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Type | Label | Description |
---|---|---|
Statement | ||
Theorem | mulsucdiv2z 16401 | 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) | ||
Theorem | sqoddm1div8z 16402 | 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) | ||
Theorem | 2teven 16403 | A number which is twice an integer is even. (Contributed by AV, 16-Jul-2021.) |
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 = (2 · 𝐴)) → 2 ∥ 𝐵) | ||
Theorem | zeo5 16404 | An integer is either even or odd, version of zeo3 16385 avoiding the negation of the representation of an odd number. (Proposed by BJ, 21-Jun-2021.) (Contributed by AV, 26-Jun-2021.) |
⊢ (𝑁 ∈ ℤ → (2 ∥ 𝑁 ∨ 2 ∥ (𝑁 + 1))) | ||
Theorem | evend2 16405 | 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 12729 and zeo2 12730. (Contributed by AV, 22-Jun-2021.) |
⊢ (𝑁 ∈ ℤ → (2 ∥ 𝑁 ↔ (𝑁 / 2) ∈ ℤ)) | ||
Theorem | oddp1d2 16406 | 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 12729 and zeo2 12730. (Contributed by AV, 22-Jun-2021.) |
⊢ (𝑁 ∈ ℤ → (¬ 2 ∥ 𝑁 ↔ ((𝑁 + 1) / 2) ∈ ℤ)) | ||
Theorem | zob 16407 | Alternate characterizations of an odd number. (Contributed by AV, 7-Jun-2020.) |
⊢ (𝑁 ∈ ℤ → (((𝑁 + 1) / 2) ∈ ℤ ↔ ((𝑁 − 1) / 2) ∈ ℤ)) | ||
Theorem | oddm1d2 16408 | 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)) | ||
Theorem | ltoddhalfle 16409 | 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 ∥ 𝑁 ∧ 𝑀 ∈ ℤ) → (𝑀 < (𝑁 / 2) ↔ 𝑀 ≤ ((𝑁 − 1) / 2))) | ||
Theorem | halfleoddlt 16410 | 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 ∥ 𝑁 ∧ 𝑀 ∈ ℤ) → ((𝑁 / 2) ≤ 𝑀 ↔ (𝑁 / 2) < 𝑀)) | ||
Theorem | opoe 16411 | 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)) | ||
Theorem | omoe 16412 | 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 16413 | 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 16414 | 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 16415 | 2 divides 0. That means 0 is even. (Contributed by AV, 11-Feb-2020.) (Revised by AV, 23-Jun-2021.) |
⊢ 2 ∥ 0 | ||
Theorem | n2dvds1 16416 | 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 16417 | 2 does not divide -1. That means -1 is odd. (Contributed by AV, 15-Aug-2021.) |
⊢ ¬ 2 ∥ -1 | ||
Theorem | z2even 16418 | 2 divides 2. That means 2 is even. (Contributed by AV, 12-Feb-2020.) (Revised by AV, 23-Jun-2021.) |
⊢ 2 ∥ 2 | ||
Theorem | n2dvds3 16419 | 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 16420 | 2 divides 4. That means 4 is even. (Contributed by AV, 23-Jul-2020.) (Revised by AV, 4-Jul-2021.) |
⊢ 2 ∥ 4 | ||
Theorem | 4dvdseven 16421 | 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 16422 | Exponentiation of -1 by an even power. Variant of m1expeven 14160. (Contributed by AV, 25-Jun-2021.) |
⊢ (2 ∥ 𝑁 → (-1↑𝑁) = 1) | ||
Theorem | m1expo 16423 | Exponentiation of -1 by an odd power. (Contributed by AV, 26-Jun-2021.) |
⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → (-1↑𝑁) = -1) | ||
Theorem | m1exp1 16424 | 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 16425 | 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 16426 | 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 16427 | 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 16428 | An odd nonnegative integer is positive. (Contributed by Steven Nguyen, 25-Mar-2023.) |
⊢ ((𝑁 ∈ ℕ0 ∧ ¬ 2 ∥ 𝑁) → 𝑁 ∈ ℕ) | ||
Theorem | nn0o1gt2 16429 | 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 16430 | 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 16431 | 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 16432 | 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 16433 | 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 16434 | 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 16435* | 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 16436* | 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 16437* | 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 16438* | 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 16439* | 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 16440* | 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)))) | ||
Theorem | divalglem0 16441 | Lemma for divalg 16451. (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 16442 | Lemma for divalg 16451. (Contributed by Paul Chapman, 21-Mar-2011.) |
⊢ 𝑁 ∈ ℤ & ⊢ 𝐷 ∈ ℤ & ⊢ 𝐷 ≠ 0 ⇒ ⊢ 0 ≤ (𝑁 + (abs‘(𝑁 · 𝐷))) | ||
Theorem | divalglem2 16443* | Lemma for divalg 16451. (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 16444* | Lemma for divalg 16451. (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 16445* | Lemma for divalg 16451. (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 16446 | Lemma for divalg 16451. (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 16447 | Lemma for divalg 16451. (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 16448* | Lemma for divalg 16451. (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 16449* | Lemma for divalg 16451. (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 16450* | Lemma for divalg 16451. (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 16451* | 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 16452* | Express the division algorithm as stated in divalg 16451 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 16453* | 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 16454 | The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor (compare divalg2 16453 and divalgb 16452). 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 16455 | The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor. Variant of divalgmod 16454. (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 16456* | 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 16457 | 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 16458 | 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 16459 | Special case of ndvdsadd 16458. 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 16460 | A quick test for non-divisibility. (Contributed by Mario Carneiro, 18-Feb-2014.) |
