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Theorem List for Intuitionistic Logic Explorer - 8301-8400   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremzdivadd 8301 Property of divisibility: if 𝐷 divides 𝐴 and 𝐵 then it divides 𝐴 + 𝐵. (Contributed by NM, 3-Oct-2008.)
(((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) ∧ ((𝐴 / 𝐷) ∈ ℤ ∧ (𝐵 / 𝐷) ∈ ℤ)) → ((𝐴 + 𝐵) / 𝐷) ∈ ℤ)
 
Theoremzdivmul 8302 Property of divisibility: if 𝐷 divides 𝐴 then it divides 𝐵 · 𝐴. (Contributed by NM, 3-Oct-2008.)
(((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) ∧ (𝐴 / 𝐷) ∈ ℤ) → ((𝐵 · 𝐴) / 𝐷) ∈ ℤ)
 
Theoremzextle 8303* An extensionality-like property for integer ordering. (Contributed by NM, 29-Oct-2005.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ ∀𝑘 ∈ ℤ (𝑘𝑀𝑘𝑁)) → 𝑀 = 𝑁)
 
Theoremzextlt 8304* An extensionality-like property for integer ordering. (Contributed by NM, 29-Oct-2005.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ ∀𝑘 ∈ ℤ (𝑘 < 𝑀𝑘 < 𝑁)) → 𝑀 = 𝑁)
 
Theoremrecnz 8305 The reciprocal of a number greater than 1 is not an integer. (Contributed by NM, 3-May-2005.)
((𝐴 ∈ ℝ ∧ 1 < 𝐴) → ¬ (1 / 𝐴) ∈ ℤ)
 
Theorembtwnnz 8306 A number between an integer and its successor is not an integer. (Contributed by NM, 3-May-2005.)
((𝐴 ∈ ℤ ∧ 𝐴 < 𝐵𝐵 < (𝐴 + 1)) → ¬ 𝐵 ∈ ℤ)
 
Theoremgtndiv 8307 A larger number does not divide a smaller positive integer. (Contributed by NM, 3-May-2005.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℕ ∧ 𝐵 < 𝐴) → ¬ (𝐵 / 𝐴) ∈ ℤ)
 
Theoremhalfnz 8308 One-half is not an integer. (Contributed by NM, 31-Jul-2004.)
¬ (1 / 2) ∈ ℤ
 
Theoremprime 8309* Two ways to express "𝐴 is a prime number (or 1)." (Contributed by NM, 4-May-2005.)
(𝐴 ∈ ℕ → (∀𝑥 ∈ ℕ ((𝐴 / 𝑥) ∈ ℕ → (𝑥 = 1 ∨ 𝑥 = 𝐴)) ↔ ∀𝑥 ∈ ℕ ((1 < 𝑥𝑥𝐴 ∧ (𝐴 / 𝑥) ∈ ℕ) → 𝑥 = 𝐴)))
 
Theoremmsqznn 8310 The square of a nonzero integer is a positive integer. (Contributed by NM, 2-Aug-2004.)
((𝐴 ∈ ℤ ∧ 𝐴 ≠ 0) → (𝐴 · 𝐴) ∈ ℕ)
 
Theoremzneo 8311 No even integer equals an odd integer (i.e. no integer can be both even and odd). Exercise 10(a) of [Apostol] p. 28. (Contributed by NM, 31-Jul-2004.) (Proof shortened by Mario Carneiro, 18-May-2014.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → (2 · 𝐴) ≠ ((2 · 𝐵) + 1))
 
Theoremnneoor 8312 A positive integer is even or odd. (Contributed by Jim Kingdon, 15-Mar-2020.)
(𝑁 ∈ ℕ → ((𝑁 / 2) ∈ ℕ ∨ ((𝑁 + 1) / 2) ∈ ℕ))
 
Theoremnneo 8313 A positive integer is even or odd but not both. (Contributed by NM, 1-Jan-2006.) (Proof shortened by Mario Carneiro, 18-May-2014.)
(𝑁 ∈ ℕ → ((𝑁 / 2) ∈ ℕ ↔ ¬ ((𝑁 + 1) / 2) ∈ ℕ))
 
Theoremnneoi 8314 A positive integer is even or odd but not both. (Contributed by NM, 20-Aug-2001.)
𝑁 ∈ ℕ       ((𝑁 / 2) ∈ ℕ ↔ ¬ ((𝑁 + 1) / 2) ∈ ℕ)
 
