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	nnoddm1d2 16401	A positive integer is odd iff its successor divided by 2 is a positive integer. (Contributed by AV, 28-Jun-2021.)
⊢ (𝑁 ∈ ℕ → (¬ 2 ∥ 𝑁 ↔ ((𝑁 + 1) / 2) ∈ ℕ))
Theorem	sumeven 16402*	If every term in a sum is even, then so is the sum. (Contributed by AV, 14-Aug-2021.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℤ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 2 ∥ 𝐵)    ⇒   ⊢ (𝜑 → 2 ∥ Σ𝑘 ∈ 𝐴 𝐵)
Theorem	sumodd 16403*	If every term in a sum is odd, then the sum is even iff the number of terms in the sum is even. (Contributed by AV, 14-Aug-2021.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℤ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → ¬ 2 ∥ 𝐵)    ⇒   ⊢ (𝜑 → (2 ∥ (♯‘𝐴) ↔ 2 ∥ Σ𝑘 ∈ 𝐴 𝐵))
Theorem	evensumodd 16404*	If every term in a sum with an even number of terms is odd, then the sum is even. (Contributed by AV, 14-Aug-2021.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℤ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → ¬ 2 ∥ 𝐵)    &   ⊢ (𝜑 → 2 ∥ (♯‘𝐴))    ⇒   ⊢ (𝜑 → 2 ∥ Σ𝑘 ∈ 𝐴 𝐵)
Theorem	oddsumodd 16405*	If every term in a sum with an odd number of terms is odd, then the sum is odd. (Contributed by AV, 14-Aug-2021.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℤ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → ¬ 2 ∥ 𝐵)    &   ⊢ (𝜑 → ¬ 2 ∥ (♯‘𝐴))    ⇒   ⊢ (𝜑 → ¬ 2 ∥ Σ𝑘 ∈ 𝐴 𝐵)
Theorem	pwp1fsum 16406*	The n-th power of a number increased by 1 expressed by a product with a finite sum. (Contributed by AV, 15-Aug-2021.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    ⇒   ⊢ (𝜑 → (((-1↑(𝑁 − 1)) · (𝐴↑𝑁)) + 1) = ((𝐴 + 1) · Σ𝑘 ∈ (0...(𝑁 − 1))((-1↑𝑘) · (𝐴↑𝑘))))
Theorem	oddpwp1fsum 16407*	An odd power of a number increased by 1 expressed by a product with a finite sum. (Contributed by AV, 15-Aug-2021.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝑁)    ⇒   ⊢ (𝜑 → ((𝐴↑𝑁) + 1) = ((𝐴 + 1) · Σ𝑘 ∈ (0...(𝑁 − 1))((-1↑𝑘) · (𝐴↑𝑘))))
6.1.5  The division algorithm
Theorem	divalglem0 16408	Lemma for divalg 16418. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ 𝑁 ∈ ℤ    &   ⊢ 𝐷 ∈ ℤ    ⇒   ⊢ ((𝑅 ∈ ℤ ∧ 𝐾 ∈ ℤ) → (𝐷 ∥ (𝑁 − 𝑅) → 𝐷 ∥ (𝑁 − (𝑅 − (𝐾 · (abs‘𝐷))))))
Theorem	divalglem1 16409	Lemma for divalg 16418. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ 𝑁 ∈ ℤ    &   ⊢ 𝐷 ∈ ℤ    &   ⊢ 𝐷 ≠ 0    ⇒   ⊢ 0 ≤ (𝑁 + (abs‘(𝑁 · 𝐷)))
Theorem	divalglem2 16410*	Lemma for divalg 16418. (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 16411*	Lemma for divalg 16418. (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 16412*	Lemma for divalg 16418. (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 16413	Lemma for divalg 16418. (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 16414	Lemma for divalg 16418. (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 16415*	Lemma for divalg 16418. (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 16416*	Lemma for divalg 16418. (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 16417*	Lemma for divalg 16418. (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 16418*	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 16419*	Express the division algorithm as stated in divalg 16418 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 16420*	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 16421	The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor (compare divalg2 16420 and divalgb 16419). 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 16422	The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor. Variant of divalgmod 16421. (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 16423*	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 16424	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 16425	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 16426	Special case of ndvdsadd 16425. 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 16427	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	5ndvds3 16428	5 does not divide 3. (Contributed by AV, 8-Sep-2025.)
