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| Type | Label | Description |
|---|---|---|
| Statement | ||
| Theorem | facp1 14201 | The factorial of a successor. (Contributed by NM, 2-Dec-2004.) (Revised by Mario Carneiro, 13-Jul-2013.) |
| ⊢ (𝑁 ∈ ℕ0 → (!‘(𝑁 + 1)) = ((!‘𝑁) · (𝑁 + 1))) | ||
| Theorem | fac2 14202 | The factorial of 2. (Contributed by NM, 17-Mar-2005.) |
| ⊢ (!‘2) = 2 | ||
| Theorem | fac3 14203 | The factorial of 3. (Contributed by NM, 17-Mar-2005.) |
| ⊢ (!‘3) = 6 | ||
| Theorem | fac4 14204 | The factorial of 4. (Contributed by Mario Carneiro, 18-Jun-2015.) |
| ⊢ (!‘4) = ;24 | ||
| Theorem | facnn2 14205 | Value of the factorial function expressed recursively. (Contributed by NM, 2-Dec-2004.) |
| ⊢ (𝑁 ∈ ℕ → (!‘𝑁) = ((!‘(𝑁 − 1)) · 𝑁)) | ||
| Theorem | faccl 14206 | Closure of the factorial function. (Contributed by NM, 2-Dec-2004.) |
| ⊢ (𝑁 ∈ ℕ0 → (!‘𝑁) ∈ ℕ) | ||
| Theorem | faccld 14207 | Closure of the factorial function, deduction version of faccl 14206. (Contributed by Glauco Siliprandi, 5-Apr-2020.) |
| ⊢ (𝜑 → 𝑁 ∈ ℕ0) ⇒ ⊢ (𝜑 → (!‘𝑁) ∈ ℕ) | ||
| Theorem | facmapnn 14208 | 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 14209 | The factorial function is nonzero. (Contributed by NM, 26-Apr-2005.) |
| ⊢ (𝑁 ∈ ℕ0 → (!‘𝑁) ≠ 0) | ||
| Theorem | facdiv 14210 | 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 14211 | 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 14212 | Ordering property of factorial. (Contributed by NM, 9-Dec-2005.) |
| ⊢ ((𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0 ∧ 𝑀 ≤ 𝑁) → (!‘𝑀) ≤ (!‘𝑁)) | ||
| Theorem | faclbnd 14213 | 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 14214 | A lower bound for the factorial function. (Contributed by NM, 17-Dec-2005.) |
| ⊢ (𝑁 ∈ ℕ0 → ((2↑𝑁) / 2) ≤ (!‘𝑁)) | ||
| Theorem | faclbnd3 14215 | A lower bound for the factorial function. (Contributed by NM, 19-Dec-2005.) |
| ⊢ ((𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0) → (𝑀↑𝑁) ≤ ((𝑀↑𝑀) · (!‘𝑁))) | ||
| Theorem | faclbnd4lem1 14216 | Lemma for faclbnd4 14220. 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 14217 | Lemma for faclbnd4 14220. Use the weak deduction theorem to convert the hypotheses of faclbnd4lem1 14216 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 14218 | Lemma for faclbnd4 14220. 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 14219 | Lemma for faclbnd4 14220. 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 14220 | Variant of faclbnd5 14221 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 14221 | 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 14222 | 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 14223 | An upper bound for the factorial function. (Contributed by Mario Carneiro, 15-Apr-2016.) |
| ⊢ (𝑁 ∈ ℕ0 → (!‘𝑁) ≤ (𝑁↑𝑁)) | ||
| Theorem | facavg 14224 | 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))) | ||
| Syntax | cbc 14225 | Extend class notation to include the binomial coefficient operation (combinatorial choose operation). |
| class C | ||
| Definition | df-bc 14226* |
Define the binomial coefficient operation. For example,
(5C3) = 10 (ex-bc 30527).