Theoremzeo 8315 An integer is even or odd. (Contributed by NM, 1-Jan-2006.)
(𝑁 ∈ ℤ → ((𝑁 / 2) ∈ ℤ ∨ ((𝑁 + 1) / 2) ∈ ℤ))
 
Theoremzeo2 8316 An integer is even or odd but not both. (Contributed by Mario Carneiro, 12-Sep-2015.)
(𝑁 ∈ ℤ → ((𝑁 / 2) ∈ ℤ ↔ ¬ ((𝑁 + 1) / 2) ∈ ℤ))
 
Theorempeano2uz2 8317* Second Peano postulate for upper integers. (Contributed by NM, 3-Oct-2004.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ {𝑥 ∈ ℤ ∣ 𝐴𝑥}) → (𝐵 + 1) ∈ {𝑥 ∈ ℤ ∣ 𝐴𝑥})
 
Theorempeano5uzti 8318* Peano's inductive postulate for upper integers. (Contributed by NM, 6-Jul-2005.) (Revised by Mario Carneiro, 25-Jul-2013.)
(𝑁 ∈ ℤ → ((𝑁𝐴 ∧ ∀𝑥𝐴 (𝑥 + 1) ∈ 𝐴) → {𝑘 ∈ ℤ ∣ 𝑁𝑘} ⊆ 𝐴))
 
Theorempeano5uzi 8319* Peano's inductive postulate for upper integers. (Contributed by NM, 6-Jul-2005.) (Revised by Mario Carneiro, 3-May-2014.)
𝑁 ∈ ℤ       ((𝑁𝐴 ∧ ∀𝑥𝐴 (𝑥 + 1) ∈ 𝐴) → {𝑘 ∈ ℤ ∣ 𝑁𝑘} ⊆ 𝐴)
 
Theoremdfuzi 8320* An expression for the upper integers that start at 𝑁 that is analogous to dfnn2 7892 for positive integers. (Contributed by NM, 6-Jul-2005.) (Proof shortened by Mario Carneiro, 3-May-2014.)
𝑁 ∈ ℤ       {𝑧 ∈ ℤ ∣ 𝑁𝑧} = {𝑥 ∣ (𝑁𝑥 ∧ ∀𝑦𝑥 (𝑦 + 1) ∈ 𝑥)}
 
Theoremuzind 8321* Induction on the upper integers that start at 𝑀. The first four hypotheses give us the substitution instances we need; the last two are the basis and the induction step. (Contributed by NM, 5-Jul-2005.)
(𝑗 = 𝑀 → (𝜑𝜓))    &   (𝑗 = 𝑘 → (𝜑𝜒))    &   (𝑗 = (𝑘 + 1) → (𝜑𝜃))    &   (𝑗 = 𝑁 → (𝜑𝜏))    &   (𝑀 ∈ ℤ → 𝜓)    &   ((𝑀 ∈ ℤ ∧ 𝑘 ∈ ℤ ∧ 𝑀𝑘) → (𝜒𝜃))       ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑀𝑁) → 𝜏)
 
Theoremuzind2 8322* Induction on the upper integers that start after an integer 𝑀. The first four hypotheses give us the substitution instances we need; the last two are the basis and the induction step. (Contributed by NM, 25-Jul-2005.)
(𝑗 = (𝑀 + 1) → (𝜑𝜓))    &   (𝑗 = 𝑘 → (𝜑𝜒))    &   (𝑗 = (𝑘 + 1) → (𝜑𝜃))    &   (𝑗 = 𝑁 → (𝜑𝜏))    &   (𝑀 ∈ ℤ → 𝜓)    &   ((𝑀 ∈ ℤ ∧ 𝑘 ∈ ℤ ∧ 𝑀 < 𝑘) → (𝜒𝜃))       ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑀 < 𝑁) → 𝜏)
 
Theoremuzind3 8323* Induction on the upper integers that start at an integer 𝑀. The first four hypotheses give us the substitution instances we need, and the last two are the basis and the induction step. (Contributed by NM, 26-Jul-2005.)
(𝑗 = 𝑀 → (𝜑𝜓))    &   (𝑗 = 𝑚 → (𝜑𝜒))    &   (𝑗 = (𝑚 + 1) → (𝜑𝜃))    &   (𝑗 = 𝑁 → (𝜑𝜏))    &   (𝑀 ∈ ℤ → 𝜓)    &   ((𝑀 ∈ ℤ ∧ 𝑚 ∈ {𝑘 ∈ ℤ ∣ 𝑀𝑘}) → (𝜒𝜃))       ((𝑀 ∈ ℤ ∧ 𝑁 ∈ {𝑘 ∈ ℤ ∣ 𝑀𝑘}) → 𝜏)
 
