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Theorem List for Metamath Proof Explorer - 13001-13100   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremsq11d 13001 The square function is one-to-one for nonnegative reals. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐵 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑 → 0 ≤ 𝐵)    &   (𝜑 → (𝐴↑2) = (𝐵↑2))       (𝜑𝐴 = 𝐵)
 
Theoremmulsubdivbinom2 13002 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)))
 
Theoremmuldivbinom2 13003 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)))
 
Theoremsq10 13004 The square of 10 is 100. (Contributed by AV, 14-Jun-2021.) (Revised by AV, 1-Aug-2021.)
(10↑2) = 100
 
Theoremsq10e99m1 13005 The square of 10 is 99 plus 1. (Contributed by AV, 14-Jun-2021.) (Revised by AV, 1-Aug-2021.)
(10↑2) = (99 + 1)
 
Theorem3dec 13006 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)
 
Theoremsq10OLD 13007 Old version of sq10 13004. Obsolete as of 1-Aug-2021. (Contributed by AV, 14-Jun-2021.) (Proof modification is discouraged.) (New usage is discouraged.)
(10↑2) = 100
 
Theoremsq10e99m1OLD 13008 Old version of sq10e99m1 13005. Obsolete as of 1-Aug-2021. (Contributed by AV, 14-Jun-2021.) (Proof modification is discouraged.) (New usage is discouraged.)
(10↑2) = (99 + 1)
 
Theorem3decOLD 13009 Old version of 3dec 13006. Obsolete as of 1-Aug-2021. (Contributed by AV, 14-Jun-2021.) (Proof modification is discouraged.) (New usage is discouraged.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0       𝐴𝐵𝐶 = ((((10↑2) · 𝐴) + (10 · 𝐵)) + 𝐶)
 
5.6.8  Ordered pair theorem for nonnegative integers
 
Theoremnn0le2msqi 13010 The square function on nonnegative integers is monotonic. (Contributed by Raph Levien, 10-Dec-2002.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0       (𝐴𝐵 ↔ (𝐴 · 𝐴) ≤ (𝐵 · 𝐵))
 
Theoremnn0opthlem1 13011 A rather pretty lemma for nn0opthi 13013. (Contributed by Raph Levien, 10-Dec-2002.)
𝐴 ∈ ℕ0    &   𝐶 ∈ ℕ0       (𝐴 < 𝐶 ↔ ((𝐴 · 𝐴) + (2 · 𝐴)) < (𝐶 · 𝐶))
 
Theoremnn0opthlem2 13012 Lemma for nn0opthi 13013. (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))
 
Theoremnn0opthi 13013 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 4162 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 = 𝐶𝐵 = 𝐷))
 
Theoremnn0opth2i 13014 An ordered pair theorem for nonnegative integers. Theorem 17.3 of [Quine] p. 124. See comments for nn0opthi 13013. (Contributed by NM, 22-Jul-2004.)
𝐴 ∈ ℕ0    &   𝐵 ∈ ℕ0    &   𝐶 ∈ ℕ0    &   𝐷 ∈ ℕ0       ((((𝐴 + 𝐵)↑2) + 𝐵) = (((𝐶 + 𝐷)↑2) + 𝐷) ↔ (𝐴 = 𝐶𝐵 = 𝐷))
 
Theoremnn0opth2 13015 An ordered pair theorem for nonnegative integers. Theorem 17.3 of [Quine] p. 124. See nn0opthi 13013. (Contributed by NM, 22-Jul-2004.)
(((𝐴 ∈ ℕ0𝐵 ∈ ℕ0) ∧ (𝐶 ∈ ℕ0𝐷 ∈ ℕ0)) → ((((𝐴 + 𝐵)↑2) + 𝐵) = (((𝐶 + 𝐷)↑2) + 𝐷) ↔ (𝐴 = 𝐶𝐵 = 𝐷)))
 
5.6.9  Factorial function
 
Syntaxcfa 13016 Extend class notation to include the factorial of nonnegative integers.
class !
 
