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Theorem List for Metamath Proof Explorer - 14901-15000   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremzeneo 14901 No even integer equals an odd integer (i.e. no integer can be both even and odd). Exercise 10(a) of [Apostol] p. 28. This variant of zneo 11336 follows immediately from the fact that a contradiction implies anything, see pm2.21i 115. (Contributed by AV, 22-Jun-2021.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ((2 ∥ 𝐴 ∧ ¬ 2 ∥ 𝐵) → 𝐴𝐵))

Theoremodd2np1lem 14902* Lemma for odd2np1 14903. (Contributed by Scott Fenton, 3-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(𝑁 ∈ ℕ0 → (∃𝑛 ∈ ℤ ((2 · 𝑛) + 1) = 𝑁 ∨ ∃𝑘 ∈ ℤ (𝑘 · 2) = 𝑁))

Theoremodd2np1 14903* An integer is odd iff it is one plus twice another integer. (Contributed by Scott Fenton, 3-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(𝑁 ∈ ℤ → (¬ 2 ∥ 𝑁 ↔ ∃𝑛 ∈ ℤ ((2 · 𝑛) + 1) = 𝑁))

Theoremeven2n 14904* An integer is even iff it is twice another integer. (Contributed by AV, 25-Jun-2020.)
(2 ∥ 𝑁 ↔ ∃𝑛 ∈ ℤ (2 · 𝑛) = 𝑁)

Theoremoddm1even 14905 An integer is odd iff its predecessor is even. (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝑁 ∈ ℤ → (¬ 2 ∥ 𝑁 ↔ 2 ∥ (𝑁 − 1)))

Theoremoddp1even 14906 An integer is odd iff its successor is even. (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝑁 ∈ ℤ → (¬ 2 ∥ 𝑁 ↔ 2 ∥ (𝑁 + 1)))

Theoremoexpneg 14907 The exponential of the negative of a number, when the exponent is odd. (Contributed by Mario Carneiro, 25-Apr-2015.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ ∧ ¬ 2 ∥ 𝑁) → (-𝐴𝑁) = -(𝐴𝑁))

Theoremmod2eq0even 14908 An integer is 0 modulo 2 iff it is even (i.e. divisible by 2), see example 2 in [ApostolNT] p. 107. (Contributed by AV, 21-Jul-2021.)
(𝑁 ∈ ℤ → ((𝑁 mod 2) = 0 ↔ 2 ∥ 𝑁))

Theoremmod2eq1n2dvds 14909 An integer is 1 modulo 2 iff it is odd (i.e. not divisible by 2), see example 3 in [ApostolNT] p. 107. (Contributed by AV, 24-May-2020.) (Proof shortened by AV, 5-Jul-2020.)
(𝑁 ∈ ℤ → ((𝑁 mod 2) = 1 ↔ ¬ 2 ∥ 𝑁))

Theoremoddnn02np1 14910* A nonnegative integer is odd iff it is one plus twice another nonnegative integer. (Contributed by AV, 19-Jun-2021.)
(𝑁 ∈ ℕ0 → (¬ 2 ∥ 𝑁 ↔ ∃𝑛 ∈ ℕ0 ((2 · 𝑛) + 1) = 𝑁))

Theoremoddge22np1 14911* An integer greater than one is odd iff it is one plus twice a positive integer. (Contributed by AV, 16-Aug-2021.)
(𝑁 ∈ (ℤ‘2) → (¬ 2 ∥ 𝑁 ↔ ∃𝑛 ∈ ℕ ((2 · 𝑛) + 1) = 𝑁))

Theoremevennn02n 14912* A nonnegative integer is even iff it is twice another nonnegative integer. (Contributed by AV, 12-Aug-2021.)
(𝑁 ∈ ℕ0 → (2 ∥ 𝑁 ↔ ∃𝑛 ∈ ℕ0 (2 · 𝑛) = 𝑁))

Theoremevennn2n 14913* A positive integer is even iff it is twice another positive integer. (Contributed by AV, 12-Aug-2021.)
(𝑁 ∈ ℕ → (2 ∥ 𝑁 ↔ ∃𝑛 ∈ ℕ (2 · 𝑛) = 𝑁))

Theorem2tp1odd 14914 A number which is twice an integer increased by 1 is odd. (Contributed by AV, 16-Jul-2021.)
((𝐴 ∈ ℤ ∧ 𝐵 = ((2 · 𝐴) + 1)) → ¬ 2 ∥ 𝐵)

