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Type | Label | Description |
---|---|---|
Statement | ||
Theorem | sqge0d 13801 | A square of a real is nonnegative. (Contributed by Mario Carneiro, 28-May-2016.) |
⊢ (𝜑 → 𝐴 ∈ ℝ) ⇒ ⊢ (𝜑 → 0 ≤ (𝐴↑2)) | ||
Theorem | sqgt0d 13802 | The square of a nonzero real is positive. (Contributed by Mario Carneiro, 28-May-2016.) |
⊢ (𝜑 → 𝐴 ∈ ℝ) & ⊢ (𝜑 → 𝐴 ≠ 0) ⇒ ⊢ (𝜑 → 0 < (𝐴↑2)) | ||
Theorem | ltexp2d 13803 | Ordering relationship for exponentiation. (Contributed by Mario Carneiro, 28-May-2016.) |
⊢ (𝜑 → 𝐴 ∈ ℝ) & ⊢ (𝜑 → 𝑀 ∈ ℤ) & ⊢ (𝜑 → 𝑁 ∈ ℤ) & ⊢ (𝜑 → 1 < 𝐴) ⇒ ⊢ (𝜑 → (𝑀 < 𝑁 ↔ (𝐴↑𝑀) < (𝐴↑𝑁))) | ||
Theorem | leexp2d 13804 | Ordering law for exponentiation. (Contributed by Mario Carneiro, 28-May-2016.) |
⊢ (𝜑 → 𝐴 ∈ ℝ) & ⊢ (𝜑 → 𝑀 ∈ ℤ) & ⊢ (𝜑 → 𝑁 ∈ ℤ) & ⊢ (𝜑 → 1 < 𝐴) ⇒ ⊢ (𝜑 → (𝑀 ≤ 𝑁 ↔ (𝐴↑𝑀) ≤ (𝐴↑𝑁))) | ||
Theorem | expcand 13805 | Ordering relationship for exponentiation. (Contributed by Mario Carneiro, 28-May-2016.) |
⊢ (𝜑 → 𝐴 ∈ ℝ) & ⊢ (𝜑 → 𝑀 ∈ ℤ) & ⊢ (𝜑 → 𝑁 ∈ ℤ) & ⊢ (𝜑 → 1 < 𝐴) & ⊢ (𝜑 → (𝐴↑𝑀) = (𝐴↑𝑁)) ⇒ ⊢ (𝜑 → 𝑀 = 𝑁) | ||
Theorem | leexp2ad 13806 | Ordering relationship for exponentiation. (Contributed by Mario Carneiro, 28-May-2016.) |
⊢ (𝜑 → 𝐴 ∈ ℝ) & ⊢ (𝜑 → 1 ≤ 𝐴) & ⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀)) ⇒ ⊢ (𝜑 → (𝐴↑𝑀) ≤ (𝐴↑𝑁)) | ||
Theorem | leexp2rd 13807 | Ordering relationship for exponentiation. (Contributed by Mario Carneiro, 28-May-2016.) |
⊢ (𝜑 → 𝐴 ∈ ℝ) & ⊢ (𝜑 → 𝑀 ∈ ℕ0) & ⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀)) & ⊢ (𝜑 → 0 ≤ 𝐴) & ⊢ (𝜑 → 𝐴 ≤ 1) ⇒ ⊢ (𝜑 → (𝐴↑𝑁) ≤ (𝐴↑𝑀)) | ||
Theorem | lt2sqd 13808 | The square function on nonnegative reals is strictly monotonic. (Contributed by Mario Carneiro, 28-May-2016.) |
⊢ (𝜑 → 𝐴 ∈ ℝ) & ⊢ (𝜑 → 𝐵 ∈ ℝ) & ⊢ (𝜑 → 0 ≤ 𝐴) & ⊢ (𝜑 → 0 ≤ 𝐵) ⇒ ⊢ (𝜑 → (𝐴 < 𝐵 ↔ (𝐴↑2) < (𝐵↑2))) | ||
Theorem | le2sqd 13809 | The square function on nonnegative reals is monotonic. (Contributed by Mario Carneiro, 28-May-2016.) |
⊢ (𝜑 → 𝐴 ∈ ℝ) & ⊢ (𝜑 → 𝐵 ∈ ℝ) & ⊢ (𝜑 → 0 ≤ 𝐴) & ⊢ (𝜑 → 0 ≤ 𝐵) ⇒ ⊢ (𝜑 → (𝐴 ≤ 𝐵 ↔ (𝐴↑2) ≤ (𝐵↑2))) | ||
Theorem | sq11d 13810 | The square function is one-to-one for nonnegative reals. (Contributed by Mario Carneiro, 28-May-2016.) |
⊢ (𝜑 → 𝐴 ∈ ℝ) & ⊢ (𝜑 → 𝐵 ∈ ℝ) & ⊢ (𝜑 → 0 ≤ 𝐴) & ⊢ (𝜑 → 0 ≤ 𝐵) & ⊢ (𝜑 → (𝐴↑2) = (𝐵↑2)) ⇒ ⊢ (𝜑 → 𝐴 = 𝐵) | ||
Theorem | mulsubdivbinom2 13811 | The square of a binomial with factor minus a number divided by a nonzero number. (Contributed by AV, 19-Jul-2021.) |
⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐷 ∈ ℂ) ∧ (𝐶 ∈ ℂ ∧ 𝐶 ≠ 0)) → (((((𝐶 · 𝐴) + 𝐵)↑2) − 𝐷) / 𝐶) = (((𝐶 · (𝐴↑2)) + (2 · (𝐴 · 𝐵))) + (((𝐵↑2) − 𝐷) / 𝐶))) | ||
Theorem | muldivbinom2 13812 | The square of a binomial with factor divided by a nonzero number. (Contributed by AV, 19-Jul-2021.) |
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐶 ≠ 0)) → ((((𝐶 · 𝐴) + 𝐵)↑2) / 𝐶) = (((𝐶 · (𝐴↑2)) + (2 · (𝐴 · 𝐵))) + ((𝐵↑2) / 𝐶))) | ||
