P. 113 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 11201-11300   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	bcval 11201	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 11202 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 11202	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 11203	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 11204	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 11205	Closure of the binomial coefficient in the positive reals. (This is mostly a lemma before we have bccl2 11220.) (Contributed by Mario Carneiro, 10-Mar-2014.)
⊢ (𝐾 ∈ (0...𝑁) → (𝑁C𝐾) ∈ ℝ+)
Theorem	bccmpl 11206	"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 11207	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 11208	The binomial coefficient " 0 choose 𝐾 " is 0 for a positive integer K. Note that (0C0) = 1 (see bcn0 11207). (Contributed by Alexander van der Vekens, 1-Jan-2018.)
⊢ (𝐾 ∈ ℕ → (0C𝐾) = 0)
Theorem	bcnn 11209	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 11210	Binomial coefficient: 𝑁 choose 1. (Contributed by NM, 21-Jun-2005.) (Revised by Mario Carneiro, 8-Nov-2013.)
⊢ (𝑁 ∈ ℕ0 → (𝑁C1) = 𝑁)
Theorem	bcnp1n 11211	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 11212	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 11213	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 11214	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 11215	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 11216	Binomial coefficient: 𝑁 choose 2. (Contributed by Mario Carneiro, 22-May-2014.)
⊢ (𝑁 ∈ ℕ0 → (𝑁C2) = ((𝑁 · (𝑁 − 1)) / 2))
Theorem	bcp1m1 11217	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 11218	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 11219	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 11220	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 11221	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 11222	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 11223	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 11224	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 11225	The binomial coefficent of (𝑁 − 1) is 𝑁. (Contributed by Scott Fenton, 16-May-2014.)
⊢ (𝑁 ∈ ℕ0 → (𝑁C(𝑁 − 1)) = 𝑁)
Theorem	4bc3eq4 11226	The value of four choose three. (Contributed by Scott Fenton, 11-Jun-2016.)
⊢ (4C3) = 4
Theorem	4bc2eq6 11227	The value of four choose two. (Contributed by Scott Fenton, 9-Jan-2017.)
⊢ (4C2) = 6
4.6.10  The ` # ` (set size) function
Syntax	chash 11228	Extend the definition of a class to include the set size function.
class ♯
Definition	df-ihash 11229*	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 7213), 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 7217). Note that we use the sharp sign (♯) for this function and we use the different character octothorpe (#) for the apartness relation (see df-ap 8912). 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 11230*	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 11231	The value of the ♯ function on an infinite set. (Contributed by Jim Kingdon, 20-Feb-2022.)
⊢ (ω ≼ 𝐴 → (♯‘𝐴) = +∞)
Theorem	hashennnuni 11232*	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 11233*	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 11234	Closure of the ♯ function. (Contributed by Paul Chapman, 26-Oct-2012.) (Revised by Mario Carneiro, 13-Jul-2014.)
⊢ (𝐴 ∈ Fin → (♯‘𝐴) ∈ ℕ0)
Theorem	hashfiv01gt1 11235	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 11236	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 11237	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 11238*	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 11239*	There is no bijection between a finite set and an infinite set. By infnfi 7199 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 11240	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 11241	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 11242	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 11243	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 11244	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 11245	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 11246	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 11247	Two ways of saying a finite set is not empty. Also, "A is inhabited" would be equivalent by fin0 7189. (Contributed by Alexander van der Vekens, 23-Sep-2018.) (Intuitionized by Jim Kingdon, 23-Feb-2022.)
⊢ (𝐴 ∈ Fin → (0 < (♯‘𝐴) ↔ 𝐴 ≠ ∅))
Theorem	hashnncl 11248	Positive natural closure of the hash function. (Contributed by Mario Carneiro, 16-Jan-2015.)
