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	faclbnd3 11101	A lower bound for the factorial function. (Contributed by NM, 19-Dec-2005.)
⊢ ((𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0) → (𝑀↑𝑁) ≤ ((𝑀↑𝑀) · (!‘𝑁)))
Theorem	faclbnd6 11102	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 11103	An upper bound for the factorial function. (Contributed by Mario Carneiro, 15-Apr-2016.)
⊢ (𝑁 ∈ ℕ0 → (!‘𝑁) ≤ (𝑁↑𝑁))
Theorem	facavg 11104	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 11105	Extend class notation to include the binomial coefficient operation (combinatorial choose operation).
class C
Definition	df-bc 11106*	Define the binomial coefficient operation. For example, (5C3) = 10 (ex-bc 16484). 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 11107	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 11108 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 11108	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 11109	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 11110	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 11111	Closure of the binomial coefficient in the positive reals. (This is mostly a lemma before we have bccl2 11126.) (Contributed by Mario Carneiro, 10-Mar-2014.)
⊢ (𝐾 ∈ (0...𝑁) → (𝑁C𝐾) ∈ ℝ+)
Theorem	bccmpl 11112	"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 11113	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 11114	The binomial coefficient " 0 choose 𝐾 " is 0 for a positive integer K. Note that (0C0) = 1 (see bcn0 11113). (Contributed by Alexander van der Vekens, 1-Jan-2018.)
⊢ (𝐾 ∈ ℕ → (0C𝐾) = 0)
Theorem	bcnn 11115	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 11116	Binomial coefficient: 𝑁 choose 1. (Contributed by NM, 21-Jun-2005.) (Revised by Mario Carneiro, 8-Nov-2013.)
⊢ (𝑁 ∈ ℕ0 → (𝑁C1) = 𝑁)
Theorem	bcnp1n 11117	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 11118	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 11119	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 11120	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 11121	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 11122	Binomial coefficient: 𝑁 choose 2. (Contributed by Mario Carneiro, 22-May-2014.)
⊢ (𝑁 ∈ ℕ0 → (𝑁C2) = ((𝑁 · (𝑁 − 1)) / 2))
Theorem	bcp1m1 11123	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 11124	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 11125	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 11126	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 11127	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 11128	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 11129	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 11130	The binomial coefficent of (𝑁 − 1) is 𝑁. (Contributed by Scott Fenton, 16-May-2014.)
⊢ (𝑁 ∈ ℕ0 → (𝑁C(𝑁 − 1)) = 𝑁)
Theorem	4bc3eq4 11131	The value of four choose three. (Contributed by Scott Fenton, 11-Jun-2016.)
⊢ (4C3) = 4
Theorem	4bc2eq6 11132	The value of four choose two. (Contributed by Scott Fenton, 9-Jan-2017.)
⊢ (4C2) = 6
4.6.10  The ` # ` (set size) function
Syntax	chash 11133	Extend the definition of a class to include the set size function.
class ♯
Definition	df-ihash 11134*	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 7165), 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 7169). Note that we use the sharp sign (♯) for this function and we use the different character octothorpe (#) for the apartness relation (see df-ap 8852). 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 11135*	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 11136	The value of the ♯ function on an infinite set. (Contributed by Jim Kingdon, 20-Feb-2022.)
⊢ (ω ≼ 𝐴 → (♯‘𝐴) = +∞)
Theorem	hashennnuni 11137*	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 11138*	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 11139	Closure of the ♯ function. (Contributed by Paul Chapman, 26-Oct-2012.) (Revised by Mario Carneiro, 13-Jul-2014.)
⊢ (𝐴 ∈ Fin → (♯‘𝐴) ∈ ℕ0)
Theorem	hashfiv01gt1 11140	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 11141	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 11142	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 11143*	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 11144*	There is no bijection between a finite set and an infinite set. By infnfi 7151 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 11145	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 11146	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 11147	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 11148	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 11149	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 11150	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 11151	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 11152	Two ways of saying a finite set is not empty. Also, "A is inhabited" would be equivalent by fin0 7141. (Contributed by Alexander van der Vekens, 23-Sep-2018.) (Intuitionized by Jim Kingdon, 23-Feb-2022.)
