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Theorem List for Metamath Proof Explorer - 13101-13200   *Has distinct variable group(s)
TypeLabelDescription
Statement

Definitiondf-fac 13101 Define the factorial function on nonnegative integers. For example, (!‘5) = 120 because 1 · 2 · 3 · 4 · 5 = 120 (ex-fac 27438). In the literature, the factorial function is written as a postscript exclamation point. (Contributed by NM, 2-Dec-2004.)
! = ({⟨0, 1⟩} ∪ seq1( · , I ))

Theoremfacnn 13102 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 13103 The factorial of 0. (Contributed by NM, 2-Dec-2004.) (Revised by Mario Carneiro, 13-Jul-2013.)
(!‘0) = 1

Theoremfac1 13104 The factorial of 1. (Contributed by NM, 2-Dec-2004.) (Revised by Mario Carneiro, 13-Jul-2013.)
(!‘1) = 1

Theoremfacp1 13105 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 13106 The factorial of 2. (Contributed by NM, 17-Mar-2005.)
(!‘2) = 2

Theoremfac3 13107 The factorial of 3. (Contributed by NM, 17-Mar-2005.)
(!‘3) = 6

Theoremfac4 13108 The factorial of 4. (Contributed by Mario Carneiro, 18-Jun-2015.)
(!‘4) = 24

Theoremfacnn2 13109 Value of the factorial function expressed recursively. (Contributed by NM, 2-Dec-2004.)
(𝑁 ∈ ℕ → (!‘𝑁) = ((!‘(𝑁 − 1)) · 𝑁))

Theoremfaccl 13110 Closure of the factorial function. (Contributed by NM, 2-Dec-2004.)
(𝑁 ∈ ℕ0 → (!‘𝑁) ∈ ℕ)

Theoremfaccld 13111 Closure of the factorial function, deduction version of faccl 13110. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
(𝜑𝑁 ∈ ℕ0)       (𝜑 → (!‘𝑁) ∈ ℕ)

Theoremfacmapnn 13112 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 13113 The factorial function is nonzero. (Contributed by NM, 26-Apr-2005.)
(𝑁 ∈ ℕ0 → (!‘𝑁) ≠ 0)

Theoremfacdiv 13114 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 13115 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 13116 Ordering property of factorial. (Contributed by NM, 9-Dec-2005.)
((𝑀 ∈ ℕ0𝑁 ∈ ℕ0𝑀𝑁) → (!‘𝑀) ≤ (!‘𝑁))

Theoremfaclbnd 13117 A lower bound for the factorial function. (Contributed by NM, 17-Dec-2005.)
((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → (𝑀↑(𝑁 + 1)) ≤ ((𝑀𝑀) · (!‘𝑁)))

Theoremfaclbnd2 13118 A lower bound for the factorial function. (Contributed by NM, 17-Dec-2005.)
(𝑁 ∈ ℕ0 → ((2↑𝑁) / 2) ≤ (!‘𝑁))

Theoremfaclbnd3 13119 A lower bound for the factorial function. (Contributed by NM, 19-Dec-2005.)
((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → (𝑀𝑁) ≤ ((𝑀𝑀) · (!‘𝑁)))

Theoremfaclbnd4lem1 13120 Lemma for faclbnd4 13124. 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 13121 Lemma for faclbnd4 13124. Use the weak deduction theorem to convert the hypotheses of faclbnd4lem1 13120 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 13122 Lemma for faclbnd4 13124. 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 13123 Lemma for faclbnd4 13124. 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 13124 Variant of faclbnd5 13125 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 13125 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 13126 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 13127 An upper bound for the factorial function. (Contributed by Mario Carneiro, 15-Apr-2016.)
(𝑁 ∈ ℕ0 → (!‘𝑁) ≤ (𝑁𝑁))

Theoremfacavg 13128 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 13129 Extend class notation to include the binomial coefficient operation (combinatorial choose operation).
class C

Definitiondf-bc 13130* Define the binomial coefficient operation. For example, (5C3) = 10 (ex-bc 27439).

