The Coq Proof Assistant

# 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.
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.
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 (only parsing).
Notation beq_nat_true_iff := Nat.eqb_eq (only parsing).
Notation beq_nat_false_iff := Nat.eqb_neq (only parsing).
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.