Counting on ℕ

This file defines the count function, which gives, for any predicate on the natural numbers, "how many numbers under k satisfy this predicate?". We then prove several expected lemmas about count, relating it to the cardinality of other objects, and helping to evaluate it for specific k.

def Nat.count (p : ℕ → Prop) [DecidablePred p] (n : ℕ) : ℕ

Count the number of naturals k < n satisfying p k.

Equations

• Nat.count p n = List.countP (fun (b : ℕ) => decide (p b)) (List.range n)

@[simp]

theorem Nat.count_zero (p : ℕ → Prop) [DecidablePred p] : count p 0 = 0

def Nat.CountSet.fintype (p : ℕ → Prop) [DecidablePred p] (n : ℕ) : Fintype { i : ℕ // i < n ∧ p i }

A fintype instance for the set relevant to Nat.count. Locally an instance in locale count

Equations

• Nat.CountSet.fintype p n = Fintype.ofFinset ({x ∈ Finset.range n | p x}) ⋯

theorem Nat.count_eq_card_filter_range (p : ℕ → Prop) [DecidablePred p] (n : ℕ) : count p n = {x ∈ Finset.range n | p x}.card

theorem Nat.count_eq_card_fintype (p : ℕ → Prop) [DecidablePred p] (n : ℕ) : count p n = Fintype.card { k : ℕ // k < n ∧ p k }

count p n can be expressed as the cardinality of {k // k < n ∧ p k}.

theorem Nat.count_le (p : ℕ → Prop) [DecidablePred p] {n : ℕ} : count p n ≤ n

theorem Nat.count_succ (p : ℕ → Prop) [DecidablePred p] (n : ℕ) : count p (n + 1) = count p n + if p n then 1 else 0

theorem Nat.count_monotone (p : ℕ → Prop) [DecidablePred p] : Monotone (count p)

theorem Nat.count_add (p : ℕ → Prop) [DecidablePred p] (a b : ℕ) : count p (a + b) = count p a + count (fun (k : ℕ) => p (a + k)) b

theorem Nat.count_add' (p : ℕ → Prop) [DecidablePred p] (a b : ℕ) : count p (a + b) = count (fun (k : ℕ) => p (k + b)) a + count p b

theorem Nat.count_one (p : ℕ → Prop) [DecidablePred p] : count p 1 = if p 0 then 1 else 0

theorem Nat.count_succ' (p : ℕ → Prop) [DecidablePred p] (n : ℕ) : count p (n + 1) = count (fun (k : ℕ) => p (k + 1)) n + if p 0 then 1 else 0

@[simp]

theorem Nat.count_lt_count_succ_iff {p : ℕ → Prop} [DecidablePred p] {n : ℕ} : count p n < count p (n + 1) ↔ p n

theorem Nat.count_succ_eq_succ_count_iff {p : ℕ → Prop} [DecidablePred p] {n : ℕ} : count p (n + 1) = count p n + 1 ↔ p n

theorem Nat.count_succ_eq_count_iff {p : ℕ → Prop} [DecidablePred p] {n : ℕ} : count p (n + 1) = count p n ↔ ¬p n

theorem Nat.count_succ_eq_succ_count {p : ℕ → Prop} [DecidablePred p] {n : ℕ} : p n → count p (n + 1) = count p n + 1

Alias of the reverse direction of Nat.count_succ_eq_succ_count_iff.

theorem Nat.count_succ_eq_count {p : ℕ → Prop} [DecidablePred p] {n : ℕ} : ¬p n → count p (n + 1) = count p n

Alias of the reverse direction of Nat.count_succ_eq_count_iff.

theorem Nat.lt_of_count_lt_count {p : ℕ → Prop} [DecidablePred p] {a b : ℕ} (h : count p a < count p b) : a < b

theorem Nat.count_strict_mono {p : ℕ → Prop} [DecidablePred p] {m n : ℕ} (hm : p m) (hmn : m < n) : count p m < count p n

theorem Nat.count_injective {p : ℕ → Prop} [DecidablePred p] {m n : ℕ} (hm : p m) (hn : p n) (heq : count p m = count p n) : m = n

theorem Nat.count_le_card {p : ℕ → Prop} [DecidablePred p] (hp : (setOf p).Finite) (n : ℕ) : count p n ≤ hp.toFinset.card

theorem Nat.count_lt_card {p : ℕ → Prop} [DecidablePred p] {n : ℕ} (hp : (setOf p).Finite) (hpn : p n) : count p n < hp.toFinset.card

theorem Nat.count_iff_forall {p : ℕ → Prop} [DecidablePred p] {n : ℕ} : count p n = n ↔ ∀ n' < n, p n'

theorem Nat.count_of_forall {p : ℕ → Prop} [DecidablePred p] {n : ℕ} : (∀ n' < n, p n') → count p n = n

Alias of the reverse direction of Nat.count_iff_forall.

@[simp]

theorem Nat.count_true (n : ℕ) : count (fun (x : ℕ) => True) n = n

theorem Nat.count_iff_forall_not {p : ℕ → Prop} [DecidablePred p] {n : ℕ} : count p n = 0 ↔ ∀ m < n, ¬p m

theorem Nat.count_of_forall_not {p : ℕ → Prop} [DecidablePred p] {n : ℕ} : (∀ m < n, ¬p m) → count p n = 0

Alias of the reverse direction of Nat.count_iff_forall_not.

@[simp]

theorem Nat.count_false (n : ℕ) : count (fun (x : ℕ) => False) n = 0

theorem Nat.count_mono_left {p : ℕ → Prop} [DecidablePred p] {q : ℕ → Prop} [DecidablePred q] {n : ℕ} (hpq : ∀ (k : ℕ), p k → q k) : count p n ≤ count q n