⊢ (𝐴 ∈ Fin → ((♯‘𝐴) ∈ ℕ ↔ 𝐴 ≠ ∅))
Theorem	hash0 11249	The empty set has size zero. (Contributed by Mario Carneiro, 8-Jul-2014.)
⊢ (♯‘∅) = 0
Theorem	fihashelne0d 11250	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 11251	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 11252	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 11253	A set equinumerous to the ordinal one has size 1 . (Contributed by Jim Kingdon, 11-Mar-2026.)
⊢ (𝐴 ≈ 1o → (♯‘𝐴) = 1)
Theorem	fihashfn 11254	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 11255	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 11256	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 11257	Dominance relation for the size function. (Contributed by Jim Kingdon, 24-Feb-2022.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) → ((♯‘𝐴) ≤ (♯‘𝐵) ↔ 𝐴 ≼ 𝐵))
Theorem	hashunlem 11258	Lemma for hashun 11259. 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 11259	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 11260	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 11261	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 11262	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 11263	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 11264	There is (at least) one set with two different elements: the unordered pair containing 0 and 1. In contrast to pr0hash2ex 11270, 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 11265	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 11266	Size of a natural number ordinal. (Contributed by Mario Carneiro, 5-Jan-2016.)
⊢ (♯‘1o) = 1
Theorem	hash2 11267	Size of a natural number ordinal. (Contributed by Mario Carneiro, 5-Jan-2016.)
⊢ (♯‘2o) = 2
Theorem	hash3 11268	Size of a natural number ordinal. (Contributed by Mario Carneiro, 5-Jan-2016.)
⊢ (♯‘3o) = 3
Theorem	hash4 11269	Size of a natural number ordinal. (Contributed by Mario Carneiro, 5-Jan-2016.)
⊢ (♯‘4o) = 4
Theorem	pr0hash2ex 11270	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 11271	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 11272	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 11273	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 11274	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 11275	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 11276	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 11277	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 11278	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 11279	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 11280	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 11281	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 11282	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)))
Theorem	hashpwfi 11283	The number of finite subsets of a finite set is two raised to the power of the size of the set. For a similar theorem with set size expressed using equinumerosity, see 2omapfi 7320. For the number of subsets (which need not be finite) of a set, see pw1mapen 17145. (Contributed by Jim Kingdon, 5-Jun-2026.)
⊢ (𝐴 ∈ Fin → (♯‘(𝒫 𝐴 ∩ Fin)) = (2↑(♯‘𝐴)))
Theorem	fimaxq 11284*	A finite set of rational numbers has a maximum. (Contributed by Jim Kingdon, 6-Sep-2022.)
⊢ ((𝐴 ⊆ ℚ ∧ 𝐴 ∈ Fin ∧ 𝐴 ≠ ∅) → ∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 𝑦 ≤ 𝑥)
Theorem	fiubm 11285*	Lemma for fiubz 11286 and fiubnn 11287. A general form of those theorems. (Contributed by Jim Kingdon, 29-Oct-2024.)
⊢ (𝜑 → 𝐴 ⊆ 𝐵)    &   ⊢ (𝜑 → 𝐵 ⊆ ℚ)    &   ⊢ (𝜑 → 𝐶 ∈ 𝐵)    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    ⇒   ⊢ (𝜑 → ∃𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐴 𝑦 ≤ 𝑥)
Theorem	fiubz 11286*	A finite set of integers has an upper bound which is an integer. (Contributed by Jim Kingdon, 29-Oct-2024.)
⊢ ((𝐴 ⊆ ℤ ∧ 𝐴 ∈ Fin) → ∃𝑥 ∈ ℤ ∀𝑦 ∈ 𝐴 𝑦 ≤ 𝑥)
Theorem	fiubnn 11287*	A finite set of natural numbers has an upper bound which is a a natural number. (Contributed by Jim Kingdon, 29-Oct-2024.)
