P. 144 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 14301-14400   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	facmapnn 14301	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 14302	The factorial function is nonzero. (Contributed by NM, 26-Apr-2005.)
⊢ (𝑁 ∈ ℕ0 → (!‘𝑁) ≠ 0)
Theorem	facdiv 14303	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 14304	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 14305	Ordering property of factorial. (Contributed by NM, 9-Dec-2005.)
⊢ ((𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0 ∧ 𝑀 ≤ 𝑁) → (!‘𝑀) ≤ (!‘𝑁))
Theorem	faclbnd 14306	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 14307	A lower bound for the factorial function. (Contributed by NM, 17-Dec-2005.)
⊢ (𝑁 ∈ ℕ0 → ((2↑𝑁) / 2) ≤ (!‘𝑁))
Theorem	faclbnd3 14308	A lower bound for the factorial function. (Contributed by NM, 19-Dec-2005.)
⊢ ((𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0) → (𝑀↑𝑁) ≤ ((𝑀↑𝑀) · (!‘𝑁)))
Theorem	faclbnd4lem1 14309	Lemma for faclbnd4 14313. 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 14310	Lemma for faclbnd4 14313. Use the weak deduction theorem to convert the hypotheses of faclbnd4lem1 14309 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 14311	Lemma for faclbnd4 14313. 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 14312	Lemma for faclbnd4 14313. 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 14313	Variant of faclbnd5 14314 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 14314	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 14315	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 14316	An upper bound for the factorial function. (Contributed by Mario Carneiro, 15-Apr-2016.)
⊢ (𝑁 ∈ ℕ0 → (!‘𝑁) ≤ (𝑁↑𝑁))
Theorem	facavg 14317	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)))
5.6.10  The binomial coefficient operation
Syntax	cbc 14318	Extend class notation to include the binomial coefficient operation (combinatorial choose operation).
class C
Definition	df-bc 14319*	Define the binomial coefficient operation. For example, (5C3) = 10 (ex-bc 30379). 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 14320	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 14321 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 14321	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 14322	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 14323	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 14324	Closure of the binomial coefficient in the positive reals. (This is mostly a lemma before we have bccl2 14339.) (Contributed by Mario Carneiro, 10-Mar-2014.)
⊢ (𝐾 ∈ (0...𝑁) → (𝑁C𝐾) ∈ ℝ+)
Theorem	bccmpl 14325	"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 14326	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 14327	The binomial coefficient " 0 choose 𝐾 " is 0 for a positive integer K. Note that (0C0) = 1 (see bcn0 14326). (Contributed by Alexander van der Vekens, 1-Jan-2018.)
⊢ (𝐾 ∈ ℕ → (0C𝐾) = 0)
Theorem	bcnn 14328	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 14329	Binomial coefficient: 𝑁 choose 1. (Contributed by NM, 21-Jun-2005.) (Revised by Mario Carneiro, 8-Nov-2013.)
⊢ (𝑁 ∈ ℕ0 → (𝑁C1) = 𝑁)
Theorem	bcnp1n 14330	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 14331	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 14332	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 14333	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 14334	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 14335	Binomial coefficient: 𝑁 choose 2. (Contributed by Mario Carneiro, 22-May-2014.)
⊢ (𝑁 ∈ ℕ0 → (𝑁C2) = ((𝑁 · (𝑁 − 1)) / 2))
Theorem	bcp1m1 14336	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 14337	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 14338	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 14339	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 14340	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 14341	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 14342	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 14343	The binomial coefficient of (𝑁 − 1) is 𝑁. (Contributed by Scott Fenton, 16-May-2014.)
⊢ (𝑁 ∈ ℕ0 → (𝑁C(𝑁 − 1)) = 𝑁)
Theorem	4bc3eq4 14344	The value of four choose three. (Contributed by Scott Fenton, 11-Jun-2016.)
⊢ (4C3) = 4
Theorem	4bc2eq6 14345	The value of four choose two. (Contributed by Scott Fenton, 9-Jan-2017.)
⊢ (4C2) = 6
5.6.11  The ` # ` (set size) function
Syntax	chash 14346	Extend the definition of a class to include the set size function.
class ♯
Definition	df-hash 14347	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 30380). (Contributed by Paul Chapman, 22-Jun-2011.)
⊢ ♯ = (((rec((𝑥 ∈ V ↦ (𝑥 + 1)), 0) ↾ ω) ∘ card) ∪ ((V ∖ Fin) × {+∞}))
Theorem	hashkf 14348	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 14349*	The value of the ♯ function in terms of the mapping 𝐺 from ω to ℕ0. The proof avoids the use of ax-ac 10471. (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 14350*	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 14351	The value of the ♯ function on an infinite set. (Contributed by Mario Carneiro, 13-Jul-2014.)
⊢ ((𝐴 ∈ 𝑉 ∧ ¬ 𝐴 ∈ Fin) → (♯‘𝐴) = +∞)
Theorem	hashbnd 14352	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 14353	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 14354	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 14353? (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 14355	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 14356	Restriction of the domain of the size function. (Contributed by Thierry Arnoux, 31-Jan-2017.)
