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Theorem List for Intuitionistic Logic Explorer - 12201-12300   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
5.2.6  Arithmetic modulo a prime number
 
Theoremmodprm1div 12201 A prime number divides an integer minus 1 iff the integer modulo the prime number is 1. (Contributed by Alexander van der Vekens, 17-May-2018.) (Proof shortened by AV, 30-May-2023.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → ((𝐴 mod 𝑃) = 1 ↔ 𝑃 ∥ (𝐴 − 1)))
 
Theoremm1dvdsndvds 12202 If an integer minus 1 is divisible by a prime number, the integer itself is not divisible by this prime number. (Contributed by Alexander van der Vekens, 30-Aug-2018.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → (𝑃 ∥ (𝐴 − 1) → ¬ 𝑃𝐴))
 
Theoremmodprminv 12203 Show an explicit expression for the modular inverse of 𝐴 mod 𝑃. This is an application of prmdiv 12189. (Contributed by Alexander van der Vekens, 15-May-2018.)
𝑅 = ((𝐴↑(𝑃 − 2)) mod 𝑃)       ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ ¬ 𝑃𝐴) → (𝑅 ∈ (1...(𝑃 − 1)) ∧ ((𝐴 · 𝑅) mod 𝑃) = 1))
 
Theoremmodprminveq 12204 The modular inverse of 𝐴 mod 𝑃 is unique. (Contributed by Alexander van der Vekens, 17-May-2018.)
𝑅 = ((𝐴↑(𝑃 − 2)) mod 𝑃)       ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ ¬ 𝑃𝐴) → ((𝑆 ∈ (0...(𝑃 − 1)) ∧ ((𝐴 · 𝑆) mod 𝑃) = 1) ↔ 𝑆 = 𝑅))
 
Theoremvfermltl 12205 Variant of Fermat's little theorem if 𝐴 is not a multiple of 𝑃, see theorem 5.18 in [ApostolNT] p. 113. (Contributed by AV, 21-Aug-2020.) (Proof shortened by AV, 5-Sep-2020.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ ¬ 𝑃𝐴) → ((𝐴↑(𝑃 − 1)) mod 𝑃) = 1)
 
Theorempowm2modprm 12206 If an integer minus 1 is divisible by a prime number, then the integer to the power of the prime number minus 2 is 1 modulo the prime number. (Contributed by Alexander van der Vekens, 30-Aug-2018.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → (𝑃 ∥ (𝐴 − 1) → ((𝐴↑(𝑃 − 2)) mod 𝑃) = 1))
 
Theoremreumodprminv 12207* For any prime number and for any positive integer less than this prime number, there is a unique modular inverse of this positive integer. (Contributed by Alexander van der Vekens, 12-May-2018.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ (1..^𝑃)) → ∃!𝑖 ∈ (1...(𝑃 − 1))((𝑁 · 𝑖) mod 𝑃) = 1)
 
Theoremmodprm0 12208* For two positive integers less than a given prime number there is always a nonnegative integer (less than the given prime number) so that the sum of one of the two positive integers and the other of the positive integers multiplied by the nonnegative integer is 0 ( modulo the given prime number). (Contributed by Alexander van der Vekens, 17-May-2018.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ (1..^𝑃) ∧ 𝐼 ∈ (1..^𝑃)) → ∃𝑗 ∈ (0..^𝑃)((𝐼 + (𝑗 · 𝑁)) mod 𝑃) = 0)
 
Theoremnnnn0modprm0 12209* For a positive integer and a nonnegative integer both less than a given prime number there is always a second nonnegative integer (less than the given prime number) so that the sum of this second nonnegative integer multiplied with the positive integer and the first nonnegative integer is 0 ( modulo the given prime number). (Contributed by Alexander van der Vekens, 8-Nov-2018.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ (1..^𝑃) ∧ 𝐼 ∈ (0..^𝑃)) → ∃𝑗 ∈ (0..^𝑃)((𝐼 + (𝑗 · 𝑁)) mod 𝑃) = 0)
 
Theoremmodprmn0modprm0 12210* For an integer not being 0 modulo a given prime number and a nonnegative integer less than the prime number, there is always a second nonnegative integer (less than the given prime number) so that the sum of this second nonnegative integer multiplied with the integer and the first nonnegative integer is 0 ( modulo the given prime number). (Contributed by Alexander van der Vekens, 10-Nov-2018.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℤ ∧ (𝑁 mod 𝑃) ≠ 0) → (𝐼 ∈ (0..^𝑃) → ∃𝑗 ∈ (0..^𝑃)((𝐼 + (𝑗 · 𝑁)) mod 𝑃) = 0))
 