Theoremnn0ind 8324* Principle of Mathematical Induction (inference schema) on nonnegative integers. The first four hypotheses give us the substitution instances we need; the last two are the basis and the induction step. (Contributed by NM, 13-May-2004.)
(𝑥 = 0 → (𝜑𝜓))    &   (𝑥 = 𝑦 → (𝜑𝜒))    &   (𝑥 = (𝑦 + 1) → (𝜑𝜃))    &   (𝑥 = 𝐴 → (𝜑𝜏))    &   𝜓    &   (𝑦 ∈ ℕ0 → (𝜒𝜃))       (𝐴 ∈ ℕ0𝜏)
 
Theoremfzind 8325* Induction on the integers from 𝑀 to 𝑁 inclusive . The first four hypotheses give us the substitution instances we need; the last two are the basis and the induction step. (Contributed by Paul Chapman, 31-Mar-2011.)
(𝑥 = 𝑀 → (𝜑𝜓))    &   (𝑥 = 𝑦 → (𝜑𝜒))    &   (𝑥 = (𝑦 + 1) → (𝜑𝜃))    &   (𝑥 = 𝐾 → (𝜑𝜏))    &   ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑀𝑁) → 𝜓)    &   (((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ (𝑦 ∈ ℤ ∧ 𝑀𝑦𝑦 < 𝑁)) → (𝜒𝜃))       (((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ (𝐾 ∈ ℤ ∧ 𝑀𝐾𝐾𝑁)) → 𝜏)
 
Theoremfnn0ind 8326* Induction on the integers from 0 to 𝑁 inclusive . The first four hypotheses give us the substitution instances we need; the last two are the basis and the induction step. (Contributed by Paul Chapman, 31-Mar-2011.)
(𝑥 = 0 → (𝜑𝜓))    &   (𝑥 = 𝑦 → (𝜑𝜒))    &   (𝑥 = (𝑦 + 1) → (𝜑𝜃))    &   (𝑥 = 𝐾 → (𝜑𝜏))    &   (𝑁 ∈ ℕ0𝜓)    &   ((𝑁 ∈ ℕ0𝑦 ∈ ℕ0𝑦 < 𝑁) → (𝜒𝜃))       ((𝑁 ∈ ℕ0𝐾 ∈ ℕ0𝐾𝑁) → 𝜏)
 
Theoremnn0ind-raph 8327* Principle of Mathematical Induction (inference schema) on nonnegative integers. The first four hypotheses give us the substitution instances we need; the last two are the basis and the induction step. Raph Levien remarks: "This seems a bit painful. I wonder if an explicit substitution version would be easier." (Contributed by Raph Levien, 10-Apr-2004.)
(𝑥 = 0 → (𝜑𝜓))    &   (𝑥 = 𝑦 → (𝜑𝜒))    &   (𝑥 = (𝑦 + 1) → (𝜑𝜃))    &   (𝑥 = 𝐴 → (𝜑𝜏))    &   𝜓    &   (𝑦 ∈ ℕ0 → (𝜒𝜃))       (𝐴 ∈ ℕ0𝜏)
 
Theoremzindd 8328* Principle of Mathematical Induction on all integers, deduction version. The first five hypotheses give the substitutions; the last three are the basis, the induction, and the extension to negative numbers. (Contributed by Paul Chapman, 17-Apr-2009.) (Proof shortened by Mario Carneiro, 4-Jan-2017.)
(𝑥 = 0 → (𝜑𝜓))    &   (𝑥 = 𝑦 → (𝜑𝜒))    &   (𝑥 = (𝑦 + 1) → (𝜑𝜏))    &   (𝑥 = -𝑦 → (𝜑𝜃))    &   (𝑥 = 𝐴 → (𝜑𝜂))    &   (𝜁𝜓)    &   (𝜁 → (𝑦 ∈ ℕ0 → (𝜒𝜏)))    &   (𝜁 → (𝑦 ∈ ℕ → (𝜒𝜃)))       (𝜁 → (𝐴 ∈ ℤ → 𝜂))
 