⊢ (♯ ↾ 𝐴) Fn 𝐴
Theorem	dmhashres 14357	Restriction of the domain of the size function. (Contributed by Thierry Arnoux, 12-Jan-2017.)
⊢ dom (♯ ↾ 𝐴) = 𝐴
Theorem	hashnn0pnf 14358	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 14355? (Contributed by Alexander van der Vekens, 6-Dec-2017.)
⊢ (𝑀 ∈ 𝑉 → ((♯‘𝑀) ∈ ℕ0 ∨ (♯‘𝑀) = +∞))
Theorem	hashnnn0genn0 14359	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 14360	The size of a set is never minus infinity. (Contributed by Alexander van der Vekens, 21-Dec-2017.)
⊢ (𝐴 ∈ 𝑉 → (♯‘𝐴) ≠ -∞)
Theorem	hashv01gt1 14361	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 14362	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 14363	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 14364	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 14365*	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 14366*	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	hasheqf1oi 14367*	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.)
⊢ (𝐴 ∈ 𝑉 → (∃𝑓 𝑓:𝐴–1-1-onto→𝐵 → (♯‘𝐴) = (♯‘𝐵)))
Theorem	hashf1rn 14368	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.)
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐹:𝐴–1-1→𝐵) → (♯‘𝐹) = (♯‘ran 𝐹))
Theorem	hasheqf1od 14369	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.)
⊢ (𝜑 → 𝐴 ∈ 𝑈)    &   ⊢ (𝜑 → 𝐹:𝐴–1-1-onto→𝐵)    ⇒   ⊢ (𝜑 → (♯‘𝐴) = (♯‘𝐵))
Theorem	fz1eqb 14370	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 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0) → ((1...𝑀) = (1...𝑁) ↔ 𝑀 = 𝑁))
Theorem	hashcard 14371	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))
Theorem	hashcl 14372	Closure of the ♯ function. (Contributed by Paul Chapman, 26-Oct-2012.) (Revised by Mario Carneiro, 13-Jul-2014.)
⊢ (𝐴 ∈ Fin → (♯‘𝐴) ∈ ℕ0)
Theorem	hashxrcl 14373	Extended real closure of the ♯ function. (Contributed by Mario Carneiro, 22-Apr-2015.)
⊢ (𝐴 ∈ 𝑉 → (♯‘𝐴) ∈ ℝ*)
Theorem	hashclb 14374	Reverse closure of the ♯ function. (Contributed by Mario Carneiro, 15-Jan-2015.)
⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ Fin ↔ (♯‘𝐴) ∈ ℕ0))
Theorem	nfile 14375	The size of any infinite set is always greater than or equal to the size of any set. (Contributed by AV, 13-Nov-2020.)
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ ¬ 𝐵 ∈ Fin) → (♯‘𝐴) ≤ (♯‘𝐵))
Theorem	hashvnfin 14376	A set of finite size is a finite set. (Contributed by Alexander van der Vekens, 8-Dec-2017.)
⊢ ((𝑆 ∈ 𝑉 ∧ 𝑁 ∈ ℕ0) → ((♯‘𝑆) = 𝑁 → 𝑆 ∈ Fin))
Theorem	hashnfinnn0 14377	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.)
⊢ ((𝐴 ∈ 𝑉 ∧ ¬ 𝐴 ∈ Fin) → (♯‘𝐴) ∉ ℕ0)
Theorem	isfinite4 14378	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)
Theorem	hasheq0 14379	Two ways of saying a set is empty. (Contributed by Paul Chapman, 26-Oct-2012.) (Revised by Mario Carneiro, 27-Jul-2014.)
⊢ (𝐴 ∈ 𝑉 → ((♯‘𝐴) = 0 ↔ 𝐴 = ∅))
Theorem	hashneq0 14380	Two ways of saying a set is not empty. (Contributed by Alexander van der Vekens, 23-Sep-2018.)
⊢ (𝐴 ∈ 𝑉 → (0 < (♯‘𝐴) ↔ 𝐴 ≠ ∅))
Theorem	hashgt0n0 14381	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.)
⊢ ((𝐴 ∈ 𝑉 ∧ 0 < (♯‘𝐴)) → 𝐴 ≠ ∅)
Theorem	hashnncl 14382	Positive natural closure of the hash function. (Contributed by Mario Carneiro, 16-Jan-2015.)
⊢ (𝐴 ∈ Fin → ((♯‘𝐴) ∈ ℕ ↔ 𝐴 ≠ ∅))
Theorem	hash0 14383	The empty set has size zero. (Contributed by Mario Carneiro, 8-Jul-2014.)
⊢ (♯‘∅) = 0
Theorem	hashelne0d 14384	A set with an element has nonzero size. (Contributed by Rohan Ridenour, 3-Aug-2023.)