Theoremmulsucdiv2z 14915 An integer multiplied with its successor divided by 2 yields an integer, i.e. an integer multiplied with its successor is even. (Contributed by AV, 19-Jul-2021.)
(𝑁 ∈ ℤ → ((𝑁 · (𝑁 + 1)) / 2) ∈ ℤ)

Theoremsqoddm1div8z 14916 A squared odd number minus 1 divided by 8 is an integer. (Contributed by AV, 19-Jul-2021.)
((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → (((𝑁↑2) − 1) / 8) ∈ ℤ)

Theorem2teven 14917 A number which is twice an integer is even. (Contributed by AV, 16-Jul-2021.)
((𝐴 ∈ ℤ ∧ 𝐵 = (2 · 𝐴)) → 2 ∥ 𝐵)

Theoremzeo5 14918 An integer is either even or odd, version of zeo3 14899 avoiding the negation of the representation of an odd number. (Proposed by BJ, 21-Jun-2021.) (Contributed by AV, 26-Jun-2020.)
(𝑁 ∈ ℤ → (2 ∥ 𝑁 ∨ 2 ∥ (𝑁 + 1)))

Theoremevend2 14919 An integer is even iff its quotient with 2 is an integer. This is a representation of even numbers without using the divides relation, see zeo 11339 and zeo2 11340. (Contributed by AV, 22-Jun-2021.)
(𝑁 ∈ ℤ → (2 ∥ 𝑁 ↔ (𝑁 / 2) ∈ ℤ))

Theoremoddp1d2 14920 An integer is odd iff its successor divided by 2 is an integer. This is a representation of odd numbers without using the divides relation, see zeo 11339 and zeo2 11340. (Contributed by AV, 22-Jun-2021.)
(𝑁 ∈ ℤ → (¬ 2 ∥ 𝑁 ↔ ((𝑁 + 1) / 2) ∈ ℤ))

Theoremzob 14921 Alternate characterizations of an odd number. (Contributed by AV, 7-Jun-2020.)
(𝑁 ∈ ℤ → (((𝑁 + 1) / 2) ∈ ℤ ↔ ((𝑁 − 1) / 2) ∈ ℤ))

Theoremoddm1d2 14922 An integer is odd iff its predecessor divided by 2 is an integer. This is another representation of odd numbers without using the divides relation. (Contributed by AV, 18-Jun-2021.) (Proof shortened by AV, 22-Jun-2021.)
(𝑁 ∈ ℤ → (¬ 2 ∥ 𝑁 ↔ ((𝑁 − 1) / 2) ∈ ℤ))

Theoremltoddhalfle 14923 An integer is less than half of an odd number iff it is less than or equal to the half of the predecessor of the odd number (which is an even number). (Contributed by AV, 29-Jun-2021.)
((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁𝑀 ∈ ℤ) → (𝑀 < (𝑁 / 2) ↔ 𝑀 ≤ ((𝑁 − 1) / 2)))

Theoremhalfleoddlt 14924 An integer is greater than half of an odd number iff it is greater than or equal to the half of the odd number. (Contributed by AV, 1-Jul-2021.)
((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁𝑀 ∈ ℤ) → ((𝑁 / 2) ≤ 𝑀 ↔ (𝑁 / 2) < 𝑀))

Theoremopoe 14925 The sum 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))

Theoremomoe 14926 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 ∥ (𝐴𝐵))

Theoremopeo 14927 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))

Theoremomeo 14928 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 ∥ (𝐴𝐵))

Theoremm1expe 14929 Exponentiation of -1 by an even power. Variant of m1expeven 12769. (Contributed by AV, 25-Jun-2021.)
(2 ∥ 𝑁 → (-1↑𝑁) = 1)

Theoremm1expo 14930 Exponentiation of -1 by an odd power. (Contributed by AV, 26-Jun-2021.)
((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → (-1↑𝑁) = -1)

Theoremm1exp1 14931 Exponentiation of negative one is one iff the exponent is even. (Contributed by AV, 20-Jun-2021.)
(𝑁 ∈ ℤ → ((-1↑𝑁) = 1 ↔ 2 ∥ 𝑁))

Theoremnn0enne 14932 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))

Theoremnn0ehalf 14933 The half of an even nonnegative integer is a nonnegative integer. (Contributed by AV, 22-Jun-2020.) (Revised by AV, 28-Jun-2021.)
((𝑁 ∈ ℕ0 ∧ 2 ∥ 𝑁) → (𝑁 / 2) ∈ ℕ0)