⊢ ((𝐴 ⊆ ℕ ∧ 𝐴 ∈ Fin) → ∃𝑥 ∈ ℕ ∀𝑦 ∈ 𝐴 𝑦 ≤ 𝑥)
Theorem	resunimafz0 11288	The union of a restriction by an image over an open range of nonnegative integers and a singleton of an ordered pair is a restriction by an image over an interval of nonnegative integers. (Contributed by Mario Carneiro, 8-Apr-2015.) (Revised by AV, 20-Feb-2021.)
⊢ (𝜑 → Fun 𝐼)    &   ⊢ (𝜑 → 𝐹:(0..^(♯‘𝐹))⟶dom 𝐼)    &   ⊢ (𝜑 → 𝑁 ∈ (0..^(♯‘𝐹)))    ⇒   ⊢ (𝜑 → (𝐼 ↾ (𝐹 “ (0...𝑁))) = ((𝐼 ↾ (𝐹 “ (0..^𝑁))) ∪ {⟨(𝐹‘𝑁), (𝐼‘(𝐹‘𝑁))⟩}))
Theorem	fnfz0hash 11289	The size of a function on a finite set of sequential nonnegative integers. (Contributed by Alexander van der Vekens, 25-Jun-2018.)
⊢ ((𝑁 ∈ ℕ0 ∧ 𝐹 Fn (0...𝑁)) → (♯‘𝐹) = (𝑁 + 1))
Theorem	ffz0hash 11290	The size of a function on a finite set of sequential nonnegative integers equals the upper bound of the sequence increased by 1. (Contributed by Alexander van der Vekens, 15-Mar-2018.) (Proof shortened by AV, 11-Apr-2021.)
⊢ ((𝑁 ∈ ℕ0 ∧ 𝐹:(0...𝑁)⟶𝐵) → (♯‘𝐹) = (𝑁 + 1))
Theorem	ffzo0hash 11291	The size of a function on a half-open range of nonnegative integers. (Contributed by Alexander van der Vekens, 25-Mar-2018.)
⊢ ((𝑁 ∈ ℕ0 ∧ 𝐹 Fn (0..^𝑁)) → (♯‘𝐹) = 𝑁)
Theorem	fnfzo0hash 11292	The size of a function on a half-open range of nonnegative integers equals the upper bound of this range. (Contributed by Alexander van der Vekens, 26-Jan-2018.) (Proof shortened by AV, 11-Apr-2021.)
⊢ ((𝑁 ∈ ℕ0 ∧ 𝐹:(0..^𝑁)⟶𝐵) → (♯‘𝐹) = 𝑁)
Theorem	sseqn 11293*	Two ways to express the subsets of a class of a given size. It might seem that {𝑥 ∈ 𝒫 𝐴 ∣ (♯‘𝑥) = 𝑁} would suffice, but that would require the converse of hashcl 11234 or something similar. Although each side of the equality would be well defined if we changed 𝑁 ∈ ℕ0 to 𝑁 ∈ ℤ, they would give different results for the (degenerate) case of a negative size, as shown at ssenneg 11294 and sshashneg 11295. (Contributed by Jim Kingdon, 22-May-2026.)
⊢ (𝑁 ∈ ℕ0 → {𝑥 ∈ 𝒫 𝐴 ∣ 𝑥 ≈ (1...𝑁)} = {𝑥 ∈ (𝒫 𝐴 ∩ Fin) ∣ (♯‘𝑥) = 𝑁})
Theorem	ssenneg 11294*	Subsets of a class of a negative size (a degenerate case). Together with sshashneg 11295 this shows that sseqn 11293 could not be extended beyond 𝑁 ∈ ℕ0. (Contributed by Jim Kingdon, 22-May-2026.)
⊢ ((𝑁 ∈ ℤ ∧ 𝑁 < 0) → {𝑥 ∈ 𝒫 𝐴 ∣ 𝑥 ≈ (1...𝑁)} = {∅})
Theorem	sshashneg 11295*	Subsets of a class of a negative size (a degenerate case). Together with ssenneg 11294 this shows that sseqn 11293 could not be extended beyond 𝑁 ∈ ℕ0. (Contributed by Jim Kingdon, 22-May-2026.)
