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Theorem List for Metamath Proof Explorer - 12901-13000   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremleexp2d 12901 Ordering law for exponentiation. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)    &   (𝜑 → 1 < 𝐴)       (𝜑 → (𝑀𝑁 ↔ (𝐴𝑀) ≤ (𝐴𝑁)))

Theoremexpcand 12902 Ordering relationship for exponentiation. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)    &   (𝜑 → 1 < 𝐴)    &   (𝜑 → (𝐴𝑀) = (𝐴𝑁))       (𝜑𝑀 = 𝑁)

Theoremleexp2ad 12903 Ordering relationship for exponentiation. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑 → 1 ≤ 𝐴)    &   (𝜑𝑁 ∈ (ℤ𝑀))       (𝜑 → (𝐴𝑀) ≤ (𝐴𝑁))

Theoremleexp2rd 12904 Ordering relationship for exponentiation. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝑀 ∈ ℕ0)    &   (𝜑𝑁 ∈ (ℤ𝑀))    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑𝐴 ≤ 1)       (𝜑 → (𝐴𝑁) ≤ (𝐴𝑀))

Theoremlt2sqd 12905 The square function on nonnegative reals is strictly monotonic. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐵 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑 → 0 ≤ 𝐵)       (𝜑 → (𝐴 < 𝐵 ↔ (𝐴↑2) < (𝐵↑2)))

Theoremle2sqd 12906 The square function on nonnegative reals is monotonic. (Contributed by Mario Carneiro, 28-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐵 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑 → 0 ≤ 𝐵)       (𝜑 → (𝐴𝐵 ↔ (𝐴↑2) ≤ (𝐵↑2)))

Theoremsq11d 12907 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 12908 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 12909 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 12910 The square of 10 is 100. (Contributed by AV, 14-Jun-2021.) (Revised by AV, 1-Aug-2021.)
(10↑2) = 100

Theoremsq10e99m1 12911 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 12912 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 12913 Old version of sq10 12910. Obsolete as of 1-Aug-2021. (Contributed by AV, 14-Jun-2021.) (Proof modification is discouraged.) (New usage is discouraged.)
(10↑2) = 100

Theoremsq10e99m1OLD 12914 Old version of sq10e99m1 12911. 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 12915 Old version of 3dec 12912. 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 12916 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 12917 A rather pretty lemma for nn0opthi 12919. (Contributed by Raph Levien, 10-Dec-2002.)
𝐴 ∈ ℕ0    &   𝐶 ∈ ℕ0       (𝐴 < 𝐶 ↔ ((𝐴 · 𝐴) + (2 · 𝐴)) < (𝐶 · 𝐶))

Theoremnn0opthlem2 12918 Lemma for nn0opthi 12919. (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 12919 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 4132 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 12920 An ordered pair theorem for nonnegative integers. Theorem 17.3 of [Quine] p. 124. See comments for nn0opthi 12919. (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 12921 An ordered pair theorem for nonnegative integers. Theorem 17.3 of [Quine] p. 124. See nn0opthi 12919. (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 12922 Extend class notation to include the factorial of nonnegative integers.
class !

Definitiondf-fac 12923 Define the factorial function on nonnegative integers. For example, (!‘5) = 120 because 1 · 2 · 3 · 4 · 5 = 120 (ex-fac 26700). In the literature, the factorial function is written as a postscript exclamation point. (Contributed by NM, 2-Dec-2004.)
! = ({⟨0, 1⟩} ∪ seq1( · , I ))

Theoremfacnn 12924 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 12925 The factorial of 0. (Contributed by NM, 2-Dec-2004.) (Revised by Mario Carneiro, 13-Jul-2013.)
(!‘0) = 1

Theoremfac1 12926 The factorial of 1. (Contributed by NM, 2-Dec-2004.) (Revised by Mario Carneiro, 13-Jul-2013.)
(!‘1) = 1

Theoremfacp1 12927 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 12928 The factorial of 2. (Contributed by NM, 17-Mar-2005.)
(!‘2) = 2

Theoremfac3 12929 The factorial of 3. (Contributed by NM, 17-Mar-2005.)
(!‘3) = 6

Theoremfac4 12930 The factorial of 4. (Contributed by Mario Carneiro, 18-Jun-2015.)
(!‘4) = 24

Theoremfacnn2 12931 Value of the factorial function expressed recursively. (Contributed by NM, 2-Dec-2004.)
(𝑁 ∈ ℕ → (!‘𝑁) = ((!‘(𝑁 − 1)) · 𝑁))

Theoremfaccl 12932 Closure of the factorial function. (Contributed by NM, 2-Dec-2004.)
(𝑁 ∈ ℕ0 → (!‘𝑁) ∈ ℕ)

Theoremfaccld 12933 Closure of the factorial function, deduction version of faccl 12932. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
(𝜑𝑁 ∈ ℕ0)       (𝜑 → (!‘𝑁) ∈ ℕ)

Theoremfacmapnn 12934 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 12935 The factorial function is nonzero. (Contributed by NM, 26-Apr-2005.)
(𝑁 ∈ ℕ0 → (!‘𝑁) ≠ 0)

Theoremfacdiv 12936 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 12937 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 12938 Ordering property of factorial. (Contributed by NM, 9-Dec-2005.)
((𝑀 ∈ ℕ0𝑁 ∈ ℕ0𝑀𝑁) → (!‘𝑀) ≤ (!‘𝑁))

Theoremfaclbnd 12939 A lower bound for the factorial function. (Contributed by NM, 17-Dec-2005.)
((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → (𝑀↑(𝑁 + 1)) ≤ ((𝑀𝑀) · (!‘𝑁)))