Definitiondf-fac 13017 Define the factorial function on nonnegative integers. For example, (!‘5) = 120 because 1 · 2 · 3 · 4 · 5 = 120 (ex-fac 27196). In the literature, the factorial function is written as a postscript exclamation point. (Contributed by NM, 2-Dec-2004.)
! = ({⟨0, 1⟩} ∪ seq1( · , I ))
 
Theoremfacnn 13018 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))
 
Theoremfac0 13019 The factorial of 0. (Contributed by NM, 2-Dec-2004.) (Revised by Mario Carneiro, 13-Jul-2013.)
(!‘0) = 1
 
Theoremfac1 13020 The factorial of 1. (Contributed by NM, 2-Dec-2004.) (Revised by Mario Carneiro, 13-Jul-2013.)
(!‘1) = 1
 
Theoremfacp1 13021 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)))
 
Theoremfac2 13022 The factorial of 2. (Contributed by NM, 17-Mar-2005.)
(!‘2) = 2
 
Theoremfac3 13023 The factorial of 3. (Contributed by NM, 17-Mar-2005.)
(!‘3) = 6
 
Theoremfac4 13024 The factorial of 4. (Contributed by Mario Carneiro, 18-Jun-2015.)
(!‘4) = 24
 
Theoremfacnn2 13025 Value of the factorial function expressed recursively. (Contributed by NM, 2-Dec-2004.)
(𝑁 ∈ ℕ → (!‘𝑁) = ((!‘(𝑁 − 1)) · 𝑁))
 
Theoremfaccl 13026 Closure of the factorial function. (Contributed by NM, 2-Dec-2004.)
(𝑁 ∈ ℕ0 → (!‘𝑁) ∈ ℕ)
 
Theoremfaccld 13027 Closure of the factorial function, deduction version of faccl 13026. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
(𝜑𝑁 ∈ ℕ0)       (𝜑 → (!‘𝑁) ∈ ℕ)
 
Theoremfacmapnn 13028 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)
 
Theoremfacne0 13029 The factorial function is nonzero. (Contributed by NM, 26-Apr-2005.)
(𝑁 ∈ ℕ0 → (!‘𝑁) ≠ 0)
 
Theoremfacdiv 13030 A positive integer divides the factorial of an equal or larger number. (Contributed by NM, 2-May-2005.)
((𝑀 ∈ ℕ0𝑁 ∈ ℕ ∧ 𝑁𝑀) → ((!‘𝑀) / 𝑁) ∈ ℕ)
 
Theoremfacndiv 13031 No positive integer (greater than one) divides the factorial plus one of an equal or larger number. (Contributed by NM, 3-May-2005.)
(((𝑀 ∈ ℕ0𝑁 ∈ ℕ) ∧ (1 < 𝑁𝑁𝑀)) → ¬ (((!‘𝑀) + 1) / 𝑁) ∈ ℤ)
 
Theoremfacwordi 13032 Ordering property of factorial. (Contributed by NM, 9-Dec-2005.)
((𝑀 ∈ ℕ0𝑁 ∈ ℕ0𝑀𝑁) → (!‘𝑀) ≤ (!‘𝑁))
 
Theoremfaclbnd 13033 A lower bound for the factorial function. (Contributed by NM, 17-Dec-2005.)
((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → (𝑀↑(𝑁 + 1)) ≤ ((𝑀𝑀) · (!‘𝑁)))
 
Theoremfaclbnd2 13034 A lower bound for the factorial function. (Contributed by NM, 17-Dec-2005.)
(𝑁 ∈ ℕ0 → ((2↑𝑁) / 2) ≤ (!‘𝑁))
 
Theoremfaclbnd3 13035 A lower bound for the factorial function. (Contributed by NM, 19-Dec-2005.)
((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → (𝑀𝑁) ≤ ((𝑀𝑀) · (!‘𝑁)))
 