⊢ ((𝑁 ∈ ℤ ∧ 𝑁 < 0) → {𝑥 ∈ (𝒫 𝐴 ∩ Fin) ∣ (♯‘𝑥) = 𝑁} = ∅)
Theorem	hashfibclem 11296*	Lemma for hashfibc 11297: inductive step. (Contributed by Mario Carneiro, 13-Jul-2014.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ (𝜑 → ¬ 𝑧 ∈ 𝐴)    &   ⊢ (𝜑 → ∀𝑗 ∈ ℤ ((♯‘𝐴)C𝑗) = (♯‘{𝑥 ∈ (𝒫 𝐴 ∩ Fin) ∣ (♯‘𝑥) = 𝑗}))    &   ⊢ (𝜑 → 𝐾 ∈ ℤ)    ⇒   ⊢ (𝜑 → ((♯‘(𝐴 ∪ {𝑧}))C𝐾) = (♯‘{𝑥 ∈ (𝒫 (𝐴 ∪ {𝑧}) ∩ Fin) ∣ (♯‘𝑥) = 𝐾}))
Theorem	hashfibc 11297*	The binomial coefficient counts the number of subsets of a finite set of a given size. This is Metamath 100 proof #58 (formula for the number of combinations). For more on the notation for subsets of a given size, see sseqn 11293. (Contributed by Mario Carneiro, 13-Jul-2014.)
⊢ ((𝐴 ∈ Fin ∧ 𝐾 ∈ ℤ) → ((♯‘𝐴)C𝐾) = (♯‘{𝑥 ∈ (𝒫 𝐴 ∩ Fin) ∣ (♯‘𝑥) = 𝐾}))
Theorem	hashfacen 11298*	The number of bijections between two sets is a cardinal invariant. (Contributed by Mario Carneiro, 21-Jan-2015.)
⊢ ((𝐴 ≈ 𝐵 ∧ 𝐶 ≈ 𝐷) → {𝑓 ∣ 𝑓:𝐴–1-1-onto→𝐶} ≈ {𝑓 ∣ 𝑓:𝐵–1-1-onto→𝐷})
Theorem	hashf1lem1 11299*	Lemma for hashf1 11301. (Contributed by Mario Carneiro, 17-Apr-2015.) (Proof shortened by AV, 14-Aug-2024.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ (𝜑 → 𝐵 ∈ Fin)    &   ⊢ (𝜑 → ¬ 𝑧 ∈ 𝐴)    &   ⊢ (𝜑 → ((♯‘𝐴) + 1) ≤ (♯‘𝐵))    &   ⊢ (𝜑 → 𝐹:𝐴–1-1→𝐵)    ⇒   ⊢ (𝜑 → {𝑓 ∣ ((𝑓 ↾ 𝐴) = 𝐹 ∧ 𝑓:(𝐴 ∪ {𝑧})–1-1→𝐵)} ≈ (𝐵 ∖ ran 𝐹))
Theorem	hashf1lem2 11300*	Lemma for hashf1 11301. (Contributed by Mario Carneiro, 17-Apr-2015.)
⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ (𝜑 → 𝐵 ∈ Fin)    &   ⊢ (𝜑 → ¬ 𝑧 ∈ 𝐴)    &   ⊢ (𝜑 → ((♯‘𝐴) + 1) ≤ (♯‘𝐵))    ⇒   ⊢ (𝜑 → (♯‘{𝑓 ∣ 𝑓:(𝐴 ∪ {𝑧})–1-1→𝐵}) = (((♯‘𝐵) − (♯‘𝐴)) · (♯‘{𝑓 ∣ 𝑓:𝐴–1-1→𝐵})))