P. 166 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 16501-16600   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	omoe 16501	The difference of two odds is even. (Contributed by Scott Fenton, 7-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ (((𝐴 ∈ ℤ ∧ ¬ 2 ∥ 𝐴) ∧ (𝐵 ∈ ℤ ∧ ¬ 2 ∥ 𝐵)) → 2 ∥ (𝐴 − 𝐵))
Theorem	opeo 16502	The sum of an odd and an even is odd. (Contributed by Scott Fenton, 7-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ (((𝐴 ∈ ℤ ∧ ¬ 2 ∥ 𝐴) ∧ (𝐵 ∈ ℤ ∧ 2 ∥ 𝐵)) → ¬ 2 ∥ (𝐴 + 𝐵))
Theorem	omeo 16503	The difference of an odd and an even is odd. (Contributed by Scott Fenton, 7-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ (((𝐴 ∈ ℤ ∧ ¬ 2 ∥ 𝐴) ∧ (𝐵 ∈ ℤ ∧ 2 ∥ 𝐵)) → ¬ 2 ∥ (𝐴 − 𝐵))
Theorem	z0even 16504	2 divides 0. That means 0 is even. (Contributed by AV, 11-Feb-2020.) (Revised by AV, 23-Jun-2021.)
⊢ 2 ∥ 0
Theorem	n2dvds1 16505	2 does not divide 1. That means 1 is odd. (Contributed by David A. Wheeler, 8-Dec-2018.) (Proof shortened by Steven Nguyen, 3-May-2023.)
⊢ ¬ 2 ∥ 1
Theorem	n2dvdsm1 16506	2 does not divide -1. That means -1 is odd. (Contributed by AV, 15-Aug-2021.)
⊢ ¬ 2 ∥ -1
Theorem	z2even 16507	2 divides 2. That means 2 is even. (Contributed by AV, 12-Feb-2020.) (Revised by AV, 23-Jun-2021.)
⊢ 2 ∥ 2
Theorem	n2dvds3 16508	2 does not divide 3. That means 3 is odd. (Contributed by AV, 28-Feb-2021.) (Proof shortened by Steven Nguyen, 3-May-2023.)
⊢ ¬ 2 ∥ 3
Theorem	z4even 16509	2 divides 4. That means 4 is even. (Contributed by AV, 23-Jul-2020.) (Revised by AV, 4-Jul-2021.)
⊢ 2 ∥ 4
Theorem	4dvdseven 16510	An integer which is divisible by 4 is divisible by 2, that is, is even. (Contributed by AV, 4-Jul-2021.)
⊢ (4 ∥ 𝑁 → 2 ∥ 𝑁)
Theorem	m1expe 16511	Exponentiation of -1 by an even power. Variant of m1expeven 14220. (Contributed by AV, 25-Jun-2021.)
⊢ (2 ∥ 𝑁 → (-1↑𝑁) = 1)
Theorem	m1expo 16512	Exponentiation of -1 by an odd power. (Contributed by AV, 26-Jun-2021.)
⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → (-1↑𝑁) = -1)
Theorem	m1exp1 16513	Exponentiation of negative one is one iff the exponent is even. (Contributed by AV, 20-Jun-2021.)
⊢ (𝑁 ∈ ℤ → ((-1↑𝑁) = 1 ↔ 2 ∥ 𝑁))
Theorem	nn0enne 16514	A positive integer is an even nonnegative integer iff it is an even positive integer. (Contributed by AV, 30-May-2020.)
⊢ (𝑁 ∈ ℕ → ((𝑁 / 2) ∈ ℕ0 ↔ (𝑁 / 2) ∈ ℕ))
Theorem	nn0ehalf 16515	The half of an even nonnegative integer is a nonnegative integer. (Contributed by AV, 22-Jun-2020.) (Revised by AV, 28-Jun-2021.) (Proof shortened by AV, 10-Jul-2022.)
⊢ ((𝑁 ∈ ℕ0 ∧ 2 ∥ 𝑁) → (𝑁 / 2) ∈ ℕ0)
Theorem	nnehalf 16516	The half of an even positive integer is a positive integer. (Contributed by AV, 28-Jun-2021.)
⊢ ((𝑁 ∈ ℕ ∧ 2 ∥ 𝑁) → (𝑁 / 2) ∈ ℕ)
Theorem	nn0onn 16517	An odd nonnegative integer is positive. (Contributed by Steven Nguyen, 25-Mar-2023.)
⊢ ((𝑁 ∈ ℕ0 ∧ ¬ 2 ∥ 𝑁) → 𝑁 ∈ ℕ)
Theorem	nn0o1gt2 16518	An odd nonnegative integer is either 1 or greater than 2. (Contributed by AV, 2-Jun-2020.)
⊢ ((𝑁 ∈ ℕ0 ∧ ((𝑁 + 1) / 2) ∈ ℕ0) → (𝑁 = 1 ∨ 2 < 𝑁))
Theorem	nno 16519	An alternate characterization of an odd integer greater than 1. (Contributed by AV, 2-Jun-2020.) (Proof shortened by AV, 10-Jul-2022.)
⊢ ((𝑁 ∈ (ℤ≥‘2) ∧ ((𝑁 + 1) / 2) ∈ ℕ0) → ((𝑁 − 1) / 2) ∈ ℕ)
Theorem	nn0o 16520	An alternate characterization of an odd nonnegative integer. (Contributed by AV, 28-May-2020.) (Proof shortened by AV, 2-Jun-2020.)
⊢ ((𝑁 ∈ ℕ0 ∧ ((𝑁 + 1) / 2) ∈ ℕ0) → ((𝑁 − 1) / 2) ∈ ℕ0)
Theorem	nn0ob 16521	Alternate characterizations of an odd nonnegative integer. (Contributed by AV, 4-Jun-2020.)