5.2.7  Pythagorean Triples
 
Theoremcoprimeprodsq 12211 If three numbers are coprime, and the square of one is the product of the other two, then there is a formula for the other two in terms of gcd and square. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(((𝐴 ∈ ℕ0𝐵 ∈ ℤ ∧ 𝐶 ∈ ℕ0) ∧ ((𝐴 gcd 𝐵) gcd 𝐶) = 1) → ((𝐶↑2) = (𝐴 · 𝐵) → 𝐴 = ((𝐴 gcd 𝐶)↑2)))
 
Theoremcoprimeprodsq2 12212 If three numbers are coprime, and the square of one is the product of the other two, then there is a formula for the other two in terms of gcd and square. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℕ0𝐶 ∈ ℕ0) ∧ ((𝐴 gcd 𝐵) gcd 𝐶) = 1) → ((𝐶↑2) = (𝐴 · 𝐵) → 𝐵 = ((𝐵 gcd 𝐶)↑2)))
 
Theoremoddprm 12213 A prime not equal to 2 is odd. (Contributed by Mario Carneiro, 4-Feb-2015.) (Proof shortened by AV, 10-Jul-2022.)
(𝑁 ∈ (ℙ ∖ {2}) → ((𝑁 − 1) / 2) ∈ ℕ)
 
Theoremnnoddn2prm 12214 A prime not equal to 2 is an odd positive integer. (Contributed by AV, 28-Jun-2021.)
(𝑁 ∈ (ℙ ∖ {2}) → (𝑁 ∈ ℕ ∧ ¬ 2 ∥ 𝑁))
 
Theoremoddn2prm 12215 A prime not equal to 2 is odd. (Contributed by AV, 28-Jun-2021.)
(𝑁 ∈ (ℙ ∖ {2}) → ¬ 2 ∥ 𝑁)
 
Theoremnnoddn2prmb 12216 A number is a prime number not equal to 2 iff it is an odd prime number. Conversion theorem for two representations of odd primes. (Contributed by AV, 14-Jul-2021.)
(𝑁 ∈ (ℙ ∖ {2}) ↔ (𝑁 ∈ ℙ ∧ ¬ 2 ∥ 𝑁))
 
Theoremprm23lt5 12217 A prime less than 5 is either 2 or 3. (Contributed by AV, 5-Jul-2021.)
((𝑃 ∈ ℙ ∧ 𝑃 < 5) → (𝑃 = 2 ∨ 𝑃 = 3))
 
Theoremprm23ge5 12218 A prime is either 2 or 3 or greater than or equal to 5. (Contributed by AV, 5-Jul-2021.)
(𝑃 ∈ ℙ → (𝑃 = 2 ∨ 𝑃 = 3 ∨ 𝑃 ∈ (ℤ‘5)))
 
Theorempythagtriplem1 12219* Lemma for pythagtrip 12237. Prove a weaker version of one direction of the theorem. (Contributed by Scott Fenton, 28-Mar-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(∃𝑛 ∈ ℕ ∃𝑚 ∈ ℕ ∃𝑘 ∈ ℕ (𝐴 = (𝑘 · ((𝑚↑2) − (𝑛↑2))) ∧ 𝐵 = (𝑘 · (2 · (𝑚 · 𝑛))) ∧ 𝐶 = (𝑘 · ((𝑚↑2) + (𝑛↑2)))) → ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2))
 
Theorempythagtriplem2 12220* Lemma for pythagtrip 12237. Prove the full version of one direction of the theorem. (Contributed by Scott Fenton, 28-Mar-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (∃𝑛 ∈ ℕ ∃𝑚 ∈ ℕ ∃𝑘 ∈ ℕ ({𝐴, 𝐵} = {(𝑘 · ((𝑚↑2) − (𝑛↑2))), (𝑘 · (2 · (𝑚 · 𝑛)))} ∧ 𝐶 = (𝑘 · ((𝑚↑2) + (𝑛↑2)))) → ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2)))
 
Theorempythagtriplem3 12221 Lemma for pythagtrip 12237. Show that 𝐶 and 𝐵 are relatively prime under some conditions. (Contributed by Scott Fenton, 8-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (𝐵 gcd 𝐶) = 1)
 
Theorempythagtriplem4 12222 Lemma for pythagtrip 12237. Show that 𝐶𝐵 and 𝐶 + 𝐵 are relatively prime. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → ((𝐶𝐵) gcd (𝐶 + 𝐵)) = 1)
 
