Created
May 18, 2017 22:09
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Proof that subtraction is not commutative using Coq.
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| Require Import Coq.Arith.Arith. | |
| Require Import Omega. | |
| (* from http://stackoverflow.com/a/44039996/2370606 *) | |
| Lemma subtraction_does_not_commute : | |
| forall a b : nat, a <> b -> a - b <> b - a. | |
| Proof. | |
| induction a. intros b. | |
| - now rewrite Nat.sub_0_r. | |
| - destruct b. | |
| + trivial. | |
| + repeat rewrite Nat.sub_succ; auto. | |
| Qed. | |
| Lemma subtraction_does_not_commute' : | |
| forall a b : nat, a <> b -> a - b <> b - a. | |
| Proof. | |
| intros; omega. | |
| Qed. |
See also the talk I made out of this https://github.com/MikeMKH/talks/tree/master/on-the-commutability-of-subtraction
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Thanks to Anton Trunov on StackOverflow! http://stackoverflow.com/a/44039996/2370606