Theoremnnehalf 14934 The half of an even positive integer is a positive integer. (Contributed by AV, 28-Jun-2021.)
((𝑁 ∈ ℕ ∧ 2 ∥ 𝑁) → (𝑁 / 2) ∈ ℕ)

Theoremnn0o1gt2 14935 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))

Theoremnno 14936 An alternate characterization of an odd integer greater than 1. (Contributed by AV, 2-Jun-2020.)
((𝑁 ∈ (ℤ‘2) ∧ ((𝑁 + 1) / 2) ∈ ℕ0) → ((𝑁 − 1) / 2) ∈ ℕ)

Theoremnn0o 14937 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)

Theoremnn0ob 14938 Alternate characterizations of an odd nonnegative integer. (Contributed by AV, 4-Jun-2020.)
(𝑁 ∈ ℕ0 → (((𝑁 + 1) / 2) ∈ ℕ0 ↔ ((𝑁 − 1) / 2) ∈ ℕ0))

Theoremnn0oddm1d2 14939 A positive integer is odd iff its predecessor divided by 2 is a positive integer. (Contributed by AV, 28-Jun-2021.)
(𝑁 ∈ ℕ0 → (¬ 2 ∥ 𝑁 ↔ ((𝑁 − 1) / 2) ∈ ℕ0))

Theoremnnoddm1d2 14940 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))

Theoremz0even 14941 0 is even. (Contributed by AV, 11-Feb-2020.) (Revised by AV, 23-Jun-2021.)
2 ∥ 0

Theoremn2dvds1 14942 2 does not divide 1 (common case). That means 1 is odd. (Contributed by David A. Wheeler, 8-Dec-2018.)
¬ 2 ∥ 1

Theoremn2dvdsm1 14943 2 does not divide -1. That means -1 is odd. (Contributed by AV, 15-Aug-2021.)
¬ 2 ∥ -1

Theoremz2even 14944 2 is even. (Contributed by AV, 12-Feb-2020.) (Revised by AV, 23-Jun-2021.)
2 ∥ 2

Theoremn2dvds3 14945 2 does not divide 3, i.e. 3 is an odd number. (Contributed by AV, 28-Feb-2021.)
¬ 2 ∥ 3

Theoremz4even 14946 4 is an even number. (Contributed by AV, 23-Jul-2020.) (Revised by AV, 4-Jul-2021.)
2 ∥ 4

Theorem4dvdseven 14947 An integer which is divisible by 4 is an even integer. (Contributed by AV, 4-Jul-2021.)
(4 ∥ 𝑁 → 2 ∥ 𝑁)

Theoremsumeven 14948* If every term in a sum is even, then so is the sum. (Contributed by AV, 14-Aug-2021.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℤ)    &   ((𝜑𝑘𝐴) → 2 ∥ 𝐵)       (𝜑 → 2 ∥ Σ𝑘𝐴 𝐵)

Theoremsumodd 14949* 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.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℤ)    &   ((𝜑𝑘𝐴) → ¬ 2 ∥ 𝐵)       (𝜑 → (2 ∥ (#‘𝐴) ↔ 2 ∥ Σ𝑘𝐴 𝐵))

Theoremevensumodd 14950* 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.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℤ)    &   ((𝜑𝑘𝐴) → ¬ 2 ∥ 𝐵)    &   (𝜑 → 2 ∥ (#‘𝐴))       (𝜑 → 2 ∥ Σ𝑘𝐴 𝐵)

Theoremoddsumodd 14951* 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.)
(𝜑𝐴 ∈ Fin)    &   ((𝜑𝑘𝐴) → 𝐵 ∈ ℤ)    &   ((𝜑𝑘𝐴) → ¬ 2 ∥ 𝐵)    &   (𝜑 → ¬ 2 ∥ (#‘𝐴))       (𝜑 → ¬ 2 ∥ Σ𝑘𝐴 𝐵)

Theorempwp1fsum 14952* The n-th 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 → (((-1↑(𝑁 − 1)) · (𝐴𝑁)) + 1) = ((𝐴 + 1) · Σ𝑘 ∈ (0...(𝑁 − 1))((-1↑𝑘) · (𝐴𝑘))))

