P. 170 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 16901-17000   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	powm2modprm 16901	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 16902*	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 16903*	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 16904*	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 16905*	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
Theorem	coprimeprodsq 16906	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 16907	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 16908	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 16909	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 16910	A prime not equal to 2 is odd. (Contributed by AV, 28-Jun-2021.)
⊢ (𝑁 ∈ (ℙ ∖ {2}) → ¬ 2 ∥ 𝑁)
Theorem	nnoddn2prmb 16911	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 16912	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 16913	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 16914*	Lemma for pythagtrip 16932. 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 16915*	Lemma for pythagtrip 16932. 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 16916	Lemma for pythagtrip 16932. 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 16917	Lemma for pythagtrip 16932. 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 16918	Lemma for pythagtrip 16932. 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 16919	Lemma for pythagtrip 16932. 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 16920	Lemma for pythagtrip 16932. 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 16921	Lemma for pythagtrip 16932. 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 16922	Lemma for pythagtrip 16932. 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 16923	Lemma for pythagtrip 16932. 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 16924	Lemma for pythagtrip 16932. 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 16925	Lemma for pythagtrip 16932. 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 16926	Lemma for pythagtrip 16932. 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 16927	Lemma for pythagtrip 16932. 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 16928	Lemma for pythagtrip 16932. 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 16929	Lemma for pythagtrip 16932. 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 16930*	Lemma for pythagtrip 16932. 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 16931*	Lemma for pythagtrip 16932. 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 16932*	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))))))
Theorem	iserodd 16933*	Collect the odd terms in a sequence. (Contributed by Mario Carneiro, 7-Apr-2015.) (Proof shortened by AV, 10-Jul-2022.)
⊢ ((𝜑 ∧ 𝑘 ∈ ℕ0) → 𝐶 ∈ ℂ)    &   ⊢ (𝑛 = ((2 · 𝑘) + 1) → 𝐵 = 𝐶)    ⇒   ⊢ (𝜑 → (seq0( + , (𝑘 ∈ ℕ0 ↦ 𝐶)) ⇝ 𝐴 ↔ seq1( + , (𝑛 ∈ ℕ ↦ if(2 ∥ 𝑛, 0, 𝐵))) ⇝ 𝐴))
6.2.7  The prime count function
Syntax	cpc 16934	Extend class notation with the prime count function.
class pCnt
Definition	df-pc 16935*	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	pclem 16936*	- Lemma for the prime power pre-function's properties. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ 𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑁}    ⇒   ⊢ ((𝑃 ∈ (ℤ≥‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝐴 ⊆ ℤ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℤ ∀𝑦 ∈ 𝐴 𝑦 ≤ 𝑥))
Theorem	pcprecl 16937*	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 16938*	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 16939*	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 16940*	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 16941*	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	pcval 16942*	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	pceulem 16943*	Lemma for pceu 16944. (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 16944*	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	pczpre 16945*	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 16946	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 16947	Closure of the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → (𝑃 pCnt 𝑁) ∈ ℕ0)
Theorem	pccld 16948	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 16949	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 16950	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 16951	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 16952	The value of the prime power function at zero. (Contributed by Mario Carneiro, 3-Oct-2014.)
⊢ (𝑃 ∈ ℙ → (𝑃 pCnt 0) = +∞)
Theorem	pc1 16953	Value of the prime count function at 1. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ (𝑃 ∈ ℙ → (𝑃 pCnt 1) = 0)
Theorem	pcqcl 16954	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 16955	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 16956	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 16957	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 16958	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 16959	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	pcge0 16960	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 16961	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 16962	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 16963	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 16964	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 16965	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 16966	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))))
Theorem	pcdvdsb 16967	𝑃↑𝐴 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) ↔ (𝑃↑𝐴) ∥ 𝑁))
Theorem	pcelnn 16968	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 ↔ 𝑃 ∥ 𝑁))
Theorem	pceq0 16969	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 ↔ ¬ 𝑃 ∥ 𝑁))
Theorem	pcidlem 16970	The prime count of a prime power. (Contributed by Mario Carneiro, 12-Mar-2014.)
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ0) → (𝑃 pCnt (𝑃↑𝐴)) = 𝐴)
Theorem	pcid 16971	The prime count of a prime power. (Contributed by Mario Carneiro, 9-Sep-2014.)
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → (𝑃 pCnt (𝑃↑𝐴)) = 𝐴)
Theorem	pcneg 16972	The prime count of a negative number. (Contributed by Mario Carneiro, 13-Mar-2014.)
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℚ) → (𝑃 pCnt -𝐴) = (𝑃 pCnt 𝐴))
Theorem	pcabs 16973	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))
Theorem	pcdvdstr 16974	The prime count increases under the divisibility relation. (Contributed by Mario Carneiro, 13-Mar-2014.)