Theorem | sq10 13813 | The square of 10 is 100. (Contributed by AV, 14-Jun-2021.) (Revised by AV, 1-Aug-2021.) |
⊢ (;10↑2) = ;;100 | ||
Theorem | sq10e99m1 13814 | The square of 10 is 99 plus 1. (Contributed by AV, 14-Jun-2021.) (Revised by AV, 1-Aug-2021.) |
⊢ (;10↑2) = (;99 + 1) | ||
Theorem | 3dec 13815 | A "decimal constructor" which is used to build up "decimal integers" or "numeric terms" in base 10 with 3 "digits". (Contributed by AV, 14-Jun-2021.) (Revised by AV, 1-Aug-2021.) |
⊢ 𝐴 ∈ ℕ0 & ⊢ 𝐵 ∈ ℕ0 ⇒ ⊢ ;;𝐴𝐵𝐶 = ((((;10↑2) · 𝐴) + (;10 · 𝐵)) + 𝐶) | ||
Theorem | nn0le2msqi 13816 | The square function on nonnegative integers is monotonic. (Contributed by Raph Levien, 10-Dec-2002.) |
⊢ 𝐴 ∈ ℕ0 & ⊢ 𝐵 ∈ ℕ0 ⇒ ⊢ (𝐴 ≤ 𝐵 ↔ (𝐴 · 𝐴) ≤ (𝐵 · 𝐵)) | ||
Theorem | nn0opthlem1 13817 | A rather pretty lemma for nn0opthi 13819. (Contributed by Raph Levien, 10-Dec-2002.) |
⊢ 𝐴 ∈ ℕ0 & ⊢ 𝐶 ∈ ℕ0 ⇒ ⊢ (𝐴 < 𝐶 ↔ ((𝐴 · 𝐴) + (2 · 𝐴)) < (𝐶 · 𝐶)) | ||
Theorem | nn0opthlem2 13818 | Lemma for nn0opthi 13819. (Contributed by Raph Levien, 10-Dec-2002.) (Revised by Scott Fenton, 8-Sep-2010.) |
⊢ 𝐴 ∈ ℕ0 & ⊢ 𝐵 ∈ ℕ0 & ⊢ 𝐶 ∈ ℕ0 & ⊢ 𝐷 ∈ ℕ0 ⇒ ⊢ ((𝐴 + 𝐵) < 𝐶 → ((𝐶 · 𝐶) + 𝐷) ≠ (((𝐴 + 𝐵) · (𝐴 + 𝐵)) + 𝐵)) | ||
Theorem | nn0opthi 13819 | An ordered pair theorem for nonnegative integers. Theorem 17.3 of [Quine] p. 124. We can represent an ordered pair of nonnegative integers 𝐴 and 𝐵 by (((𝐴 + 𝐵) · (𝐴 + 𝐵)) + 𝐵). If two such ordered pairs are equal, their first elements are equal and their second elements are equal. Contrast this ordered pair representation with the standard one df-op 4538 that works for any set. (Contributed by Raph Levien, 10-Dec-2002.) (Proof shortened by Scott Fenton, 8-Sep-2010.) |
⊢ 𝐴 ∈ ℕ0 & ⊢ 𝐵 ∈ ℕ0 & ⊢ 𝐶 ∈ ℕ0 & ⊢ 𝐷 ∈ ℕ0 ⇒ ⊢ ((((𝐴 + 𝐵) · (𝐴 + 𝐵)) + 𝐵) = (((𝐶 + 𝐷) · (𝐶 + 𝐷)) + 𝐷) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)) | ||
Theorem | nn0opth2i 13820 | An ordered pair theorem for nonnegative integers. Theorem 17.3 of [Quine] p. 124. See comments for nn0opthi 13819. (Contributed by NM, 22-Jul-2004.) |
⊢ 𝐴 ∈ ℕ0 & ⊢ 𝐵 ∈ ℕ0 & ⊢ 𝐶 ∈ ℕ0 & ⊢ 𝐷 ∈ ℕ0 ⇒ ⊢ ((((𝐴 + 𝐵)↑2) + 𝐵) = (((𝐶 + 𝐷)↑2) + 𝐷) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)) | ||
Theorem | nn0opth2 13821 | An ordered pair theorem for nonnegative integers. Theorem 17.3 of [Quine] p. 124. See nn0opthi 13819. (Contributed by NM, 22-Jul-2004.) |
⊢ (((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ0) ∧ (𝐶 ∈ ℕ0 ∧ 𝐷 ∈ ℕ0)) → ((((𝐴 + 𝐵)↑2) + 𝐵) = (((𝐶 + 𝐷)↑2) + 𝐷) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))) | ||
Syntax | cfa 13822 | Extend class notation to include the factorial of nonnegative integers. |
class ! | ||
Definition | df-fac 13823 | Define the factorial function on nonnegative integers. For example, (!‘5) = 120 because 1 · 2 · 3 · 4 · 5 = 120 (ex-fac 28506). In the literature, the factorial function is written as a postscript exclamation point. (Contributed by NM, 2-Dec-2004.) |