⊢ (𝐴 ∈ Fin → (0 < (♯‘𝐴) ↔ 𝐴 ≠ ∅))
Theorem	hashnncl 11153	Positive natural closure of the hash function. (Contributed by Mario Carneiro, 16-Jan-2015.)
⊢ (𝐴 ∈ Fin → ((♯‘𝐴) ∈ ℕ ↔ 𝐴 ≠ ∅))
Theorem	hash0 11154	The empty set has size zero. (Contributed by Mario Carneiro, 8-Jul-2014.)
⊢ (♯‘∅) = 0
Theorem	fihashelne0d 11155	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 11156	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 11157	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 11158	A set equinumerous to the ordinal one has size 1 . (Contributed by Jim Kingdon, 11-Mar-2026.)
⊢ (𝐴 ≈ 1o → (♯‘𝐴) = 1)
Theorem	fihashfn 11159	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 11160	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 11161	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 11162	Dominance relation for the size function. (Contributed by Jim Kingdon, 24-Feb-2022.)
⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) → ((♯‘𝐴) ≤ (♯‘𝐵) ↔ 𝐴 ≼ 𝐵))
Theorem	hashunlem 11163	Lemma for hashun 11164. 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 11164	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 11165	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 11166	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 11167	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 11168	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 11169	There is (at least) one set with two different elements: the unordered pair containing 0 and 1. In contrast to pr0hash2ex 11175, 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 11170	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 11171	Size of a natural number ordinal. (Contributed by Mario Carneiro, 5-Jan-2016.)
⊢ (♯‘1o) = 1
Theorem	hash2 11172	Size of a natural number ordinal. (Contributed by Mario Carneiro, 5-Jan-2016.)
⊢ (♯‘2o) = 2
Theorem	hash3 11173	Size of a natural number ordinal. (Contributed by Mario Carneiro, 5-Jan-2016.)
⊢ (♯‘3o) = 3
Theorem	hash4 11174	Size of a natural number ordinal. (Contributed by Mario Carneiro, 5-Jan-2016.)
⊢ (♯‘4o) = 4
Theorem	pr0hash2ex 11175	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 11176	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 11177	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 11178	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 11179	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 11180	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 11181	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 11182	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 11183	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 11184	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 11185	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 11186	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	fimaxq 11187*	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 11188*	Lemma for fiubz 11189 and fiubnn 11190. 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 11189*	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 11190*	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 11191	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 11192	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 11193	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 11194	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 11195	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 11196*	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 11139 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 11197 and sshashneg 11198. (Contributed by Jim Kingdon, 22-May-2026.)
⊢ (𝑁 ∈ ℕ0 → {𝑥 ∈ 𝒫 𝐴 ∣ 𝑥 ≈ (1...𝑁)} = {𝑥 ∈ (𝒫 𝐴 ∩ Fin) ∣ (♯‘𝑥) = 𝑁})
Theorem	ssenneg 11197*	Subsets of a class of a negative size (a degenerate case). Together with sshashneg 11198 this shows that sseqn 11196 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 11198*	Subsets of a class of a negative size (a degenerate case). Together with ssenneg 11197 this shows that sseqn 11196 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 11199*	Lemma for hashfibc 11200: 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 11200*	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 11196. (Contributed by Mario Carneiro, 13-Jul-2014.)
⊢ ((𝐴 ∈ Fin ∧ 𝐾 ∈ ℤ) → ((♯‘𝐴)C𝐾) = (♯‘{𝑥 ∈ (𝒫 𝐴 ∩ Fin) ∣ (♯‘𝑥) = 𝐾}))