⊢ ((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐴 ∥ 𝐵)) → (𝑃 pCnt 𝐴) ≤ (𝑃 pCnt 𝐵))
Theorem	pcgcd1 16975	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))
Theorem	pcgcd 16976	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)))
Theorem	pc2dvds 16977*	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) → (𝐴 ∥ 𝐵 ↔ ∀𝑝 ∈ ℙ (𝑝 pCnt 𝐴) ≤ (𝑝 pCnt 𝐵)))
Theorem	pc11 16978*	The prime count function, viewed as a function from ℕ to (ℕ ↑m ℙ), is one-to-one. (Contributed by Mario Carneiro, 23-Feb-2014.)
⊢ ((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ0) → (𝐴 = 𝐵 ↔ ∀𝑝 ∈ ℙ (𝑝 pCnt 𝐴) = (𝑝 pCnt 𝐵)))
Theorem	pcz 16979*	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)))
Theorem	pcprmpw2 16980*	Self-referential expression for a prime power. (Contributed by Mario Carneiro, 16-Jan-2015.)
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ) → (∃𝑛 ∈ ℕ0 𝐴 ∥ (𝑃↑𝑛) ↔ 𝐴 = (𝑃↑(𝑃 pCnt 𝐴))))
Theorem	pcprmpw 16981*	Self-referential expression for a prime power. (Contributed by Mario Carneiro, 16-Jan-2015.)
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ) → (∃𝑛 ∈ ℕ0 𝐴 = (𝑃↑𝑛) ↔ 𝐴 = (𝑃↑(𝑃 pCnt 𝐴))))
Theorem	dvdsprmpweq 16982*	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 ∥ (𝑃↑𝑁) → ∃𝑛 ∈ ℕ0 𝐴 = (𝑃↑𝑛)))
Theorem	dvdsprmpweqnn 16983*	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 ∥ (𝑃↑𝑁) → ∃𝑛 ∈ ℕ 𝐴 = (𝑃↑𝑛)))
Theorem	dvdsprmpweqle 16984*	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 ∥ (𝑃↑𝑁) → ∃𝑛 ∈ ℕ0 (𝑛 ≤ 𝑁 ∧ 𝐴 = (𝑃↑𝑛))))
Theorem	difsqpwdvds 16985	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 ∈ ℕ0 ∧ 𝐵 ∈ ℕ0 ∧ (𝐵 + 1) < 𝐴) ∧ (𝐶 ∈ ℙ ∧ 𝐷 ∈ ℕ0)) → ((𝐶↑𝐷) = ((𝐴↑2) − (𝐵↑2)) → 𝐶 ∥ (2 · 𝐵)))
Theorem	pcaddlem 16986	Lemma for pcadd 16987. The original numbers 𝐴 and 𝐵 have been decomposed using the prime count function as (𝑃↑𝑀) · (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.)
⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → 𝐴 = ((𝑃↑𝑀) · (𝑅 / 𝑆)))    &   ⊢ (𝜑 → 𝐵 = ((𝑃↑𝑁) · (𝑇 / 𝑈)))    &   ⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀))    &   ⊢ (𝜑 → (𝑅 ∈ ℤ ∧ ¬ 𝑃 ∥ 𝑅))    &   ⊢ (𝜑 → (𝑆 ∈ ℕ ∧ ¬ 𝑃 ∥ 𝑆))    &   ⊢ (𝜑 → (𝑇 ∈ ℤ ∧ ¬ 𝑃 ∥ 𝑇))    &   ⊢ (𝜑 → (𝑈 ∈ ℕ ∧ ¬ 𝑃 ∥ 𝑈))    ⇒   ⊢ (𝜑 → 𝑀 ≤ (𝑃 pCnt (𝐴 + 𝐵)))
Theorem	pcadd 16987	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.)
⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → 𝐴 ∈ ℚ)    &   ⊢ (𝜑 → 𝐵 ∈ ℚ)    &   ⊢ (𝜑 → (𝑃 pCnt 𝐴) ≤ (𝑃 pCnt 𝐵))    ⇒   ⊢ (𝜑 → (𝑃 pCnt 𝐴) ≤ (𝑃 pCnt (𝐴 + 𝐵)))
Theorem	pcadd2 16988	The inequality of pcadd 16987 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.)
⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝜑 → 𝐴 ∈ ℚ)    &   ⊢ (𝜑 → 𝐵 ∈ ℚ)    &   ⊢ (𝜑 → (𝑃 pCnt 𝐴) < (𝑃 pCnt 𝐵))    ⇒   ⊢ (𝜑 → (𝑃 pCnt 𝐴) = (𝑃 pCnt (𝐴 + 𝐵)))
Theorem	pcmptcl 16989	Closure for the prime power map. (Contributed by Mario Carneiro, 12-Mar-2014.)