⊢ ! = ({〈0, 1〉} ∪ seq1( · , I )) | ||
Theorem | facnn 13824 | Value of the factorial function for positive integers. (Contributed by NM, 2-Dec-2004.) (Revised by Mario Carneiro, 13-Jul-2013.) |
⊢ (𝑁 ∈ ℕ → (!‘𝑁) = (seq1( · , I )‘𝑁)) | ||
Theorem | fac0 13825 | The factorial of 0. (Contributed by NM, 2-Dec-2004.) (Revised by Mario Carneiro, 13-Jul-2013.) |
⊢ (!‘0) = 1 | ||
Theorem | fac1 13826 | The factorial of 1. (Contributed by NM, 2-Dec-2004.) (Revised by Mario Carneiro, 13-Jul-2013.) |
⊢ (!‘1) = 1 | ||
Theorem | facp1 13827 | The factorial of a successor. (Contributed by NM, 2-Dec-2004.) (Revised by Mario Carneiro, 13-Jul-2013.) |
⊢ (𝑁 ∈ ℕ0 → (!‘(𝑁 + 1)) = ((!‘𝑁) · (𝑁 + 1))) | ||
Theorem | fac2 13828 | The factorial of 2. (Contributed by NM, 17-Mar-2005.) |
⊢ (!‘2) = 2 | ||
Theorem | fac3 13829 | The factorial of 3. (Contributed by NM, 17-Mar-2005.) |
⊢ (!‘3) = 6 | ||
Theorem | fac4 13830 | The factorial of 4. (Contributed by Mario Carneiro, 18-Jun-2015.) |
⊢ (!‘4) = ;24 | ||
Theorem | facnn2 13831 | Value of the factorial function expressed recursively. (Contributed by NM, 2-Dec-2004.) |
⊢ (𝑁 ∈ ℕ → (!‘𝑁) = ((!‘(𝑁 − 1)) · 𝑁)) | ||
Theorem | faccl 13832 | Closure of the factorial function. (Contributed by NM, 2-Dec-2004.) |
⊢ (𝑁 ∈ ℕ0 → (!‘𝑁) ∈ ℕ) | ||
Theorem | faccld 13833 | Closure of the factorial function, deduction version of faccl 13832. (Contributed by Glauco Siliprandi, 5-Apr-2020.) |
⊢ (𝜑 → 𝑁 ∈ ℕ0) ⇒ ⊢ (𝜑 → (!‘𝑁) ∈ ℕ) | ||
Theorem | facmapnn 13834 | The factorial function restricted to positive integers is a mapping from the positive integers to the positive integers. (Contributed by AV, 8-Aug-2020.) |
⊢ (𝑛 ∈ ℕ ↦ (!‘𝑛)) ∈ (ℕ ↑m ℕ) | ||
Theorem | facne0 13835 | The factorial function is nonzero. (Contributed by NM, 26-Apr-2005.) |
⊢ (𝑁 ∈ ℕ0 → (!‘𝑁) ≠ 0) | ||
Theorem | facdiv 13836 | A positive integer divides the factorial of an equal or larger number. (Contributed by NM, 2-May-2005.) |
⊢ ((𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ ∧ 𝑁 ≤ 𝑀) → ((!‘𝑀) / 𝑁) ∈ ℕ) | ||
Theorem | facndiv 13837 | No positive integer (greater than one) divides the factorial plus one of an equal or larger number. (Contributed by NM, 3-May-2005.) |
⊢ (((𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ) ∧ (1 < 𝑁 ∧ 𝑁 ≤ 𝑀)) → ¬ (((!‘𝑀) + 1) / 𝑁) ∈ ℤ) | ||
Theorem | facwordi 13838 | Ordering property of factorial. (Contributed by NM, 9-Dec-2005.) |
⊢ ((𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0 ∧ 𝑀 ≤ 𝑁) → (!‘𝑀) ≤ (!‘𝑁)) | ||
Theorem | faclbnd 13839 | A lower bound for the factorial function. (Contributed by NM, 17-Dec-2005.) |
⊢ ((𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0) → (𝑀↑(𝑁 + 1)) ≤ ((𝑀↑𝑀) · (!‘𝑁))) | ||
Theorem | faclbnd2 13840 | A lower bound for the factorial function. (Contributed by NM, 17-Dec-2005.) |