Theoremfaclbnd4lem1 13036 Lemma for faclbnd4 13040. 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)) · (𝑀𝑁)) ≤ (((2↑((𝐾 + 1)↑2)) · (𝑀↑(𝑀 + (𝐾 + 1)))) · (!‘𝑁)))
 
Theoremfaclbnd4lem2 13037 Lemma for faclbnd4 13040. Use the weak deduction theorem to convert the hypotheses of faclbnd4lem1 13036 to antecedents. (Contributed by NM, 23-Dec-2005.)
((𝑀 ∈ ℕ0𝐾 ∈ ℕ0𝑁 ∈ ℕ) → ((((𝑁 − 1)↑𝐾) · (𝑀↑(𝑁 − 1))) ≤ (((2↑(𝐾↑2)) · (𝑀↑(𝑀 + 𝐾))) · (!‘(𝑁 − 1))) → ((𝑁↑(𝐾 + 1)) · (𝑀𝑁)) ≤ (((2↑((𝐾 + 1)↑2)) · (𝑀↑(𝑀 + (𝐾 + 1)))) · (!‘𝑁))))
 
Theoremfaclbnd4lem3 13038 Lemma for faclbnd4 13040. The 𝑁 = 0 case. (Contributed by NM, 23-Dec-2005.)
(((𝑀 ∈ ℕ0𝐾 ∈ ℕ0) ∧ 𝑁 = 0) → ((𝑁𝐾) · (𝑀𝑁)) ≤ (((2↑(𝐾↑2)) · (𝑀↑(𝑀 + 𝐾))) · (!‘𝑁)))
 
Theoremfaclbnd4lem4 13039 Lemma for faclbnd4 13040. Prove the 0 < 𝑁 case by induction on 𝐾. (Contributed by NM, 19-Dec-2005.)
((𝑁 ∈ ℕ ∧ 𝐾 ∈ ℕ0𝑀 ∈ ℕ0) → ((𝑁𝐾) · (𝑀𝑁)) ≤ (((2↑(𝐾↑2)) · (𝑀↑(𝑀 + 𝐾))) · (!‘𝑁)))
 
Theoremfaclbnd4 13040 Variant of faclbnd5 13041 providing a non-strict lower bound. (Contributed by NM, 23-Dec-2005.)
((𝑁 ∈ ℕ0𝐾 ∈ ℕ0𝑀 ∈ ℕ0) → ((𝑁𝐾) · (𝑀𝑁)) ≤ (((2↑(𝐾↑2)) · (𝑀↑(𝑀 + 𝐾))) · (!‘𝑁)))
 
Theoremfaclbnd5 13041 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 ∈ ℕ0𝐾 ∈ ℕ0𝑀 ∈ ℕ) → ((𝑁𝐾) · (𝑀𝑁)) < ((2 · ((2↑(𝐾↑2)) · (𝑀↑(𝑀 + 𝐾)))) · (!‘𝑁)))
 
Theoremfaclbnd6 13042 Geometric lower bound for the factorial function, where N is usually held constant. (Contributed by Paul Chapman, 28-Dec-2007.)
((𝑁 ∈ ℕ0𝑀 ∈ ℕ0) → ((!‘𝑁) · ((𝑁 + 1)↑𝑀)) ≤ (!‘(𝑁 + 𝑀)))
 
Theoremfacubnd 13043 An upper bound for the factorial function. (Contributed by Mario Carneiro, 15-Apr-2016.)
(𝑁 ∈ ℕ0 → (!‘𝑁) ≤ (𝑁𝑁))
 
Theoremfacavg 13044 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 ∈ ℕ0𝑁 ∈ ℕ0) → (!‘(⌊‘((𝑀 + 𝑁) / 2))) ≤ ((!‘𝑀) · (!‘𝑁)))
 
5.6.10  The binomial coefficient operation
 
Syntaxcbc 13045 Extend class notation to include the binomial coefficient operation (combinatorial choose operation).
class C
 
Definitiondf-bc 13046* Define the binomial coefficient operation. For example, (5C3) = 10 (ex-bc 27197).

