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Theorem List for Intuitionistic Logic Explorer - 12801-12900   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremprmdiveq 12801 The modular inverse of 𝐴 mod 𝑃 is unique. (Contributed by Mario Carneiro, 24-Jan-2015.)
𝑅 = ((𝐴↑(𝑃 − 2)) mod 𝑃)       ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ ¬ 𝑃𝐴) → ((𝑆 ∈ (0...(𝑃 − 1)) ∧ 𝑃 ∥ ((𝐴 · 𝑆) − 1)) ↔ 𝑆 = 𝑅))
 
Theoremprmdivdiv 12802 The (modular) inverse of the inverse of a number is itself. (Contributed by Mario Carneiro, 24-Jan-2015.)
𝑅 = ((𝐴↑(𝑃 − 2)) mod 𝑃)       ((𝑃 ∈ ℙ ∧ 𝐴 ∈ (1...(𝑃 − 1))) → 𝐴 = ((𝑅↑(𝑃 − 2)) mod 𝑃))
 
Theoremhashgcdlem 12803* A correspondence between elements of specific GCD and relative primes in a smaller ring. (Contributed by Stefan O'Rear, 12-Sep-2015.)
𝐴 = {𝑦 ∈ (0..^(𝑀 / 𝑁)) ∣ (𝑦 gcd (𝑀 / 𝑁)) = 1}    &   𝐵 = {𝑧 ∈ (0..^𝑀) ∣ (𝑧 gcd 𝑀) = 𝑁}    &   𝐹 = (𝑥𝐴 ↦ (𝑥 · 𝑁))       ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ∧ 𝑁𝑀) → 𝐹:𝐴1-1-onto𝐵)
 
Theoremdvdsfi 12804* A natural number has finitely many divisors. (Contributed by Jim Kingdon, 9-Oct-2025.)
(𝑁 ∈ ℕ → {𝑥 ∈ ℕ ∣ 𝑥𝑁} ∈ Fin)
 
Theoremhashgcdeq 12805* Number of initial positive integers with specified divisors. (Contributed by Stefan O'Rear, 12-Sep-2015.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (♯‘{𝑥 ∈ (0..^𝑀) ∣ (𝑥 gcd 𝑀) = 𝑁}) = if(𝑁𝑀, (ϕ‘(𝑀 / 𝑁)), 0))
 
Theoremphisum 12806* The divisor sum identity of the totient function. Theorem 2.2 in [ApostolNT] p. 26. (Contributed by Stefan O'Rear, 12-Sep-2015.)
(𝑁 ∈ ℕ → Σ𝑑 ∈ {𝑥 ∈ ℕ ∣ 𝑥𝑁} (ϕ‘𝑑) = 𝑁)
 
Theoremodzval 12807* Value of the order function. This is a function of functions; the inner argument selects the base (i.e., mod 𝑁 for some 𝑁, often prime) and the outer argument selects the integer or equivalence class (if you want to think about it that way) from the integers mod 𝑁. In order to ensure the supremum is well-defined, we only define the expression when 𝐴 and 𝑁 are coprime. (Contributed by Mario Carneiro, 23-Feb-2014.) (Revised by AV, 26-Sep-2020.)
((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) → ((od𝑁)‘𝐴) = inf({𝑛 ∈ ℕ ∣ 𝑁 ∥ ((𝐴𝑛) − 1)}, ℝ, < ))
 
Theoremodzcllem 12808 - Lemma for odzcl 12809, showing existence of a recurrent point for the exponential. (Contributed by Mario Carneiro, 28-Feb-2014.) (Proof shortened by AV, 26-Sep-2020.)
((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) → (((od𝑁)‘𝐴) ∈ ℕ ∧ 𝑁 ∥ ((𝐴↑((od𝑁)‘𝐴)) − 1)))
 
Theoremodzcl 12809 The order of a group element is an integer. (Contributed by Mario Carneiro, 28-Feb-2014.)
((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) → ((od𝑁)‘𝐴) ∈ ℕ)
 
Theoremodzid 12810 Any element raised to the power of its order is 1. (Contributed by Mario Carneiro, 28-Feb-2014.)
((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) → 𝑁 ∥ ((𝐴↑((od𝑁)‘𝐴)) − 1))
 
Theoremodzdvds 12811 The only powers of 𝐴 that are congruent to 1 are the multiples of the order of 𝐴. (Contributed by Mario Carneiro, 28-Feb-2014.) (Proof shortened by AV, 26-Sep-2020.)
(((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) ∧ 𝐾 ∈ ℕ0) → (𝑁 ∥ ((𝐴𝐾) − 1) ↔ ((od𝑁)‘𝐴) ∥ 𝐾))
 
