P. 130 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 12901-13000   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	divnumden 12901	Calculate the reduced form of a quotient using gcd. (Contributed by Stefan O'Rear, 13-Sep-2014.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℕ) → ((numer‘(𝐴 / 𝐵)) = (𝐴 / (𝐴 gcd 𝐵)) ∧ (denom‘(𝐴 / 𝐵)) = (𝐵 / (𝐴 gcd 𝐵))))
Theorem	divdenle 12902	Reducing a quotient never increases the denominator. (Contributed by Stefan O'Rear, 13-Sep-2014.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℕ) → (denom‘(𝐴 / 𝐵)) ≤ 𝐵)
Theorem	qnumgt0 12903	A rational is positive iff its canonical numerator is. (Contributed by Stefan O'Rear, 15-Sep-2014.)
⊢ (𝐴 ∈ ℚ → (0 < 𝐴 ↔ 0 < (numer‘𝐴)))
Theorem	qgt0numnn 12904	A rational is positive iff its canonical numerator is a positive integer. (Contributed by Stefan O'Rear, 15-Sep-2014.)
⊢ ((𝐴 ∈ ℚ ∧ 0 < 𝐴) → (numer‘𝐴) ∈ ℕ)
Theorem	nn0gcdsq 12905	Squaring commutes with GCD, in particular two coprime numbers have coprime squares. (Contributed by Stefan O'Rear, 15-Sep-2014.)
⊢ ((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ0) → ((𝐴 gcd 𝐵)↑2) = ((𝐴↑2) gcd (𝐵↑2)))
Theorem	zgcdsq 12906	nn0gcdsq 12905 extended to integers by symmetry. (Contributed by Stefan O'Rear, 15-Sep-2014.)
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ((𝐴 gcd 𝐵)↑2) = ((𝐴↑2) gcd (𝐵↑2)))
Theorem	numdensq 12907	Squaring a rational squares its canonical components. (Contributed by Stefan O'Rear, 15-Sep-2014.)
⊢ (𝐴 ∈ ℚ → ((numer‘(𝐴↑2)) = ((numer‘𝐴)↑2) ∧ (denom‘(𝐴↑2)) = ((denom‘𝐴)↑2)))
Theorem	numsq 12908	Square commutes with canonical numerator. (Contributed by Stefan O'Rear, 15-Sep-2014.)
⊢ (𝐴 ∈ ℚ → (numer‘(𝐴↑2)) = ((numer‘𝐴)↑2))
Theorem	densq 12909	Square commutes with canonical denominator. (Contributed by Stefan O'Rear, 15-Sep-2014.)
⊢ (𝐴 ∈ ℚ → (denom‘(𝐴↑2)) = ((denom‘𝐴)↑2))
Theorem	qden1elz 12910	A rational is an integer iff it has denominator 1. (Contributed by Stefan O'Rear, 15-Sep-2014.)
⊢ (𝐴 ∈ ℚ → ((denom‘𝐴) = 1 ↔ 𝐴 ∈ ℤ))
Theorem	nn0sqrtelqelz 12911	If a nonnegative integer has a rational square root, that root must be an integer. (Contributed by Jim Kingdon, 24-May-2022.)
⊢ ((𝐴 ∈ ℕ0 ∧ (√‘𝐴) ∈ ℚ) → (√‘𝐴) ∈ ℤ)
Theorem	nonsq 12912	Any integer strictly between two adjacent squares has a non-rational square root. (Contributed by Stefan O'Rear, 15-Sep-2014.)
⊢ (((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ0) ∧ ((𝐵↑2) < 𝐴 ∧ 𝐴 < ((𝐵 + 1)↑2))) → ¬ (√‘𝐴) ∈ ℚ)
5.2.5  Euler's theorem
Syntax	codz 12913	Extend class notation with the order function on the class of integers modulo N.
class odℤ
Syntax	cphi 12914	Extend class notation with the Euler phi function.
class ϕ
Definition	df-odz 12915*	Define the order function on the class of integers modulo N. (Contributed by Mario Carneiro, 23-Feb-2014.) (Revised by AV, 26-Sep-2020.)
⊢ odℤ = (𝑛 ∈ ℕ ↦ (𝑥 ∈ {𝑥 ∈ ℤ ∣ (𝑥 gcd 𝑛) = 1} ↦ inf({𝑚 ∈ ℕ ∣ 𝑛 ∥ ((𝑥↑𝑚) − 1)}, ℝ, < )))
Definition	df-phi 12916*	Define the Euler phi function (also called "Euler totient function"), which counts the number of integers less than 𝑛 and coprime to it, see definition in [ApostolNT] p. 25. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ ϕ = (𝑛 ∈ ℕ ↦ (♯‘{𝑥 ∈ (1...𝑛) ∣ (𝑥 gcd 𝑛) = 1}))
Theorem	phivalfi 12917*	Finiteness of an expression used to define the Euler ϕ function. (Contributed by Jim Kingon, 28-May-2022.)
