P. 131 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 13001-13100   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	hashdvds 13001*	The number of numbers in a given residue class in a finite set of integers. (Contributed by Mario Carneiro, 12-Mar-2014.) (Proof shortened by Mario Carneiro, 7-Jun-2016.)
⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝐴 ∈ ℤ)    &   ⊢ (𝜑 → 𝐵 ∈ (ℤ≥‘(𝐴 − 1)))    &   ⊢ (𝜑 → 𝐶 ∈ ℤ)    ⇒   ⊢ (𝜑 → (♯‘{𝑥 ∈ (𝐴...𝐵) ∣ 𝑁 ∥ (𝑥 − 𝐶)}) = ((⌊‘((𝐵 − 𝐶) / 𝑁)) − (⌊‘(((𝐴 − 1) − 𝐶) / 𝑁))))
Theorem	phiprmpw 13002	Value of the Euler ϕ function at a prime power. Theorem 2.5(a) in [ApostolNT] p. 28. (Contributed by Mario Carneiro, 24-Feb-2014.)
⊢ ((𝑃 ∈ ℙ ∧ 𝐾 ∈ ℕ) → (ϕ‘(𝑃↑𝐾)) = ((𝑃↑(𝐾 − 1)) · (𝑃 − 1)))
Theorem	phiprm 13003	Value of the Euler ϕ function at a prime. (Contributed by Mario Carneiro, 28-Feb-2014.)
⊢ (𝑃 ∈ ℙ → (ϕ‘𝑃) = (𝑃 − 1))
Theorem	crth 13004*	The Chinese Remainder Theorem: the function that maps 𝑥 to its remainder classes mod 𝑀 and mod 𝑁 is 1-1 and onto when 𝑀 and 𝑁 are coprime. (Contributed by Mario Carneiro, 24-Feb-2014.) (Proof shortened by Mario Carneiro, 2-May-2016.)
⊢ 𝑆 = (0..^(𝑀 · 𝑁))    &   ⊢ 𝑇 = ((0..^𝑀) × (0..^𝑁))    &   ⊢ 𝐹 = (𝑥 ∈ 𝑆 ↦ ⟨(𝑥 mod 𝑀), (𝑥 mod 𝑁)⟩)    &   ⊢ (𝜑 → (𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ∧ (𝑀 gcd 𝑁) = 1))    ⇒   ⊢ (𝜑 → 𝐹:𝑆–1-1-onto→𝑇)
Theorem	phimullem 13005*	Lemma for phimul 13006. (Contributed by Mario Carneiro, 24-Feb-2014.)
⊢ 𝑆 = (0..^(𝑀 · 𝑁))    &   ⊢ 𝑇 = ((0..^𝑀) × (0..^𝑁))    &   ⊢ 𝐹 = (𝑥 ∈ 𝑆 ↦ ⟨(𝑥 mod 𝑀), (𝑥 mod 𝑁)⟩)    &   ⊢ (𝜑 → (𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ∧ (𝑀 gcd 𝑁) = 1))    &   ⊢ 𝑈 = {𝑦 ∈ (0..^𝑀) ∣ (𝑦 gcd 𝑀) = 1}    &   ⊢ 𝑉 = {𝑦 ∈ (0..^𝑁) ∣ (𝑦 gcd 𝑁) = 1}    &   ⊢ 𝑊 = {𝑦 ∈ 𝑆 ∣ (𝑦 gcd (𝑀 · 𝑁)) = 1}    ⇒   ⊢ (𝜑 → (ϕ‘(𝑀 · 𝑁)) = ((ϕ‘𝑀) · (ϕ‘𝑁)))
Theorem	phimul 13006	The Euler ϕ function is a multiplicative function, meaning that it distributes over multiplication at relatively prime arguments. Theorem 2.5(c) in [ApostolNT] p. 28. (Contributed by Mario Carneiro, 24-Feb-2014.)
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ∧ (𝑀 gcd 𝑁) = 1) → (ϕ‘(𝑀 · 𝑁)) = ((ϕ‘𝑀) · (ϕ‘𝑁)))
Theorem	eulerthlem1 13007*	Lemma for eulerth 13013. (Contributed by Mario Carneiro, 8-May-2015.)