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 14227 | 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 14228 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 14228 | 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 14229 | 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 14230 | 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 14231 | Closure of the binomial coefficient in the positive reals. (This is mostly a lemma before we have bccl2 14246.) (Contributed by Mario Carneiro, 10-Mar-2014.) |
| ⊢ (𝐾 ∈ (0...𝑁) → (𝑁C𝐾) ∈ ℝ+) | ||
| Theorem | bccmpl 14232 | "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 14233 | 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 14234 | The binomial coefficient " 0 choose 𝐾 " is 0 for a positive integer K. Note that (0C0) = 1 (see bcn0 14233). (Contributed by Alexander van der Vekens, 1-Jan-2018.) |
| ⊢ (𝐾 ∈ ℕ → (0C𝐾) = 0) | ||
| Theorem | bcnn 14235 | 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 14236 | Binomial coefficient: 𝑁 choose 1. (Contributed by NM, 21-Jun-2005.) (Revised by Mario Carneiro, 8-Nov-2013.) |
| ⊢ (𝑁 ∈ ℕ0 → (𝑁C1) = 𝑁) | ||
| Theorem | bcnp1n 14237 | 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 14238 | 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 14239 | 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 14240 | 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 14241 | 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 14242 | Binomial coefficient: 𝑁 choose 2. (Contributed by Mario Carneiro, 22-May-2014.) |
| ⊢ (𝑁 ∈ ℕ0 → (𝑁C2) = ((𝑁 · (𝑁 − 1)) / 2)) | ||
| Theorem | bcp1m1 14243 | 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 14244 | 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 14245 | 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 14246 | 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 14247 | 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 14248 | 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 14249 | 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 14250 | The binomial coefficient of (𝑁 − 1) is 𝑁. (Contributed by Scott Fenton, 16-May-2014.) |
| ⊢ (𝑁 ∈ ℕ0 → (𝑁C(𝑁 − 1)) = 𝑁) | ||
| Theorem | 4bc3eq4 14251 | The value of four choose three. (Contributed by Scott Fenton, 11-Jun-2016.) |
| ⊢ (4C3) = 4 | ||
| Theorem | 4bc2eq6 14252 | The value of four choose two. (Contributed by Scott Fenton, 9-Jan-2017.) |
| ⊢ (4C2) = 6 | ||
| Syntax | chash 14253 | Extend the definition of a class to include the set size function. |
| class ♯ | ||
| Definition | df-hash 14254 | 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 30528). (Contributed by Paul Chapman, 22-Jun-2011.) |
| ⊢ ♯ = (((rec((𝑥 ∈ V ↦ (𝑥 + 1)), 0) ↾ ω) ∘ card) ∪ ((V ∖ Fin) × {+∞})) | ||
| Theorem | hashkf 14255 | 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 14256* | The value of the ♯ function in terms of the mapping 𝐺 from ω to ℕ0. The proof avoids the use of ax-ac 10369. (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 14257* | 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 14258 | The value of the ♯ function on an infinite set. (Contributed by Mario Carneiro, 13-Jul-2014.) |
| ⊢ ((𝐴 ∈ 𝑉 ∧ ¬ 𝐴 ∈ Fin) → (♯‘𝐴) = +∞) | ||
| Theorem | hashbnd 14259 | 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 14260 | 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 14261 | 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 14260? (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 14262 | 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 14263 | Restriction of the domain of the size function. (Contributed by Thierry Arnoux, 31-Jan-2017.) |
| ⊢ (♯ ↾ 𝐴) Fn 𝐴 | ||
| Theorem | dmhashres 14264 | Restriction of the domain of the size function. (Contributed by Thierry Arnoux, 12-Jan-2017.) |
| ⊢ dom (♯ ↾ 𝐴) = 𝐴 | ||
| Theorem | hashnn0pnf 14265 | 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 14262? (Contributed by Alexander van der Vekens, 6-Dec-2017.) |
| ⊢ (𝑀 ∈ 𝑉 → ((♯‘𝑀) ∈ ℕ0 ∨ (♯‘𝑀) = +∞)) | ||
| Theorem | hashnnn0genn0 14266 | 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 14267 | The size of a set is never minus infinity. (Contributed by Alexander van der Vekens, 21-Dec-2017.) |
| ⊢ (𝐴 ∈ 𝑉 → (♯‘𝐴) ≠ -∞) | ||
| Theorem | hashv01gt1 14268 | 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 14269 | 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 14270 | 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 14271 | 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 14272* | 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 14273* | 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 14274* | 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 14275 | 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 14276 | 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 14277 | 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 14278 | 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 14279 | Closure of the ♯ function. (Contributed by Paul Chapman, 26-Oct-2012.) (Revised by Mario Carneiro, 13-Jul-2014.) |
| ⊢ (𝐴 ∈ Fin → (♯‘𝐴) ∈ ℕ0) | ||
| Theorem | hashxrcl 14280 | Extended real closure of the ♯ function. (Contributed by Mario Carneiro, 22-Apr-2015.) |
| ⊢ (𝐴 ∈ 𝑉 → (♯‘𝐴) ∈ ℝ*) | ||
| Theorem | hashclb 14281 | Reverse closure of the ♯ function. (Contributed by Mario Carneiro, 15-Jan-2015.) |
| ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ Fin ↔ (♯‘𝐴) ∈ ℕ0)) | ||
| Theorem | nfile 14282 | 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 14283 | 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 14284 | 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 14285 | 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 14286 | 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 14287 | 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 14288 | 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 14289 | Positive natural closure of the hash function. (Contributed by Mario Carneiro, 16-Jan-2015.) |
| ⊢ (𝐴 ∈ Fin → ((♯‘𝐴) ∈ ℕ ↔ 𝐴 ≠ ∅)) | ||
| Theorem | hash0 14290 | The empty set has size zero. (Contributed by Mario Carneiro, 8-Jul-2014.) |
| ⊢ (♯‘∅) = 0 | ||
| Theorem | hashelne0d 14291 | 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 14292 | 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 14293 | 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 14294 | 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 14295* | 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 14296* | 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 14297* | 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 14298 | A function is equinumerous to its domain. (Contributed by Mario Carneiro, 12-Mar-2015.) |
| ⊢ (𝐹 Fn 𝐴 → (♯‘𝐹) = (♯‘𝐴)) | ||
| Theorem | fseq1hash 14299 | 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 14300 | 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))) | ||
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