P. 112 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 11101-11200   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	fac2 11101	The factorial of 2. (Contributed by NM, 17-Mar-2005.)
⊢ (!‘2) = 2
Theorem	fac3 11102	The factorial of 3. (Contributed by NM, 17-Mar-2005.)
⊢ (!‘3) = 6
Theorem	fac4 11103	The factorial of 4. (Contributed by Mario Carneiro, 18-Jun-2015.)
⊢ (!‘4) = ;24
Theorem	facnn2 11104	Value of the factorial function expressed recursively. (Contributed by NM, 2-Dec-2004.)
⊢ (𝑁 ∈ ℕ → (!‘𝑁) = ((!‘(𝑁 − 1)) · 𝑁))
Theorem	faccl 11105	Closure of the factorial function. (Contributed by NM, 2-Dec-2004.)
⊢ (𝑁 ∈ ℕ0 → (!‘𝑁) ∈ ℕ)
Theorem	faccld 11106	Closure of the factorial function, deduction version of faccl 11105. (Contributed by Glauco Siliprandi, 5-Apr-2020.)
⊢ (𝜑 → 𝑁 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (!‘𝑁) ∈ ℕ)
Theorem	facne0 11107	The factorial function is nonzero. (Contributed by NM, 26-Apr-2005.)
⊢ (𝑁 ∈ ℕ0 → (!‘𝑁) ≠ 0)
Theorem	facdiv 11108	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 11109	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 11110	Ordering property of factorial. (Contributed by NM, 9-Dec-2005.)
⊢ ((𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0 ∧ 𝑀 ≤ 𝑁) → (!‘𝑀) ≤ (!‘𝑁))
Theorem	faclbnd 11111	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 11112	A lower bound for the factorial function. (Contributed by NM, 17-Dec-2005.)
⊢ (𝑁 ∈ ℕ0 → ((2↑𝑁) / 2) ≤ (!‘𝑁))
Theorem	faclbnd3 11113	A lower bound for the factorial function. (Contributed by NM, 19-Dec-2005.)
⊢ ((𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0) → (𝑀↑𝑁) ≤ ((𝑀↑𝑀) · (!‘𝑁)))
Theorem	faclbnd6 11114	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 11115	An upper bound for the factorial function. (Contributed by Mario Carneiro, 15-Apr-2016.)
⊢ (𝑁 ∈ ℕ0 → (!‘𝑁) ≤ (𝑁↑𝑁))
Theorem	facavg 11116	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)))
4.6.9  The binomial coefficient operation
Syntax	cbc 11117	Extend class notation to include the binomial coefficient operation (combinatorial choose operation).
class C
Definition	df-bc 11118*	Define the binomial coefficient operation. For example, (5C3) = 10 (ex-bc 16546). 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 ≤ 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 11119	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 11120 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 11120	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 11121	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 11122	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 11123	Closure of the binomial coefficient in the positive reals. (This is mostly a lemma before we have bccl2 11138.) (Contributed by Mario Carneiro, 10-Mar-2014.)
⊢ (𝐾 ∈ (0...𝑁) → (𝑁C𝐾) ∈ ℝ+)
Theorem	bccmpl 11124	"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 11125	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 11126	The binomial coefficient " 0 choose 𝐾 " is 0 for a positive integer K. Note that (0C0) = 1 (see bcn0 11125). (Contributed by Alexander van der Vekens, 1-Jan-2018.)
⊢ (𝐾 ∈ ℕ → (0C𝐾) = 0)
Theorem	bcnn 11127	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 11128	Binomial coefficient: 𝑁 choose 1. (Contributed by NM, 21-Jun-2005.) (Revised by Mario Carneiro, 8-Nov-2013.)
⊢ (𝑁 ∈ ℕ0 → (𝑁C1) = 𝑁)
Theorem	bcnp1n 11129	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 11130	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 11131	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 11132	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 11133	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.) (Revised by Jim Kingdon, 23-Apr-2023.)
⊢ ((𝑁 ∈ ℕ0 ∧ 𝐾 ∈ ℕ) → (𝑁C𝐾) = ((seq((𝑁 − 𝐾) + 1)( · , I )‘𝑁) / (!‘𝐾)))
Theorem	bcn2 11134	Binomial coefficient: 𝑁 choose 2. (Contributed by Mario Carneiro, 22-May-2014.)