Theoremoddpwp1fsum 14953* 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 14954 Lemma for divalg 14964. (Contributed by Paul Chapman, 21-Mar-2011.)
𝑁 ∈ ℤ    &   𝐷 ∈ ℤ       ((𝑅 ∈ ℤ ∧ 𝐾 ∈ ℤ) → (𝐷 ∥ (𝑁𝑅) → 𝐷 ∥ (𝑁 − (𝑅 − (𝐾 · (abs‘𝐷))))))

Theoremdivalglem1 14955 Lemma for divalg 14964. (Contributed by Paul Chapman, 21-Mar-2011.)
𝑁 ∈ ℤ    &   𝐷 ∈ ℤ    &   𝐷 ≠ 0       0 ≤ (𝑁 + (abs‘(𝑁 · 𝐷)))

Theoremdivalglem2 14956* Lemma for divalg 14964. (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 14957* Lemma for divalg 14964. (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 14958* Lemma for divalg 14964. (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 14959 Lemma for divalg 14964. (Contributed by Paul Chapman, 21-Mar-2011.)
𝐴 ∈ ℕ    &   𝑋 ∈ (0...(𝐴 − 1))    &   𝐾 ∈ ℤ       (𝐾 ≠ 0 → ¬ (𝑋 + (𝐾 · 𝐴)) ∈ (0...(𝐴 − 1)))

Theoremdivalglem7 14960 Lemma for divalg 14964. (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 14961* Lemma for divalg 14964. (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 14962* Lemma for divalg 14964. (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 14963* Lemma for divalg 14964. (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 14964* 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 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 14965* Express the division algorithm as stated in divalg 14964 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 14966* 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 14967 The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor (compare divalg2 14966 and divalgb 14965). 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)))))

TheoremdivalgmodOLD 14968* Obsolete proof of divalgmod 14967 as of 21-Aug-2021. (Contributed by Paul Chapman, 31-Mar-2011.) (New usage is discouraged.) (Proof modification is discouraged.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ) → (𝑟 = (𝑁 mod 𝐷) ↔ (𝑟 ∈ ℕ0 ∧ (𝑟 < 𝐷𝐷 ∥ (𝑁𝑟)))))

Theoremdivalgmodcl 14969 The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor. Variant of divalgmod 14967. (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 14970* 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 14971 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 14972 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 14973 Special case of ndvdsadd 14972. 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 14974 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 14975 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 14976 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 14977 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 14978 The floor of an odd number divided by 4, multiplied by 2 is less than the half of the odd number. (Contributed by AV, 4-Jul-2021.)
((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → ((⌊‘(𝑁 / 4)) · 2) < (𝑁 / 2))

6.1.6  Bit sequences

Syntaxcbits 14979 Define the binary bits of an integer.
class bits

Syntaxcsmu 14981 Define the sequence multiplication on bit sequences.
class smul

Definitiondf-bits 14982* 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 14983* 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 14984 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 14985 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 14986 The set of bits of an integer is a subset of 0. (Contributed by Mario Carneiro, 5-Sep-2016.)
(bits‘𝑁) ⊆ ℕ0

Theorembitsf 14987 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 14988 Value of the zeroth bit. (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝑁 ∈ ℤ → (0 ∈ (bits‘𝑁) ↔ ¬ 2 ∥ 𝑁))

Theorembits0e 14989 The zeroth bit of an even number is zero. (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝑁 ∈ ℤ → ¬ 0 ∈ (bits‘(2 · 𝑁)))

Theorembits0o 14990 The zeroth bit of an odd number is zero. (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝑁 ∈ ℤ → 0 ∈ (bits‘((2 · 𝑁) + 1)))

Theorembitsp1 14991 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 14992 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 14993 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 14994* Lemma for bitsfzo 14995. (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 14995 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 14996 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 14997 Every number is associated with a finite set of bits. (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝑁 ∈ ℕ0 → (bits‘𝑁) ∈ Fin)

Theorembitscmp 14998 The bit complement of 𝑁 is -𝑁 − 1. (Thus, by bitsfi 14997, all negative numbers have cofinite bits representations.) (Contributed by Mario Carneiro, 5-Sep-2016.)
(𝑁 ∈ ℤ → (ℕ0 ∖ (bits‘𝑁)) = (bits‘(-𝑁 − 1)))

Theorem0bits 14999 The bits of zero. (Contributed by Mario Carneiro, 6-Sep-2016.)
(bits‘0) = ∅

Theoremm1bits 15000 The bits of negative one. (Contributed by Mario Carneiro, 5-Sep-2016.)
(bits‘-1) = ℕ0

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