Theorembtwnz 8329* Any real number can be sandwiched between two integers. Exercise 2 of [Apostol] p. 28. (Contributed by NM, 10-Nov-2004.)
(𝐴 ∈ ℝ → (∃𝑥 ∈ ℤ 𝑥 < 𝐴 ∧ ∃𝑦 ∈ ℤ 𝐴 < 𝑦))
 
Theoremnn0zd 8330 A positive integer is an integer. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℕ0)       (𝜑𝐴 ∈ ℤ)
 
Theoremnnzd 8331 A nonnegative integer is an integer. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℕ)       (𝜑𝐴 ∈ ℤ)
 
Theoremzred 8332 An integer is a real number. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℤ)       (𝜑𝐴 ∈ ℝ)
 
Theoremzcnd 8333 An integer is a complex number. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℤ)       (𝜑𝐴 ∈ ℂ)
 
Theoremznegcld 8334 Closure law for negative integers. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℤ)       (𝜑 → -𝐴 ∈ ℤ)
 
Theorempeano2zd 8335 Deduction from second Peano postulate generalized to integers. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℤ)       (𝜑 → (𝐴 + 1) ∈ ℤ)
 
Theoremzaddcld 8336 Closure of addition of integers. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)       (𝜑 → (𝐴 + 𝐵) ∈ ℤ)
 
Theoremzsubcld 8337 Closure of subtraction of integers. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)       (𝜑 → (𝐴𝐵) ∈ ℤ)
 
Theoremzmulcld 8338 Closure of multiplication of integers. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)       (𝜑 → (𝐴 · 𝐵) ∈ ℤ)
 
Theoremzadd2cl 8339 Increasing an integer by 2 results in an integer. (Contributed by Alexander van der Vekens, 16-Sep-2018.)
(𝑁 ∈ ℤ → (𝑁 + 2) ∈ ℤ)
 
3.4.9  Decimal arithmetic
 
Syntaxcdc 8340 Constant used for decimal constructor.
class 𝐴𝐵
 
Definitiondf-dec 8341 Define the "decimal constructor", which is used to build up "decimal integers" or "numeric terms" in base 10. For example, (1000 + 2000) = 3000. (Contributed by Mario Carneiro, 17-Apr-2015.)
𝐴𝐵 = ((10 · 𝐴) + 𝐵)
 
Theoremdeceq1 8342 Equality theorem for the decimal constructor. (Contributed by Mario Carneiro, 17-Apr-2015.)
(𝐴 = 𝐵𝐴𝐶 = 𝐵𝐶)
 
Theoremdeceq2 8343 Equality theorem for the decimal constructor. (Contributed by Mario Carneiro, 17-Apr-2015.)
(𝐴 = 𝐵𝐶𝐴 = 𝐶𝐵)
 
Theoremdeceq1i 8344 Equality theorem for the decimal constructor. (Contributed by Mario Carneiro, 17-Apr-2015.)
𝐴 = 𝐵       𝐴𝐶 = 𝐵𝐶
 
Theoremdeceq2i 8345 Equality theorem for the decimal constructor. (Contributed by Mario Carneiro, 17-Apr-2015.)
𝐴 = 𝐵       𝐶𝐴 = 𝐶𝐵
 
Theoremdeceq12i 8346 Equality theorem for the decimal constructor. (Contributed by Mario Carneiro, 17-Apr-2015.)
𝐴 = 𝐵    &   𝐶 = 𝐷       𝐴𝐶 = 𝐵𝐷
 
Theoremnumnncl 8347 Closure for a numeral (with units place). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 ∈ ℕ0    &   𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ       ((𝑇 · 𝐴) + 𝐵) ∈ ℕ
 
Theoremnum0u 8348 Add a zero in the units place. (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 ∈ ℕ0    &   𝐴 ∈ ℕ0       (𝑇 · 𝐴) = ((𝑇 · 𝐴) + 0)
 
Theoremnum0h 8349 Add a zero in the higher places. (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 ∈ ℕ0    &   𝐴 ∈ ℕ0       𝐴 = ((𝑇 · 0) + 𝐴)
 
Theoremnumcl 8350 Closure for a decimal integer (with units place). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 ∈ ℕ0    &   𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0       ((𝑇 · 𝐴) + 𝐵) ∈ ℕ0
 
