(************************************************************************) (* v * The Coq Proof Assistant / The Coq Development Team *) (* 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. Proof. intros P H0 H1 Hn. cut (forall n, P n /\ P (S n)). intros H'n n. elim (H'n n). auto with arith. induction n. auto with arith. intros. elim IHn; auto with arith. Qed. (** [0 n/2 < n] *) Lemma lt_div2 : forall n, 0 < n -> div2 n < n. Proof. intro n. pattern n in |- *. apply ind_0_1_SS. (* n = 0 *) inversion 1. (* n=1 *) simpl; trivial. (* n=S S n' *) intro n'; case (zerop n'). intro n'_eq_0. rewrite n'_eq_0. auto with arith. auto with arith. Qed. 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). Proof. destruct n; intro H. (* 0 *) trivial. (* S n *) inversion_clear H. apply odd_div2 in H0 as <-. trivial. destruct n; intro. (* 0 *) inversion H. (* S n *) inversion_clear H. apply even_div2 in H0 as <-. trivial. Qed. Lemma div2_even n : div2 n = div2 (S n) -> even n with div2_odd n : S (div2 n) = div2 (S n) -> odd n. Proof. { destruct n; intro H. - constructor. - constructor. apply div2_odd. rewrite H. trivial. } { destruct n; intro H. - discriminate. - constructor. apply div2_even. injection H as <-. trivial. } Qed. 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)). Proof. split; split; auto using div2_odd, div2_even, odd_div2, even_div2. Qed. (** 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)). Proof. intro. unfold double in |- *. simpl in |- *. auto with arith. Qed. Lemma double_plus : forall n (m:nat), double (n + m) = double n + double m. Proof. intros m n. unfold double in |- *. do 2 rewrite plus_assoc_reverse. rewrite (plus_permute n). reflexivity. Qed. Hint Resolve double_S: arith. Lemma even_odd_double : forall n, (even n <-> n = double (div2 n)) /\ (odd n <-> n = S (double (div2 n))). Proof. intro n. pattern n in |- *. apply ind_0_1_SS. (* n = 0 *) split; split; auto with arith. intro H. inversion H. (* n = 1 *) split; split; auto with arith. intro H. inversion H. inversion H1. (* n = (S (S n')) *) intros. destruct H as ((IH1,IH2),(IH3,IH4)). split; split. intro H. inversion H. inversion H1. simpl in |- *. rewrite (double_S (div2 n0)). auto with arith. simpl in |- *. rewrite (double_S (div2 n0)). intro H. injection H. auto with arith. intro H. inversion H. inversion H1. simpl in |- *. rewrite (double_S (div2 n0)). auto with arith. simpl in |- *. rewrite (double_S (div2 n0)). intro H. injection H. auto with arith. Qed. (** Specializations *) Lemma even_double : forall n, even n -> n = double (div2 n). Proof fun n => proj1 (proj1 (even_odd_double n)). Lemma double_even : forall n, n = double (div2 n) -> even n. Proof fun n => proj2 (proj1 (even_odd_double n)). Lemma odd_double : forall n, odd n -> n = S (double (div2 n)). Proof fun n => proj1 (proj2 (even_odd_double n)). Lemma double_odd : forall n, n = S (double (div2 n)) -> odd n. Proof fun n => proj2 (proj2 (even_odd_double 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}. Proof. intros n H. exists (div2 n). auto with arith. Defined. Lemma odd_S2n : forall n, odd n -> {p : nat | n = S (double p)}. Proof. intros n H. exists (div2 n). auto with arith. Defined. (** Doubling before dividing by two brings back to the initial number. *) Lemma div2_double : forall n:nat, div2 (2*n) = n. Proof. induction n. simpl; auto. simpl. replace (n+S(n+0)) with (S (2*n)). f_equal; auto. simpl; auto with arith. Qed. Lemma div2_double_plus_one : forall n:nat, div2 (S (2*n)) = n. Proof. induction n. simpl; auto. simpl. replace (n+S(n+0)) with (S (2*n)). f_equal; auto. simpl; auto with arith. Qed.