How to prove False from obviously contradictory assumptions

Question

Suppose I want to prove following Theorem:

Theorem succ_neq_zero : forall n m: nat, S n = m -> 0 = m -> False.

This one is trivial since m cannot be both successor and zero, as assumed. However I found it quite tricky to prove it, and I don't know how to make it without an auxiliary lemma:

Lemma succ_neq_zero_lemma : forall n : nat, O = S n -> False.
Proof.
  intros.
  inversion H.
Qed.

Theorem succ_neq_zero : forall n m: nat, S n = m -> 0 = m -> False.
Proof.
  intros.
  symmetry in H.
  apply (succ_neq_zero_lemma n).
  transitivity m.
  assumption.
  assumption.
Qed.

I am pretty sure there is a better way to prove this. What is the best way to do it?

Answer 1

You just need to substitute for m in the first equation:

Theorem succ_neq_zero : forall n m: nat, S n = m -> 0 = m -> False.
Proof.
intros n m H1 H2; rewrite <- H2 in H1; inversion H1.
Qed.

Answer 2

There's a very easy way to prove it:

Theorem succ_neq_zero : forall n m: nat, S n = m -> 0 = m -> False.
Proof.
  congruence.
Qed.

The congruence tactic is a decision procedure for ground equalities on uninterpreted symbols. It's complete for uninterpreted symbols and for constructors, so in cases like this one, it can prove that the equality 0 = m is impossible.

Answer 3

It might be useful to know how congruence works.

To prove that two terms constructed by different constructors are in fact different, just create a function that returns True in one case and False in the other cases, and then use it to prove True = False. I think this is explained in Coq'Art

Example not_congruent: 0 <> 1.
  intros C. (* now our goal is 'False' *)
  pose (fun m=>match m with 0=>True |S _=>False end) as f.
  assert (Contra: f 1 = f 0) by (rewrite C; reflexivity).
  now replace False with True by Contra.
Qed.