⊢ (𝑁 ∈ ℕ0 → (𝑁C2) = ((𝑁 · (𝑁 − 1)) / 2))
Theorem	bcp1m1 11135	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 11136	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 11137	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 11138	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	bcm1n 11139	The proportion of one binomial coefficient to another with 𝑁 decreased by 1. (Contributed by Thierry Arnoux, 9-Nov-2016.)
⊢ ((𝐾 ∈ (0...(𝑁 − 1)) ∧ 𝑁 ∈ ℕ) → (((𝑁 − 1)C𝐾) / (𝑁C𝐾)) = ((𝑁 − 𝐾) / 𝑁))
Theorem	bcn2m1 11140	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 11141	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 11142	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 11143	The binomial coefficent of (𝑁 − 1) is 𝑁. (Contributed by Scott Fenton, 16-May-2014.)
⊢ (𝑁 ∈ ℕ0 → (𝑁C(𝑁 − 1)) = 𝑁)
Theorem	4bc3eq4 11144	The value of four choose three. (Contributed by Scott Fenton, 11-Jun-2016.)
⊢ (4C3) = 4
Theorem	4bc2eq6 11145	The value of four choose two. (Contributed by Scott Fenton, 9-Jan-2017.)
⊢ (4C2) = 6
4.6.10  The ` # ` (set size) function
Syntax	chash 11146	Extend the definition of a class to include the set size function.
class ♯
Definition	df-ihash 11147*	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. Since we don't know that an arbitrary set is either finite or infinite (by inffiexmid 7168), the behavior beyond finite sets is not as useful as it might appear. For example, we wouldn't expect to be able to define this function in a meaningful way on 𝒫 1o, which cannot be shown to be finite (per pw1fin 7172). Note that we use the sharp sign (♯) for this function and we use the different character octothorpe (#) for the apartness relation (see df-ap 8861). We adopt the former notation from Corollary 8.2.4 of [AczelRathjen], p. 80 (although that work only defines it for finite sets). This definition (in terms of ∪ and ≼) is not taken directly from the literature, but for finite sets should be equivalent to the conventional definition that the size of a finite set is the unique natural number which is equinumerous to the given set. (Contributed by Jim Kingdon, 19-Feb-2022.)
⊢ ♯ = ((frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0) ∪ {⟨ω, +∞⟩}) ∘ (𝑥 ∈ V ↦ ∪ {𝑦 ∈ (ω ∪ {ω}) ∣ 𝑦 ≼ 𝑥}))
Theorem	hashinfuni 11148*	The ordinal size of an infinite set is ω. (Contributed by Jim Kingdon, 20-Feb-2022.)
⊢ (ω ≼ 𝐴 → ∪ {𝑦 ∈ (ω ∪ {ω}) ∣ 𝑦 ≼ 𝐴} = ω)
Theorem	hashinfom 11149	The value of the ♯ function on an infinite set. (Contributed by Jim Kingdon, 20-Feb-2022.)
⊢ (ω ≼ 𝐴 → (♯‘𝐴) = +∞)
Theorem	hashennnuni 11150*	The ordinal size of a set equinumerous to an element of ω is that element of ω. (Contributed by Jim Kingdon, 20-Feb-2022.)
⊢ ((𝑁 ∈ ω ∧ 𝑁 ≈ 𝐴) → ∪ {𝑦 ∈ (ω ∪ {ω}) ∣ 𝑦 ≼ 𝐴} = 𝑁)
Theorem	hashennn 11151*	The size of a set equinumerous to an element of ω. (Contributed by Jim Kingdon, 21-Feb-2022.)
⊢ ((𝑁 ∈ ω ∧ 𝑁 ≈ 𝐴) → (♯‘𝐴) = (frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)‘𝑁))
Theorem	hashcl 11152	Closure of the ♯ function. (Contributed by Paul Chapman, 26-Oct-2012.) (Revised by Mario Carneiro, 13-Jul-2014.)
⊢ (𝐴 ∈ Fin → (♯‘𝐴) ∈ ℕ0)
Theorem	hashfiv01gt1 11153	The size of a finite set is either 0 or 1 or greater than 1. (Contributed by Jim Kingdon, 21-Feb-2022.)