⊢ (𝜑 → (𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1))    &   ⊢ 𝑆 = {𝑦 ∈ (0..^𝑁) ∣ (𝑦 gcd 𝑁) = 1}    &   ⊢ 𝑇 = (1...(ϕ‘𝑁))    &   ⊢ (𝜑 → 𝐹:𝑇–1-1-onto→𝑆)    &   ⊢ 𝐺 = (𝑥 ∈ 𝑇 ↦ ((𝐴 · (𝐹‘𝑥)) mod 𝑁))    ⇒   ⊢ (𝜑 → 𝐺:𝑇⟶𝑆)
Theorem	eulerthlemfi 13008*	Lemma for eulerth 13013. The set 𝑆 is finite. (Contributed by Mario Carneiro, 28-Feb-2014.) (Revised by Jim Kingdon, 7-Sep-2024.)
⊢ (𝜑 → (𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1))    &   ⊢ 𝑆 = {𝑦 ∈ (0..^𝑁) ∣ (𝑦 gcd 𝑁) = 1}    ⇒   ⊢ (𝜑 → 𝑆 ∈ Fin)
Theorem	eulerthlemrprm 13009*	Lemma for eulerth 13013. 𝑁 and ∏𝑥 ∈ (1...(ϕ‘𝑁))(𝐹‘𝑥) are relatively prime. (Contributed by Mario Carneiro, 28-Feb-2014.) (Revised by Jim Kingdon, 2-Sep-2024.)
⊢ (𝜑 → (𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1))    &   ⊢ 𝑆 = {𝑦 ∈ (0..^𝑁) ∣ (𝑦 gcd 𝑁) = 1}    &   ⊢ (𝜑 → 𝐹:(1...(ϕ‘𝑁))–1-1-onto→𝑆)    ⇒   ⊢ (𝜑 → (𝑁 gcd ∏𝑥 ∈ (1...(ϕ‘𝑁))(𝐹‘𝑥)) = 1)
Theorem	eulerthlema 13010*	Lemma for eulerth 13013. (Contributed by Mario Carneiro, 28-Feb-2014.) (Revised by Jim Kingdon, 2-Sep-2024.)
⊢ (𝜑 → (𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1))    &   ⊢ 𝑆 = {𝑦 ∈ (0..^𝑁) ∣ (𝑦 gcd 𝑁) = 1}    &   ⊢ (𝜑 → 𝐹:(1...(ϕ‘𝑁))–1-1-onto→𝑆)    ⇒   ⊢ (𝜑 → (((𝐴↑(ϕ‘𝑁)) · ∏𝑥 ∈ (1...(ϕ‘𝑁))(𝐹‘𝑥)) mod 𝑁) = (∏𝑥 ∈ (1...(ϕ‘𝑁))((𝐴 · (𝐹‘𝑥)) mod 𝑁) mod 𝑁))
Theorem	eulerthlemh 13011*	Lemma for eulerth 13013. A permutation of (1...(ϕ‘𝑁)). (Contributed by Mario Carneiro, 28-Feb-2014.) (Revised by Jim Kingdon, 5-Sep-2024.)
⊢ (𝜑 → (𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1))    &   ⊢ 𝑆 = {𝑦 ∈ (0..^𝑁) ∣ (𝑦 gcd 𝑁) = 1}    &   ⊢ (𝜑 → 𝐹:(1...(ϕ‘𝑁))–1-1-onto→𝑆)    &   ⊢ 𝐻 = (◡𝐹 ∘ (𝑦 ∈ (1...(ϕ‘𝑁)) ↦ ((𝐴 · (𝐹‘𝑦)) mod 𝑁)))    ⇒   ⊢ (𝜑 → 𝐻:(1...(ϕ‘𝑁))–1-1-onto→(1...(ϕ‘𝑁)))
Theorem	eulerthlemth 13012*	Lemma for eulerth 13013. The result. (Contributed by Mario Carneiro, 28-Feb-2014.) (Revised by Jim Kingdon, 2-Sep-2024.)
