Lemmas about bitwise operations on natural numbers.

Possibly only of archaeological significance.

def Nat.boddDiv2 : ℕ → Bool × ℕ

boddDiv2 n returns a 2-tuple of type (Bool,Nat) where the Bool value indicates whether n is odd or not and the Nat value returns ⌊n/2⌋

Equations

• Nat.boddDiv2 0 = (false, 0)
• Nat.boddDiv2 (Nat.succ n) = match Nat.boddDiv2 n with | (false, m) => (true, m) | (true, m) => (false, Nat.succ m)

def Nat.div2 (n : ℕ) : ℕ

div2 n = ⌊n/2⌋ the greatest integer smaller than n/2

Equations

• Nat.div2 n = (Nat.boddDiv2 n).snd

def Nat.bodd (n : ℕ) : Bool

bodd n returns true if n is odd

Equations

• Nat.bodd n = (Nat.boddDiv2 n).fst

@[simp]

theorem Nat.bodd_zero : Nat.bodd 0 = false

theorem Nat.bodd_one : Nat.bodd 1 = true

theorem Nat.bodd_two : Nat.bodd 2 = false

@[simp]

theorem Nat.bodd_succ (n : ℕ) : Nat.bodd (Nat.succ n) = !Nat.bodd n

@[simp]

theorem Nat.bodd_add (m : ℕ) (n : ℕ) : Nat.bodd (m + n) = xor (Nat.bodd m) (Nat.bodd n)

@[simp]

theorem Nat.bodd_mul (m : ℕ) (n : ℕ) : Nat.bodd (m * n) = (Nat.bodd m && Nat.bodd n)

theorem Nat.mod_two_of_bodd (n : ℕ) : n % 2 = bif Nat.bodd n then 1 else 0

@[simp]

theorem Nat.div2_zero : Nat.div2 0 = 0

theorem Nat.div2_one : Nat.div2 1 = 0

theorem Nat.div2_two : Nat.div2 2 = 1

@[simp]

theorem Nat.div2_succ (n : ℕ) : Nat.div2 (Nat.succ n) = bif Nat.bodd n then Nat.succ (Nat.div2 n) else Nat.div2 n

theorem Nat.bodd_add_div2 (n : ℕ) : (bif Nat.bodd n then 1 else 0) + 2 * Nat.div2 n = n

theorem Nat.div2_val (n : ℕ) : Nat.div2 n = n / 2

def Nat.bit (b : Bool) : ℕ → ℕ

bit b appends the digit b to the binary representation of its natural number input.

Equations

• Nat.bit b = bif b then bit1 else bit0

theorem Nat.bit0_val (n : ℕ) : bit0 n = 2 * n

theorem Nat.bit1_val (n : ℕ) : bit1 n = 2 * n + 1

theorem Nat.bit_val (b : Bool) (n : ℕ) : Nat.bit b n = 2 * n + bif b then 1 else 0

theorem Nat.bit_decomp (n : ℕ) : Nat.bit (Nat.bodd n) (Nat.div2 n) = n

def Nat.bitCasesOn {C : ℕ → Sort u} (n : ℕ) (h : (b : Bool) → (n : ℕ) → C (Nat.bit b n)) : C n

For a predicate C : Nat → Sort*, if instances can be constructed for natural numbers of the form bit b n, they can be constructed for any given natural number.

Equations

• Nat.bitCasesOn n h = (_ : Nat.bit (Nat.bodd n) (Nat.div2 n) = n) ▸ h (Nat.bodd n) (Nat.div2 n)

theorem Nat.bit_zero : Nat.bit false 0 = 0

def Nat.shiftLeft' (b : Bool) (m : ℕ) : ℕ → ℕ

shiftLeft' b m n performs a left shift of m n times and adds the bit b as the least significant bit each time. Returns the corresponding natural number

Equations

• Nat.shiftLeft' b m 0 = m
• Nat.shiftLeft' b m (Nat.succ n) = Nat.bit b (Nat.shiftLeft' b m n)

@[simp]

theorem Nat.shiftLeft'_false {m : ℕ} (n : ℕ) : Nat.shiftLeft' false m n = m <<< n

@[simp]

theorem Nat.shiftLeft_eq' (m : ℕ) (n : ℕ) : Nat.shiftLeft m n = m <<< n

Std4 takes the unprimed name for Nat.shiftLeft_eq m n : m <<< n = m * 2 ^ n.

