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I'm trying to use a lemma for a bigger proof, but I can't find a way to prove one of those two things. Can somebody help me? Here is the proof so far:

Lemma less_r : (forall m n p : nat, n + m < p + n + m).

Proof.
 intros.
 apply PeanoNat.Nat.add_lt_mono_r.
 apply PeanoNat.Nat.lt_add_pos_l.
 admit.
Qed.
Anton Trunov
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Rafael Santos
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1 Answers1

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Your statement cannot be proved because it does not hold. For instance, if we take n = m = p = 0, it implies 0 < 0, a clear contradiction.

Arthur Azevedo De Amorim
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  • But there's no way to prove that, except where 0 < 0? – Rafael Santos Jun 30 '18 at 18:06
  • @RafaelSantos I do not understand what you mean. You need to modify your statement if you want to prove it. One possibility would be to add a precondition about `p`: `forall n m p, 0 < p -> n + m < p + n + m`, which you can easily prove with the `omega` tactic (after adding `Require Import Omega.` to your file). But I don't know if that lemma suits your needs. – Arthur Azevedo De Amorim Jun 30 '18 at 18:58
  • So just so we understand what you want to do, you want to get coq to disprove your claim? – Lasse V. Karlsen Jun 30 '18 at 19:38