Library Coq.Arith.EqNat


Require Import PeanoNat.
Local Open Scope nat_scope.

Equality on natural numbers

Propositional equality


Fixpoint eq_nat n m : Prop :=
  match n, m with
    | O, O => True
    | O, S _ => False
    | S _, O => False
    | S n1, S m1 => eq_nat n1 m1
  end.

Theorem eq_nat_refl n : eq_nat n n.
Hint Resolve eq_nat_refl: arith v62.

eq restricted to nat and eq_nat are equivalent

Theorem eq_nat_is_eq n m : eq_nat n m <-> n = m.

Lemma eq_eq_nat n m : n = m -> eq_nat n m.

Lemma eq_nat_eq n m : eq_nat n m -> n = m.

Hint Immediate eq_eq_nat eq_nat_eq: arith v62.

Theorem eq_nat_elim :
  forall n (P:nat -> Prop), P n -> forall m, eq_nat n m -> P m.

Theorem eq_nat_decide : forall n m, {eq_nat n m} + {~ eq_nat n m}.

Boolean equality on nat.

We reuse the one already defined in module Nat. In scope nat_scope, the notation "=?" can be used.

Notation beq_nat := Nat.eqb (compat "8.4").

Notation beq_nat_true_iff := Nat.eqb_eq (compat "8.4").
Notation beq_nat_false_iff := Nat.eqb_neq (compat "8.4").

Lemma beq_nat_refl n : true = (n =? n).

Lemma beq_nat_true n m : (n =? m) = true -> n=m.

Lemma beq_nat_false n m : (n =? m) = false -> n<>m.

TODO: is it really useful here to have a Defined ? Otherwise we could use Nat.eqb_eq

Definition beq_nat_eq : forall n m, true = (n =? m) -> n = m.