Theoremfaclbnd2 12940 A lower bound for the factorial function. (Contributed by NM, 17-Dec-2005.)
(𝑁 ∈ ℕ0 → ((2↑𝑁) / 2) ≤ (!‘𝑁))

Theoremfaclbnd3 12941 A lower bound for the factorial function. (Contributed by NM, 19-Dec-2005.)
((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → (𝑀𝑁) ≤ ((𝑀𝑀) · (!‘𝑁)))

Theoremfaclbnd4lem1 12942 Lemma for faclbnd4 12946. 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 12943 Lemma for faclbnd4 12946. Use the weak deduction theorem to convert the hypotheses of faclbnd4lem1 12942 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 12944 Lemma for faclbnd4 12946. 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 12945 Lemma for faclbnd4 12946. 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 12946 Variant of faclbnd5 12947 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 12947 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 12948 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 12949 An upper bound for the factorial function. (Contributed by Mario Carneiro, 15-Apr-2016.)
(𝑁 ∈ ℕ0 → (!‘𝑁) ≤ (𝑁𝑁))

Theoremfacavg 12950 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 12951 Extend class notation to include the binomial coefficient operation (combinatorial choose operation).
class C

Definitiondf-bc 12952* Define the binomial coefficient operation. For example, (5C3) = 10 (ex-bc 26701).

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 12953 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 12954 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 12954 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 12955 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 12956 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 12957 Closure of the binomial coefficient in the positive reals. (This is mostly a lemma before we have bccl2 12972.) (Contributed by Mario Carneiro, 10-Mar-2014.)
(𝐾 ∈ (0...𝑁) → (𝑁C𝐾) ∈ ℝ+)

Theorembccmpl 12958 "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 12959 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 12960 The binomial coefficient " 0 choose 𝐾 " is 0 for a positive integer K. Note that (0C0) = 1 (see bcn0 12959). (Contributed by Alexander van der Vekens, 1-Jan-2018.)
(𝐾 ∈ ℕ → (0C𝐾) = 0)

Theorembcnn 12961 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 12962 Binomial coefficient: 𝑁 choose 1. (Contributed by NM, 21-Jun-2005.) (Revised by Mario Carneiro, 8-Nov-2013.)
(𝑁 ∈ ℕ0 → (𝑁C1) = 𝑁)

Theorembcnp1n 12963 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 12964 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 12965 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 12966 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 12967 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 12968 Binomial coefficient: 𝑁 choose 2. (Contributed by Mario Carneiro, 22-May-2014.)
(𝑁 ∈ ℕ0 → (𝑁C2) = ((𝑁 · (𝑁 − 1)) / 2))

Theorembcp1m1 12969 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 12970 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 12971 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 12972 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 12973 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 12974 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 12975 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 12976 The binomial coefficent of (𝑁 − 1) is 𝑁. (Contributed by Scott Fenton, 16-May-2014.)
(𝑁 ∈ ℕ0 → (𝑁C(𝑁 − 1)) = 𝑁)

Theorem4bc3eq4 12977 The value of four choose three. (Contributed by Scott Fenton, 11-Jun-2016.)
(4C3) = 4

Theorem4bc2eq6 12978 The value of four choose two. (Contributed by Scott Fenton, 9-Jan-2017.)
(4C2) = 6

5.6.11  The ` # ` (set size) function

Syntaxchash 12979 Extend the definition of a class to include the set size function.
class #

Definitiondf-hash 12980 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 26702). (Contributed by Paul Chapman, 22-Jun-2011.)
# = (((rec((𝑥 ∈ V ↦ (𝑥 + 1)), 0) ↾ ω) ∘ card) ∪ ((V ∖ Fin) × {+∞}))

Theoremhashkf 12981 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 12982* The value of the # function in terms of the mapping 𝐺 from ω to 0. The proof avoids the use of ax-ac 9164. (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 12983* 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 12984 The value of the # function on an infinite set. (Contributed by Mario Carneiro, 13-Jul-2014.)
((𝐴𝑉 ∧ ¬ 𝐴 ∈ Fin) → (#‘𝐴) = +∞)

Theoremhashbnd 12985 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 12986 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 12987 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 12986? (Contributed by Mario Carneiro, 13-Sep-2013.) (Revised by Mario Carneiro, 13-Jul-2014.) (Proof shortened by AV, 24-Oct-2021.)
#:V⟶(ℕ0 ∪ {+∞})

TheoremhashfOLD 12988 Obsolete version of hashf 12987 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 12989 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 12990 Restriction of the domain of the size function. (Contributed by Thierry Arnoux, 31-Jan-2017.)
(# ↾ 𝐴) Fn 𝐴

Theoremdmhashres 12991 Restriction of the domain of the size function. (Contributed by Thierry Arnoux, 12-Jan-2017.)
dom (# ↾ 𝐴) = 𝐴

Theoremhashnn0pnf 12992 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 12989? (Contributed by Alexander van der Vekens, 6-Dec-2017.)
(𝑀𝑉 → ((#‘𝑀) ∈ ℕ0 ∨ (#‘𝑀) = +∞))

Theoremhashnnn0genn0 12993 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 12994 The size of a set is never minus infinity. (Contributed by Alexander van der Vekens, 21-Dec-2017.)
(𝐴𝑉 → (#‘𝐴) ≠ -∞)

Theoremhashv01gt1 12995 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 12996 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 12997 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 12998 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 12999* 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 13000* 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)

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