⊢ (𝑁 ∈ ℕ0 → ((2↑𝑁) / 2) ≤ (!‘𝑁)) | ||
Theorem | faclbnd3 13841 | A lower bound for the factorial function. (Contributed by NM, 19-Dec-2005.) |
⊢ ((𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0) → (𝑀↑𝑁) ≤ ((𝑀↑𝑀) · (!‘𝑁))) | ||
Theorem | faclbnd4lem1 13842 | Lemma for faclbnd4 13846. Prepare the induction step. (Contributed by NM, 20-Dec-2005.) |
⊢ 𝑁 ∈ ℕ & ⊢ 𝐾 ∈ ℕ0 & ⊢ 𝑀 ∈ ℕ0 ⇒ ⊢ ((((𝑁 − 1)↑𝐾) · (𝑀↑(𝑁 − 1))) ≤ (((2↑(𝐾↑2)) · (𝑀↑(𝑀 + 𝐾))) · (!‘(𝑁 − 1))) → ((𝑁↑(𝐾 + 1)) · (𝑀↑𝑁)) ≤ (((2↑((𝐾 + 1)↑2)) · (𝑀↑(𝑀 + (𝐾 + 1)))) · (!‘𝑁))) | ||
Theorem | faclbnd4lem2 13843 | Lemma for faclbnd4 13846. Use the weak deduction theorem to convert the hypotheses of faclbnd4lem1 13842 to antecedents. (Contributed by NM, 23-Dec-2005.) |
⊢ ((𝑀 ∈ ℕ0 ∧ 𝐾 ∈ ℕ0 ∧ 𝑁 ∈ ℕ) → ((((𝑁 − 1)↑𝐾) · (𝑀↑(𝑁 − 1))) ≤ (((2↑(𝐾↑2)) · (𝑀↑(𝑀 + 𝐾))) · (!‘(𝑁 − 1))) → ((𝑁↑(𝐾 + 1)) · (𝑀↑𝑁)) ≤ (((2↑((𝐾 + 1)↑2)) · (𝑀↑(𝑀 + (𝐾 + 1)))) · (!‘𝑁)))) | ||
Theorem | faclbnd4lem3 13844 | Lemma for faclbnd4 13846. The 𝑁 = 0 case. (Contributed by NM, 23-Dec-2005.) |
⊢ (((𝑀 ∈ ℕ0 ∧ 𝐾 ∈ ℕ0) ∧ 𝑁 = 0) → ((𝑁↑𝐾) · (𝑀↑𝑁)) ≤ (((2↑(𝐾↑2)) · (𝑀↑(𝑀 + 𝐾))) · (!‘𝑁))) | ||
Theorem | faclbnd4lem4 13845 | Lemma for faclbnd4 13846. Prove the 0 < 𝑁 case by induction on 𝐾. (Contributed by NM, 19-Dec-2005.) |
⊢ ((𝑁 ∈ ℕ ∧ 𝐾 ∈ ℕ0 ∧ 𝑀 ∈ ℕ0) → ((𝑁↑𝐾) · (𝑀↑𝑁)) ≤ (((2↑(𝐾↑2)) · (𝑀↑(𝑀 + 𝐾))) · (!‘𝑁))) | ||
Theorem | faclbnd4 13846 | Variant of faclbnd5 13847 providing a non-strict lower bound. (Contributed by NM, 23-Dec-2005.) |
⊢ ((𝑁 ∈ ℕ0 ∧ 𝐾 ∈ ℕ0 ∧ 𝑀 ∈ ℕ0) → ((𝑁↑𝐾) · (𝑀↑𝑁)) ≤ (((2↑(𝐾↑2)) · (𝑀↑(𝑀 + 𝐾))) · (!‘𝑁))) | ||
Theorem | faclbnd5 13847 | The factorial function grows faster than powers and exponentiations. If we consider 𝐾 and 𝑀 to be constants, the right-hand side of the inequality is a constant times 𝑁-factorial. (Contributed by NM, 24-Dec-2005.) |
⊢ ((𝑁 ∈ ℕ0 ∧ 𝐾 ∈ ℕ0 ∧ 𝑀 ∈ ℕ) → ((𝑁↑𝐾) · (𝑀↑𝑁)) < ((2 · ((2↑(𝐾↑2)) · (𝑀↑(𝑀 + 𝐾)))) · (!‘𝑁))) | ||
Theorem | faclbnd6 13848 | Geometric lower bound for the factorial function, where N is usually held constant. (Contributed by Paul Chapman, 28-Dec-2007.) |
⊢ ((𝑁 ∈ ℕ0 ∧ 𝑀 ∈ ℕ0) → ((!‘𝑁) · ((𝑁 + 1)↑𝑀)) ≤ (!‘(𝑁 + 𝑀))) | ||
Theorem | facubnd 13849 | An upper bound for the factorial function. (Contributed by Mario Carneiro, 15-Apr-2016.) |
⊢ (𝑁 ∈ ℕ0 → (!‘𝑁) ≤ (𝑁↑𝑁)) | ||
Theorem | facavg 13850 | The product of two factorials is greater than or equal to the factorial of (the floor of) their average. (Contributed by NM, 9-Dec-2005.) |
⊢ ((𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0) → (!‘(⌊‘((𝑀 + 𝑁) / 2))) ≤ ((!‘𝑀) · (!‘𝑁))) | ||
Syntax | cbc 13851 | Extend class notation to include the binomial coefficient operation (combinatorial choose operation). |
class C | ||
Definition | df-bc 13852* |
Define the binomial coefficient operation. For example,
(5C3) = 10 (ex-bc 28507).
