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.
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.