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| Type | Label | Description |
|---|---|---|
| Statement | ||
| Syntax | cbc 11201 | Extend class notation to include the binomial coefficient operation (combinatorial choose operation). |
| class C | ||
| Definition | df-bc 11202* |
Define the binomial coefficient operation. For example,
(5C3) = 10 (ex-bc 16914).
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 11203 | 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 11204 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 11204 | 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 11205 | 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 11206 | 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 11207 | Closure of the binomial coefficient in the positive reals. (This is mostly a lemma before we have bccl2 11222.) (Contributed by Mario Carneiro, 10-Mar-2014.) |
| ⊢ (𝐾 ∈ (0...𝑁) → (𝑁C𝐾) ∈ ℝ+) | ||
| Theorem | bccmpl 11208 | "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 11209 | 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 11210 | The binomial coefficient " 0 choose 𝐾 " is 0 for a positive integer K. Note that (0C0) = 1 (see bcn0 11209). (Contributed by Alexander van der Vekens, 1-Jan-2018.) |
| ⊢ (𝐾 ∈ ℕ → (0C𝐾) = 0) | ||
| Theorem | bcnn 11211 | 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 11212 | Binomial coefficient: 𝑁 choose 1. (Contributed by NM, 21-Jun-2005.) (Revised by Mario Carneiro, 8-Nov-2013.) |
| ⊢ (𝑁 ∈ ℕ0 → (𝑁C1) = 𝑁) | ||
| Theorem | bcnp1n 11213 | 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 11214 | 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 11215 | 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 11216 | 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 11217 | 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 11218 | Binomial coefficient: 𝑁 choose 2. (Contributed by Mario Carneiro, 22-May-2014.) |
| ⊢ (𝑁 ∈ ℕ0 → (𝑁C2) = ((𝑁 · (𝑁 − 1)) / 2)) | ||
| Theorem | bcp1m1 11219 | 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 11220 | 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 11221 | 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 11222 | 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 11223 | 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 11224 | 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 11225 | 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 11226 | 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 11227 | The binomial coefficent of (𝑁 − 1) is 𝑁. (Contributed by Scott Fenton, 16-May-2014.) |
| ⊢ (𝑁 ∈ ℕ0 → (𝑁C(𝑁 − 1)) = 𝑁) | ||
| Theorem | 4bc3eq4 11228 | The value of four choose three. (Contributed by Scott Fenton, 11-Jun-2016.) |
| ⊢ (4C3) = 4 | ||
| Theorem | 4bc2eq6 11229 | The value of four choose two. (Contributed by Scott Fenton, 9-Jan-2017.) |
| ⊢ (4C2) = 6 | ||
| Syntax | chash 11230 | Extend the definition of a class to include the set size function. |
| class ♯ | ||
| Definition | df-ihash 11231* |
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 8913). 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 11232* | 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 11233 | The value of the ♯ function on an infinite set. (Contributed by Jim Kingdon, 20-Feb-2022.) |
| ⊢ (ω ≼ 𝐴 → (♯‘𝐴) = +∞) | ||
| Theorem | hashennnuni 11234* | 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 11235* | 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 11236 | Closure of the ♯ function. (Contributed by Paul Chapman, 26-Oct-2012.) (Revised by Mario Carneiro, 13-Jul-2014.) |
| ⊢ (𝐴 ∈ Fin → (♯‘𝐴) ∈ ℕ0) | ||
| Theorem | hashfiv01gt1 11237 | 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 11238 | 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 11239 | 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 11240* | 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 11241* | 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 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.) |
| ⊢ ((𝐴 ∈ Fin ∧ 𝐹:𝐴–1-1-onto→𝐵) → (♯‘𝐴) = (♯‘𝐵)) | ||
| Theorem | fihashf1rn 11243 | 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 11244 | 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 11245 | 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 11246 | 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 11247 | 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 11248 | 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 11249 | 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 11250 | Positive natural closure of the hash function. (Contributed by Mario Carneiro, 16-Jan-2015.) |
| ⊢ (𝐴 ∈ Fin → ((♯‘𝐴) ∈ ℕ ↔ 𝐴 ≠ ∅)) | ||