Theorempythagtriplem10 12223 Lemma for pythagtrip 12237. Show that 𝐶𝐵 is positive. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2)) → 0 < (𝐶𝐵))
 
Theorempythagtriplem6 12224 Lemma for pythagtrip 12237. Calculate (√‘(𝐶𝐵)). (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (√‘(𝐶𝐵)) = ((𝐶𝐵) gcd 𝐴))
 
Theorempythagtriplem7 12225 Lemma for pythagtrip 12237. Calculate (√‘(𝐶 + 𝐵)). (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (√‘(𝐶 + 𝐵)) = ((𝐶 + 𝐵) gcd 𝐴))
 
Theorempythagtriplem8 12226 Lemma for pythagtrip 12237. Show that (√‘(𝐶𝐵)) is a positive integer. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (√‘(𝐶𝐵)) ∈ ℕ)
 
Theorempythagtriplem9 12227 Lemma for pythagtrip 12237. Show that (√‘(𝐶 + 𝐵)) is a positive integer. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (√‘(𝐶 + 𝐵)) ∈ ℕ)
 
Theorempythagtriplem11 12228 Lemma for pythagtrip 12237. Show that 𝑀 (which will eventually be closely related to the 𝑚 in the final statement) is a natural. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
𝑀 = (((√‘(𝐶 + 𝐵)) + (√‘(𝐶𝐵))) / 2)       (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝑀 ∈ ℕ)
 
Theorempythagtriplem12 12229 Lemma for pythagtrip 12237. Calculate the square of 𝑀. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
𝑀 = (((√‘(𝐶 + 𝐵)) + (√‘(𝐶𝐵))) / 2)       (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (𝑀↑2) = ((𝐶 + 𝐴) / 2))
 
Theorempythagtriplem13 12230 Lemma for pythagtrip 12237. Show that 𝑁 (which will eventually be closely related to the 𝑛 in the final statement) is a natural. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶𝐵))) / 2)       (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝑁 ∈ ℕ)
 
Theorempythagtriplem14 12231 Lemma for pythagtrip 12237. Calculate the square of 𝑁. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶𝐵))) / 2)       (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (𝑁↑2) = ((𝐶𝐴) / 2))
 
Theorempythagtriplem15 12232 Lemma for pythagtrip 12237. Show the relationship between 𝑀, 𝑁, and 𝐴. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
𝑀 = (((√‘(𝐶 + 𝐵)) + (√‘(𝐶𝐵))) / 2)    &   𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶𝐵))) / 2)       (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝐴 = ((𝑀↑2) − (𝑁↑2)))
 
Theorempythagtriplem16 12233 Lemma for pythagtrip 12237. Show the relationship between 𝑀, 𝑁, and 𝐵. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
𝑀 = (((√‘(𝐶 + 𝐵)) + (√‘(𝐶𝐵))) / 2)    &   𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶𝐵))) / 2)       (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝐵 = (2 · (𝑀 · 𝑁)))
 
Theorempythagtriplem17 12234 Lemma for pythagtrip 12237. Show the relationship between 𝑀, 𝑁, and 𝐶. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
𝑀 = (((√‘(𝐶 + 𝐵)) + (√‘(𝐶𝐵))) / 2)    &   𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶𝐵))) / 2)       (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝐶 = ((𝑀↑2) + (𝑁↑2)))
 
Theorempythagtriplem18 12235* Lemma for pythagtrip 12237. Wrap the previous 𝑀 and 𝑁 up in quantifiers. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → ∃𝑛 ∈ ℕ ∃𝑚 ∈ ℕ (𝐴 = ((𝑚↑2) − (𝑛↑2)) ∧ 𝐵 = (2 · (𝑚 · 𝑛)) ∧ 𝐶 = ((𝑚↑2) + (𝑛↑2))))
 
Theorempythagtriplem19 12236* Lemma for pythagtrip 12237. Introduce 𝑘 and remove the relative primality requirement. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ¬ 2 ∥ (𝐴 / (𝐴 gcd 𝐵))) → ∃𝑛 ∈ ℕ ∃𝑚 ∈ ℕ ∃𝑘 ∈ ℕ (𝐴 = (𝑘 · ((𝑚↑2) − (𝑛↑2))) ∧ 𝐵 = (𝑘 · (2 · (𝑚 · 𝑛))) ∧ 𝐶 = (𝑘 · ((𝑚↑2) + (𝑛↑2)))))
 