⊢ (𝑀 ∈ Fin → ((♯‘𝑀) = 0 ∨ (♯‘𝑀) = 1 ∨ 1 < (♯‘𝑀)))
Theorem	hashfz1 11154	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 11155	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	hasheqf1o 11156*	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 11157*	There is no bijection between a finite set and an infinite set. By infnfi 7154 the theorem would also hold if "infinite" were expressed as ω ≼ 𝐵. (Contributed by Alexander van der Vekens, 25-Dec-2017.)
⊢ ((𝐴 ∈ Fin ∧ ¬ 𝐵 ∈ Fin) → ¬ ∃𝑓 𝑓:𝐴–1-1-onto→𝐵)
Theorem	fihasheqf1oi 11158	The size of two finite sets is equal if there is a bijection mapping one of the sets onto the other. (Contributed by Jim Kingdon, 21-Feb-2022.)
⊢ ((𝐴 ∈ Fin ∧ 𝐹:𝐴–1-1-onto→𝐵) → (♯‘𝐴) = (♯‘𝐵))
Theorem	fihashf1rn 11159	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 Jim Kingdon, 21-Feb-2022.)
⊢ ((𝐴 ∈ Fin ∧ 𝐹:𝐴–1-1→𝐵) → (♯‘𝐹) = (♯‘ran 𝐹))
Theorem	fihasheqf1od 11160	The size of two finite sets is equal if there is a bijection mapping one of the sets onto the other. (Contributed by Jim Kingdon, 21-Feb-2022.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ (𝜑 → 𝐹:𝐴–1-1-onto→𝐵)    ⇒   ⊢ (𝜑 → (♯‘𝐴) = (♯‘𝐵))
Theorem	fz1eqb 11161	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	filtinf 11162	The size of an infinite set is greater than the size of a finite set. (Contributed by Jim Kingdon, 21-Feb-2022.)
⊢ ((𝐴 ∈ Fin ∧ ω ≼ 𝐵) → (♯‘𝐴) < (♯‘𝐵))
Theorem	isfinite4im 11163	A finite set is equinumerous to the range of integers from one up to the hash value of the set. (Contributed by Jim Kingdon, 22-Feb-2022.)
⊢ (𝐴 ∈ Fin → (1...(♯‘𝐴)) ≈ 𝐴)
Theorem	fihasheq0 11164	Two ways of saying a finite set is empty. (Contributed by Paul Chapman, 26-Oct-2012.) (Revised by Mario Carneiro, 27-Jul-2014.) (Intuitionized by Jim Kingdon, 23-Feb-2022.)
⊢ (𝐴 ∈ Fin → ((♯‘𝐴) = 0 ↔ 𝐴 = ∅))
Theorem	fihashneq0 11165	Two ways of saying a finite set is not empty. Also, "A is inhabited" would be equivalent by fin0 7144. (Contributed by Alexander van der Vekens, 23-Sep-2018.) (Intuitionized by Jim Kingdon, 23-Feb-2022.)
⊢ (𝐴 ∈ Fin → (0 < (♯‘𝐴) ↔ 𝐴 ≠ ∅))
Theorem	hashnncl 11166	Positive natural closure of the hash function. (Contributed by Mario Carneiro, 16-Jan-2015.)
⊢ (𝐴 ∈ Fin → ((♯‘𝐴) ∈ ℕ ↔ 𝐴 ≠ ∅))
Theorem	hash0 11167	The empty set has size zero. (Contributed by Mario Carneiro, 8-Jul-2014.)
⊢ (♯‘∅) = 0
Theorem	fihashelne0d 11168	A finite set with an element has nonzero size. (Contributed by Rohan Ridenour, 3-Aug-2023.)
⊢ (𝜑 → 𝐵 ∈ 𝐴)    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    ⇒   ⊢ (𝜑 → ¬ (♯‘𝐴) = 0)
Theorem	hashsng 11169	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	fihashen1 11170	A finite set has size 1 if and only if it is equinumerous to the ordinal 1. (Contributed by AV, 14-Apr-2019.) (Intuitionized by Jim Kingdon, 23-Feb-2022.)
⊢ (𝐴 ∈ Fin → ((♯‘𝐴) = 1 ↔ 𝐴 ≈ 1o))
Theorem	en1hash 11171	A set equinumerous to the ordinal one has size 1 . (Contributed by Jim Kingdon, 11-Mar-2026.)
⊢ (𝐴 ≈ 1o → (♯‘𝐴) = 1)
Theorem	fihashfn 11172	A function on a finite set is equinumerous to its domain. (Contributed by Mario Carneiro, 12-Mar-2015.) (Intuitionized by Jim Kingdon, 24-Feb-2022.)