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛↑𝐴), 1))    &   ⊢ (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈ ℕ0)    ⇒   ⊢ (𝜑 → (𝐹:ℕ⟶ℕ ∧ seq1( · , 𝐹):ℕ⟶ℕ))
Theorem	pcmpt 16990*	Construct a function with given prime count characteristics. (Contributed by Mario Carneiro, 12-Mar-2014.)
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛↑𝐴), 1))    &   ⊢ (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈ ℕ0)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝑛 = 𝑃 → 𝐴 = 𝐵)    ⇒   ⊢ (𝜑 → (𝑃 pCnt (seq1( · , 𝐹)‘𝑁)) = if(𝑃 ≤ 𝑁, 𝐵, 0))
Theorem	pcmpt2 16991*	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, (𝑛↑𝐴), 1))    &   ⊢ (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈ ℕ0)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑃 ∈ ℙ)    &   ⊢ (𝑛 = 𝑃 → 𝐴 = 𝐵)    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘𝑁))    ⇒   ⊢ (𝜑 → (𝑃 pCnt ((seq1( · , 𝐹)‘𝑀) / (seq1( · , 𝐹)‘𝑁))) = if((𝑃 ≤ 𝑀 ∧ ¬ 𝑃 ≤ 𝑁), 𝐵, 0))
Theorem	pcmptdvds 16992	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, (𝑛↑𝐴), 1))    &   ⊢ (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈ ℕ0)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘𝑁))    ⇒   ⊢ (𝜑 → (seq1( · , 𝐹)‘𝑁) ∥ (seq1( · , 𝐹)‘𝑀))
Theorem	pcprod 16993*	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)
Theorem	sumhash 16994*	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 ∧ 𝐴 ⊆ 𝐵) → Σ𝑘 ∈ 𝐵 if(𝑘 ∈ 𝐴, 1, 0) = (♯‘𝐴))
Theorem	fldivp1 16995	The difference between the floors of adjacent fractions is either 1 or 0. (Contributed by Mario Carneiro, 8-Mar-2014.)
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ) → ((⌊‘((𝑀 + 1) / 𝑁)) − (⌊‘(𝑀 / 𝑁))) = if(𝑁 ∥ (𝑀 + 1), 1, 0))
Theorem	pcfaclem 16996	Lemma for pcfac 16997. (Contributed by Mario Carneiro, 20-May-2014.)
⊢ ((𝑁 ∈ ℕ0 ∧ 𝑀 ∈ (ℤ≥‘𝑁) ∧ 𝑃 ∈ ℙ) → (⌊‘(𝑁 / (𝑃↑𝑀))) = 0)
Theorem	pcfac 16997*	Calculate the prime count of a factorial. (Contributed by Mario Carneiro, 11-Mar-2014.) (Revised by Mario Carneiro, 21-May-2014.)
⊢ ((𝑁 ∈ ℕ0 ∧ 𝑀 ∈ (ℤ≥‘𝑁) ∧ 𝑃 ∈ ℙ) → (𝑃 pCnt (!‘𝑁)) = Σ𝑘 ∈ (1...𝑀)(⌊‘(𝑁 / (𝑃↑𝑘))))
Theorem	pcbc 16998*	Calculate the prime count of a binomial coefficient. (Contributed by Mario Carneiro, 11-Mar-2014.) (Revised by Mario Carneiro, 21-May-2014.)
⊢ ((𝑁 ∈ ℕ ∧ 𝐾 ∈ (0...𝑁) ∧ 𝑃 ∈ ℙ) → (𝑃 pCnt (𝑁C𝐾)) = Σ𝑘 ∈ (1...𝑁)((⌊‘(𝑁 / (𝑃↑𝑘))) − ((⌊‘((𝑁 − 𝐾) / (𝑃↑𝑘))) + (⌊‘(𝐾 / (𝑃↑𝑘))))))
Theorem	qexpz 16999	If a power of a rational number is an integer, then the number is an integer. In other words, all n-th roots are irrational unless they are integers (so that the original number is an n-th power). (Contributed by Mario Carneiro, 10-Aug-2015.)
⊢ ((𝐴 ∈ ℚ ∧ 𝑁 ∈ ℕ ∧ (𝐴↑𝑁) ∈ ℤ) → 𝐴 ∈ ℤ)
Theorem	expnprm 17000	A second or higher power of a rational number is not a prime number. Or by contraposition, the n-th root of a prime number is irrational. Suggested by Norm Megill. (Contributed by Mario Carneiro, 10-Aug-2015.)
⊢ ((𝐴 ∈ ℚ ∧ 𝑁 ∈ (ℤ≥‘2)) → ¬ (𝐴↑𝑁) ∈ ℙ)