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". (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 ≤ 𝑘𝑛 does not hold. (Contributed by NM, 10-Jul-2005.)

C = (𝑛 ∈ ℕ0, 𝑘 ∈ ℤ ↦ if(𝑘 ∈ (0...𝑛), ((!‘𝑛) / ((!‘(𝑛𝑘)) · (!‘𝑘))), 0))
 
Theorembcval 13047 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 ≤ 𝐾𝑁 does not hold. See bcval2 13048 for the value in the standard domain. (Contributed by NM, 10-Jul-2005.) (Revised by Mario Carneiro, 7-Nov-2013.)
((𝑁 ∈ ℕ0𝐾 ∈ ℤ) → (𝑁C𝐾) = if(𝐾 ∈ (0...𝑁), ((!‘𝑁) / ((!‘(𝑁𝐾)) · (!‘𝐾))), 0))
 
Theorembcval2 13048 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) / ((!‘(𝑁𝐾)) · (!‘𝐾))))
 
Theorembcval3 13049 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 ∈ ℕ0𝐾 ∈ ℤ ∧ ¬ 𝐾 ∈ (0...𝑁)) → (𝑁C𝐾) = 0)
 
Theorembcval4 13050 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 ∈ ℕ0𝐾 ∈ ℤ ∧ (𝐾 < 0 ∨ 𝑁 < 𝐾)) → (𝑁C𝐾) = 0)
 
Theorembcrpcl 13051 Closure of the binomial coefficient in the positive reals. (This is mostly a lemma before we have bccl2 13066.) (Contributed by Mario Carneiro, 10-Mar-2014.)
(𝐾 ∈ (0...𝑁) → (𝑁C𝐾) ∈ ℝ+)
 
Theorembccmpl 13052 "Complementing" its second argument doesn't change a binary coefficient. (Contributed by NM, 21-Jun-2005.) (Revised by Mario Carneiro, 5-Mar-2014.)
((𝑁 ∈ ℕ0𝐾 ∈ ℤ) → (𝑁C𝐾) = (𝑁C(𝑁𝐾)))
 
Theorembcn0 13053 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)
 
Theorembc0k 13054 The binomial coefficient " 0 choose 𝐾 " is 0 for a positive integer K. Note that (0C0) = 1 (see bcn0 13053). (Contributed by Alexander van der Vekens, 1-Jan-2018.)
(𝐾 ∈ ℕ → (0C𝐾) = 0)
 
Theorembcnn 13055 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)
 
Theorembcn1 13056 Binomial coefficient: 𝑁 choose 1. (Contributed by NM, 21-Jun-2005.) (Revised by Mario Carneiro, 8-Nov-2013.)
(𝑁 ∈ ℕ0 → (𝑁C1) = 𝑁)
 
Theorembcnp1n 13057 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))
 
Theorembcm1k 13058 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)))
 
Theorembcp1n 13059 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))))
 
Theorembcp1nk 13060 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))))
 
Theorembcval5 13061 Write out the top and bottom parts of the binomial coefficient (𝑁C𝐾) = (𝑁 · (𝑁 − 1) · ... · ((𝑁𝐾) + 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 ∈ ℕ0𝐾 ∈ ℕ) → (𝑁C𝐾) = ((seq((𝑁𝐾) + 1)( · , I )‘𝑁) / (!‘𝐾)))
 
Theorembcn2 13062 Binomial coefficient: 𝑁 choose 2. (Contributed by Mario Carneiro, 22-May-2014.)
(𝑁 ∈ ℕ0 → (𝑁C2) = ((𝑁 · (𝑁 − 1)) / 2))
 
Theorembcp1m1 13063 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))
 
Theorembcpasc 13064 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 ∈ ℕ0𝐾 ∈ ℤ) → ((𝑁C𝐾) + (𝑁C(𝐾 − 1))) = ((𝑁 + 1)C𝐾))
 