| Theorem | hash0 11251 | The empty set has size zero. (Contributed by Mario Carneiro, 8-Jul-2014.) |
| ⊢ (♯‘∅) = 0 | ||
| Theorem | fihashelne0d 11252 | 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 11253 | 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 11254 | 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 11255 | A set equinumerous to the ordinal one has size 1 . (Contributed by Jim Kingdon, 11-Mar-2026.) |
| ⊢ (𝐴 ≈ 1o → (♯‘𝐴) = 1) | ||
| Theorem | fihashfn 11256 | 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 11257 | 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 11258 | 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 11259 | Dominance relation for the size function. (Contributed by Jim Kingdon, 24-Feb-2022.) |
| ⊢ ((𝐴 ∈ Fin ∧ 𝐵 ∈ Fin) → ((♯‘𝐴) ≤ (♯‘𝐵) ↔ 𝐴 ≼ 𝐵)) | ||
| Theorem | hashunlem 11260 | Lemma for hashun 11261. 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 11261 | 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 11262 | 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 11263 | 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 11264 | 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 11265 | 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 11266 | There is (at least) one set with two different elements: the unordered pair containing 0 and 1. In contrast to pr0hash2ex 11272, 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 11267 | 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 11268 | Size of a natural number ordinal. (Contributed by Mario Carneiro, 5-Jan-2016.) |
| ⊢ (♯‘1o) = 1 | ||
| Theorem | hash2 11269 | Size of a natural number ordinal. (Contributed by Mario Carneiro, 5-Jan-2016.) |
| ⊢ (♯‘2o) = 2 | ||
| Theorem | hash3 11270 | Size of a natural number ordinal. (Contributed by Mario Carneiro, 5-Jan-2016.) |
| ⊢ (♯‘3o) = 3 | ||
| Theorem | hash4 11271 | Size of a natural number ordinal. (Contributed by Mario Carneiro, 5-Jan-2016.) |
| ⊢ (♯‘4o) = 4 | ||
| Theorem | pr0hash2ex 11272 | 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 11273 | 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 11274 | 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 11275 | 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 11276 | 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 11277 | 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 11278 | 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 11279 | 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 11280 | 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 11281 | 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 11282 | 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 11283 | 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 11284 | 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 11285 | 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 7321. For the number of subsets (which need not be finite) of a set, see pw1mapen 17197. (Contributed by Jim Kingdon, 5-Jun-2026.) |
| ⊢ (𝐴 ∈ Fin → (♯‘(𝒫 𝐴 ∩ Fin)) = (2↑(♯‘𝐴))) | ||
| Theorem | fimaxq 11286* | 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 11287* | Lemma for fiubz 11288 and fiubnn 11289. 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 11288* | 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 11289* | 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 11290 | 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 11291 | 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 11292 | 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 11293 | 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 11294 | 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 11295* | 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 11236 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 11296 and sshashneg 11297. (Contributed by Jim Kingdon, 22-May-2026.) |
| ⊢ (𝑁 ∈ ℕ0 → {𝑥 ∈ 𝒫 𝐴 ∣ 𝑥 ≈ (1...𝑁)} = {𝑥 ∈ (𝒫 𝐴 ∩ Fin) ∣ (♯‘𝑥) = 𝑁}) | ||
| Theorem | ssenneg 11296* | Subsets of a class of a negative size (a degenerate case). Together with sshashneg 11297 this shows that sseqn 11295 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 11297* | Subsets of a class of a negative size (a degenerate case). Together with ssenneg 11296 this shows that sseqn 11295 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 11298* | Lemma for hashfibc 11299: 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 11299* | 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 11295. (Contributed by Mario Carneiro, 13-Jul-2014.) |
| ⊢ ((𝐴 ∈ Fin ∧ 𝐾 ∈ ℤ) → ((♯‘𝐴)C𝐾) = (♯‘{𝑥 ∈ (𝒫 𝐴 ∩ Fin) ∣ (♯‘𝑥) = 𝐾})) | ||
| Theorem | hashfacen 11300* | 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}) | ||
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