@[simp]

theorem Nat.shiftRight_eq (m : ℕ) (n : ℕ) : Nat.shiftRight m n = m >>> n

theorem Nat.binaryRec_decreasing {n : ℕ} (h : n ≠ 0) : Nat.div2 n < n

def Nat.binaryRec {C : ℕ → Sort u} (z : C 0) (f : (b : Bool) → (n : ℕ) → C n → C (Nat.bit b n)) (n : ℕ) : C n

A recursion principle for bit representations of natural numbers. For a predicate C : Nat → Sort*, if instances can be constructed for natural numbers of the form bit b n, they can be constructed for all natural numbers.

Equations

• One or more equations did not get rendered due to their size.

def Nat.size : ℕ → ℕ

size n : Returns the size of a natural number in bits i.e. the length of its binary representation

Equations

• Nat.size = Nat.binaryRec 0 fun (x : Bool) (x : ℕ) => Nat.succ

def Nat.bits : ℕ → List Bool

bits n returns a list of Bools which correspond to the binary representation of n, where the head of the list represents the least significant bit

Equations

• Nat.bits = Nat.binaryRec [] fun (b : Bool) (x : ℕ) (IH : List Bool) => b :: IH

def Nat.ldiff : ℕ → ℕ → ℕ

ldiff a b performs bitwise set difference. For each corresponding pair of bits taken as booleans, say aᵢ and bᵢ, it applies the boolean operation aᵢ ∧ ¬bᵢ to obtain the iᵗʰ bit of the result.

Equations

• Nat.ldiff = Nat.bitwise fun (a b : Bool) => a && !b

@[simp]

theorem Nat.binaryRec_zero {C : ℕ → Sort u} (z : C 0) (f : (b : Bool) → (n : ℕ) → C n → C (Nat.bit b n)) : Nat.binaryRec z f 0 = z

bitwise ops

theorem Nat.bodd_bit (b : Bool) (n : ℕ) : Nat.bodd (Nat.bit b n) = b

theorem Nat.div2_bit (b : Bool) (n : ℕ) : Nat.div2 (Nat.bit b n) = n

theorem Nat.shiftLeft'_add (b : Bool) (m : ℕ) (n : ℕ) (k : ℕ) : Nat.shiftLeft' b m (n + k) = Nat.shiftLeft' b (Nat.shiftLeft' b m n) k

theorem Nat.shiftLeft_add (m : ℕ) (n : ℕ) (k : ℕ) : m <<< (n + k) = m <<< n <<< k

theorem Nat.shiftLeft'_sub (b : Bool) (m : ℕ) {n : ℕ} {k : ℕ} : k ≤ n → Nat.shiftLeft' b m (n - k) = Nat.shiftLeft' b m n >>> k

theorem Nat.shiftLeft_sub (m : ℕ) {n : ℕ} {k : ℕ} : k ≤ n → m <<< (n - k) = m <<< n >>> k

@[simp]

theorem Nat.testBit_zero (b : Bool) (n : ℕ) : Nat.testBit (Nat.bit b n) 0 = b

theorem Nat.bodd_eq_and_one_ne_zero (n : ℕ) : Nat.bodd n = (n &&& 1 != 0)

theorem Nat.testBit_succ (m : ℕ) (b : Bool) (n : ℕ) : Nat.testBit (Nat.bit b n) (Nat.succ m) = Nat.testBit n m

theorem Nat.binaryRec_eq {C : ℕ → Sort u} {z : C 0} {f : (b : Bool) → (n : ℕ) → C n → C (Nat.bit b n)} (h : f false 0 z = z) (b : Bool) (n : ℕ) : Nat.binaryRec z f (Nat.bit b n) = f b n (Nat.binaryRec z f n)