Sep-2016.)
(𝑁 ∈ ℕ0 → Σ𝑛 ∈ (bits‘𝑁)(2↑𝑛) = 𝑁)
 
Theorembitsinv2 16150* 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‘Σ𝑛𝐴 (2↑𝑛)) = 𝐴)
 
Theorembitsf1ocnv 16151* 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 15540. (Contributed by Mario Carneiro, 8-Sep-2016.)
((bits ↾ ℕ0):ℕ01-1-onto→(𝒫 ℕ0 ∩ Fin) ∧ (bits ↾ ℕ0) = (𝑥 ∈ (𝒫 ℕ0 ∩ Fin) ↦ Σ𝑛𝑥 (2↑𝑛)))
 
Theorembitsf1o 16152 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 15540. (Contributed by Mario Carneiro, 8-Sep-2016.)
(bits ↾ ℕ0):ℕ01-1-onto→(𝒫 ℕ0 ∩ Fin)
 
Theorembitsf1 16153 The bits function is an injection from to 𝒫 ℕ0. It is obviously not a bijection (by Cantor's theorem canth2 8917), and in fact its range is the set of finite and cofinite subsets of 0. (Contributed by Mario Carneiro, 22-Sep-2016.)
bits:ℤ–1-1→𝒫 ℕ0
 
Theorem2ebits 16154 The bits of a power of two. (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝑁 ∈ ℕ0 → (bits‘(2↑𝑁)) = {𝑁})
 
Theorembitsinv 16155* The inverse of the bits function. (Contributed by Mario Carneiro, 8-Sep-2016.)
𝐾 = (bits ↾ ℕ0)       (𝐴 ∈ (𝒫 ℕ0 ∩ Fin) → (𝐾𝐴) = Σ𝑘𝐴 (2↑𝑘))
 
Theorembitsinvp1 16156 Recursive definition of the inverse of the bits function. (Contributed by Mario Carneiro, 8-Sep-2016.)
𝐾 = (bits ↾ ℕ0)       ((𝐴 ⊆ ℕ0𝑁 ∈ ℕ0) → (𝐾‘(𝐴 ∩ (0..^(𝑁 + 1)))) = ((𝐾‘(𝐴 ∩ (0..^𝑁))) + if(𝑁𝐴, (2↑𝑁), 0)))
 
Theoremsadadd2lem2 16157 The core of the proof of sadadd2 16167. 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)))
 
Definitiondf-sad 16158* Define the addition of two bit sequences, using df-had 1595 and df-cad 1609 bit operations. (Contributed by Mario Carneiro, 5-Sep-2016.)
sadd = (𝑥 ∈ 𝒫 ℕ0, 𝑦 ∈ 𝒫 ℕ0 ↦ {𝑘 ∈ ℕ0 ∣ hadd(𝑘𝑥, 𝑘𝑦, ∅ ∈ (seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝑥, 𝑚𝑦, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘𝑘))})
 
Theoremsadfval 16159* Define the addition of two bit sequences, using df-had 1595 and df-cad 1609 bit operations. (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))       (𝜑 → (𝐴 sadd 𝐵) = {𝑘 ∈ ℕ0 ∣ hadd(𝑘𝐴, 𝑘𝐵, ∅ ∈ (𝐶𝑘))})
 
Theoremsadcf 16160* The carry sequence is a sequence of elements of 2o encoding a "sequence of wffs". (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))       (𝜑𝐶:ℕ0⟶2o)
 
Theoremsadc0 16161* The initial element of the carry sequence is . (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))       (𝜑 → ¬ ∅ ∈ (𝐶‘0))
 
Theoremsadcp1 16162* 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.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (∅ ∈ (𝐶‘(𝑁 + 1)) ↔ cadd(𝑁𝐴, 𝑁𝐵, ∅ ∈ (𝐶𝑁))))
 
Theoremsadval 16163* The full adder sequence is the half adder function applied to the inputs and the carry sequence. (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (𝑁 ∈ (𝐴 sadd 𝐵) ↔ hadd(𝑁𝐴, 𝑁𝐵, ∅ ∈ (𝐶𝑁))))
 
Theoremsadcaddlem 16164* Lemma for sadcadd 16165. (Contributed by Mario Carneiro, 8-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)    &   𝐾 = (bits ↾ ℕ0)    &   (𝜑 → (∅ ∈ (𝐶𝑁) ↔ (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)))))))
 
Theoremsadcadd 16165* Non-recursive definition of the carry sequence. (Contributed by Mario Carneiro, 8-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)    &   𝐾 = (bits ↾ ℕ0)       (𝜑 → (∅ ∈ (𝐶𝑁) ↔ (2↑𝑁) ≤ ((𝐾‘(𝐴 ∩ (0..^𝑁))) + (𝐾‘(𝐵 ∩ (0..^𝑁))))))
 