Theoremnumsuc 8351 The successor of a decimal integer (no carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 ∈ ℕ0    &   𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   (𝐵 + 1) = 𝐶    &   𝑁 = ((𝑇 · 𝐴) + 𝐵)       (𝑁 + 1) = ((𝑇 · 𝐴) + 𝐶)
 
Theoremdecnncl 8352 Closure for a numeral. (Contributed by Mario Carneiro, 17-Apr-2015.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ       𝐴𝐵 ∈ ℕ
 
Theoremdeccl 8353 Closure for a numeral. (Contributed by Mario Carneiro, 17-Apr-2015.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0       𝐴𝐵 ∈ ℕ0
 
Theoremdec0u 8354 Add a zero in the units place. (Contributed by Mario Carneiro, 17-Apr-2015.)
𝐴 ∈ ℕ0       (10 · 𝐴) = 𝐴0
 
Theoremdec0h 8355 Add a zero in the higher places. (Contributed by Mario Carneiro, 17-Apr-2015.)
𝐴 ∈ ℕ0       𝐴 = 0𝐴
 
Theoremnumnncl2 8356 Closure for a decimal integer (zero units place). (Contributed by Mario Carneiro, 9-Mar-2015.)
𝑇 ∈ ℕ    &   𝐴 ∈ ℕ       ((𝑇 · 𝐴) + 0) ∈ ℕ
 
Theoremdecnncl2 8357 Closure for a decimal integer (zero units place). (Contributed by Mario Carneiro, 17-Apr-2015.)
𝐴 ∈ ℕ       𝐴0 ∈ ℕ
 
Theoremnumlt 8358 Comparing two decimal integers (equal higher places). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 ∈ ℕ    &   𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝐶 ∈ ℕ    &   𝐵 < 𝐶       ((𝑇 · 𝐴) + 𝐵) < ((𝑇 · 𝐴) + 𝐶)
 
Theoremnumltc 8359 Comparing two decimal integers (unequal higher places). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 ∈ ℕ    &   𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝐶 ∈ ℕ0    &   𝐷 ∈ ℕ0    &   𝐶 < 𝑇    &   𝐴 < 𝐵       ((𝑇 · 𝐴) + 𝐶) < ((𝑇 · 𝐵) + 𝐷)
 
Theoremdeclt 8360 Comparing two decimal integers (equal higher places). (Contributed by Mario Carneiro, 17-Apr-2015.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝐶 ∈ ℕ    &   𝐵 < 𝐶       𝐴𝐵 < 𝐴𝐶
 
Theoremdecltc 8361 Comparing two decimal integers (unequal higher places). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝐶 ∈ ℕ0    &   𝐷 ∈ ℕ0    &   𝐶 < 10    &   𝐴 < 𝐵       𝐴𝐶 < 𝐵𝐷
 
Theoremdecsuc 8362 The successor of a decimal integer (no carry). (Contributed by Mario Carneiro, 17-Apr-2015.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   (𝐵 + 1) = 𝐶    &   𝑁 = 𝐴𝐵       (𝑁 + 1) = 𝐴𝐶
 
Theoremnumlti 8363 Comparing a digit to a decimal integer. (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 ∈ ℕ    &   𝐴 ∈ ℕ    &   𝐵 ∈ ℕ0    &   𝐶 ∈ ℕ0    &   𝐶 < 𝑇       𝐶 < ((𝑇 · 𝐴) + 𝐵)
 
Theoremdeclti 8364 Comparing a digit to a decimal integer. (Contributed by Mario Carneiro, 18-Feb-2014.)
𝐴 ∈ ℕ    &   𝐵 ∈ ℕ0    &   𝐶 ∈ ℕ0    &   𝐶 < 10       𝐶 < 𝐴𝐵
 
Theoremnumsucc 8365 The successor of a decimal integer (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑌 ∈ ℕ0    &   𝑇 = (𝑌 + 1)    &   𝐴 ∈ ℕ0    &   (𝐴 + 1) = 𝐵    &   𝑁 = ((𝑇 · 𝐴) + 𝑌)       (𝑁 + 1) = ((𝑇 · 𝐵) + 0)
 
Theoremdecsucc 8366 The successor of a decimal integer (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝐴 ∈ ℕ0    &   (𝐴 + 1) = 𝐵    &   𝑁 = 𝐴9       (𝑁 + 1) = 𝐵0
 
Theorem1e0p1 8367 The successor of zero. (Contributed by Mario Carneiro, 18-Feb-2014.)
1 = (0 + 1)
 
