| Metamath
Proof Explorer Theorem List (p. 144 of 505) | < Previous Next > | |
| Bad symbols? Try the
GIF version. |
||
|
Mirrors > Metamath Home Page > MPE Home Page > Theorem List Contents > Recent Proofs This page: Page List |
||
| Color key: | (1-31179) |
(31180-32702) |
(32703-50434) |
| Type | Label | Description |
|---|---|---|
| Statement | ||
| Definition | df-fac 14301 | Define the factorial function on nonnegative integers. For example, (!‘5) = 120 because 1 · 2 · 3 · 4 · 5 = 120 (ex-fac 30711). In the literature, the factorial function is written as a postscript exclamation point. (Contributed by NM, 2-Dec-2004.) |
| ⊢ ! = ({〈0, 1〉} ∪ seq1( · , I )) | ||
| Theorem | facnn 14302 | Value of the factorial function for positive integers. (Contributed by NM, 2-Dec-2004.) (Revised by Mario Carneiro, 13-Jul-2013.) |
| ⊢ (𝑁 ∈ ℕ → (!‘𝑁) = (seq1( · , I )‘𝑁)) | ||
| Theorem | fac0 14303 | The factorial of 0. (Contributed by NM, 2-Dec-2004.) (Revised by Mario Carneiro, 13-Jul-2013.) |
| ⊢ (!‘0) = 1 | ||
| Theorem | fac1 14304 | The factorial of 1. (Contributed by NM, 2-Dec-2004.) (Revised by Mario Carneiro, 13-Jul-2013.) |
| ⊢ (!‘1) = 1 | ||
| Theorem | facp1 14305 | 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 14306 | The factorial of 2. (Contributed by NM, 17-Mar-2005.) |
| ⊢ (!‘2) = 2 | ||
| Theorem | fac3 14307 | The factorial of 3. (Contributed by NM, 17-Mar-2005.) |
| ⊢ (!‘3) = 6 | ||
| Theorem | fac4 14308 | The factorial of 4. (Contributed by Mario Carneiro, 18-Jun-2015.) |
| ⊢ (!‘4) = ;24 | ||
| Theorem | facnn2 14309 | Value of the factorial function expressed recursively. (Contributed by NM, 2-Dec-2004.) |
| ⊢ (𝑁 ∈ ℕ → (!‘𝑁) = ((!‘(𝑁 − 1)) · 𝑁)) | ||
| Theorem | faccl 14310 | Closure of the factorial function. (Contributed by NM, 2-Dec-2004.) |
| ⊢ (𝑁 ∈ ℕ0 → (!‘𝑁) ∈ ℕ) | ||
| Theorem | faccld 14311 | Closure of the factorial function, deduction version of faccl 14310. (Contributed by Glauco Siliprandi, 5-Apr-2020.) |
| ⊢ (𝜑 → 𝑁 ∈ ℕ0) ⇒ ⊢ (𝜑 → (!‘𝑁) ∈ ℕ) | ||
| Theorem | facmapnn 14312 | 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 14313 | The factorial function is nonzero. (Contributed by NM, 26-Apr-2005.) |
| ⊢ (𝑁 ∈ ℕ0 → (!‘𝑁) ≠ 0) | ||
| Theorem | facdiv 14314 | 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 14315 | 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 14316 | Ordering property of factorial. (Contributed by NM, 9-Dec-2005.) |
| ⊢ ((𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0 ∧ 𝑀 ≤ 𝑁) → (!‘𝑀) ≤ (!‘𝑁)) | ||
| Theorem | faclbnd 14317 | 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 14318 | A lower bound for the factorial function. (Contributed by NM, 17-Dec-2005.) |
| ⊢ (𝑁 ∈ ℕ0 → ((2↑𝑁) / 2) ≤ (!‘𝑁)) | ||
| Theorem | faclbnd3 14319 | A lower bound for the factorial function. (Contributed by NM, 19-Dec-2005.) |
| ⊢ ((𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0) → (𝑀↑𝑁) ≤ ((𝑀↑𝑀) · (!‘𝑁))) | ||
| Theorem | faclbnd4lem1 14320 | Lemma for faclbnd4 14324. 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 14321 | Lemma for faclbnd4 14324. Use the weak deduction theorem to convert the hypotheses of faclbnd4lem1 14320 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 14322 | Lemma for faclbnd4 14324. 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 14323 | Lemma for faclbnd4 14324. 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 14324 | Variant of faclbnd5 14325 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 14325 | 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 14326 | 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 14327 | An upper bound for the factorial function. (Contributed by Mario Carneiro, 15-Apr-2016.) |
| ⊢ (𝑁 ∈ ℕ0 → (!‘𝑁) ≤ (𝑁↑𝑁)) | ||
| Theorem | facavg 14328 | 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 14329 | Extend class notation to include the binomial coefficient operation (combinatorial choose operation). |
| class C | ||
| Definition | df-bc 14330* |
Define the binomial coefficient operation. For example,
(5C3) = 10 (ex-bc 30712).