⊢ ((𝐹 Fn 𝐴 ∧ 𝐴 ∈ Fin) → (♯‘𝐹) = (♯‘𝐴))
Theorem	fseq1hash 11173	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	omgadd 11174	Mapping ordinal addition to integer addition. (Contributed by Jim Kingdon, 24-Feb-2022.)
⊢ 𝐺 = frec((𝑥 ∈ ℤ ↦ (𝑥 + 1)), 0)    ⇒   ⊢ ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (𝐺‘(𝐴 +o 𝐵)) = ((𝐺‘𝐴) + (𝐺‘𝐵)))
Theorem	fihashdom 11175	Dominance relation for the size function. (Contributed by Jim Kingdon, 24-Feb-2022.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) → ((♯‘𝐴) ≤ (♯‘𝐵) ↔ 𝐴 ≼ 𝐵))
Theorem	hashunlem 11176	Lemma for hashun 11177. Ordinal size of the union. (Contributed by Jim Kingdon, 25-Feb-2022.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ (𝜑 → 𝐵 ∈ Fin)    &   ⊢ (𝜑 → (𝐴 ∩ 𝐵) = ∅)    &   ⊢ (𝜑 → 𝑁 ∈ ω)    &   ⊢ (𝜑 → 𝑀 ∈ ω)    &   ⊢ (𝜑 → 𝐴 ≈ 𝑁)    &   ⊢ (𝜑 → 𝐵 ≈ 𝑀)    ⇒   ⊢ (𝜑 → (𝐴 ∪ 𝐵) ≈ (𝑁 +o 𝑀))
Theorem	hashun 11177	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	fihashgt0 11178	The cardinality of a finite nonempty set is greater than zero. (Contributed by Thierry Arnoux, 2-Mar-2017.)
⊢ ((𝐴 ∈ Fin ∧ 𝐴 ≠ ∅) → 0 < (♯‘𝐴))
Theorem	1elfz0hash 11179	1 is an element of the finite set of sequential nonnegative integers bounded by the size of a nonempty finite set. (Contributed by AV, 9-May-2020.)
⊢ ((𝐴 ∈ Fin ∧ 𝐴 ≠ ∅) → 1 ∈ (0...(♯‘𝐴)))
Theorem	hashunsng 11180	The size of the union of a finite set with a disjoint singleton is one more than the size of the set. (Contributed by Paul Chapman, 30-Nov-2012.)
⊢ (𝐵 ∈ 𝑉 → ((𝐴 ∈ Fin ∧ ¬ 𝐵 ∈ 𝐴) → (♯‘(𝐴 ∪ {𝐵})) = ((♯‘𝐴) + 1)))
Theorem	hashprg 11181	The size of an unordered pair. (Contributed by Mario Carneiro, 27-Sep-2013.) (Revised by Mario Carneiro, 5-May-2016.) (Revised by AV, 18-Sep-2021.)
⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (𝐴 ≠ 𝐵 ↔ (♯‘{𝐴, 𝐵}) = 2))
Theorem	prhash2ex 11182	There is (at least) one set with two different elements: the unordered pair containing 0 and 1. In contrast to pr0hash2ex 11188, numbers are used instead of sets because their representation is shorter (and more comprehensive). (Contributed by AV, 29-Jan-2020.)
⊢ (♯‘{0, 1}) = 2
Theorem	hashp1i 11183	Size of a natural number ordinal. (Contributed by Mario Carneiro, 5-Jan-2016.)
⊢ 𝐴 ∈ ω    &   ⊢ 𝐵 = suc 𝐴    &   ⊢ (♯‘𝐴) = 𝑀    &   ⊢ (𝑀 + 1) = 𝑁    ⇒   ⊢ (♯‘𝐵) = 𝑁
Theorem	hash1 11184	Size of a natural number ordinal. (Contributed by Mario Carneiro, 5-Jan-2016.)
⊢ (♯‘1o) = 1
Theorem	hash2 11185	Size of a natural number ordinal. (Contributed by Mario Carneiro, 5-Jan-2016.)
⊢ (♯‘2o) = 2
Theorem	hash3 11186	Size of a natural number ordinal. (Contributed by Mario Carneiro, 5-Jan-2016.)