Theorempythagtrip 12237* Parameterize the Pythagorean triples. If 𝐴, 𝐵, and 𝐶 are naturals, then they obey the Pythagorean triple formula iff they are parameterized by three naturals. This proof follows the Isabelle proof at http://afp.sourceforge.net/entries/Fermat3_4.shtml. This is Metamath 100 proof #23. (Contributed by Scott Fenton, 19-Apr-2014.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) → (((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ↔ ∃𝑛 ∈ ℕ ∃𝑚 ∈ ℕ ∃𝑘 ∈ ℕ ({𝐴, 𝐵} = {(𝑘 · ((𝑚↑2) − (𝑛↑2))), (𝑘 · (2 · (𝑚 · 𝑛)))} ∧ 𝐶 = (𝑘 · ((𝑚↑2) + (𝑛↑2))))))
 
5.2.8  The prime count function
 
Syntaxcpc 12238 Extend class notation with the prime count function.
class pCnt
 
Definitiondf-pc 12239* Define the prime count function, which returns the largest exponent of a given prime (or other positive integer) that divides the number. For rational numbers, it returns negative values according to the power of a prime in the denominator. (Contributed by Mario Carneiro, 23-Feb-2014.)
pCnt = (𝑝 ∈ ℙ, 𝑟 ∈ ℚ ↦ if(𝑟 = 0, +∞, (℩𝑧𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑟 = (𝑥 / 𝑦) ∧ 𝑧 = (sup({𝑛 ∈ ℕ0 ∣ (𝑝𝑛) ∥ 𝑥}, ℝ, < ) − sup({𝑛 ∈ ℕ0 ∣ (𝑝𝑛) ∥ 𝑦}, ℝ, < ))))))
 
Theorempclem0 12240* Lemma for the prime power pre-function's properties. (Contributed by Mario Carneiro, 23-Feb-2014.) (Revised by Jim Kingdon, 7-Oct-2024.)
𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑁}       ((𝑃 ∈ (ℤ‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → 0 ∈ 𝐴)
 
Theorempclemub 12241* Lemma for the prime power pre-function's properties. (Contributed by Mario Carneiro, 23-Feb-2014.) (Revised by Jim Kingdon, 7-Oct-2024.)
𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑁}       ((𝑃 ∈ (ℤ‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ∃𝑥 ∈ ℤ ∀𝑦𝐴 𝑦𝑥)
 
Theorempclemdc 12242* Lemma for the prime power pre-function's properties. (Contributed by Jim Kingdon, 8-Oct-2024.)
𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑁}       ((𝑃 ∈ (ℤ‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ∀𝑥 ∈ ℤ DECID 𝑥𝐴)
 
Theorempcprecl 12243* Closure of the prime power pre-function. (Contributed by Mario Carneiro, 23-Feb-2014.)
𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑁}    &   𝑆 = sup(𝐴, ℝ, < )       ((𝑃 ∈ (ℤ‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑆 ∈ ℕ0 ∧ (𝑃𝑆) ∥ 𝑁))
 
Theorempcprendvds 12244* Non-divisibility property of the prime power pre-function. (Contributed by Mario Carneiro, 23-Feb-2014.)
𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑁}    &   𝑆 = sup(𝐴, ℝ, < )       ((𝑃 ∈ (ℤ‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ¬ (𝑃↑(𝑆 + 1)) ∥ 𝑁)
 
Theorempcprendvds2 12245* Non-divisibility property of the prime power pre-function. (Contributed by Mario Carneiro, 23-Feb-2014.)
𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑁}    &   𝑆 = sup(𝐴, ℝ, < )       ((𝑃 ∈ (ℤ‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ¬ 𝑃 ∥ (𝑁 / (𝑃𝑆)))
 
Theorempcpre1 12246* Value of the prime power pre-function at 1. (Contributed by Mario Carneiro, 23-Feb-2014.) (Revised by Mario Carneiro, 26-Apr-2016.)
𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑁}    &   𝑆 = sup(𝐴, ℝ, < )       ((𝑃 ∈ (ℤ‘2) ∧ 𝑁 = 1) → 𝑆 = 0)
 
Theorempcpremul 12247* Multiplicative property of the prime count pre-function. Note that the primality of 𝑃 is essential for this property; (4 pCnt 2) = 0 but (4 pCnt (2 · 2)) = 1 ≠ 2 · (4 pCnt 2) = 0. Since this is needed to show uniqueness for the real prime count function (over ), we don't bother to define it off the primes. (Contributed by Mario Carneiro, 23-Feb-2014.)
𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑀}, ℝ, < )    &   𝑇 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑁}, ℝ, < )    &   𝑈 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ (𝑀 · 𝑁)}, ℝ, < )       ((𝑃 ∈ ℙ ∧ (𝑀 ∈ ℤ ∧ 𝑀 ≠ 0) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑆 + 𝑇) = 𝑈)
 
