Induction principles involving factorizations #
Definitions #
Given P 0, P 1
and a way to extend P a
to P (p ^ n * a)
for prime p
not dividing a
,
we can define P
for all natural numbers.
Equations
- One or more equations did not get rendered due to their size.
Instances For
Given P 0
, P 1
, and P (p ^ n)
for positive prime powers, and a way to extend P a
and
P b
to P (a * b)
when a, b
are positive coprime, we can define P
for all natural numbers.
Equations
- One or more equations did not get rendered due to their size.
Instances For
Given P 0
, P (p ^ n)
for all prime powers, and a way to extend P a
and P b
to
P (a * b)
when a, b
are positive coprime, we can define P
for all natural numbers.
Equations
- Nat.recOnPrimeCoprime h0 hp h a = Nat.recOnPosPrimePosCoprime (fun (p n : ℕ) (h : Nat.Prime p) (x : 0 < n) => hp p n h) h0 (hp 2 0 Nat.prime_two) h a
Instances For
Given P 0
, P 1
, P p
for all primes, and a way to extend P a
and P b
to
P (a * b)
, we can define P
for all natural numbers.
Equations
- Nat.recOnMul h0 h1 hp h a = Nat.recOnPrimeCoprime h0 (Nat.recOnMul.hp'' h1 hp h) (fun (a b : ℕ) (x : 1 < a) (x : 1 < b) (x : a.Coprime b) => h a b) a
Instances For
The predicate holds on prime powers
Equations
- Nat.recOnMul.hp'' h1 hp h p 0 hp' = h1
- Nat.recOnMul.hp'' h1 hp h p n_2.succ hp' = h (p.pow n_2) p (Nat.recOnMul.hp'' h1 hp h p n_2 hp') (hp p hp')
Instances For
Lemmas on multiplicative functions #
For any multiplicative function f
with f 1 = 1
and any n ≠ 0
,
we can evaluate f n
by evaluating f
at p ^ k
over the factorization of n
For any multiplicative function f
with f 1 = 1
and f 0 = 1
,
we can evaluate f n
by evaluating f
at p ^ k
over the factorization of n