⊢ ¬ 5 ∥ 3
Theorem	5ndvds6 16429	5 does not divide 6. (Contributed by AV, 8-Sep-2025.)
⊢ ¬ 5 ∥ 6
Theorem	flodddiv4 16430	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 16431	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 16432	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 16433	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 16434	Define the binary bits of an integer.
class bits
Syntax	csad 16435	Define the sequence addition on bit sequences.
class sadd
Syntax	csmu 16436	Define the sequence multiplication on bit sequences.
class smul
Definition	df-bits 16437*	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 16438*	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 16439	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 16440	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 16441	The set of bits of an integer is a subset of ℕ0. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (bits‘𝑁) ⊆ ℕ0
Theorem	bitsf 16442	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 16443	Value of the zeroth bit. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℤ → (0 ∈ (bits‘𝑁) ↔ ¬ 2 ∥ 𝑁))
Theorem	bits0e 16444	The zeroth bit of an even number is zero. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℤ → ¬ 0 ∈ (bits‘(2 · 𝑁)))
Theorem	bits0o 16445	The zeroth bit of an odd number is one. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℤ → 0 ∈ (bits‘((2 · 𝑁) + 1)))
Theorem	bitsp1 16446	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 16447	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 16448	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 16449*	Lemma for bitsfzo 16450. (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 16450	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 16451	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 16452	Every number is associated with a finite set of bits. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℕ0 → (bits‘𝑁) ∈ Fin)
Theorem	bitscmp 16453	The bit complement of 𝑁 is -𝑁 − 1. (Thus, by bitsfi 16452, all negative numbers have cofinite bits representations.) (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℤ → (ℕ0 ∖ (bits‘𝑁)) = (bits‘(-𝑁 − 1)))
Theorem	0bits 16454	The bits of zero. (Contributed by Mario Carneiro, 6-Sep-2016.)
⊢ (bits‘0) = ∅
Theorem	m1bits 16455	The bits of negative one. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (bits‘-1) = ℕ0
Theorem	bitsinv1lem 16456	Lemma for bitsinv1 16457. (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 16457*	There is an explicit inverse to the bits function for nonnegative integers (which can be extended to negative integers using bitscmp 16453), part 1. (Contributed by Mario Carneiro, 7-Sep-2016.)
⊢ (𝑁 ∈ ℕ0 → Σ𝑛 ∈ (bits‘𝑁)(2↑𝑛) = 𝑁)
Theorem	bitsinv2 16458*	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 16459*	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 15839. (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 16460	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 15839. (Contributed by Mario Carneiro, 8-Sep-2016.)
⊢ (bits ↾ ℕ0):ℕ0–1-1-onto→(𝒫 ℕ0 ∩ Fin)
Theorem	bitsf1 16461	The bits function is an injection from ℤ to 𝒫 ℕ0. It is obviously not a bijection (by Cantor's theorem canth2 9096), 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 16462	The bits of a power of two. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℕ0 → (bits‘(2↑𝑁)) = {𝑁})
Theorem	bitsinv 16463*	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 16464	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 16465	The core of the proof of sadadd2 16475. 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 16466*	Define the addition of two bit sequences, using df-had 1613 and df-cad 1626 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 16467*	Define the addition of two bit sequences, using df-had 1613 and df-cad 1626 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 16468*	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 16469*	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 16470*	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 16471*	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 16472*	Lemma for sadcadd 16473. (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 16473*	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 16474*	Lemma for sadadd2 16475. (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 16475*	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 16476*	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 16477	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 16478	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 16479*	Lemma for sadadd 16482. (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 16480	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 16481*	Lemma for sadadd 16482. (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 16482	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 1613 and df-cad 1626. 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 16483	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 16484	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 16485	Lemma for sadass 16486. (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 16486	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 16487	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 16488	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 16489	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 16490*	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 16491*	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 16492*	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 16493*	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 16494*	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 16495*	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 16496*	Define the addition of two bit sequences, using df-had 1613 and df-cad 1626 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 16497*	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 16498*	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 16499	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 16500*	Lemma for smu01 16501 and smu02 16502. (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) = ∅)