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Theorem List for Metamath Proof Explorer - 15501-15600   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremodzphi 15501 The order of any group element is a divisor of the Euler ϕ function. (Contributed by Mario Carneiro, 28-Feb-2014.)
((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) → ((od𝑁)‘𝐴) ∥ (ϕ‘𝑁))
 
6.2.5  Arithmetic modulo a prime number
 
Theoremmodprm1div 15502 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.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → ((𝐴 mod 𝑃) = 1 ↔ 𝑃 ∥ (𝐴 − 1)))
 
Theoremm1dvdsndvds 15503 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 15504 Show an explicit expression for the modular inverse of 𝐴 mod 𝑃. This is an application of prmdiv 15490. (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 15505 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 15506 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)
 
TheoremvfermltlALT 15507 Alternate proof of vfermltl 15506, not using Euler's theorem. (Contributed by AV, 21-Aug-2020.) (New usage is discouraged.) (Proof modification is discouraged.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ ¬ 𝑃𝐴) → ((𝐴↑(𝑃 − 1)) mod 𝑃) = 1)
 
Theorempowm2modprm 15508 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 15509* 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 15510* 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 15511* 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 15512* 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))
 
6.2.6  Pythagorean Triples
 
Theoremcoprimeprodsq 15513 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 15514 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 15515 A prime not equal to 2 is odd. (Contributed by Mario Carneiro, 4-Feb-2015.)
(𝑁 ∈ (ℙ ∖ {2}) → ((𝑁 − 1) / 2) ∈ ℕ)
 
Theoremnnoddn2prm 15516 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 15517 A prime not equal to 2 is odd. (Contributed by AV, 28-Jun-2021.)
(𝑁 ∈ (ℙ ∖ {2}) → ¬ 2 ∥ 𝑁)
 
Theoremnnoddn2prmb 15518 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 15519 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 15520 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 15521* Lemma for pythagtrip 15539. 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 15522* Lemma for pythagtrip 15539. 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 15523 Lemma for pythagtrip 15539. 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 15524 Lemma for pythagtrip 15539. 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 15525 Lemma for pythagtrip 15539. 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 15526 Lemma for pythagtrip 15539. 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 15527 Lemma for pythagtrip 15539. 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 15528 Lemma for pythagtrip 15539. 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 15529 Lemma for pythagtrip 15539. 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 15530 Lemma for pythagtrip 15539. 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 15531 Lemma for pythagtrip 15539. 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 15532 Lemma for pythagtrip 15539. 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 15533 Lemma for pythagtrip 15539. 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 15534 Lemma for pythagtrip 15539. 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 15535 Lemma for pythagtrip 15539. 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 15536 Lemma for pythagtrip 15539. 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 15537* Lemma for pythagtrip 15539. 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 15538* Lemma for pythagtrip 15539. 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 15539* 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))))))
 
Theoremiserodd 15540* Collect the odd terms in a sequence. (Contributed by Mario Carneiro, 7-Apr-2015.)
((𝜑𝑘 ∈ ℕ0) → 𝐶 ∈ ℂ)    &   (𝑛 = ((2 · 𝑘) + 1) → 𝐵 = 𝐶)       (𝜑 → (seq0( + , (𝑘 ∈ ℕ0𝐶)) ⇝ 𝐴 ↔ seq1( + , (𝑛 ∈ ℕ ↦ if(2 ∥ 𝑛, 0, 𝐵))) ⇝ 𝐴))
 
6.2.7  The prime count function
 
Syntaxcpc 15541 Extend class notation with the prime count function.
class pCnt
 
Definitiondf-pc 15542* 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, < ))))))
 
Theorempclem 15543* - Lemma for the prime power pre-function's properties. (Contributed by Mario Carneiro, 23-Feb-2014.)
𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑁}       ((𝑃 ∈ (ℤ‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝐴 ⊆ ℤ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℤ ∀𝑦𝐴 𝑦𝑥))
 
Theorempcprecl 15544* 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 15545* 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 15546* 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 15547* 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 15548* 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)
 
Theorempcval 15549* 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))))
 
Theorempceulem 15550* Lemma for pceu 15551. (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 15551* 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)))
 
Theorempczpre 15552* 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 15553 Closure of the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) ∈ ℕ0)
 
Theorempccl 15554 Closure of the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → (𝑃 pCnt 𝑁) ∈ ℕ0)
 
Theorempccld 15555 Closure of the prime power function. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝑃 ∈ ℙ)    &   (𝜑𝑁 ∈ ℕ)       (𝜑 → (𝑃 pCnt 𝑁) ∈ ℕ0)
 
Theorempcmul 15556 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 15557 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 15558 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 15559 The value of the prime power function at zero. (Contributed by Mario Carneiro, 3-Oct-2014.)
(𝑃 ∈ ℙ → (𝑃 pCnt 0) = +∞)
 
Theorempc1 15560 Value of the prime count function at 1. (Contributed by Mario Carneiro, 23-Feb-2014.)
(𝑃 ∈ ℙ → (𝑃 pCnt 1) = 0)
 
Theorempcqcl 15561 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 15562 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 15563 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 15564 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)))
 
Theorempcxcl 15565 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 15566 The prime count of an integer is greater or equal to zero. (Contributed by Mario Carneiro, 3-Oct-2014.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℤ) → 0 ≤ (𝑃 pCnt 𝑁))
 
Theorempczdvds 15567 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 15568 Defining property of the prime count function. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → (𝑃↑(𝑃 pCnt 𝑁)) ∥ 𝑁)
 
Theorempczndvds 15569 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 15570 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 15571 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 15572 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 15573 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 15574 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 15575 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 15576 The prime count of a prime power. (Contributed by Mario Carneiro, 12-Mar-2014.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ0) → (𝑃 pCnt (𝑃𝐴)) = 𝐴)
 
Theorempcid 15577 The prime count of a prime power. (Contributed by Mario Carneiro, 9-Sep-2014.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → (𝑃 pCnt (𝑃𝐴)) = 𝐴)
 
Theorempcneg 15578 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 15579 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 15580 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 15581 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 15582 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 15583* 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 15584* 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 15585* 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 15586* 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 15587* 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 15588* 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 15589* 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 15590* 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 15591 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 15592 Lemma for pcadd 15593. 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 15593 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)))
 
Theorempcadd2 15594 The inequality of pcadd 15593 becomes an equality when one of the factors has prime count strictly less than the other. (Contributed by Mario Carneiro, 16-Jan-2015.) (Revised by Mario Carneiro, 26-Jun-2015.)
(𝜑𝑃 ∈ ℙ)    &   (𝜑𝐴 ∈ ℚ)    &   (𝜑𝐵 ∈ ℚ)    &   (𝜑 → (𝑃 pCnt 𝐴) < (𝑃 pCnt 𝐵))       (𝜑 → (𝑃 pCnt 𝐴) = (𝑃 pCnt (𝐴 + 𝐵)))
 
Theorempcmptcl 15595 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 15596* 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 15597* 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 15598 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 15599* 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)
 
Theoremsumhash 15600* 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) → Σ𝑘𝐵 if(𝑘𝐴, 1, 0) = (#‘𝐴))
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