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 13131 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 13132 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 13132 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 13133 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 13134 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 13135 Closure of the binomial coefficient in the positive reals. (This is mostly a lemma before we have bccl2 13150.) (Contributed by Mario Carneiro, 10-Mar-2014.)
(𝐾 ∈ (0...𝑁) → (𝑁C𝐾) ∈ ℝ+)

Theorembccmpl 13136 "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 13137 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 13138 The binomial coefficient " 0 choose 𝐾 " is 0 for a positive integer K. Note that (0C0) = 1 (see bcn0 13137). (Contributed by Alexander van der Vekens, 1-Jan-2018.)
(𝐾 ∈ ℕ → (0C𝐾) = 0)

Theorembcnn 13139 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 13140 Binomial coefficient: 𝑁 choose 1. (Contributed by NM, 21-Jun-2005.) (Revised by Mario Carneiro, 8-Nov-2013.)
(𝑁 ∈ ℕ0 → (𝑁C1) = 𝑁)

Theorembcnp1n 13141 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 13142 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 13143 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 13144 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 13145 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 13146 Binomial coefficient: 𝑁 choose 2. (Contributed by Mario Carneiro, 22-May-2014.)
(𝑁 ∈ ℕ0 → (𝑁C2) = ((𝑁 · (𝑁 − 1)) / 2))

Theorembcp1m1 13147 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 13148 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 13149 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 13150 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 13151 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 13152 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 13153 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 13154 The binomial coefficent of (𝑁 − 1) is 𝑁. (Contributed by Scott Fenton, 16-May-2014.)
(𝑁 ∈ ℕ0 → (𝑁C(𝑁 − 1)) = 𝑁)

Theorem4bc3eq4 13155 The value of four choose three. (Contributed by Scott Fenton, 11-Jun-2016.)
(4C3) = 4

Theorem4bc2eq6 13156 The value of four choose two. (Contributed by Scott Fenton, 9-Jan-2017.)
(4C2) = 6

5.6.11  The ` # ` (set size) function

Syntaxchash 13157 Extend the definition of a class to include the set size function.
class #

Definitiondf-hash 13158 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 27440). (Contributed by Paul Chapman, 22-Jun-2011.)
# = (((rec((𝑥 ∈ V ↦ (𝑥 + 1)), 0) ↾ ω) ∘ card) ∪ ((V ∖ Fin) × {+∞}))

Theoremhashkf 13159 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 13160* The value of the # function in terms of the mapping 𝐺 from ω to 0. The proof avoids the use of ax-ac 9319. (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 13161* 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 13162 The value of the # function on an infinite set. (Contributed by Mario Carneiro, 13-Jul-2014.)
((𝐴𝑉 ∧ ¬ 𝐴 ∈ Fin) → (#‘𝐴) = +∞)

Theoremhashbnd 13163 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 13164 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 13165 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 13164? (Contributed by Mario Carneiro, 13-Sep-2013.) (Revised by Mario Carneiro, 13-Jul-2014.) (Proof shortened by AV, 24-Oct-2021.)
#:V⟶(ℕ0 ∪ {+∞})

TheoremhashfOLD 13166 Obsolete version of hashf 13165 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 13167 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 13168 Restriction of the domain of the size function. (Contributed by Thierry Arnoux, 31-Jan-2017.)
(# ↾ 𝐴) Fn 𝐴

Theoremdmhashres 13169 Restriction of the domain of the size function. (Contributed by Thierry Arnoux, 12-Jan-2017.)
dom (# ↾ 𝐴) = 𝐴

Theoremhashnn0pnf 13170 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 13167? (Contributed by Alexander van der Vekens, 6-Dec-2017.)
(𝑀𝑉 → ((#‘𝑀) ∈ ℕ0 ∨ (#‘𝑀) = +∞))

Theoremhashnnn0genn0 13171 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 13172 The size of a set is never minus infinity. (Contributed by Alexander van der Vekens, 21-Dec-2017.)
(𝐴𝑉 → (#‘𝐴) ≠ -∞)

Theoremhashv01gt1 13173 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 13174 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 13175 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 13176 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 13177* 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 13178* 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 13179 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 13180* 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)))

Theoremhashf1rn 13181 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))

Theoremhasheqf1od 13182 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))

Theoremfz1eqb 13183 Two possibly-empty 1-based finite sets of sequential integers are equal iff their endpoints are equal. (Contributed by Paul Chapman, 22-Jun-2011.) (Proof shortened by Mario Carneiro, 29-Mar-2014.)
((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → ((1...𝑀) = (1...𝑁) ↔ 𝑀 = 𝑁))

Theoremhashcard 13184 The size function of the cardinality function. (Contributed by Mario Carneiro, 19-Sep-2013.) (Revised by Mario Carneiro, 4-Nov-2013.)
(𝐴 ∈ Fin → (#‘(card‘𝐴)) = (#‘𝐴))