In the literature, this function is often written as a column vector of the two arguments, or with the arguments as subscripts before and after the letter "C". The expression (𝑁C𝐾) is read "𝑁 choose 𝐾". Definition of binomial coefficient in [Gleason] p. 295. As suggested by Gleason, we define it to be 0 when 0 ≤ 𝑘 ≤ 𝑛 does not hold. (Contributed by NM, 10-Jul-2005.) |
⊢ C = (𝑛 ∈ ℕ0, 𝑘 ∈ ℤ ↦ if(𝑘 ∈ (0...𝑛), ((!‘𝑛) / ((!‘(𝑛 − 𝑘)) · (!‘𝑘))), 0)) | ||
Theorem | bcval 13853 | Value of the binomial coefficient, 𝑁 choose 𝐾. Definition of binomial coefficient in [Gleason] p. 295. As suggested by Gleason, we define it to be 0 when 0 ≤ 𝐾 ≤ 𝑁 does not hold. See bcval2 13854 for the value in the standard domain. (Contributed by NM, 10-Jul-2005.) (Revised by Mario Carneiro, 7-Nov-2013.) |
⊢ ((𝑁 ∈ ℕ0 ∧ 𝐾 ∈ ℤ) → (𝑁C𝐾) = if(𝐾 ∈ (0...𝑁), ((!‘𝑁) / ((!‘(𝑁 − 𝐾)) · (!‘𝐾))), 0)) | ||
Theorem | bcval2 13854 | Value of the binomial coefficient, 𝑁 choose 𝐾, in its standard domain. (Contributed by NM, 9-Jun-2005.) (Revised by Mario Carneiro, 7-Nov-2013.) |
⊢ (𝐾 ∈ (0...𝑁) → (𝑁C𝐾) = ((!‘𝑁) / ((!‘(𝑁 − 𝐾)) · (!‘𝐾)))) | ||
Theorem | bcval3 13855 | Value of the binomial coefficient, 𝑁 choose 𝐾, outside of its standard domain. Remark in [Gleason] p. 295. (Contributed by NM, 14-Jul-2005.) (Revised by Mario Carneiro, 8-Nov-2013.) |
⊢ ((𝑁 ∈ ℕ0 ∧ 𝐾 ∈ ℤ ∧ ¬ 𝐾 ∈ (0...𝑁)) → (𝑁C𝐾) = 0) | ||
Theorem | bcval4 13856 | Value of the binomial coefficient, 𝑁 choose 𝐾, outside of its standard domain. Remark in [Gleason] p. 295. (Contributed by NM, 14-Jul-2005.) (Revised by Mario Carneiro, 7-Nov-2013.) |
⊢ ((𝑁 ∈ ℕ0 ∧ 𝐾 ∈ ℤ ∧ (𝐾 < 0 ∨ 𝑁 < 𝐾)) → (𝑁C𝐾) = 0) | ||
Theorem | bcrpcl 13857 | Closure of the binomial coefficient in the positive reals. (This is mostly a lemma before we have bccl2 13872.) (Contributed by Mario Carneiro, 10-Mar-2014.) |
⊢ (𝐾 ∈ (0...𝑁) → (𝑁C𝐾) ∈ ℝ+) | ||
Theorem | bccmpl 13858 | "Complementing" its second argument doesn't change a binary coefficient. (Contributed by NM, 21-Jun-2005.) (Revised by Mario Carneiro, 5-Mar-2014.) |
⊢ ((𝑁 ∈ ℕ0 ∧ 𝐾 ∈ ℤ) → (𝑁C𝐾) = (𝑁C(𝑁 − 𝐾))) | ||
Theorem | bcn0 13859 | 𝑁 choose 0 is 1. Remark in [Gleason] p. 296. (Contributed by NM, 17-Jun-2005.) (Revised by Mario Carneiro, 8-Nov-2013.) |
⊢ (𝑁 ∈ ℕ0 → (𝑁C0) = 1) | ||
Theorem | bc0k 13860 | The binomial coefficient " 0 choose 𝐾 " is 0 for a positive integer K. Note that (0C0) = 1 (see bcn0 13859). (Contributed by Alexander van der Vekens, 1-Jan-2018.) |
⊢ (𝐾 ∈ ℕ → (0C𝐾) = 0) | ||
Theorem | bcnn 13861 | 𝑁 choose 𝑁 is 1. Remark in [Gleason] p. 296. (Contributed by NM, 17-Jun-2005.) (Revised by Mario Carneiro, 8-Nov-2013.) |
⊢ (𝑁 ∈ ℕ0 → (𝑁C𝑁) = 1) | ||
Theorem | bcn1 13862 | Binomial coefficient: 𝑁 choose 1. (Contributed by NM, 21-Jun-2005.) (Revised by Mario Carneiro, 8-Nov-2013.) |