Theoremdec10p 8368 Ten plus an integer. (Contributed by Mario Carneiro, 19-Apr-2015.)
(10 + 𝐴) = 1𝐴
 
Theoremdec10 8369 The decimal form of 10. NB: In our presentations of large numbers later on, we will use our symbol for 10 at the highest digits when advantageous, because we can use this theorem to convert back to "long form" (where each digit is in the range 0-9) with no extra effort. However, we cannot do this for lower digits while maintaining the ease of use of the decimal system, since it requires nontrivial number knowledge (more than just equality theorems) to convert back. (Contributed by Mario Carneiro, 18-Feb-2014.)
10 = 10
 
Theoremnumma 8370 Perform a multiply-add of two decimal integers 𝑀 and 𝑁 against a fixed multiplicand 𝑃 (no carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 ∈ ℕ0    &   𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝐶 ∈ ℕ0    &   𝐷 ∈ ℕ0    &   𝑀 = ((𝑇 · 𝐴) + 𝐵)    &   𝑁 = ((𝑇 · 𝐶) + 𝐷)    &   𝑃 ∈ ℕ0    &   ((𝐴 · 𝑃) + 𝐶) = 𝐸    &   ((𝐵 · 𝑃) + 𝐷) = 𝐹       ((𝑀 · 𝑃) + 𝑁) = ((𝑇 · 𝐸) + 𝐹)
 
Theoremnummac 8371 Perform a multiply-add of two decimal integers 𝑀 and 𝑁 against a fixed multiplicand 𝑃 (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 ∈ ℕ0    &   𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝐶 ∈ ℕ0    &   𝐷 ∈ ℕ0    &   𝑀 = ((𝑇 · 𝐴) + 𝐵)    &   𝑁 = ((𝑇 · 𝐶) + 𝐷)    &   𝑃 ∈ ℕ0    &   𝐹 ∈ ℕ0    &   𝐺 ∈ ℕ0    &   ((𝐴 · 𝑃) + (𝐶 + 𝐺)) = 𝐸    &   ((𝐵 · 𝑃) + 𝐷) = ((𝑇 · 𝐺) + 𝐹)       ((𝑀 · 𝑃) + 𝑁) = ((𝑇 · 𝐸) + 𝐹)
 
Theoremnumma2c 8372 Perform a multiply-add of two decimal integers 𝑀 and 𝑁 against a fixed multiplicand 𝑃 (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 ∈ ℕ0    &   𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝐶 ∈ ℕ0    &   𝐷 ∈ ℕ0    &   𝑀 = ((𝑇 · 𝐴) + 𝐵)    &   𝑁 = ((𝑇 · 𝐶) + 𝐷)    &   𝑃 ∈ ℕ0    &   𝐹 ∈ ℕ0    &   𝐺 ∈ ℕ0    &   ((𝑃 · 𝐴) + (𝐶 + 𝐺)) = 𝐸    &   ((𝑃 · 𝐵) + 𝐷) = ((𝑇 · 𝐺) + 𝐹)       ((𝑃 · 𝑀) + 𝑁) = ((𝑇 · 𝐸) + 𝐹)
 
Theoremnumadd 8373 Add two decimal integers 𝑀 and 𝑁 (no carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 ∈ ℕ0    &   𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝐶 ∈ ℕ0    &   𝐷 ∈ ℕ0    &   𝑀 = ((𝑇 · 𝐴) + 𝐵)    &   𝑁 = ((𝑇 · 𝐶) + 𝐷)    &   (𝐴 + 𝐶) = 𝐸    &   (𝐵 + 𝐷) = 𝐹       (𝑀 + 𝑁) = ((𝑇 · 𝐸) + 𝐹)
 
Theoremnumaddc 8374 Add two decimal integers 𝑀 and 𝑁 (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 ∈ ℕ0    &   𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝐶 ∈ ℕ0    &   𝐷 ∈ ℕ0    &   𝑀 = ((𝑇 · 𝐴) + 𝐵)    &   𝑁 = ((𝑇 · 𝐶) + 𝐷)    &   𝐹 ∈ ℕ0    &   ((𝐴 + 𝐶) + 1) = 𝐸    &   (𝐵 + 𝐷) = ((𝑇 · 1) + 𝐹)       (𝑀 + 𝑁) = ((𝑇 · 𝐸) + 𝐹)
 
