如何在Coq中证明反对称

Question

Define relation [<=] on natural numbers by saying that [m <= n] holds if there is a number [k] such that [m = k + n].

Reflexive and transitive have been proved.

reflexive: ref: forall n:nat, n <= n.
transitive: trans: forall l m n:nat, l <= m -> m <= n -> l <= n

Now, how to prove forall lm : nat, l <= m -> m <= l -> m = l ?

Answer 1

I think you meant " m <= n holds if there is k such that n = m + k ".

The only additional lemma I needed to prove antisymmetry is:

Lemma le_S_n : forall m n, le (S m) (S n) -> le m n.

Then my proof of antisymmetry goes by induction. Here is the full script:

Require Import Arith.

Definition le (m n :nat) := exists k, n = m + k.

Lemma le_refl : forall m, le m m.
Proof.
intro m; exists 0.
now rewrite <- plus_n_O.
Qed.

Lemma le_trans: forall m n p, le m n -> le n p -> le m p.
Proof.
intros m n p [k1 hk1] [k2 hk2]; exists (k1 + k2).
now rewrite plus_assoc, <- hk1.
Qed.

Lemma le_S_n : forall m n, le (S m) (S n) -> le m n.
Proof.
intros m n [k hk]; exists k.
simpl in hk.
now injection hk; intros.
Qed.

Lemma le_antisym: forall m n, le m n -> le n m -> m = n.
Proof.
induction m as [ | m hi]; destruct n as [ | n ]; simpl in *; intros h1 h2.
- reflexivity.
- destruct h2 as [k hk].
  simpl in hk; discriminate hk.
- destruct h1 as [k hk].
  simpl in hk; discriminate hk.
- now rewrite (hi n); [ reflexivity | apply le_S_n | apply le_S_n ].
Qed.

Answer 2

Here is a short solution:

Require Omega.

Definition le (n m:nat)  := exists k, n + k = m.

Theorem le_nm_mn_eq: forall n m, le n m -> le m n -> n = m.
Proof.
  intros n m Hnm Hmn.
  inversion Hnm; inversion Hmn.
  omega.
Qed.

It uses omega so that I don't have to do all the fiddling with nat equalities.