⊢ (𝜑 → (𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1))    &   ⊢ 𝑆 = {𝑦 ∈ (0..^𝑁) ∣ (𝑦 gcd 𝑁) = 1}    &   ⊢ (𝜑 → 𝐹:(1...(ϕ‘𝑁))–1-1-onto→𝑆)    ⇒   ⊢ (𝜑 → ((𝐴↑(ϕ‘𝑁)) mod 𝑁) = (1 mod 𝑁))
Theorem	eulerth 13013	Euler's theorem, a generalization of Fermat's little theorem. If 𝐴 and 𝑁 are coprime, then 𝐴↑ϕ(𝑁)≡1 (mod 𝑁). This is Metamath 100 proof #10. Also called Euler-Fermat theorem, see theorem 5.17 in [ApostolNT] p. 113. (Contributed by Mario Carneiro, 28-Feb-2014.)
⊢ ((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) → ((𝐴↑(ϕ‘𝑁)) mod 𝑁) = (1 mod 𝑁))
Theorem	fermltl 13014	Fermat's little theorem. When 𝑃 is prime, 𝐴↑𝑃≡𝐴 (mod 𝑃) for any 𝐴, see theorem 5.19 in [ApostolNT] p. 114. (Contributed by Mario Carneiro, 28-Feb-2014.) (Proof shortened by AV, 19-Mar-2022.)
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → ((𝐴↑𝑃) mod 𝑃) = (𝐴 mod 𝑃))
Theorem	prmdiv 13015	Show an explicit expression for the modular inverse of 𝐴 mod 𝑃. (Contributed by Mario Carneiro, 24-Jan-2015.)
⊢ 𝑅 = ((𝐴↑(𝑃 − 2)) mod 𝑃)    ⇒   ⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ ¬ 𝑃 ∥ 𝐴) → (𝑅 ∈ (1...(𝑃 − 1)) ∧ 𝑃 ∥ ((𝐴 · 𝑅) − 1)))
Theorem	prmdiveq 13016	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 ∧ ¬ 𝑃 ∥ 𝐴) → ((𝑆 ∈ (0...(𝑃 − 1)) ∧ 𝑃 ∥ ((𝐴 · 𝑆) − 1)) ↔ 𝑆 = 𝑅))
Theorem	prmdivdiv 13017	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))
Theorem	hashgcdlem 13018*	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 = (𝑥 ∈ 𝐴 ↦ (𝑥 · 𝑁))    ⇒   ⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ∧ 𝑁 ∥ 𝑀) → 𝐹:𝐴–1-1-onto→𝐵)
Theorem	dvdsfi 13019*	A natural number has finitely many divisors. (Contributed by Jim Kingdon, 9-Oct-2025.)
⊢ (𝑁 ∈ ℕ → {𝑥 ∈ ℕ ∣ 𝑥 ∥ 𝑁} ∈ Fin)
Theorem	hashgcdeq 13020*	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(𝑁 ∥ 𝑀, (ϕ‘(𝑀 / 𝑁)), 0))
Theorem	phisum 13021*	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 ∣ 𝑥 ∥ 𝑁} (ϕ‘𝑑) = 𝑁)
Theorem	odzval 13022*	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) → ((odℤ‘𝑁)‘𝐴) = inf({𝑛 ∈ ℕ ∣ 𝑁 ∥ ((𝐴↑𝑛) − 1)}, ℝ, < ))
Theorem	odzcllem 13023	- Lemma for odzcl 13024, 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) → (((odℤ‘𝑁)‘𝐴) ∈ ℕ ∧ 𝑁 ∥ ((𝐴↑((odℤ‘𝑁)‘𝐴)) − 1)))
Theorem	odzcl 13024	The order of a group element is an integer. (Contributed by Mario Carneiro, 28-Feb-2014.)
⊢ ((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) → ((odℤ‘𝑁)‘𝐴) ∈ ℕ)
Theorem	odzid 13025	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↑((odℤ‘𝑁)‘𝐴)) − 1))
Theorem	odzdvds 13026	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 ∥ ((𝐴↑𝐾) − 1) ↔ ((odℤ‘𝑁)‘𝐴) ∥ 𝐾))
Theorem	odzphi 13027	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) → ((odℤ‘𝑁)‘𝐴) ∥ (ϕ‘𝑁))
5.2.6  Arithmetic modulo a prime number
Theorem	modprm1div 13028	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)))
Theorem	m1dvdsndvds 13029	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) → ¬ 𝑃 ∥ 𝐴))
Theorem	modprminv 13030	Show an explicit expression for the modular inverse of 𝐴 mod 𝑃. This is an application of prmdiv 13015. (Contributed by Alexander van der Vekens, 15-May-2018.)
