Coq - proving something which has already been defined?

Question

Taking the very straightforward proof of "the sum of two naturals is odd if one of them is even and the other odd":

Require Import Arith.
Require Import Coq.omega.Omega.

Definition even (n: nat) := exists k, n = 2 * k.
Definition odd  (n: nat) := exists k, n = 2 * k + 1.

Lemma sum_odd_even : forall n m, odd (n + m) -> odd n /\ even m \/ even n /\ odd m.
Proof.
  intros n. intros m. left.
  destruct H. firstorder.

The state at the end of this block of code is:

2 subgoals
n, m, x : nat
H : n + m = 2 * x + 1
______________________________________(1/2)
odd n
______________________________________(2/2)
even m

To my understanding, it is telling me that I need to prove to it that I have an odd number n and an even number m through the hypothesis? Even though I have already stated than n is odd and m is even? How do I proceed from here?

UPDATE:

After a bit of fidgeting around (in light of the comments), I guess I would have to do something like this?

Lemma even_or_odd: forall (n: nat), even n \/ odd n.
Proof.
  induction n as [|n IHn].
  (* Base Case *)
  left. unfold even. exists 0. firstorder.
  (* step case *)
  destruct IHn as [IHeven | IHodd].
  right. unfold even in IHeven. destruct IHeven as [k Heq].
  unfold odd. exists k. firstorder.
  left. unfold odd in IHodd. destruct IHodd as [k Heq].
  unfold even. exists (k + 1). firstorder.
Qed.

Which means that now:

Lemma sum_odd : forall n m, odd (n + m) -> odd n /\ even m \/ even n /\ odd m.
Proof.
  intros n. intros m. left. destruct H. firstorder.
  pose proof (even_or_odd n). pose proof (even_or_odd m).

Result:

    2 subgoals
n, m, x : nat
H : n + m = 2 * x + 1
H0 : even n \/ odd n
H1 : even m \/ odd m
______________________________________(1/2)
odd n
______________________________________(2/2)
even m

Intuitively, all that I have done is saying that every number is either even or odd. Now I have to tell coq that my odd and even numbers are indeed odd and even (I guess?).

UPDATE 2:

As an aside, the problem is solvable with just firstorder:

Lemma sum_odd : forall n m, odd (n + m) -> odd n /\ even m \/ even n /\ odd m.
Proof.
  intros n. intros m. firstorder.
  pose proof (even_or_odd n). pose proof (even_or_odd m).
  destruct H0 as [Even_n | Odd_n]. destruct H1 as [Even_m | Odd_m].
  exfalso. firstorder.
  right. auto.
  destruct H1. left. auto.
  exfalso. firstorder.
Qed.

Where do you think you stated that n is odd and m is even? You used `left` to indicate that you wanted to prove those two things (instead of `even n /\ odd m`), but before you do that you need to do some induction/case analysis. — Tej Chajed, May 04 '18 at 14:08
You're saying that you need to prove this by induction? Why? The classical way of proving this is simply by stating: `n = 2a + 1`, `m = 2b` therefore `n + m = 2a + 1 + 2b` => `n + m = 2(a + b) + 1`. I cannot find an apparent way of doing this proof by definition/substitution in coq. I suppose I just (erroneously) assumed that coq would assume n is odd and m is even, since to prove `odd (n + m) -> odd n /\ even m` requires the assumption that n is odd and m is even. — R. Rengold, May 04 '18 at 15:02
The argument "`n = 2a + 1`, `m = 2b` therefore `n + m = 2a + 1 + 2b`" proves `odd n /\ even m -> odd (n + m)`, not the inverse that you are trying to prove. — Isabelle Newbie, May 04 '18 at 18:51

score 5 · Accepted Answer · answered May 06 '18 at 11:15

Your use of left is still incorrect and keeps you from completing the proof. You apply it to the following goal:

odd (n + m) -> odd n /\ even m \/ even n /\ odd m

and it gives:

H : odd (n + m)
______________________________________(1/1)
odd n /\ even m

You are committing to proving that if n + m is odd, then n is odd and m is even. But this is not true: n might be odd and m might be even. Only apply left or right once you have enough information in the context to be sure which one you want to prove.

So let's restart without left:

Lemma sum_odd : forall n m, odd (n + m) -> odd n /\ even m \/ even n /\ odd m.
Proof.
  intros n. intros m. firstorder.
  pose proof (even_or_odd n). pose proof (even_or_odd m).

At this point we are at:

H : n + m = 2 * x + 1
H0 : even n \/ odd n
H1 : even m \/ odd m
______________________________________(1/1)
odd n /\ even m \/ even n /\ odd m

Now you want to prove something from disjunctions. In order to prove something of the form A \/ B -> C in Coq's constructive logic, you must prove both A -> C and B -> C. You do this by case analysis on the A \/ B (using destruct or other tactics). In this case we have two disjunctions to decompose:

  destruct H0 as [Even_n | Odd_n], H1 as [Even_m | Odd_m].

This gives four cases. I'll show you the first two, the last two are symmetric.

Fist case:

H : n + m = 2 * x + 1
Even_n : even n
Even_m : even m
______________________________________(1/1)
odd n /\ even m \/ even n /\ odd m

The assumptions are contradictory: If both n and m are even, then H cannot hold. We can prove this as follows:

  - exfalso. destruct Even_n, Even_m. omega.

(Step through this to understand what happens!) The exfalso is not really necessary, but it's good documentation that we are doing a proof by showing that the assumptions contradict.

Second case:

H : n + m = 2 * x + 1
Even_n : even n
Odd_m : odd m
______________________________________(1/1)
odd n /\ even m \/ even n /\ odd m

Now, knowing assumptions that apply in this case, we can commit to the right disjunct. This is why your left kept you from making progress!

  - right.

All that remains to be proved is:

Even_n : even n
Odd_m : odd m
______________________________________(1/1)
even n /\ odd m

And auto can handle this.

Coq - proving something which has already been defined?

1 Answers1