Theoremnummul1c 8375 The product of a decimal integer with a number. (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 ∈ ℕ0    &   𝑃 ∈ ℕ0    &   𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝑁 = ((𝑇 · 𝐴) + 𝐵)    &   𝐷 ∈ ℕ0    &   𝐸 ∈ ℕ0    &   ((𝐴 · 𝑃) + 𝐸) = 𝐶    &   (𝐵 · 𝑃) = ((𝑇 · 𝐸) + 𝐷)       (𝑁 · 𝑃) = ((𝑇 · 𝐶) + 𝐷)
 
Theoremnummul2c 8376 The product of a decimal integer with a number (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑇 ∈ ℕ0    &   𝑃 ∈ ℕ0    &   𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝑁 = ((𝑇 · 𝐴) + 𝐵)    &   𝐷 ∈ ℕ0    &   𝐸 ∈ ℕ0    &   ((𝑃 · 𝐴) + 𝐸) = 𝐶    &   (𝑃 · 𝐵) = ((𝑇 · 𝐸) + 𝐷)       (𝑃 · 𝑁) = ((𝑇 · 𝐶) + 𝐷)
 
Theoremdecma 8377 Perform a multiply-add of two numerals 𝑀 and 𝑁 against a fixed multiplicand 𝑃 (no carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝐶 ∈ ℕ0    &   𝐷 ∈ ℕ0    &   𝑀 = 𝐴𝐵    &   𝑁 = 𝐶𝐷    &   𝑃 ∈ ℕ0    &   ((𝐴 · 𝑃) + 𝐶) = 𝐸    &   ((𝐵 · 𝑃) + 𝐷) = 𝐹       ((𝑀 · 𝑃) + 𝑁) = 𝐸𝐹
 
Theoremdecmac 8378 Perform a multiply-add of two numerals 𝑀 and 𝑁 against a fixed multiplicand 𝑃 (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝐶 ∈ ℕ0    &   𝐷 ∈ ℕ0    &   𝑀 = 𝐴𝐵    &   𝑁 = 𝐶𝐷    &   𝑃 ∈ ℕ0    &   𝐹 ∈ ℕ0    &   𝐺 ∈ ℕ0    &   ((𝐴 · 𝑃) + (𝐶 + 𝐺)) = 𝐸    &   ((𝐵 · 𝑃) + 𝐷) = 𝐺𝐹       ((𝑀 · 𝑃) + 𝑁) = 𝐸𝐹
 
Theoremdecma2c 8379 Perform a multiply-add of two numerals 𝑀 and 𝑁 against a fixed multiplicand 𝑃 (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝐶 ∈ ℕ0    &   𝐷 ∈ ℕ0    &   𝑀 = 𝐴𝐵    &   𝑁 = 𝐶𝐷    &   𝑃 ∈ ℕ0    &   𝐹 ∈ ℕ0    &   𝐺 ∈ ℕ0    &   ((𝑃 · 𝐴) + (𝐶 + 𝐺)) = 𝐸    &   ((𝑃 · 𝐵) + 𝐷) = 𝐺𝐹       ((𝑃 · 𝑀) + 𝑁) = 𝐸𝐹
 
Theoremdecadd 8380 Add two numerals 𝑀 and 𝑁 (no carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝐶 ∈ ℕ0    &   𝐷 ∈ ℕ0    &   𝑀 = 𝐴𝐵    &   𝑁 = 𝐶𝐷    &   (𝐴 + 𝐶) = 𝐸    &   (𝐵 + 𝐷) = 𝐹       (𝑀 + 𝑁) = 𝐸𝐹
 
Theoremdecaddc 8381 Add two numerals 𝑀 and 𝑁 (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝐶 ∈ ℕ0    &   𝐷 ∈ ℕ0    &   𝑀 = 𝐴𝐵    &   𝑁 = 𝐶𝐷    &   ((𝐴 + 𝐶) + 1) = 𝐸    &   𝐹 ∈ ℕ0    &   (𝐵 + 𝐷) = 1𝐹       (𝑀 + 𝑁) = 𝐸𝐹
 
Theoremdecaddc2 8382 Add two numerals 𝑀 and 𝑁 (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝐶 ∈ ℕ0    &   𝐷 ∈ ℕ0    &   𝑀 = 𝐴𝐵    &   𝑁 = 𝐶𝐷    &   ((𝐴 + 𝐶) + 1) = 𝐸    &   (𝐵 + 𝐷) = 10       (𝑀 + 𝑁) = 𝐸0
 
