a + b = 0 -> a = 0 and b = 0 in Coq

Question

I want to prove the following :

1 subgoals
a : nat
b : nat
H0 : a + b = 0
______________________________________(1/1)
a = 0 /\ b = 0

It seems very easy, even trivial, but I dont know how to do it. I tried induction , case , unsuccessfully. Any idea ?

Thanks for your help.

Answer 1

You can prove forall n1 n2, n1 + n2 = 0 -> n1 = 0 by case analysis on n1 .

If n1 is 0 , the conclusion is 0 = 0 , which you can prove because = is reflexive.

If there's an n3 such that n1 = S n3 , then the hypothesis S n3 + n2 = 0 reduces to S (n3 + n2) = 0 and implies False because 0 and S are different constructors. False implies anything so you're done.

You can prove forall n1 n2, n1 + n2 = 0 -> n2 = 0 by using the previous fact and commutativity of addition. Then you're able to prove forall n1 n2, n1 + n2 = 0 -> n1 = 0 /\\ n2 = 0 .

Check eq_refl.
Check O_S.
Check False_rect.
Conjecture plus_comm : forall n1 n2, n1 + n2 = n2 + n1.
Check conj.

It's probably better to try to automate the proof as much as possible though.

Require Import Coq.Setoids.Setoid.

Set Firstorder Depth 0.

Create HintDb Hints.

Ltac simplify := firstorder || autorewrite with Hints in *.

Conjecture C1 : forall t1 (x1 : t1), x1 = x1 <-> True.
Conjecture C2 : forall n1, S n1 = 0 <-> False.
Conjecture C3 : forall n1, 0 = S n1 <-> False.
Conjecture C4 : forall n1 n2, S n1 = S n2 <-> n1 = n2.
Conjecture C5 : forall n1, 0 + n1 = n1.
Conjecture C6 : forall n1 n2, S n1 + n2 = S (n1 + n2).

Hint Rewrite C1 C2 C3 C4 C5 C6 : Hints.

Theorem T1 : forall n1 n2, n1 + n2 = 0 <-> n1 = 0 /\ n2 = 0.
Proof. destruct n1; repeat simplify. Qed.

Hint Rewrite T1 : Hints.

Either way, this fact has already been proved in the standard library.

Require Import Coq.Arith.Arith.

Check plus_is_O.