Library Coq.Arith.Div2



Require Import Lt.
Require Import Plus.
Require Import Compare_dec.
Require Import Even.

Open Local Scope nat_scope.

Implicit Type n : nat.

Here we define n/2 and prove some of its properties

Fixpoint div2 n : nat :=
  match n with
  | O => 0
  | S O => 0
  | S (S n') => S (div2 n')
  end.

Since div2 is recursively defined on 0, 1 and (S (S n)), it is useful to prove the corresponding induction principle

Lemma ind_0_1_SS :
  forall P:nat -> Prop,
    P 0 -> P 1 -> (forall n, P n -> P (S (S n))) -> forall n, P n.

0 <n => n/2 < n

Lemma lt_div2 : forall n, 0 < n -> div2 n < n.

Hint Resolve lt_div2: arith.

Properties related to the parity

Lemma even_div2 : forall n, even n -> div2 n = div2 (S n)
with odd_div2 : forall n, odd n -> S (div2 n) = div2 (S n).

Lemma div2_even : forall n, div2 n = div2 (S n) -> even n
with div2_odd : forall n, S (div2 n) = div2 (S n) -> odd n.

Hint Resolve even_div2 div2_even odd_div2 div2_odd: arith.

Lemma even_odd_div2 :
  forall n,
    (even n <-> div2 n = div2 (S n)) /\ (odd n <-> S (div2 n) = div2 (S n)).

Properties related to the double (2n)

Definition double n := n + n.

Hint Unfold double: arith.

Lemma double_S : forall n, double (S n) = S (S (double n)).

Lemma double_plus : forall n (m:nat), double (n + m) = double n + double m.

Hint Resolve double_S: arith.

Lemma even_odd_double :
  forall n,
    (even n <-> n = double (div2 n)) /\ (odd n <-> n = S (double (div2 n))).
Specializations

Lemma even_double : forall n, even n -> n = double (div2 n).
Lemma double_even : forall n, n = double (div2 n) -> even n.
Lemma odd_double : forall n, odd n -> n = S (double (div2 n)).
Lemma double_odd : forall n, n = S (double (div2 n)) -> odd n.
Hint Resolve even_double double_even odd_double double_odd: arith.
Application:
  • if n is even then there is a p such that n = 2p
  • if n is odd then there is a p such that n = 2p+1
(Immediate: it is n/2)
Lemma even_2n : forall n, even n -> {p : nat | n = double p}.

Lemma odd_S2n : forall n, odd n -> {p : nat | n = S (double p)}.

Doubling before dividing by two brings back to the initial number.

Lemma div2_double : forall n:nat, div2 (2*n) = n.

Lemma div2_double_plus_one : forall n:nat, div2 (S (2*n)) = n.