Theoremdecaddi 8383 Add two numerals 𝑀 and 𝑁 (no carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝑁 ∈ ℕ0    &   𝑀 = 𝐴𝐵    &   (𝐵 + 𝑁) = 𝐶       (𝑀 + 𝑁) = 𝐴𝐶
 
Theoremdecaddci 8384 Add two numerals 𝑀 and 𝑁 (no carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝑁 ∈ ℕ0    &   𝑀 = 𝐴𝐵    &   (𝐴 + 1) = 𝐷    &   𝐶 ∈ ℕ0    &   (𝐵 + 𝑁) = 1𝐶       (𝑀 + 𝑁) = 𝐷𝐶
 
Theoremdecaddci2 8385 Add two numerals 𝑀 and 𝑁 (no carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝑁 ∈ ℕ0    &   𝑀 = 𝐴𝐵    &   (𝐴 + 1) = 𝐷    &   (𝐵 + 𝑁) = 10       (𝑀 + 𝑁) = 𝐷0
 
Theoremdecmul1c 8386 The product of a numeral with a number. (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑃 ∈ ℕ0    &   𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝑁 = 𝐴𝐵    &   𝐷 ∈ ℕ0    &   𝐸 ∈ ℕ0    &   ((𝐴 · 𝑃) + 𝐸) = 𝐶    &   (𝐵 · 𝑃) = 𝐸𝐷       (𝑁 · 𝑃) = 𝐶𝐷
 
Theoremdecmul2c 8387 The product of a numeral with a number (with carry). (Contributed by Mario Carneiro, 18-Feb-2014.)
𝑃 ∈ ℕ0    &   𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝑁 = 𝐴𝐵    &   𝐷 ∈ ℕ0    &   𝐸 ∈ ℕ0    &   ((𝑃 · 𝐴) + 𝐸) = 𝐶    &   (𝑃 · 𝐵) = 𝐸𝐷       (𝑃 · 𝑁) = 𝐶𝐷
 
Theorem6p5lem 8388 Lemma for 6p5e11 8389 and related theorems. (Contributed by Mario Carneiro, 19-Apr-2015.)
𝐴 ∈ ℕ0    &   𝐷 ∈ ℕ0    &   𝐸 ∈ ℕ0    &   𝐵 = (𝐷 + 1)    &   𝐶 = (𝐸 + 1)    &   (𝐴 + 𝐷) = 1𝐸       (𝐴 + 𝐵) = 1𝐶
 
Theorem6p5e11 8389 6 + 5 = 11. (Contributed by Mario Carneiro, 19-Apr-2015.)
(6 + 5) = 11
 
Theorem6p6e12 8390 6 + 6 = 12. (Contributed by Mario Carneiro, 19-Apr-2015.)
(6 + 6) = 12
 
Theorem7p4e11 8391 7 + 4 = 11. (Contributed by Mario Carneiro, 19-Apr-2015.)
(7 + 4) = 11
 
Theorem7p5e12 8392 7 + 5 = 12. (Contributed by Mario Carneiro, 19-Apr-2015.)
(7 + 5) = 12
 
Theorem7p6e13 8393 7 + 6 = 13. (Contributed by Mario Carneiro, 19-Apr-2015.)
(7 + 6) = 13
 
Theorem7p7e14 8394 7 + 7 = 14. (Contributed by Mario Carneiro, 19-Apr-2015.)
(7 + 7) = 14
 
Theorem8p3e11 8395 8 + 3 = 11. (Contributed by Mario Carneiro, 19-Apr-2015.)
(8 + 3) = 11
 
Theorem8p4e12 8396 8 + 4 = 12. (Contributed by Mario Carneiro, 19-Apr-2015.)
(8 + 4) = 12
 
Theorem8p5e13 8397 8 + 5 = 13. (Contributed by Mario Carneiro, 19-Apr-2015.)
(8 + 5) = 13
 
Theorem8p6e14 8398 8 + 6 = 14. (Contributed by Mario Carneiro, 19-Apr-2015.)
(8 + 6) = 14
 
Theorem8p7e15 8399 8 + 7 = 15. (Contributed by Mario Carneiro, 19-Apr-2015.)
(8 + 7) = 15
 
Theorem8p8e16 8400 8 + 8 = 16. (Contributed by Mario Carneiro, 19-Apr-2015.)
(8 + 8) = 16
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