⊢ (𝜑 → 𝐵 ∈ 𝐴)    &   ⊢ (𝜑 → 𝐴 ∈ 𝑉)    ⇒   ⊢ (𝜑 → ¬ (♯‘𝐴) = 0)
Theorem	hashsng 14385	The size of a singleton. (Contributed by Paul Chapman, 26-Oct-2012.) (Proof shortened by Mario Carneiro, 13-Feb-2013.)
⊢ (𝐴 ∈ 𝑉 → (♯‘{𝐴}) = 1)
Theorem	hashen1 14386	A set has size 1 if and only if it is equinumerous to the ordinal 1. (Contributed by AV, 14-Apr-2019.)
⊢ (𝐴 ∈ 𝑉 → ((♯‘𝐴) = 1 ↔ 𝐴 ≈ 1o))
Theorem	hash1elsn 14387	A set of size 1 with a known element is the singleton of that element. (Contributed by Rohan Ridenour, 3-Aug-2023.)
⊢ (𝜑 → (♯‘𝐴) = 1)    &   ⊢ (𝜑 → 𝐵 ∈ 𝐴)    &   ⊢ (𝜑 → 𝐴 ∈ 𝑉)    ⇒   ⊢ (𝜑 → 𝐴 = {𝐵})
Theorem	hashrabrsn 14388*	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
Theorem	hashrabsn01 14389*	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))
Theorem	hashrabsn1 14390*	If the size of a restricted class abstraction restricted to a singleton is 1, the condition of the class abstraction must hold for the singleton. (Contributed by Alexander van der Vekens, 3-Sep-2018.)
⊢ ((♯‘{𝑥 ∈ {𝐴} ∣ 𝜑}) = 1 → [𝐴 / 𝑥]𝜑)
Theorem	hashfn 14391	A function is equinumerous to its domain. (Contributed by Mario Carneiro, 12-Mar-2015.)
⊢ (𝐹 Fn 𝐴 → (♯‘𝐹) = (♯‘𝐴))
Theorem	fseq1hash 14392	The value of the size function on a finite 1-based sequence. (Contributed by Paul Chapman, 26-Oct-2012.) (Proof shortened by Mario Carneiro, 12-Mar-2015.)
⊢ ((𝑁 ∈ ℕ0 ∧ 𝐹 Fn (1...𝑁)) → (♯‘𝐹) = 𝑁)
Theorem	hashgadd 14393	𝐺 maps ordinal addition to integer addition. (Contributed by Paul Chapman, 30-Nov-2012.) (Revised by Mario Carneiro, 15-Sep-2013.)
⊢ 𝐺 = (rec((𝑥 ∈ V ↦ (𝑥 + 1)), 0) ↾ ω)    ⇒   ⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (𝐺‘(𝐴 +o 𝐵)) = ((𝐺‘𝐴) + (𝐺‘𝐵)))
Theorem	hashgval2 14394	A short expression for the 𝐺 function of hashgf1o 13987. (Contributed by Mario Carneiro, 24-Jan-2015.)
⊢ (♯ ↾ ω) = (rec((𝑥 ∈ V ↦ (𝑥 + 1)), 0) ↾ ω)
Theorem	hashdom 14395	Dominance relation for the size function. (Contributed by Mario Carneiro, 22-Sep-2013.) (Revised by Mario Carneiro, 22-Apr-2015.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ 𝑉) → ((♯‘𝐴) ≤ (♯‘𝐵) ↔ 𝐴 ≼ 𝐵))
Theorem	hashdomi 14396	Non-strict order relation of the ♯ function on the full cardinal poset. (Contributed by Stefan O'Rear, 12-Sep-2015.)
⊢ (𝐴 ≼ 𝐵 → (♯‘𝐴) ≤ (♯‘𝐵))
Theorem	hashsdom 14397	Strict dominance relation for the size function. (Contributed by Mario Carneiro, 18-Aug-2014.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) → ((♯‘𝐴) < (♯‘𝐵) ↔ 𝐴 ≺ 𝐵))
Theorem	hashun 14398	The size of the union of disjoint finite sets is the sum of their sizes. (Contributed by Paul Chapman, 30-Nov-2012.) (Revised by Mario Carneiro, 15-Sep-2013.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin ∧ (𝐴 ∩ 𝐵) = ∅) → (♯‘(𝐴 ∪ 𝐵)) = ((♯‘𝐴) + (♯‘𝐵)))
Theorem	hashun2 14399	The size of the union of finite sets is less than or equal to the sum of their sizes. (Contributed by Mario Carneiro, 23-Sep-2013.) (Proof shortened by Mario Carneiro, 27-Jul-2014.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) → (♯‘(𝐴 ∪ 𝐵)) ≤ ((♯‘𝐴) + (♯‘𝐵)))
Theorem	hashun3 14400	The size of the union of finite sets is the sum of their sizes minus the size of the intersection. (Contributed by Mario Carneiro, 6-Aug-2017.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) → (♯‘(𝐴 ∪ 𝐵)) = (((♯‘𝐴) + (♯‘𝐵)) − (♯‘(𝐴 ∩ 𝐵))))