⊢ 𝐴 ∈ ℕ & ⊢ 𝑄 ∈ ℕ0 & ⊢ 𝑅 ∈ ℕ & ⊢ ((𝐴 · 𝑄) + 𝑅) = 𝐵 & ⊢ 𝑅 < 𝐴 ⇒ ⊢ ¬ 𝐴 ∥ 𝐵 | ||
Theorem | flodddiv4 16461 | 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 16462 | 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 16463 | 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 16464 | 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)) | ||
Syntax | cbits 16465 | Define the binary bits of an integer. |
class bits | ||
Syntax | csad 16466 | Define the sequence addition on bit sequences. |
class sadd | ||
Syntax | csmu 16467 | Define the sequence multiplication on bit sequences. |
class smul | ||
Definition | df-bits 16468* | 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 16469* | 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 16470 | 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 16471 | 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 16472 | The set of bits of an integer is a subset of ℕ0. (Contributed by Mario Carneiro, 5-Sep-2016.) |
⊢ (bits‘𝑁) ⊆ ℕ0 | ||
Theorem | bitsf 16473 | 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 16474 | Value of the zeroth bit. (Contributed by Mario Carneiro, 5-Sep-2016.) |
⊢ (𝑁 ∈ ℤ → (0 ∈ (bits‘𝑁) ↔ ¬ 2 ∥ 𝑁)) | ||
Theorem | bits0e 16475 | The zeroth bit of an even number is zero. (Contributed by Mario Carneiro, 5-Sep-2016.) |
⊢ (𝑁 ∈ ℤ → ¬ 0 ∈ (bits‘(2 · 𝑁))) | ||
Theorem | bits0o 16476 | The zeroth bit of an odd number is zero. (Contributed by Mario Carneiro, 5-Sep-2016.) |
⊢ (𝑁 ∈ ℤ → 0 ∈ (bits‘((2 · 𝑁) + 1))) | ||
Theorem | bitsp1 16477 | 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 16478 | 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 16479 | 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 16480* | Lemma for bitsfzo 16481. (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 16481 | The bits of a number are all less than 𝑀 iff the number is nonnegative and less than 2↑𝑀. (Contributed by Mario Carneiro, 5-Sep-2016.) (Proof shortened by AV, 1-Oct-2020.) |
⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℕ0) → (𝑁 ∈ (0..^(2↑𝑀)) ↔ (bits‘𝑁) ⊆ (0..^𝑀))) | ||
Theorem | bitsmod 16482 | 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 16483 | Every number is associated with a finite set of bits. (Contributed by Mario Carneiro, 5-Sep-2016.) |
⊢ (𝑁 ∈ ℕ0 → (bits‘𝑁) ∈ Fin) | ||
Theorem | bitscmp 16484 | The bit complement of 𝑁 is -𝑁 − 1. (Thus, by bitsfi 16483, all negative numbers have cofinite bits representations.) (Contributed by Mario Carneiro, 5-Sep-2016.) |
⊢ (𝑁 ∈ ℤ → (ℕ0 ∖ (bits‘𝑁)) = (bits‘(-𝑁 − 1))) | ||
Theorem | 0bits 16485 | The bits of zero. (Contributed by Mario Carneiro, 6-Sep-2016.) |
⊢ (bits‘0) = ∅ | ||
Theorem | m1bits 16486 | The bits of negative one. (Contributed by Mario Carneiro, 5-Sep-2016.) |
⊢ (bits‘-1) = ℕ0 | ||
Theorem | bitsinv1lem 16487 | Lemma for bitsinv1 16488. (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 16488* | There is an explicit inverse to the bits function for nonnegative integers (which can be extended to negative integers using bitscmp 16484), part 1. (Contributed by Mario Carneiro, 7-Sep-2016.) |
⊢ (𝑁 ∈ ℕ0 → Σ𝑛 ∈ (bits‘𝑁)(2↑𝑛) = 𝑁) | ||
Theorem | bitsinv2 16489* | 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 16490* | 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 15876. (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 16491 | 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 15876. (Contributed by Mario Carneiro, 8-Sep-2016.) |
⊢ (bits ↾ ℕ0):ℕ0–1-1-onto→(𝒫 ℕ0 ∩ Fin) | ||
Theorem | bitsf1 16492 | The bits function is an injection from ℤ to 𝒫 ℕ0. It is obviously not a bijection (by Cantor's theorem canth2 9196), 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 16493 | The bits of a power of two. (Contributed by Mario Carneiro, 5-Sep-2016.) |
⊢ (𝑁 ∈ ℕ0 → (bits‘(2↑𝑁)) = {𝑁}) | ||
Theorem | bitsinv 16494* | 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 16495 | 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 16496 | The core of the proof of sadadd2 16506. 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 16497* | Define the addition of two bit sequences, using df-had 1591 and df-cad 1604 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 16498* | Define the addition of two bit sequences, using df-had 1591 and df-cad 1604 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 16499* | 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 16500* | 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)) |
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