Theorempceulem 12248* Lemma for pceu 12249. (Contributed by Mario Carneiro, 23-Feb-2014.)
𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑥}, ℝ, < )    &   𝑇 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑦}, ℝ, < )    &   𝑈 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑠}, ℝ, < )    &   𝑉 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑡}, ℝ, < )    &   (𝜑𝑃 ∈ ℙ)    &   (𝜑𝑁 ≠ 0)    &   (𝜑 → (𝑥 ∈ ℤ ∧ 𝑦 ∈ ℕ))    &   (𝜑𝑁 = (𝑥 / 𝑦))    &   (𝜑 → (𝑠 ∈ ℤ ∧ 𝑡 ∈ ℕ))    &   (𝜑𝑁 = (𝑠 / 𝑡))       (𝜑 → (𝑆𝑇) = (𝑈𝑉))
 
Theorempceu 12249* Uniqueness for the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.)
𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑥}, ℝ, < )    &   𝑇 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑦}, ℝ, < )       ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → ∃!𝑧𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆𝑇)))
 
Theorempcval 12250* The value of the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.) (Revised by Mario Carneiro, 3-Oct-2014.)
𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑥}, ℝ, < )    &   𝑇 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑦}, ℝ, < )       ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) = (℩𝑧𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆𝑇))))
 
Theorempczpre 12251* Connect the prime count pre-function to the actual prime count function, when restricted to the integers. (Contributed by Mario Carneiro, 23-Feb-2014.) (Proof shortened by Mario Carneiro, 24-Dec-2016.)
𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑁}, ℝ, < )       ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) = 𝑆)
 
Theorempczcl 12252 Closure of the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) ∈ ℕ0)
 
Theorempccl 12253 Closure of the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → (𝑃 pCnt 𝑁) ∈ ℕ0)
 
Theorempccld 12254 Closure of the prime power function. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝑃 ∈ ℙ)    &   (𝜑𝑁 ∈ ℕ)       (𝜑 → (𝑃 pCnt 𝑁) ∈ ℕ0)
 
Theorempcmul 12255 Multiplication property of the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℤ ∧ 𝐴 ≠ 0) ∧ (𝐵 ∈ ℤ ∧ 𝐵 ≠ 0)) → (𝑃 pCnt (𝐴 · 𝐵)) = ((𝑃 pCnt 𝐴) + (𝑃 pCnt 𝐵)))
 
Theorempcdiv 12256 Division property of the prime power function. (Contributed by Mario Carneiro, 1-Mar-2014.)
((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℤ ∧ 𝐴 ≠ 0) ∧ 𝐵 ∈ ℕ) → (𝑃 pCnt (𝐴 / 𝐵)) = ((𝑃 pCnt 𝐴) − (𝑃 pCnt 𝐵)))
 
Theorempcqmul 12257 Multiplication property of the prime power function. (Contributed by Mario Carneiro, 9-Sep-2014.)
((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℚ ∧ 𝐴 ≠ 0) ∧ (𝐵 ∈ ℚ ∧ 𝐵 ≠ 0)) → (𝑃 pCnt (𝐴 · 𝐵)) = ((𝑃 pCnt 𝐴) + (𝑃 pCnt 𝐵)))
 
Theorempc0 12258 The value of the prime power function at zero. (Contributed by Mario Carneiro, 3-Oct-2014.)
(𝑃 ∈ ℙ → (𝑃 pCnt 0) = +∞)
 
Theorempc1 12259 Value of the prime count function at 1. (Contributed by Mario Carneiro, 23-Feb-2014.)
(𝑃 ∈ ℙ → (𝑃 pCnt 1) = 0)
 
Theorempcqcl 12260 Closure of the general prime count function. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) ∈ ℤ)
 
Theorempcqdiv 12261 Division property of the prime power function. (Contributed by Mario Carneiro, 10-Aug-2015.)
((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℚ ∧ 𝐴 ≠ 0) ∧ (𝐵 ∈ ℚ ∧ 𝐵 ≠ 0)) → (𝑃 pCnt (𝐴 / 𝐵)) = ((𝑃 pCnt 𝐴) − (𝑃 pCnt 𝐵)))
 