Theoremsadadd2lem 16166* Lemma for sadadd2 16167. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)    &   𝐾 = (bits ↾ ℕ0)    &   (𝜑 → ((𝐾‘((𝐴 sadd 𝐵) ∩ (0..^𝑁))) + if(∅ ∈ (𝐶𝑁), (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))))))
 
Theoremsadadd2 16167* Sum of initial segments of the sadd sequence. (Contributed by Mario Carneiro, 8-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)    &   𝐾 = (bits ↾ ℕ0)       (𝜑 → ((𝐾‘((𝐴 sadd 𝐵) ∩ (0..^𝑁))) + if(∅ ∈ (𝐶𝑁), (2↑𝑁), 0)) = ((𝐾‘(𝐴 ∩ (0..^𝑁))) + (𝐾‘(𝐵 ∩ (0..^𝑁)))))
 
Theoremsadadd3 16168* Sum of initial segments of the sadd sequence. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ 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)))
 
Theoremsadcl 16169 The sum of two sequences is a sequence. (Contributed by Mario Carneiro, 5-Sep-2016.)
((𝐴 ⊆ ℕ0𝐵 ⊆ ℕ0) → (𝐴 sadd 𝐵) ⊆ ℕ0)
 
Theoremsadcom 16170 The adder sequence function is commutative. (Contributed by Mario Carneiro, 5-Sep-2016.)
((𝐴 ⊆ ℕ0𝐵 ⊆ ℕ0) → (𝐴 sadd 𝐵) = (𝐵 sadd 𝐴))
 
Theoremsaddisjlem 16171* Lemma for sadadd 16174. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   (𝜑 → (𝐴𝐵) = ∅)    &   𝐶 = seq0((𝑐 ∈ 2o, 𝑚 ∈ ℕ0 ↦ if(cadd(𝑚𝐴, 𝑚𝐵, ∅ ∈ 𝑐), 1o, ∅)), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (𝑁 ∈ (𝐴 sadd 𝐵) ↔ 𝑁 ∈ (𝐴𝐵)))
 
Theoremsaddisj 16172 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.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   (𝜑 → (𝐴𝐵) = ∅)       (𝜑 → (𝐴 sadd 𝐵) = (𝐴𝐵))
 
Theoremsadaddlem 16173* Lemma for sadadd 16174. (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)    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (((bits‘𝐴) sadd (bits‘𝐵)) ∩ (0..^𝑁)) = (bits‘((𝐴 + 𝐵) mod (2↑𝑁))))
 
Theoremsadadd 16174 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 1595 and df-cad 1609.

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)))
 
Theoremsadid1 16175 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)
 
Theoremsadid2 16176 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)
 
Theoremsadasslem 16177 Lemma for sadass 16178. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   (𝜑𝐶 ⊆ ℕ0)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (((𝐴 sadd 𝐵) sadd 𝐶) ∩ (0..^𝑁)) = ((𝐴 sadd (𝐵 sadd 𝐶)) ∩ (0..^𝑁)))
 
Theoremsadass 16178 Sequence addition is associative. (Contributed by Mario Carneiro, 9-Sep-2016.)
((𝐴 ⊆ ℕ0𝐵 ⊆ ℕ0𝐶 ⊆ ℕ0) → ((𝐴 sadd 𝐵) sadd 𝐶) = (𝐴 sadd (𝐵 sadd 𝐶)))
 
Theoremsadeq 16179 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.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → ((𝐴 sadd 𝐵) ∩ (0..^𝑁)) = (((𝐴 ∩ (0..^𝑁)) sadd (𝐵 ∩ (0..^𝑁))) ∩ (0..^𝑁)))
 
Theorembitsres 16180 Restrict the bits of a number to an upper integer set. (Contributed by Mario Carneiro, 5-Sep-2016.)
((𝐴 ∈ ℤ ∧ 𝑁 ∈ ℕ0) → ((bits‘𝐴) ∩ (ℤ𝑁)) = (bits‘((⌊‘(𝐴 / (2↑𝑁))) · (2↑𝑁))))
 
Theorembitsuz 16181 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) ⊆ (ℤ𝑁)))
 
Theorembitsshft 16182* 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 ∣ (𝑛𝑁) ∈ (bits‘𝐴)} = (bits‘(𝐴 · (2↑𝑁))))
 
Definitiondf-smu 16183* 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 ∈ ℕ0𝑘 ∈ (seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝑥 ∧ (𝑛𝑚) ∈ 𝑦)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))‘(𝑘 + 1))})
 
Theoremsmufval 16184* The multiplication of two bit sequences as repeated sequence addition. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))       (𝜑 → (𝐴 smul 𝐵) = {𝑘 ∈ ℕ0𝑘 ∈ (𝑃‘(𝑘 + 1))})
 