Theoremodzphi 12812 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))
 
5.2.6  Arithmetic modulo a prime number
 
Theoremmodprm1div 12813 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 12814 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 12815 Show an explicit expression for the modular inverse of 𝐴 mod 𝑃. This is an application of prmdiv 12800. (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 12816 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 12817 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 12818 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 12819* 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 12820* 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 12821* 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 12822* 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 12823 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 12824 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 12825 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 12826 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 12827 A prime not equal to 2 is odd. (Contributed by AV, 28-Jun-2021.)
(𝑁 ∈ (ℙ ∖ {2}) → ¬ 2 ∥ 𝑁)
 
Theoremnnoddn2prmb 12828 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 12829 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 12830 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 12831* Lemma for pythagtrip 12849. 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 12832* Lemma for pythagtrip 12849. 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 12833 Lemma for pythagtrip 12849. 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 12834 Lemma for pythagtrip 12849. 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 12835 Lemma for pythagtrip 12849. 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 12836 Lemma for pythagtrip 12849. 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 12837 Lemma for pythagtrip 12849. 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 12838 Lemma for pythagtrip 12849. 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 12839 Lemma for pythagtrip 12849. 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 12840 Lemma for pythagtrip 12849. 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 12841 Lemma for pythagtrip 12849. 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 12842 Lemma for pythagtrip 12849. 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 12843 Lemma for pythagtrip 12849. 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 12844 Lemma for pythagtrip 12849. 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 12845 Lemma for pythagtrip 12849. 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 12846 Lemma for pythagtrip 12849. 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 12847* Lemma for pythagtrip 12849. 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 12848* Lemma for pythagtrip 12849. 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 12849* 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 12850 Extend class notation with the prime count function.
class pCnt
 
Definitiondf-pc 12851* 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 12852* 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 12853* 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 12854* 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 12855* 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 12856* 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 12857* 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 12858* 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 12859* 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 12860* Lemma for pceu 12861. (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 12861* 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 12862* 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 12863* 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 12864 Closure of the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) ∈ ℕ0)
 
Theorempccl 12865 Closure of the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → (𝑃 pCnt 𝑁) ∈ ℕ0)
 
Theorempccld 12866 Closure of the prime power function. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝑃 ∈ ℙ)    &   (𝜑𝑁 ∈ ℕ)       (𝜑 → (𝑃 pCnt 𝑁) ∈ ℕ0)
 
Theorempcmul 12867 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 12868 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 12869 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 12870 The value of the prime power function at zero. (Contributed by Mario Carneiro, 3-Oct-2014.)
(𝑃 ∈ ℙ → (𝑃 pCnt 0) = +∞)
 
Theorempc1 12871 Value of the prime count function at 1. (Contributed by Mario Carneiro, 23-Feb-2014.)
(𝑃 ∈ ℙ → (𝑃 pCnt 1) = 0)
 
Theorempcqcl 12872 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 12873 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 12874 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 12875 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 12876 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 12877 Extended real closure of the general prime count function. (Contributed by Mario Carneiro, 3-Oct-2014.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℚ) → (𝑃 pCnt 𝑁) ∈ ℝ*)
 
Theorempcxqcl 12878 The general prime count function is an integer or infinite. (Contributed by Jim Kingdon, 6-Jun-2025.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℚ) → ((𝑃 pCnt 𝑁) ∈ ℤ ∨ (𝑃 pCnt 𝑁) = +∞))
 
Theorempcge0 12879 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 12880 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 12881 Defining property of the prime count function. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → (𝑃↑(𝑃 pCnt 𝑁)) ∥ 𝑁)
 
Theorempczndvds 12882 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 12883 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 12884 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 12885 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 12886 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 12887 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 12888 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 12889 The prime count of a prime power. (Contributed by Mario Carneiro, 12-Mar-2014.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ0) → (𝑃 pCnt (𝑃𝐴)) = 𝐴)
 
Theorempcid 12890 The prime count of a prime power. (Contributed by Mario Carneiro, 9-Sep-2014.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → (𝑃 pCnt (𝑃𝐴)) = 𝐴)
 
Theorempcneg 12891 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 12892 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 12893 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 12894 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 12895 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 12896* 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 12897* 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 12898* 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 12899* 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 12900* 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))))
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