Library Coq.Arith.Div2
Nota : this file is OBSOLETE, and left only for compatibility.
Please consider using Nat.div2 directly, and results about it
(see file PeanoNat).
Here we define n/2 and prove some of its properties
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
Properties related to the parity
Lemma even_div2 n : even n -> div2 n = div2 (S n).
Lemma odd_div2 n : odd n -> S (div2 n) = div2 (S n).
Lemma div2_even n : div2 n = div2 (S n) -> even n.
Lemma div2_odd n : S (div2 n) = div2 (S n) -> odd n.
Hint Resolve even_div2 div2_even odd_div2 div2_odd: arith.
Lemma even_odd_div2 n :
(even n <-> div2 n = div2 (S n)) /\
(odd n <-> S (div2 n) = div2 (S n)).
Properties related to the double (2n)
Notation double := Nat.double (only parsing).
Hint Unfold double Nat.double: arith.
Lemma double_S n : double (S n) = S (S (double n)).
Lemma double_plus n m : double (n + m) = double n + double m.
Hint Resolve double_S: arith.
Lemma even_odd_double n :
(even n <-> n = double (div2 n)) /\ (odd n <-> n = S (double (div2 n))).
Specializations
Lemma even_double n : even n -> n = double (div2 n).
Lemma double_even n : n = double (div2 n) -> even n.
Lemma odd_double n : odd n -> n = S (double (div2 n)).
Lemma double_odd n : n = S (double (div2 n)) -> odd n.
Hint Resolve even_double double_even odd_double double_odd: arith.
Application:
(Immediate: it is n/2)
- 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
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.