⊢ 𝑅 = ((𝐴↑(𝑃 − 2)) mod 𝑃)    ⇒   ⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ ¬ 𝑃 ∥ 𝐴) → (𝑅 ∈ (1...(𝑃 − 1)) ∧ ((𝐴 · 𝑅) mod 𝑃) = 1))
Theorem	modprminveq 13031	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 ∧ ¬ 𝑃 ∥ 𝐴) → ((𝑆 ∈ (0...(𝑃 − 1)) ∧ ((𝐴 · 𝑆) mod 𝑃) = 1) ↔ 𝑆 = 𝑅))
Theorem	vfermltl 13032	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 ∧ ¬ 𝑃 ∥ 𝐴) → ((𝐴↑(𝑃 − 1)) mod 𝑃) = 1)
Theorem	powm2modprm 13033	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))
Theorem	reumodprminv 13034*	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)
Theorem	modprm0 13035*	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)
Theorem	nnnn0modprm0 13036*	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)
Theorem	modprmn0modprm0 13037*	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
Theorem	coprimeprodsq 13038	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 ∈ ℕ0 ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℕ0) ∧ ((𝐴 gcd 𝐵) gcd 𝐶) = 1) → ((𝐶↑2) = (𝐴 · 𝐵) → 𝐴 = ((𝐴 gcd 𝐶)↑2)))
Theorem	coprimeprodsq2 13039	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 ∈ ℕ0 ∧ 𝐶 ∈ ℕ0) ∧ ((𝐴 gcd 𝐵) gcd 𝐶) = 1) → ((𝐶↑2) = (𝐴 · 𝐵) → 𝐵 = ((𝐵 gcd 𝐶)↑2)))
Theorem	oddprm 13040	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)
Theorem	nnoddn2prm 13041	A prime not equal to 2 is an odd positive integer. (Contributed by AV, 28-Jun-2021.)
⊢ (𝑁 ∈ (ℙ ∖ {2}) → (𝑁 ∈ ℕ ∧ ¬ 2 ∥ 𝑁))
Theorem	oddn2prm 13042	A prime not equal to 2 is odd. (Contributed by AV, 28-Jun-2021.)
⊢ (𝑁 ∈ (ℙ ∖ {2}) → ¬ 2 ∥ 𝑁)
Theorem	nnoddn2prmb 13043	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))
Theorem	prm23lt5 13044	A prime less than 5 is either 2 or 3. (Contributed by AV, 5-Jul-2021.)
⊢ ((𝑃 ∈ ℙ ∧ 𝑃 < 5) → (𝑃 = 2 ∨ 𝑃 = 3))
Theorem	prm23ge5 13045	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)))
Theorem	pythagtriplem1 13046*	Lemma for pythagtrip 13064. 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))
Theorem	pythagtriplem2 13047*	Lemma for pythagtrip 13064. 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)))
Theorem	pythagtriplem3 13048	Lemma for pythagtrip 13064. 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)
Theorem	pythagtriplem4 13049	Lemma for pythagtrip 13064. Show that 𝐶 − 𝐵 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)) → ((𝐶 − 𝐵) gcd (𝐶 + 𝐵)) = 1)
Theorem	pythagtriplem10 13050	Lemma for pythagtrip 13064. Show that 𝐶 − 𝐵 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 < (𝐶 − 𝐵))
Theorem	pythagtriplem6 13051	Lemma for pythagtrip 13064. 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))
Theorem	pythagtriplem7 13052	Lemma for pythagtrip 13064. 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))
Theorem	pythagtriplem8 13053	Lemma for pythagtrip 13064. 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)
Theorem	pythagtriplem9 13054	Lemma for pythagtrip 13064. 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)
Theorem	pythagtriplem11 13055	Lemma for pythagtrip 13064. 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)) + (√‘(𝐶 − 𝐵))) / 2)    ⇒   ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝑀 ∈ ℕ)
Theorem	pythagtriplem12 13056	Lemma for pythagtrip 13064. Calculate the square of 𝑀. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ 𝑀 = (((√‘(𝐶 + 𝐵)) + (√‘(𝐶 − 𝐵))) / 2)    ⇒   ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (𝑀↑2) = ((𝐶 + 𝐴) / 2))
Theorem	pythagtriplem13 13057	Lemma for pythagtrip 13064. 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)) − (√‘(𝐶 − 𝐵))) / 2)    ⇒   ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝑁 ∈ ℕ)
Theorem	pythagtriplem14 13058	Lemma for pythagtrip 13064. Calculate the square of 𝑁. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
⊢ 𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶 − 𝐵))) / 2)    ⇒   ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (𝑁↑2) = ((𝐶 − 𝐴) / 2))