⊢ (𝑁 ∈ ℕ0 → (𝑁C1) = 𝑁) | ||
Theorem | bcnp1n 13863 | Binomial coefficient: 𝑁 + 1 choose 𝑁. (Contributed by NM, 20-Jun-2005.) (Revised by Mario Carneiro, 8-Nov-2013.) |
⊢ (𝑁 ∈ ℕ0 → ((𝑁 + 1)C𝑁) = (𝑁 + 1)) | ||
Theorem | bcm1k 13864 | The proportion of one binomial coefficient to another with 𝐾 decreased by 1. (Contributed by Mario Carneiro, 10-Mar-2014.) |
⊢ (𝐾 ∈ (1...𝑁) → (𝑁C𝐾) = ((𝑁C(𝐾 − 1)) · ((𝑁 − (𝐾 − 1)) / 𝐾))) | ||
Theorem | bcp1n 13865 | The proportion of one binomial coefficient to another with 𝑁 increased by 1. (Contributed by Mario Carneiro, 10-Mar-2014.) |
⊢ (𝐾 ∈ (0...𝑁) → ((𝑁 + 1)C𝐾) = ((𝑁C𝐾) · ((𝑁 + 1) / ((𝑁 + 1) − 𝐾)))) | ||
Theorem | bcp1nk 13866 | The proportion of one binomial coefficient to another with 𝑁 and 𝐾 increased by 1. (Contributed by Mario Carneiro, 16-Jan-2015.) |
⊢ (𝐾 ∈ (0...𝑁) → ((𝑁 + 1)C(𝐾 + 1)) = ((𝑁C𝐾) · ((𝑁 + 1) / (𝐾 + 1)))) | ||
Theorem | bcval5 13867 | Write out the top and bottom parts of the binomial coefficient (𝑁C𝐾) = (𝑁 · (𝑁 − 1) · ... · ((𝑁 − 𝐾) + 1)) / 𝐾! explicitly. In this form, it is valid even for 𝑁 < 𝐾, although it is no longer valid for nonpositive 𝐾. (Contributed by Mario Carneiro, 22-May-2014.) |
⊢ ((𝑁 ∈ ℕ0 ∧ 𝐾 ∈ ℕ) → (𝑁C𝐾) = ((seq((𝑁 − 𝐾) + 1)( · , I )‘𝑁) / (!‘𝐾))) | ||
Theorem | bcn2 13868 | Binomial coefficient: 𝑁 choose 2. (Contributed by Mario Carneiro, 22-May-2014.) |
⊢ (𝑁 ∈ ℕ0 → (𝑁C2) = ((𝑁 · (𝑁 − 1)) / 2)) | ||
Theorem | bcp1m1 13869 | Compute the binomial coefficient of (𝑁 + 1) over (𝑁 − 1) (Contributed by Scott Fenton, 11-May-2014.) (Revised by Mario Carneiro, 22-May-2014.) |
⊢ (𝑁 ∈ ℕ0 → ((𝑁 + 1)C(𝑁 − 1)) = (((𝑁 + 1) · 𝑁) / 2)) | ||
Theorem | bcpasc 13870 | Pascal's rule for the binomial coefficient, generalized to all integers 𝐾. Equation 2 of [Gleason] p. 295. (Contributed by NM, 13-Jul-2005.) (Revised by Mario Carneiro, 10-Mar-2014.) |
⊢ ((𝑁 ∈ ℕ0 ∧ 𝐾 ∈ ℤ) → ((𝑁C𝐾) + (𝑁C(𝐾 − 1))) = ((𝑁 + 1)C𝐾)) | ||
Theorem | bccl 13871 | A binomial coefficient, in its extended domain, is a nonnegative integer. (Contributed by NM, 10-Jul-2005.) (Revised by Mario Carneiro, 9-Nov-2013.) |
⊢ ((𝑁 ∈ ℕ0 ∧ 𝐾 ∈ ℤ) → (𝑁C𝐾) ∈ ℕ0) | ||
Theorem | bccl2 13872 | A binomial coefficient, in its standard domain, is a positive integer. (Contributed by NM, 3-Jan-2006.) (Revised by Mario Carneiro, 10-Mar-2014.) |
⊢ (𝐾 ∈ (0...𝑁) → (𝑁C𝐾) ∈ ℕ) | ||
Theorem | bcn2m1 13873 | Compute the binomial coefficient "𝑁 choose 2 " from "(𝑁 − 1) choose 2 ": (N-1) + ( (N-1) 2 ) = ( N 2 ). (Contributed by Alexander van der Vekens, 7-Jan-2018.) |
⊢ (𝑁 ∈ ℕ → ((𝑁 − 1) + ((𝑁 − 1)C2)) = (𝑁C2)) | ||
Theorem | bcn2p1 13874 | Compute the binomial coefficient "(𝑁 + 1) choose 2 " from "𝑁 choose 2 ": N + ( N 2 ) = ( (N+1) 2 ). (Contributed by Alexander van der Vekens, 8-Jan-2018.) |
⊢ (𝑁 ∈ ℕ0 → (𝑁 + (𝑁C2)) = ((𝑁 + 1)C2)) | ||
