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(* $Id$ *)
(**************************************************************************)
(* Equality on natural numbers *)
(**************************************************************************)
Fixpoint eq_nat [n:nat] : nat -> Prop :=
[m:nat]Cases n m of
O O => True
| O (S _) => False
| (S _) O => False
| (S n1) (S m1) => (eq_nat n1 m1)
end.
Theorem eq_nat_refl : (n:nat)(eq_nat n n).
Induction n; Simpl; Auto.
Qed.
Hints Resolve eq_nat_refl : arith v62.
Theorem eq_eq_nat : (n,m:nat)(n=m)->(eq_nat n m).
Induction 1; Trivial with arith.
Qed.
Hints Immediate eq_eq_nat : arith v62.
Theorem eq_nat_eq : (n,m:nat)(eq_nat n m)->(n=m).
Induction n; Induction m; Simpl; Contradiction Orelse Auto with arith.
Qed.
Hints Immediate eq_nat_eq : arith v62.
Theorem eq_nat_elim : (n:nat)(P:nat->Prop)(P n)->(m:nat)(eq_nat n m)->(P m).
Intros; Replace m with n; Auto with arith.
Qed.
Theorem eq_nat_decide : (n,m:nat){(eq_nat n m)}+{~(eq_nat n m)}.
Induction n.
Destruct m.
Auto with arith.
Intro; Right; Red; Trivial with arith.
Destruct m.
Right; Red; Auto with arith.
Intros.
Simpl.
Apply H.
Defined.
Fixpoint beq_nat [n:nat] : nat -> bool :=
[m:nat]Cases n m of
O O => true
| O (S _) => false
| (S _) O => false
| (S n1) (S m1) => (beq_nat n1 m1)
end.
Lemma beq_nat_refl : (x:nat)true=(beq_nat x x).
Proof.
Induction x; Simpl; Auto.
Qed.
Definition beq_nat_eq : (x,y:nat)true=(beq_nat x y)->x=y.
Proof.
Double Induction 1 2; Simpl.
Reflexivity.
Intros; Discriminate H0.
Intros; Discriminate H0.
Intros; Case (H0 ? H1); Reflexivity.
Defined.
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