Theorem	pythagtriplem15 13059	Lemma for pythagtrip 13064. 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)) + (√‘(𝐶 − 𝐵))) / 2)    &   ⊢ 𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶 − 𝐵))) / 2)    ⇒   ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝐴 = ((𝑀↑2) − (𝑁↑2)))
Theorem	pythagtriplem16 13060	Lemma for pythagtrip 13064. 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)) + (√‘(𝐶 − 𝐵))) / 2)    &   ⊢ 𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶 − 𝐵))) / 2)    ⇒   ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝐵 = (2 · (𝑀 · 𝑁)))
Theorem	pythagtriplem17 13061	Lemma for pythagtrip 13064. 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)) + (√‘(𝐶 − 𝐵))) / 2)    &   ⊢ 𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶 − 𝐵))) / 2)    ⇒   ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝐶 = ((𝑀↑2) + (𝑁↑2)))
Theorem	pythagtriplem18 13062*	Lemma for pythagtrip 13064. 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))))
Theorem	pythagtriplem19 13063*	Lemma for pythagtrip 13064. 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)))))
Theorem	pythagtrip 13064*	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
Syntax	cpc 13065	Extend class notation with the prime count function.
class pCnt
Definition	df-pc 13066*	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, +∞, (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑟 = (𝑥 / 𝑦) ∧ 𝑧 = (sup({𝑛 ∈ ℕ0 ∣ (𝑝↑𝑛) ∥ 𝑥}, ℝ, < ) − sup({𝑛 ∈ ℕ0 ∣ (𝑝↑𝑛) ∥ 𝑦}, ℝ, < ))))))
Theorem	pclem0 13067*	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 ∣ (𝑃↑𝑛) ∥ 𝑁}    ⇒   ⊢ ((𝑃 ∈ (ℤ≥‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → 0 ∈ 𝐴)
Theorem	pclemub 13068*	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 ∣ (𝑃↑𝑛) ∥ 𝑁}    ⇒   ⊢ ((𝑃 ∈ (ℤ≥‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ∃𝑥 ∈ ℤ ∀𝑦 ∈ 𝐴 𝑦 ≤ 𝑥)
Theorem	pclemdc 13069*	Lemma for the prime power pre-function's properties. (Contributed by Jim Kingdon, 8-Oct-2024.)
⊢ 𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑁}    ⇒   ⊢ ((𝑃 ∈ (ℤ≥‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ∀𝑥 ∈ ℤ DECID 𝑥 ∈ 𝐴)
Theorem	pcprecl 13070*	Closure of the prime power pre-function. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ 𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑁}    &   ⊢ 𝑆 = sup(𝐴, ℝ, < )    ⇒   ⊢ ((𝑃 ∈ (ℤ≥‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑆 ∈ ℕ0 ∧ (𝑃↑𝑆) ∥ 𝑁))
Theorem	pcprendvds 13071*	Non-divisibility property of the prime power pre-function. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ 𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑁}    &   ⊢ 𝑆 = sup(𝐴, ℝ, < )    ⇒   ⊢ ((𝑃 ∈ (ℤ≥‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ¬ (𝑃↑(𝑆 + 1)) ∥ 𝑁)
Theorem	pcprendvds2 13072*	Non-divisibility property of the prime power pre-function. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ 𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑁}    &   ⊢ 𝑆 = sup(𝐴, ℝ, < )    ⇒   ⊢ ((𝑃 ∈ (ℤ≥‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ¬ 𝑃 ∥ (𝑁 / (𝑃↑𝑆)))
Theorem	pcpre1 13073*	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 ∣ (𝑃↑𝑛) ∥ 𝑁}    &   ⊢ 𝑆 = sup(𝐴, ℝ, < )    ⇒   ⊢ ((𝑃 ∈ (ℤ≥‘2) ∧ 𝑁 = 1) → 𝑆 = 0)
Theorem	pcpremul 13074*	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 ℚ), we don't bother to define it off the primes. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ 𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑀}, ℝ, < )    &   ⊢ 𝑇 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑁}, ℝ, < )    &   ⊢ 𝑈 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ (𝑀 · 𝑁)}, ℝ, < )    ⇒   ⊢ ((𝑃 ∈ ℙ ∧ (𝑀 ∈ ℤ ∧ 𝑀 ≠ 0) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑆 + 𝑇) = 𝑈)