Theorembccl 13065 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 ∈ ℕ0𝐾 ∈ ℤ) → (𝑁C𝐾) ∈ ℕ0)
 
Theorembccl2 13066 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)
 
Theorembcn2m1 13067 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))
 
Theorembcn2p1 13068 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))
 
Theorempermnn 13069 The number of permutations of 𝑁𝑅 objects from a collection of 𝑁 objects is a positive integer. (Contributed by Jason Orendorff, 24-Jan-2007.)
(𝑅 ∈ (0...𝑁) → ((!‘𝑁) / (!‘𝑅)) ∈ ℕ)
 
Theorembcnm1 13070 The binomial coefficent of (𝑁 − 1) is 𝑁. (Contributed by Scott Fenton, 16-May-2014.)
(𝑁 ∈ ℕ0 → (𝑁C(𝑁 − 1)) = 𝑁)
 
Theorem4bc3eq4 13071 The value of four choose three. (Contributed by Scott Fenton, 11-Jun-2016.)
(4C3) = 4
 
Theorem4bc2eq6 13072 The value of four choose two. (Contributed by Scott Fenton, 9-Jan-2017.)
(4C2) = 6
 
5.6.11  The ` # ` (set size) function
 
Syntaxchash 13073 Extend the definition of a class to include the set size function.
class #
 
Definitiondf-hash 13074 Define the set size function #, which gives the cardinality of a finite set as a member of 0, and assigns all infinite sets the value +∞. For example, (#‘{0, 1, 2}) = 3 (ex-hash 27198). (Contributed by Paul Chapman, 22-Jun-2011.)
# = (((rec((𝑥 ∈ V ↦ (𝑥 + 1)), 0) ↾ ω) ∘ card) ∪ ((V ∖ Fin) × {+∞}))
 
Theoremhashkf 13075 The finite part of the size function maps all finite sets to their cardinality, as members of 0. (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
 
Theoremhashgval 13076* The value of the # function in terms of the mapping 𝐺 from ω to 0. The proof avoids the use of ax-ac 9241. (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))
 
Theoremhashginv 13077* 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))
 
Theoremhashinf 13078 The value of the # function on an infinite set. (Contributed by Mario Carneiro, 13-Jul-2014.)
((𝐴𝑉 ∧ ¬ 𝐴 ∈ Fin) → (#‘𝐴) = +∞)
 
Theoremhashbnd 13079 If 𝐴 has size bounded by an integer 𝐵, then 𝐴 is finite. (Contributed by Mario Carneiro, 14-Jun-2015.)
((𝐴𝑉𝐵 ∈ ℕ0 ∧ (#‘𝐴) ≤ 𝐵) → 𝐴 ∈ Fin)
 
Theoremhashfxnn0 13080 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*
 
Theoremhashf 13081 The size function maps all finite sets to their cardinality, as members of 0, and infinite sets to +∞. TODO-AV: mark as OBSOLETE and replace it by hashfxnn0 13080? (Contributed by Mario Carneiro, 13-Sep-2013.) (Revised by Mario Carneiro, 13-Jul-2014.) (Proof shortened by AV, 24-Oct-2021.)
#:V⟶(ℕ0 ∪ {+∞})
 
TheoremhashfOLD 13082 Obsolete version of hashf 13081 as of 24-Oct-2021. The size function maps all finite sets to their cardinality, as members of 0, and infinite sets to +∞. (Contributed by Mario Carneiro, 13-Sep-2013.) (Revised by Mario Carneiro, 13-Jul-2014.) (New usage is discouraged.) (Proof modification is discouraged.)
#:V⟶(ℕ0 ∪ {+∞})
 
Theoremhashxnn0 13083 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.)
(𝑀𝑉 → (#‘𝑀) ∈ ℕ0*)
 
Theoremhashresfn 13084 Restriction of the domain of the size function. (Contributed by Thierry Arnoux, 31-Jan-2017.)
(# ↾ 𝐴) Fn 𝐴
 