⊢ (𝑁 ∈ ℕ → {𝑥 ∈ (1...𝑁) ∣ (𝑥 gcd 𝑁) = 1} ∈ Fin)
Theorem	phival 12918*	Value of the Euler ϕ function. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ (𝑁 ∈ ℕ → (ϕ‘𝑁) = (♯‘{𝑥 ∈ (1...𝑁) ∣ (𝑥 gcd 𝑁) = 1}))
Theorem	phicl2 12919	Bounds and closure for the value of the Euler ϕ function. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ (𝑁 ∈ ℕ → (ϕ‘𝑁) ∈ (1...𝑁))
Theorem	phicl 12920	Closure for the value of the Euler ϕ function. (Contributed by Mario Carneiro, 28-Feb-2014.)
⊢ (𝑁 ∈ ℕ → (ϕ‘𝑁) ∈ ℕ)
Theorem	phibndlem 12921*	Lemma for phibnd 12922. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ (𝑁 ∈ (ℤ≥‘2) → {𝑥 ∈ (1...𝑁) ∣ (𝑥 gcd 𝑁) = 1} ⊆ (1...(𝑁 − 1)))
Theorem	phibnd 12922	A slightly tighter bound on the value of the Euler ϕ function. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ (𝑁 ∈ (ℤ≥‘2) → (ϕ‘𝑁) ≤ (𝑁 − 1))
Theorem	phicld 12923	Closure for the value of the Euler ϕ function. (Contributed by Mario Carneiro, 29-May-2016.)
⊢ (𝜑 → 𝑁 ∈ ℕ)    ⇒   ⊢ (𝜑 → (ϕ‘𝑁) ∈ ℕ)
Theorem	phi1 12924	Value of the Euler ϕ function at 1. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ (ϕ‘1) = 1
Theorem	dfphi2 12925*	Alternate definition of the Euler ϕ function. (Contributed by Mario Carneiro, 23-Feb-2014.) (Revised by Mario Carneiro, 2-May-2016.)
⊢ (𝑁 ∈ ℕ → (ϕ‘𝑁) = (♯‘{𝑥 ∈ (0..^𝑁) ∣ (𝑥 gcd 𝑁) = 1}))
Theorem	hashdvds 12926*	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 12927	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 12928	Value of the Euler ϕ function at a prime. (Contributed by Mario Carneiro, 28-Feb-2014.)
⊢ (𝑃 ∈ ℙ → (ϕ‘𝑃) = (𝑃 − 1))
Theorem	crth 12929*	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 12930*	Lemma for phimul 12931. (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 12931	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 12932*	Lemma for eulerth 12938. (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 12933*	Lemma for eulerth 12938. 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 12934*	Lemma for eulerth 12938. 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 12935*	Lemma for eulerth 12938. (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 12936*	Lemma for eulerth 12938. 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 12937*	Lemma for eulerth 12938. 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 12938	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 12939	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 12940	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 12941	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 12942	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 12943*	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 12944*	A natural number has finitely many divisors. (Contributed by Jim Kingdon, 9-Oct-2025.)
⊢ (𝑁 ∈ ℕ → {𝑥 ∈ ℕ ∣ 𝑥 ∥ 𝑁} ∈ Fin)
Theorem	hashgcdeq 12945*	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 12946*	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 12947*	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 12948	- Lemma for odzcl 12949, 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 12949	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 12950	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 12951	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 12952	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 12953	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 12954	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 12955	Show an explicit expression for the modular inverse of 𝐴 mod 𝑃. This is an application of prmdiv 12940. (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 12956	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 12957	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 12958	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 12959*	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 12960*	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 12961*	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 12962*	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 12963	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 12964	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 12965	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 12966	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 12967	A prime not equal to 2 is odd. (Contributed by AV, 28-Jun-2021.)
⊢ (𝑁 ∈ (ℙ ∖ {2}) → ¬ 2 ∥ 𝑁)
Theorem	nnoddn2prmb 12968	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 12969	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 12970	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 12971*	Lemma for pythagtrip 12989. 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 12972*	Lemma for pythagtrip 12989. 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 12973	Lemma for pythagtrip 12989. 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 12974	Lemma for pythagtrip 12989. 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 12975	Lemma for pythagtrip 12989. 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 12976	Lemma for pythagtrip 12989. 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 12977	Lemma for pythagtrip 12989. 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 12978	Lemma for pythagtrip 12989. 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 12979	Lemma for pythagtrip 12989. 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 12980	Lemma for pythagtrip 12989. 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 12981	Lemma for pythagtrip 12989. 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 12982	Lemma for pythagtrip 12989. 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 12983	Lemma for pythagtrip 12989. 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 12984	Lemma for pythagtrip 12989. 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 12985	Lemma for pythagtrip 12989. 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 12986	Lemma for pythagtrip 12989. 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 12987*	Lemma for pythagtrip 12989. 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 12988*	Lemma for pythagtrip 12989. 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 12989*	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 12990	Extend class notation with the prime count function.
class pCnt
Definition	df-pc 12991*	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 12992*	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 12993*	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 12994*	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 12995*	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 12996*	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 12997*	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 12998*	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 12999*	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 13000*	Lemma for pceu 13001. (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))