Theoremsmupf 16185* The sequence of partial sums of the sequence multiplication. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))       (𝜑𝑃:ℕ0⟶𝒫 ℕ0)
 
Theoremsmup0 16186* The initial element of the partial sum sequence. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))       (𝜑 → (𝑃‘0) = ∅)
 
Theoremsmupp1 16187* The initial element of the partial sum sequence. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (𝑃‘(𝑁 + 1)) = ((𝑃𝑁) sadd {𝑛 ∈ ℕ0 ∣ (𝑁𝐴 ∧ (𝑛𝑁) ∈ 𝐵)}))
 
Theoremsmuval 16188* Define the addition of two bit sequences, using df-had 1595 and df-cad 1609 bit operations. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (𝑁 ∈ (𝐴 smul 𝐵) ↔ 𝑁 ∈ (𝑃‘(𝑁 + 1))))
 
Theoremsmuval2 16189* 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.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)    &   (𝜑𝑀 ∈ (ℤ‘(𝑁 + 1)))       (𝜑 → (𝑁 ∈ (𝐴 smul 𝐵) ↔ 𝑁 ∈ (𝑃𝑀)))
 
Theoremsmupvallem 16190* 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.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)    &   (𝜑𝐴 ⊆ (0..^𝑁))    &   (𝜑𝑀 ∈ (ℤ𝑁))       (𝜑 → (𝑃𝑀) = (𝐴 smul 𝐵))
 
Theoremsmucl 16191 The product of two sequences is a sequence. (Contributed by Mario Carneiro, 19-Sep-2016.)
((𝐴 ⊆ ℕ0𝐵 ⊆ ℕ0) → (𝐴 smul 𝐵) ⊆ ℕ0)
 
Theoremsmu01lem 16192* Lemma for smu01 16193 and smu02 16194. (Contributed by Mario Carneiro, 19-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   ((𝜑 ∧ (𝑘 ∈ ℕ0𝑛 ∈ ℕ0)) → ¬ (𝑘𝐴 ∧ (𝑛𝑘) ∈ 𝐵))       (𝜑 → (𝐴 smul 𝐵) = ∅)
 
Theoremsmu01 16193 Multiplication of a sequence by 0 on the right. (Contributed by Mario Carneiro, 19-Sep-2016.)
(𝐴 ⊆ ℕ0 → (𝐴 smul ∅) = ∅)
 
Theoremsmu02 16194 Multiplication of a sequence by 0 on the left. (Contributed by Mario Carneiro, 9-Sep-2016.)
(𝐴 ⊆ ℕ0 → (∅ smul 𝐴) = ∅)
 
Theoremsmupval 16195* Rewrite the elements of the partial sum sequence in terms of sequence multiplication. (Contributed by Mario Carneiro, 20-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (𝑃𝑁) = ((𝐴 ∩ (0..^𝑁)) smul 𝐵))
 
Theoremsmup1 16196* Rewrite smupp1 16187 using only smul instead of the internal recursive function 𝑃. (Contributed by Mario Carneiro, 20-Sep-2016.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → ((𝐴 ∩ (0..^(𝑁 + 1))) smul 𝐵) = (((𝐴 ∩ (0..^𝑁)) smul 𝐵) sadd {𝑛 ∈ ℕ0 ∣ (𝑁𝐴 ∧ (𝑛𝑁) ∈ 𝐵)}))
 
Theoremsmueqlem 16197* 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.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   (𝜑𝑁 ∈ ℕ0)    &   𝑃 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ 𝐵)})), (𝑛 ∈ ℕ0 ↦ if(𝑛 = 0, ∅, (𝑛 − 1))))    &   𝑄 = seq0((𝑝 ∈ 𝒫 ℕ0, 𝑚 ∈ ℕ0 ↦ (𝑝 sadd {𝑛 ∈ ℕ0 ∣ (𝑚𝐴 ∧ (𝑛𝑚) ∈ (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)))
 
Theoremsmueq 16198 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.)
(𝜑𝐴 ⊆ ℕ0)    &   (𝜑𝐵 ⊆ ℕ0)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → ((𝐴 smul 𝐵) ∩ (0..^𝑁)) = (((𝐴 ∩ (0..^𝑁)) smul (𝐵 ∩ (0..^𝑁))) ∩ (0..^𝑁)))
 
Theoremsmumullem 16199 Lemma for smumul 16200. (Contributed by Mario Carneiro, 22-Sep-2016.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (((bits‘𝐴) ∩ (0..^𝑁)) smul (bits‘𝐵)) = (bits‘((𝐴 mod (2↑𝑁)) · 𝐵)))
 
Theoremsmumul 16200 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 16158, whose correctness is verified in sadadd 16174.

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)))
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