Theoremdmhashres 13085 Restriction of the domain of the size function. (Contributed by Thierry Arnoux, 12-Jan-2017.)
dom (# ↾ 𝐴) = 𝐴
 
Theoremhashnn0pnf 13086 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 13083? (Contributed by Alexander van der Vekens, 6-Dec-2017.)
(𝑀𝑉 → ((#‘𝑀) ∈ ℕ0 ∨ (#‘𝑀) = +∞))
 
Theoremhashnnn0genn0 13087 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.)
((𝑀𝑉 ∧ (#‘𝑀) ∉ ℕ0𝑁 ∈ ℕ0) → 𝑁 ≤ (#‘𝑀))
 
Theoremhashnemnf 13088 The size of a set is never minus infinity. (Contributed by Alexander van der Vekens, 21-Dec-2017.)
(𝐴𝑉 → (#‘𝐴) ≠ -∞)
 
Theoremhashv01gt1 13089 The size of a set is either 0 or 1 or greater than 1. (Contributed by Alexander van der Vekens, 29-Dec-2017.)
(𝑀𝑉 → ((#‘𝑀) = 0 ∨ (#‘𝑀) = 1 ∨ 1 < (#‘𝑀)))
 
Theoremhashfz1 13090 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)
 
Theoremhashen 13091 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) ↔ 𝐴𝐵))
 
Theoremhasheni 13092 Equinumerous sets have the same number of elements (even if they are not finite). (Contributed by Mario Carneiro, 15-Apr-2015.)
(𝐴𝐵 → (#‘𝐴) = (#‘𝐵))
 
Theoremhasheqf1o 13093* 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:𝐴1-1-onto𝐵))
 
Theoremfiinfnf1o 13094* 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:𝐴1-1-onto𝐵)
 
Theoremfocdmex 13095 The codomain of an onto function is a set if its domain is a set. (Contributed by AV, 4-May-2021.)
((𝐴𝑉𝐹:𝐴onto𝐵) → 𝐵 ∈ V)
 
Theoremhasheqf1oi 13096* The size of two sets is equal if there is a bijection mapping one of the sets onto the other. (Contributed by Alexander van der Vekens, 25-Dec-2017.) (Revised by AV, 4-May-2021.)
(𝐴𝑉 → (∃𝑓 𝑓:𝐴1-1-onto𝐵 → (#‘𝐴) = (#‘𝐵)))
 
Theoremhasheqf1oiOLD 13097* Obsolete version of hasheqf1oi 13096 as of 4-May-2021. (Contributed by Alexander van der Vekens, 25-Dec-2017.) (New usage is discouraged.) (Proof modification is discouraged.)
((𝐴𝑉𝐵𝑊) → (∃𝑓 𝑓:𝐴1-1-onto𝐵 → (#‘𝐴) = (#‘𝐵)))
 
Theoremhashf1rn 13098 The size of a finite set which is a one-to-one function is equal to the size of the function's range. (Contributed by Alexander van der Vekens, 12-Jan-2018.) (Revised by AV, 4-May-2021.)
((𝐴𝑉𝐹:𝐴1-1𝐵) → (#‘𝐹) = (#‘ran 𝐹))
 
Theoremhashf1rnOLD 13099 Obsolete version of hashf1rn 13098 as of 4-May-2021. (Contributed by Alexander van der Vekens, 12-Jan-2018.) (Revised by Alexander van der Vekens, 4-Feb-2018.) (New usage is discouraged.) (Proof modification is discouraged.)
((𝐴𝑉𝐵𝑊) → (𝐹:𝐴1-1𝐵 → (#‘𝐹) = (#‘ran 𝐹)))
 
Theoremhasheqf1od 13100 The size of two sets is equal if there is a bijection mapping one of the sets onto the other. (Contributed by AV, 4-May-2021.)
(𝜑𝐴𝑈)    &   (𝜑𝐹:𝐴1-1-onto𝐵)       (𝜑 → (#‘𝐴) = (#‘𝐵))
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