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 14331 | 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 14332 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 14332 | 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 14333 | 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 14334 | 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 14335 | Closure of the binomial coefficient in the positive reals. (This is mostly a lemma before we have bccl2 14350.) (Contributed by Mario Carneiro, 10-Mar-2014.) |
| ⊢ (𝐾 ∈ (0...𝑁) → (𝑁C𝐾) ∈ ℝ+) | ||
| Theorem | bccmpl 14336 | "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 14337 | 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 14338 | The binomial coefficient " 0 choose 𝐾 " is 0 for a positive integer K. Note that (0C0) = 1 (see bcn0 14337). (Contributed by Alexander van der Vekens, 1-Jan-2018.) |
| ⊢ (𝐾 ∈ ℕ → (0C𝐾) = 0) | ||
| Theorem | bcnn 14339 | 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 14340 | Binomial coefficient: 𝑁 choose 1. (Contributed by NM, 21-Jun-2005.) (Revised by Mario Carneiro, 8-Nov-2013.) |
| ⊢ (𝑁 ∈ ℕ0 → (𝑁C1) = 𝑁) | ||
| Theorem | bcnp1n 14341 | 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 14342 | 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 14343 | 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 14344 | 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 14345 | 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 14346 | Binomial coefficient: 𝑁 choose 2. (Contributed by Mario Carneiro, 22-May-2014.) |
| ⊢ (𝑁 ∈ ℕ0 → (𝑁C2) = ((𝑁 · (𝑁 − 1)) / 2)) | ||
| Theorem | bcp1m1 14347 | 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 14348 | 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 14349 | 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 14350 | 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 14351 | 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 14352 | 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 14353 | 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 14354 | The binomial coefficient of (𝑁 − 1) is 𝑁. (Contributed by Scott Fenton, 16-May-2014.) |
| ⊢ (𝑁 ∈ ℕ0 → (𝑁C(𝑁 − 1)) = 𝑁) | ||
| Theorem | 4bc3eq4 14355 | The value of four choose three. (Contributed by Scott Fenton, 11-Jun-2016.) |
| ⊢ (4C3) = 4 | ||
| Theorem | 4bc2eq6 14356 | The value of four choose two. (Contributed by Scott Fenton, 9-Jan-2017.) |
| ⊢ (4C2) = 6 | ||
| Syntax | chash 14357 | Extend the definition of a class to include the set size function. |
| class ♯ | ||
| Definition | df-hash 14358 | 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 30713). (Contributed by Paul Chapman, 22-Jun-2011.) |
| ⊢ ♯ = (((rec((𝑥 ∈ V ↦ (𝑥 + 1)), 0) ↾ ω) ∘ card) ∪ ((V ∖ Fin) × {+∞})) | ||
| Theorem | hashkf 14359 | 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 14360* | The value of the ♯ function in terms of the mapping 𝐺 from ω to ℕ0. The proof avoids the use of ax-ac 10431. (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 14361* | 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 14362 | The value of the ♯ function on an infinite set. (Contributed by Mario Carneiro, 13-Jul-2014.) |
| ⊢ ((𝐴 ∈ 𝑉 ∧ ¬ 𝐴 ∈ Fin) → (♯‘𝐴) = +∞) | ||
| Theorem | hashbnd 14363 | 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 14364 | 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 14365 | 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 14364? (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 14366 | 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 14367 | Restriction of the domain of the size function. (Contributed by Thierry Arnoux, 31-Jan-2017.) |
| ⊢ (♯ ↾ 𝐴) Fn 𝐴 | ||
| Theorem | dmhashres 14368 | Restriction of the domain of the size function. (Contributed by Thierry Arnoux, 12-Jan-2017.) |
| ⊢ dom (♯ ↾ 𝐴) = 𝐴 | ||
| Theorem | hashnn0pnf 14369 | 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 14366? (Contributed by Alexander van der Vekens, 6-Dec-2017.) |
| ⊢ (𝑀 ∈ 𝑉 → ((♯‘𝑀) ∈ ℕ0 ∨ (♯‘𝑀) = +∞)) | ||
| Theorem | hashnnn0genn0 14370 | 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 14371 | The size of a set is never minus infinity. (Contributed by Alexander van der Vekens, 21-Dec-2017.) |
| ⊢ (𝐴 ∈ 𝑉 → (♯‘𝐴) ≠ -∞) | ||
| Theorem | hashv01gt1 14372 | 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 14373 | 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 14374 | 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 14375 | 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 14376* | 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 14377* | 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 14378* | 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 14379 | 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 14380 | 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 14381 | 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 14382 | 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 14383 | Closure of the ♯ function. (Contributed by Paul Chapman, 26-Oct-2012.) (Revised by Mario Carneiro, 13-Jul-2014.) |
| ⊢ (𝐴 ∈ Fin → (♯‘𝐴) ∈ ℕ0) | ||
| Theorem | hashxrcl 14384 | Extended real closure of the ♯ function. (Contributed by Mario Carneiro, 22-Apr-2015.) |
| ⊢ (𝐴 ∈ 𝑉 → (♯‘𝐴) ∈ ℝ*) | ||
| Theorem | hashclb 14385 | Reverse closure of the ♯ function. (Contributed by Mario Carneiro, 15-Jan-2015.) |
| ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ Fin ↔ (♯‘𝐴) ∈ ℕ0)) | ||
| Theorem | nfile 14386 | 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 14387 | 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 14388 | 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 14389 | 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 14390 | 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 14391 | 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 14392 | 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 14393 | Positive natural closure of the hash function. (Contributed by Mario Carneiro, 16-Jan-2015.) |
| ⊢ (𝐴 ∈ Fin → ((♯‘𝐴) ∈ ℕ ↔ 𝐴 ≠ ∅)) | ||
| Theorem | hash0 14394 | The empty set has size zero. (Contributed by Mario Carneiro, 8-Jul-2014.) |
| ⊢ (♯‘∅) = 0 | ||
| Theorem | hashelne0d 14395 | 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 14396 | 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 14397 | 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 14398 | 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 14399* | 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 14400* | 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)) | ||
| < Previous Next > |
| Copyright terms: Public domain | < Previous Next > |