⊢ (𝑁 ∈ ℕ0 → (((𝑁 + 1) / 2) ∈ ℕ0 ↔ ((𝑁 − 1) / 2) ∈ ℕ0))
Theorem	nn0oddm1d2 16522	A positive integer is odd iff its predecessor divided by 2 is a positive integer. (Contributed by AV, 28-Jun-2021.) (Proof shortened by AV, 10-Jul-2022.)
⊢ (𝑁 ∈ ℕ0 → (¬ 2 ∥ 𝑁 ↔ ((𝑁 − 1) / 2) ∈ ℕ0))
Theorem	nnoddm1d2 16523	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 16524*	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 16525*	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 16526*	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 16527*	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 16528*	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 16529*	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 16530	Lemma for divalg 16540. (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 16531	Lemma for divalg 16540. (Contributed by Paul Chapman, 21-Mar-2011.)
⊢ 𝑁 ∈ ℤ    &   ⊢ 𝐷 ∈ ℤ    &   ⊢ 𝐷 ≠ 0    ⇒   ⊢ 0 ≤ (𝑁 + (abs‘(𝑁 · 𝐷)))
Theorem	divalglem2 16532*	Lemma for divalg 16540. (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 16533*	Lemma for divalg 16540. (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 16534*	Lemma for divalg 16540. (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 16535	Lemma for divalg 16540. (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 16536	Lemma for divalg 16540. (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 16537*	Lemma for divalg 16540. (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 16538*	Lemma for divalg 16540. (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 16539*	Lemma for divalg 16540. (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 16540*	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 16541*	Express the division algorithm as stated in divalg 16540 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 16542*	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 16543	The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor (compare divalg2 16542 and divalgb 16541). 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 16544	The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor. Variant of divalgmod 16543. (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 16545*	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 16546	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 16547	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 16548	Special case of ndvdsadd 16547. 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 16549	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 16550	5 does not divide 3. (Contributed by AV, 8-Sep-2025.)
⊢ ¬ 5 ∥ 3
Theorem	5ndvds6 16551	5 does not divide 6. (Contributed by AV, 8-Sep-2025.)
⊢ ¬ 5 ∥ 6
Theorem	flodddiv4 16552	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 16553	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 16554	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 16555	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 16556	Define the binary bits of an integer.
class bits
Syntax	csad 16557	Define the sequence addition on bit sequences.
class sadd
Syntax	csmu 16558	Define the sequence multiplication on bit sequences.
class smul
Definition	df-bits 16559*	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 16560*	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 16561	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 16562	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 16563	The set of bits of an integer is a subset of ℕ0. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (bits‘𝑁) ⊆ ℕ0
Theorem	bitsf 16564	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 16565	Value of the zeroth bit. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℤ → (0 ∈ (bits‘𝑁) ↔ ¬ 2 ∥ 𝑁))
Theorem	bits0e 16566	The zeroth bit of an even number is zero. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℤ → ¬ 0 ∈ (bits‘(2 · 𝑁)))
Theorem	bits0o 16567	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 16568	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 16569	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 16570	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 16571*	Lemma for bitsfzo 16572. (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 16572	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 16573	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 16574	Every number is associated with a finite set of bits. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℕ0 → (bits‘𝑁) ∈ Fin)
Theorem	bitscmp 16575	The bit complement of 𝑁 is -𝑁 − 1. (Thus, by bitsfi 16574, all negative numbers have cofinite bits representations.) (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℤ → (ℕ0 ∖ (bits‘𝑁)) = (bits‘(-𝑁 − 1)))
Theorem	0bits 16576	The bits of zero. (Contributed by Mario Carneiro, 6-Sep-2016.)
⊢ (bits‘0) = ∅
Theorem	m1bits 16577	The bits of negative one. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (bits‘-1) = ℕ0
Theorem	bitsinv1lem 16578	Lemma for bitsinv1 16579. (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 16579*	There is an explicit inverse to the bits function for nonnegative integers (which can be extended to negative integers using bitscmp 16575), part 1. (Contributed by Mario Carneiro, 7-Sep-2016.)
⊢ (𝑁 ∈ ℕ0 → Σ𝑛 ∈ (bits‘𝑁)(2↑𝑛) = 𝑁)
Theorem	bitsinv2 16580*	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 16581*	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 15964. (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 16582	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 15964. (Contributed by Mario Carneiro, 8-Sep-2016.)
⊢ (bits ↾ ℕ0):ℕ0–1-1-onto→(𝒫 ℕ0 ∩ Fin)
Theorem	bitsf1 16583	The bits function is an injection from ℤ to 𝒫 ℕ0. It is obviously not a bijection (by Cantor's theorem canth2 9127), 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 16584	The bits of a power of two. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ (𝑁 ∈ ℕ0 → (bits‘(2↑𝑁)) = {𝑁})
Theorem	bitsinv 16585*	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 16586	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 16587	The core of the proof of sadadd2 16597. 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 16588*	Define the addition of two bit sequences, using df-had 1624 and df-cad 1640 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 16589*	Define the addition of two bit sequences, using df-had 1624 and df-cad 1640 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 16590*	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 16591*	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 16592*	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 16593*	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 16594*	Lemma for sadcadd 16595. (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 16595*	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 16596*	Lemma for sadadd2 16597. (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 16597*	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 16598*	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 16599	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 16600	The adder sequence function is commutative. (Contributed by Mario Carneiro, 5-Sep-2016.)
⊢ ((𝐴 ⊆ ℕ0 ∧ 𝐵 ⊆ ℕ0) → (𝐴 sadd 𝐵) = (𝐵 sadd 𝐴))