Theorem | permnn 13875 | The number of permutations of 𝑁 − 𝑅 objects from a collection of 𝑁 objects is a positive integer. (Contributed by Jason Orendorff, 24-Jan-2007.) |
⊢ (𝑅 ∈ (0...𝑁) → ((!‘𝑁) / (!‘𝑅)) ∈ ℕ) | ||
Theorem | bcnm1 13876 | The binomial coefficent of (𝑁 − 1) is 𝑁. (Contributed by Scott Fenton, 16-May-2014.) |
⊢ (𝑁 ∈ ℕ0 → (𝑁C(𝑁 − 1)) = 𝑁) | ||
Theorem | 4bc3eq4 13877 | The value of four choose three. (Contributed by Scott Fenton, 11-Jun-2016.) |
⊢ (4C3) = 4 | ||
Theorem | 4bc2eq6 13878 | The value of four choose two. (Contributed by Scott Fenton, 9-Jan-2017.) |
⊢ (4C2) = 6 | ||
Syntax | chash 13879 | Extend the definition of a class to include the set size function. |
class ♯ | ||
Definition | df-hash 13880 | Define the set size function ♯, which gives the cardinality of a finite set as a member of ℕ0, and assigns all infinite sets the value +∞. For example, (♯‘{0, 1, 2}) = 3 (ex-hash 28508). (Contributed by Paul Chapman, 22-Jun-2011.) |
⊢ ♯ = (((rec((𝑥 ∈ V ↦ (𝑥 + 1)), 0) ↾ ω) ∘ card) ∪ ((V ∖ Fin) × {+∞})) | ||
Theorem | hashkf 13881 | The finite part of the size function maps all finite sets to their cardinality, as members of ℕ0. (Contributed by Mario Carneiro, 13-Sep-2013.) (Revised by Mario Carneiro, 26-Dec-2014.) |
⊢ 𝐺 = (rec((𝑥 ∈ V ↦ (𝑥 + 1)), 0) ↾ ω) & ⊢ 𝐾 = (𝐺 ∘ card) ⇒ ⊢ 𝐾:Fin⟶ℕ0 | ||
Theorem | hashgval 13882* | The value of the ♯ function in terms of the mapping 𝐺 from ω to ℕ0. The proof avoids the use of ax-ac 10056. (Contributed by Paul Chapman, 22-Jun-2011.) (Revised by Mario Carneiro, 26-Dec-2014.) |
⊢ 𝐺 = (rec((𝑥 ∈ V ↦ (𝑥 + 1)), 0) ↾ ω) ⇒ ⊢ (𝐴 ∈ Fin → (𝐺‘(card‘𝐴)) = (♯‘𝐴)) | ||
Theorem | hashginv 13883* | The converse of 𝐺 maps the size function's value to card. (Contributed by Paul Chapman, 22-Jun-2011.) (Revised by Mario Carneiro, 15-Sep-2013.) |
⊢ 𝐺 = (rec((𝑥 ∈ V ↦ (𝑥 + 1)), 0) ↾ ω) ⇒ ⊢ (𝐴 ∈ Fin → (◡𝐺‘(♯‘𝐴)) = (card‘𝐴)) | ||
Theorem | hashinf 13884 | The value of the ♯ function on an infinite set. (Contributed by Mario Carneiro, 13-Jul-2014.) |
⊢ ((𝐴 ∈ 𝑉 ∧ ¬ 𝐴 ∈ Fin) → (♯‘𝐴) = +∞) | ||
Theorem | hashbnd 13885 | If 𝐴 has size bounded by an integer 𝐵, then 𝐴 is finite. (Contributed by Mario Carneiro, 14-Jun-2015.) |
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ ℕ0 ∧ (♯‘𝐴) ≤ 𝐵) → 𝐴 ∈ Fin) | ||
Theorem | hashfxnn0 13886 | The size function is a function into the extended nonnegative integers. (Contributed by Mario Carneiro, 13-Sep-2013.) (Revised by AV, 10-Dec-2020.) |
⊢ ♯:V⟶ℕ0* | ||
Theorem | hashf 13887 | The size function maps all finite sets to their cardinality, as members of ℕ0, and infinite sets to +∞. TODO-AV: mark as OBSOLETE and replace it by hashfxnn0 13886? (Contributed by Mario Carneiro, 13-Sep-2013.) (Revised by Mario Carneiro, 13-Jul-2014.) (Proof shortened by AV, 24-Oct-2021.) |
⊢ ♯:V⟶(ℕ0 ∪ {+∞}) | ||