Theorempcrec 12262 Prime power of a reciprocal. (Contributed by Mario Carneiro, 10-Aug-2015.)
((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℚ ∧ 𝐴 ≠ 0)) → (𝑃 pCnt (1 / 𝐴)) = -(𝑃 pCnt 𝐴))
 
Theorempcexp 12263 Prime power of an exponential. (Contributed by Mario Carneiro, 10-Aug-2015.)
((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℚ ∧ 𝐴 ≠ 0) ∧ 𝑁 ∈ ℤ) → (𝑃 pCnt (𝐴𝑁)) = (𝑁 · (𝑃 pCnt 𝐴)))
 
Theorempcxnn0cl 12264 Extended nonnegative integer closure of the general prime count function. (Contributed by Jim Kingdon, 13-Oct-2024.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℤ) → (𝑃 pCnt 𝑁) ∈ ℕ0*)
 
Theorempcxcl 12265 Extended real closure of the general prime count function. (Contributed by Mario Carneiro, 3-Oct-2014.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℚ) → (𝑃 pCnt 𝑁) ∈ ℝ*)
 
Theorempcge0 12266 The prime count of an integer is greater than or equal to zero. (Contributed by Mario Carneiro, 3-Oct-2014.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℤ) → 0 ≤ (𝑃 pCnt 𝑁))
 
Theorempczdvds 12267 Defining property of the prime count function. (Contributed by Mario Carneiro, 9-Sep-2014.)
((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑃↑(𝑃 pCnt 𝑁)) ∥ 𝑁)
 
Theorempcdvds 12268 Defining property of the prime count function. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → (𝑃↑(𝑃 pCnt 𝑁)) ∥ 𝑁)
 
Theorempczndvds 12269 Defining property of the prime count function. (Contributed by Mario Carneiro, 3-Oct-2014.)
((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ¬ (𝑃↑((𝑃 pCnt 𝑁) + 1)) ∥ 𝑁)
 
Theorempcndvds 12270 Defining property of the prime count function. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → ¬ (𝑃↑((𝑃 pCnt 𝑁) + 1)) ∥ 𝑁)
 
Theorempczndvds2 12271 The remainder after dividing out all factors of 𝑃 is not divisible by 𝑃. (Contributed by Mario Carneiro, 9-Sep-2014.)
((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ¬ 𝑃 ∥ (𝑁 / (𝑃↑(𝑃 pCnt 𝑁))))
 
Theorempcndvds2 12272 The remainder after dividing out all factors of 𝑃 is not divisible by 𝑃. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → ¬ 𝑃 ∥ (𝑁 / (𝑃↑(𝑃 pCnt 𝑁))))
 
Theorempcdvdsb 12273 𝑃𝐴 divides 𝑁 if and only if 𝐴 is at most the count of 𝑃. (Contributed by Mario Carneiro, 3-Oct-2014.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℤ ∧ 𝐴 ∈ ℕ0) → (𝐴 ≤ (𝑃 pCnt 𝑁) ↔ (𝑃𝐴) ∥ 𝑁))
 
Theorempcelnn 12274 There are a positive number of powers of a prime 𝑃 in 𝑁 iff 𝑃 divides 𝑁. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → ((𝑃 pCnt 𝑁) ∈ ℕ ↔ 𝑃𝑁))
 
Theorempceq0 12275 There are zero powers of a prime 𝑃 in 𝑁 iff 𝑃 does not divide 𝑁. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → ((𝑃 pCnt 𝑁) = 0 ↔ ¬ 𝑃𝑁))
 
Theorempcidlem 12276 The prime count of a prime power. (Contributed by Mario Carneiro, 12-Mar-2014.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ0) → (𝑃 pCnt (𝑃𝐴)) = 𝐴)
 
Theorempcid 12277 The prime count of a prime power. (Contributed by Mario Carneiro, 9-Sep-2014.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → (𝑃 pCnt (𝑃𝐴)) = 𝐴)
 
Theorempcneg 12278 The prime count of a negative number. (Contributed by Mario Carneiro, 13-Mar-2014.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℚ) → (𝑃 pCnt -𝐴) = (𝑃 pCnt 𝐴))
 
Theorempcabs 12279 The prime count of an absolute value. (Contributed by Mario Carneiro, 13-Mar-2014.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℚ) → (𝑃 pCnt (abs‘𝐴)) = (𝑃 pCnt 𝐴))
 
Theorempcdvdstr 12280 The prime count increases under the divisibility relation. (Contributed by Mario Carneiro, 13-Mar-2014.)
((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐴𝐵)) → (𝑃 pCnt 𝐴) ≤ (𝑃 pCnt 𝐵))
 