⊢ (♯‘3o) = 3
Theorem	hash4 11187	Size of a natural number ordinal. (Contributed by Mario Carneiro, 5-Jan-2016.)
⊢ (♯‘4o) = 4
Theorem	pr0hash2ex 11188	There is (at least) one set with two different elements: the unordered pair containing the empty set and the singleton containing the empty set. (Contributed by AV, 29-Jan-2020.)
⊢ (♯‘{∅, {∅}}) = 2
Theorem	fihashss 11189	The size of a subset is less than or equal to the size of its superset. (Contributed by Alexander van der Vekens, 14-Jul-2018.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin ∧ 𝐵 ⊆ 𝐴) → (♯‘𝐵) ≤ (♯‘𝐴))
Theorem	fiprsshashgt1 11190	The size of a superset of a proper unordered pair is greater than 1. (Contributed by AV, 6-Feb-2021.)
⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝐴 ≠ 𝐵) ∧ 𝐶 ∈ Fin) → ({𝐴, 𝐵} ⊆ 𝐶 → 2 ≤ (♯‘𝐶)))
Theorem	fihashssdif 11191	The size of the difference of a finite set and a finite subset is the set's size minus the subset's. (Contributed by Jim Kingdon, 31-May-2022.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin ∧ 𝐵 ⊆ 𝐴) → (♯‘(𝐴 ∖ 𝐵)) = ((♯‘𝐴) − (♯‘𝐵)))
Theorem	hashdifsn 11192	The size of the difference of a finite set and a singleton subset is the set's size minus 1. (Contributed by Alexander van der Vekens, 6-Jan-2018.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ 𝐴) → (♯‘(𝐴 ∖ {𝐵})) = ((♯‘𝐴) − 1))
Theorem	hashdifpr 11193	The size of the difference of a finite set and a proper ordered pair subset is the set's size minus 2. (Contributed by AV, 16-Dec-2020.)
⊢ ((𝐴 ∈ Fin ∧ (𝐵 ∈ 𝐴 ∧ 𝐶 ∈ 𝐴 ∧ 𝐵 ≠ 𝐶)) → (♯‘(𝐴 ∖ {𝐵, 𝐶})) = ((♯‘𝐴) − 2))
Theorem	hashfz 11194	Value of the numeric cardinality of a nonempty integer range. (Contributed by Stefan O'Rear, 12-Sep-2014.) (Proof shortened by Mario Carneiro, 15-Apr-2015.)
⊢ (𝐵 ∈ (ℤ≥‘𝐴) → (♯‘(𝐴...𝐵)) = ((𝐵 − 𝐴) + 1))
Theorem	hashfzo 11195	Cardinality of a half-open set of integers. (Contributed by Stefan O'Rear, 15-Aug-2015.)
⊢ (𝐵 ∈ (ℤ≥‘𝐴) → (♯‘(𝐴..^𝐵)) = (𝐵 − 𝐴))
Theorem	hashfzo0 11196	Cardinality of a half-open set of integers based at zero. (Contributed by Stefan O'Rear, 15-Aug-2015.)
⊢ (𝐵 ∈ ℕ0 → (♯‘(0..^𝐵)) = 𝐵)
Theorem	hashfzp1 11197	Value of the numeric cardinality of a (possibly empty) integer range. (Contributed by AV, 19-Jun-2021.)
⊢ (𝐵 ∈ (ℤ≥‘𝐴) → (♯‘((𝐴 + 1)...𝐵)) = (𝐵 − 𝐴))
Theorem	hashfz0 11198	Value of the numeric cardinality of a nonempty range of nonnegative integers. (Contributed by Alexander van der Vekens, 21-Jul-2018.)
⊢ (𝐵 ∈ ℕ0 → (♯‘(0...𝐵)) = (𝐵 + 1))
Theorem	hashxp 11199	The size of the Cartesian product of two finite sets is the product of their sizes. (Contributed by Paul Chapman, 30-Nov-2012.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) → (♯‘(𝐴 × 𝐵)) = ((♯‘𝐴) · (♯‘𝐵)))
Theorem	hashmap 11200	The size of the set exponential of two finite sets is the exponential of their sizes. (This is the original motivation behind the notation for set exponentiation.) (Contributed by Mario Carneiro, 5-Aug-2014.) (Proof shortened by AV, 18-Jul-2022.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) → (♯‘(𝐴 ↑𝑚 𝐵)) = ((♯‘𝐴)↑(♯‘𝐵)))