Theorem	pceulem 13075*	Lemma for pceu 13076. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ 𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑥}, ℝ, < )    &   ⊢ 𝑇 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑦}, ℝ, < )    &   ⊢ 𝑈 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑠}, ℝ, < )    &   ⊢ 𝑉 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑡}, ℝ, < )    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → 𝑁 ≠ 0)    &   ⊢ (𝜑 → (𝑥 ∈ ℤ ∧ 𝑦 ∈ ℕ))    &   ⊢ (𝜑 → 𝑁 = (𝑥 / 𝑦))    &   ⊢ (𝜑 → (𝑠 ∈ ℤ ∧ 𝑡 ∈ ℕ))    &   ⊢ (𝜑 → 𝑁 = (𝑠 / 𝑡))    ⇒   ⊢ (𝜑 → (𝑆 − 𝑇) = (𝑈 − 𝑉))
Theorem	pceu 13076*	Uniqueness for the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ 𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑥}, ℝ, < )    &   ⊢ 𝑇 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑦}, ℝ, < )    ⇒   ⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → ∃!𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇)))
Theorem	pcval 13077*	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 ∣ (𝑃↑𝑛) ∥ 𝑥}, ℝ, < )    &   ⊢ 𝑇 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑦}, ℝ, < )    ⇒   ⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) = (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇))))
Theorem	pczpre 13078*	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 ∣ (𝑃↑𝑛) ∥ 𝑁}, ℝ, < )    ⇒   ⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) = 𝑆)
Theorem	pczcl 13079	Closure of the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) ∈ ℕ0)
Theorem	pccl 13080	Closure of the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → (𝑃 pCnt 𝑁) ∈ ℕ0)
Theorem	pccld 13081	Closure of the prime power function. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    ⇒   ⊢ (𝜑 → (𝑃 pCnt 𝑁) ∈ ℕ0)
Theorem	pcmul 13082	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)))
Theorem	pcdiv 13083	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)))
Theorem	pcqmul 13084	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)))
Theorem	pc0 13085	The value of the prime power function at zero. (Contributed by Mario Carneiro, 3-Oct-2014.)
⊢ (𝑃 ∈ ℙ → (𝑃 pCnt 0) = +∞)
Theorem	pc1 13086	Value of the prime count function at 1. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ (𝑃 ∈ ℙ → (𝑃 pCnt 1) = 0)
Theorem	pcqcl 13087	Closure of the general prime count function. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) ∈ ℤ)
Theorem	pcqdiv 13088	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)))
Theorem	pcrec 13089	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))
Theorem	pcexp 13090	Prime power of an exponential. (Contributed by Mario Carneiro, 10-Aug-2015.)
⊢ ((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℚ ∧ 𝐴 ≠ 0) ∧ 𝑁 ∈ ℤ) → (𝑃 pCnt (𝐴↑𝑁)) = (𝑁 · (𝑃 pCnt 𝐴)))
Theorem	pcxnn0cl 13091	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*)
Theorem	pcxcl 13092	Extended real closure of the general prime count function. (Contributed by Mario Carneiro, 3-Oct-2014.)
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℚ) → (𝑃 pCnt 𝑁) ∈ ℝ*)
Theorem	pcxqcl 13093	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) = +∞))
Theorem	pcge0 13094	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))
Theorem	pczdvds 13095	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)
Theorem	pcdvds 13096	Defining property of the prime count function. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → (𝑃↑(𝑃 pCnt 𝑁)) ∥ 𝑁)
Theorem	pczndvds 13097	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)
Theorem	pcndvds 13098	Defining property of the prime count function. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → ¬ (𝑃↑((𝑃 pCnt 𝑁) + 1)) ∥ 𝑁)
Theorem	pczndvds2 13099	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))))
Theorem	pcndvds2 13100	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))))