Theorem | hashxnn0 13888 | The value of the hash function for a set is an extended nonnegative integer. (Contributed by Alexander van der Vekens, 6-Dec-2017.) (Revised by AV, 10-Dec-2020.) |
⊢ (𝑀 ∈ 𝑉 → (♯‘𝑀) ∈ ℕ0*) | ||
Theorem | hashresfn 13889 | Restriction of the domain of the size function. (Contributed by Thierry Arnoux, 31-Jan-2017.) |
⊢ (♯ ↾ 𝐴) Fn 𝐴 | ||
Theorem | dmhashres 13890 | Restriction of the domain of the size function. (Contributed by Thierry Arnoux, 12-Jan-2017.) |
⊢ dom (♯ ↾ 𝐴) = 𝐴 | ||
Theorem | hashnn0pnf 13891 | The value of the hash function for a set is either a nonnegative integer or positive infinity. TODO-AV: mark as OBSOLETE and replace it by hashxnn0 13888? (Contributed by Alexander van der Vekens, 6-Dec-2017.) |
⊢ (𝑀 ∈ 𝑉 → ((♯‘𝑀) ∈ ℕ0 ∨ (♯‘𝑀) = +∞)) | ||
Theorem | hashnnn0genn0 13892 | If the size of a set is not a nonnegative integer, it is greater than or equal to any nonnegative integer. (Contributed by Alexander van der Vekens, 6-Dec-2017.) |
⊢ ((𝑀 ∈ 𝑉 ∧ (♯‘𝑀) ∉ ℕ0 ∧ 𝑁 ∈ ℕ0) → 𝑁 ≤ (♯‘𝑀)) | ||
Theorem | hashnemnf 13893 | The size of a set is never minus infinity. (Contributed by Alexander van der Vekens, 21-Dec-2017.) |
⊢ (𝐴 ∈ 𝑉 → (♯‘𝐴) ≠ -∞) | ||
Theorem | hashv01gt1 13894 | The size of a set is either 0 or 1 or greater than 1. (Contributed by Alexander van der Vekens, 29-Dec-2017.) |
⊢ (𝑀 ∈ 𝑉 → ((♯‘𝑀) = 0 ∨ (♯‘𝑀) = 1 ∨ 1 < (♯‘𝑀))) | ||
Theorem | hashfz1 13895 | The set (1...𝑁) has 𝑁 elements. (Contributed by Paul Chapman, 22-Jun-2011.) (Revised by Mario Carneiro, 15-Sep-2013.) |
⊢ (𝑁 ∈ ℕ0 → (♯‘(1...𝑁)) = 𝑁) | ||
Theorem | hashen 13896 | Two finite sets have the same number of elements iff they are equinumerous. (Contributed by Paul Chapman, 22-Jun-2011.) (Revised by Mario Carneiro, 15-Sep-2013.) |
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) → ((♯‘𝐴) = (♯‘𝐵) ↔ 𝐴 ≈ 𝐵)) | ||
Theorem | hasheni 13897 | Equinumerous sets have the same number of elements (even if they are not finite). (Contributed by Mario Carneiro, 15-Apr-2015.) |
⊢ (𝐴 ≈ 𝐵 → (♯‘𝐴) = (♯‘𝐵)) | ||
Theorem | hasheqf1o 13898* | The size of two finite sets is equal if and only if there is a bijection mapping one of the sets onto the other. (Contributed by Alexander van der Vekens, 17-Dec-2017.) |
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) → ((♯‘𝐴) = (♯‘𝐵) ↔ ∃𝑓 𝑓:𝐴–1-1-onto→𝐵)) | ||
Theorem | fiinfnf1o 13899* | There is no bijection between a finite set and an infinite set. (Contributed by Alexander van der Vekens, 25-Dec-2017.) |
⊢ ((𝐴 ∈ Fin ∧ ¬ 𝐵 ∈ Fin) → ¬ ∃𝑓 𝑓:𝐴–1-1-onto→𝐵) | ||
Theorem | focdmex 13900 | The codomain of an onto function is a set if its domain is a set. (Contributed by AV, 4-May-2021.) |
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐹:𝐴–onto→𝐵) → 𝐵 ∈ V) |
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