Theorempcgcd1 12281 The prime count of a GCD is the minimum of the prime counts of the arguments. (Contributed by Mario Carneiro, 3-Oct-2014.)
(((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) ∧ (𝑃 pCnt 𝐴) ≤ (𝑃 pCnt 𝐵)) → (𝑃 pCnt (𝐴 gcd 𝐵)) = (𝑃 pCnt 𝐴))
 
Theorempcgcd 12282 The prime count of a GCD is the minimum of the prime counts of the arguments. (Contributed by Mario Carneiro, 3-Oct-2014.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → (𝑃 pCnt (𝐴 gcd 𝐵)) = if((𝑃 pCnt 𝐴) ≤ (𝑃 pCnt 𝐵), (𝑃 pCnt 𝐴), (𝑃 pCnt 𝐵)))
 
Theorempc2dvds 12283* A characterization of divisibility in terms of prime count. (Contributed by Mario Carneiro, 23-Feb-2014.) (Revised by Mario Carneiro, 3-Oct-2014.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → (𝐴𝐵 ↔ ∀𝑝 ∈ ℙ (𝑝 pCnt 𝐴) ≤ (𝑝 pCnt 𝐵)))
 
Theorempc11 12284* The prime count function, viewed as a function from to (ℕ ↑𝑚 ℙ), is one-to-one. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝐴 ∈ ℕ0𝐵 ∈ ℕ0) → (𝐴 = 𝐵 ↔ ∀𝑝 ∈ ℙ (𝑝 pCnt 𝐴) = (𝑝 pCnt 𝐵)))
 
Theorempcz 12285* The prime count function can be used as an indicator that a given rational number is an integer. (Contributed by Mario Carneiro, 23-Feb-2014.)
(𝐴 ∈ ℚ → (𝐴 ∈ ℤ ↔ ∀𝑝 ∈ ℙ 0 ≤ (𝑝 pCnt 𝐴)))
 
Theorempcprmpw2 12286* Self-referential expression for a prime power. (Contributed by Mario Carneiro, 16-Jan-2015.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ) → (∃𝑛 ∈ ℕ0 𝐴 ∥ (𝑃𝑛) ↔ 𝐴 = (𝑃↑(𝑃 pCnt 𝐴))))
 
Theorempcprmpw 12287* Self-referential expression for a prime power. (Contributed by Mario Carneiro, 16-Jan-2015.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ) → (∃𝑛 ∈ ℕ0 𝐴 = (𝑃𝑛) ↔ 𝐴 = (𝑃↑(𝑃 pCnt 𝐴))))
 
Theoremdvdsprmpweq 12288* If a positive integer divides a prime power, it is a prime power. (Contributed by AV, 25-Jul-2021.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ ∧ 𝑁 ∈ ℕ0) → (𝐴 ∥ (𝑃𝑁) → ∃𝑛 ∈ ℕ0 𝐴 = (𝑃𝑛)))
 
Theoremdvdsprmpweqnn 12289* If an integer greater than 1 divides a prime power, it is a (proper) prime power. (Contributed by AV, 13-Aug-2021.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ (ℤ‘2) ∧ 𝑁 ∈ ℕ0) → (𝐴 ∥ (𝑃𝑁) → ∃𝑛 ∈ ℕ 𝐴 = (𝑃𝑛)))
 
Theoremdvdsprmpweqle 12290* If a positive integer divides a prime power, it is a prime power with a smaller exponent. (Contributed by AV, 25-Jul-2021.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ ∧ 𝑁 ∈ ℕ0) → (𝐴 ∥ (𝑃𝑁) → ∃𝑛 ∈ ℕ0 (𝑛𝑁𝐴 = (𝑃𝑛))))
 
Theoremdifsqpwdvds 12291 If the difference of two squares is a power of a prime, the prime divides twice the second squared number. (Contributed by AV, 13-Aug-2021.)
(((𝐴 ∈ ℕ0𝐵 ∈ ℕ0 ∧ (𝐵 + 1) < 𝐴) ∧ (𝐶 ∈ ℙ ∧ 𝐷 ∈ ℕ0)) → ((𝐶𝐷) = ((𝐴↑2) − (𝐵↑2)) → 𝐶 ∥ (2 · 𝐵)))
 