Theoremhashcl 13185 Closure of the # function. (Contributed by Paul Chapman, 26-Oct-2012.) (Revised by Mario Carneiro, 13-Jul-2014.)
(𝐴 ∈ Fin → (#‘𝐴) ∈ ℕ0)

Theoremhashxrcl 13186 Extended real closure of the # function. (Contributed by Mario Carneiro, 22-Apr-2015.)
(𝐴𝑉 → (#‘𝐴) ∈ ℝ*)

Theoremhashclb 13187 Reverse closure of the # function. (Contributed by Mario Carneiro, 15-Jan-2015.)
(𝐴𝑉 → (𝐴 ∈ Fin ↔ (#‘𝐴) ∈ ℕ0))

Theoremnfile 13188 The size of any infinite set is always greater than or equal to the size of any set. (Contributed by AV, 13-Nov-2020.)
((𝐴𝑉𝐵𝑊 ∧ ¬ 𝐵 ∈ Fin) → (#‘𝐴) ≤ (#‘𝐵))

Theoremhashvnfin 13189 A set of finite size is a finite set. (Contributed by Alexander van der Vekens, 8-Dec-2017.)
((𝑆𝑉𝑁 ∈ ℕ0) → ((#‘𝑆) = 𝑁𝑆 ∈ Fin))

Theoremhashnfinnn0 13190 The size of an infinite set is not a nonnegative integer. (Contributed by Alexander van der Vekens, 21-Dec-2017.) (Proof shortened by Alexander van der Vekens, 18-Jan-2018.)
((𝐴𝑉 ∧ ¬ 𝐴 ∈ Fin) → (#‘𝐴) ∉ ℕ0)

Theoremisfinite4 13191 A finite set is equinumerous to the range of integers from one up to the hash value of the set. In other words, counting objects with natural numbers works if and only if it is a finite collection. (Contributed by Richard Penner, 26-Feb-2020.)
(𝐴 ∈ Fin ↔ (1...(#‘𝐴)) ≈ 𝐴)

Theoremhasheq0 13192 Two ways of saying a finite set is empty. (Contributed by Paul Chapman, 26-Oct-2012.) (Revised by Mario Carneiro, 27-Jul-2014.)
(𝐴𝑉 → ((#‘𝐴) = 0 ↔ 𝐴 = ∅))

Theoremhashneq0 13193 Two ways of saying a set is not empty. (Contributed by Alexander van der Vekens, 23-Sep-2018.)
(𝐴𝑉 → (0 < (#‘𝐴) ↔ 𝐴 ≠ ∅))

Theoremhashgt0n0 13194 If the size of a set is greater than 0, the set is not empty. (Contributed by AV, 5-Aug-2018.) (Proof shortened by AV, 18-Nov-2018.)
((𝐴𝑉 ∧ 0 < (#‘𝐴)) → 𝐴 ≠ ∅)

Theoremhashnncl 13195 Positive natural closure of the hash function. (Contributed by Mario Carneiro, 16-Jan-2015.)
(𝐴 ∈ Fin → ((#‘𝐴) ∈ ℕ ↔ 𝐴 ≠ ∅))

Theoremhash0 13196 The empty set has size zero. (Contributed by Mario Carneiro, 8-Jul-2014.)
(#‘∅) = 0

Theoremhashsng 13197 The size of a singleton. (Contributed by Paul Chapman, 26-Oct-2012.) (Proof shortened by Mario Carneiro, 13-Feb-2013.)
(𝐴𝑉 → (#‘{𝐴}) = 1)

Theoremhashen1 13198 A set has size 1 if and only if it is equinumerous to the ordinal 1. (Contributed by AV, 14-Apr-2019.)
(𝐴𝑉 → ((#‘𝐴) = 1 ↔ 𝐴 ≈ 1𝑜))

Theoremhashrabrsn 13199* The size of a restricted class abstraction restricted to a singleton is a nonnegative integer. (Contributed by Alexander van der Vekens, 22-Dec-2017.)
(#‘{𝑥 ∈ {𝐴} ∣ 𝜑}) ∈ ℕ0

Theoremhashrabsn01 13200* The size of a restricted class abstraction restricted to a singleton is either 0 or 1. (Contributed by Alexander van der Vekens, 3-Sep-2018.)
((#‘{𝑥 ∈ {𝐴} ∣ 𝜑}) = 𝑁 → (𝑁 = 0 ∨ 𝑁 = 1))

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