Theorempcaddlem 12292 Lemma for pcadd 12293. The original numbers 𝐴 and 𝐵 have been decomposed using the prime count function as (𝑃𝑀) · (𝑅 / 𝑆) where 𝑅, 𝑆 are both not divisible by 𝑃 and 𝑀 = (𝑃 pCnt 𝐴), and similarly for 𝐵. (Contributed by Mario Carneiro, 9-Sep-2014.)
(𝜑𝑃 ∈ ℙ)    &   (𝜑𝐴 = ((𝑃𝑀) · (𝑅 / 𝑆)))    &   (𝜑𝐵 = ((𝑃𝑁) · (𝑇 / 𝑈)))    &   (𝜑𝑁 ∈ (ℤ𝑀))    &   (𝜑 → (𝑅 ∈ ℤ ∧ ¬ 𝑃𝑅))    &   (𝜑 → (𝑆 ∈ ℕ ∧ ¬ 𝑃𝑆))    &   (𝜑 → (𝑇 ∈ ℤ ∧ ¬ 𝑃𝑇))    &   (𝜑 → (𝑈 ∈ ℕ ∧ ¬ 𝑃𝑈))       (𝜑𝑀 ≤ (𝑃 pCnt (𝐴 + 𝐵)))
 
Theorempcadd 12293 An inequality for the prime count of a sum. This is the source of the ultrametric inequality for the p-adic metric. (Contributed by Mario Carneiro, 9-Sep-2014.)
(𝜑𝑃 ∈ ℙ)    &   (𝜑𝐴 ∈ ℚ)    &   (𝜑𝐵 ∈ ℚ)    &   (𝜑 → (𝑃 pCnt 𝐴) ≤ (𝑃 pCnt 𝐵))       (𝜑 → (𝑃 pCnt 𝐴) ≤ (𝑃 pCnt (𝐴 + 𝐵)))
 
Theorempcmptcl 12294 Closure for the prime power map. (Contributed by Mario Carneiro, 12-Mar-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛𝐴), 1))    &   (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈ ℕ0)       (𝜑 → (𝐹:ℕ⟶ℕ ∧ seq1( · , 𝐹):ℕ⟶ℕ))
 
Theorempcmpt 12295* Construct a function with given prime count characteristics. (Contributed by Mario Carneiro, 12-Mar-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛𝐴), 1))    &   (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈ ℕ0)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 ∈ ℙ)    &   (𝑛 = 𝑃𝐴 = 𝐵)       (𝜑 → (𝑃 pCnt (seq1( · , 𝐹)‘𝑁)) = if(𝑃𝑁, 𝐵, 0))
 
Theorempcmpt2 12296* Dividing two prime count maps yields a number with all dividing primes confined to an interval. (Contributed by Mario Carneiro, 14-Mar-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛𝐴), 1))    &   (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈ ℕ0)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑃 ∈ ℙ)    &   (𝑛 = 𝑃𝐴 = 𝐵)    &   (𝜑𝑀 ∈ (ℤ𝑁))       (𝜑 → (𝑃 pCnt ((seq1( · , 𝐹)‘𝑀) / (seq1( · , 𝐹)‘𝑁))) = if((𝑃𝑀 ∧ ¬ 𝑃𝑁), 𝐵, 0))
 
Theorempcmptdvds 12297 The partial products of the prime power map form a divisibility chain. (Contributed by Mario Carneiro, 12-Mar-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛𝐴), 1))    &   (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈ ℕ0)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑀 ∈ (ℤ𝑁))       (𝜑 → (seq1( · , 𝐹)‘𝑁) ∥ (seq1( · , 𝐹)‘𝑀))
 
Theorempcprod 12298* The product of the primes taken to their respective powers reconstructs the original number. (Contributed by Mario Carneiro, 12-Mar-2014.)
𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛↑(𝑛 pCnt 𝑁)), 1))       (𝑁 ∈ ℕ → (seq1( · , 𝐹)‘𝑁) = 𝑁)
 
Theoremsumhashdc 12299* The sum of 1 over a set is the size of the set. (Contributed by Mario Carneiro, 8-Mar-2014.) (Revised by Mario Carneiro, 20-May-2014.)
((𝐵 ∈ Fin ∧ 𝐴𝐵 ∧ ∀𝑥𝐵 DECID 𝑥𝐴) → Σ𝑘𝐵 if(𝑘𝐴, 1, 0) = (♯‘𝐴))
 
Theoremfldivp1 12300 The difference between the floors of adjacent fractions is either 1 or 0. (Contributed by Mario Carneiro, 8-Mar-2014.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ) → ((⌊‘((𝑀 + 1) / 𝑁)) − (⌊‘(𝑀 / 𝑁))) = if(𝑁 ∥ (𝑀 + 1), 1, 0))
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