How do I prove the simplified Chinese Remainder Theorem?

问题:

I've managed to prove

Theorem modulo_inv : forall m n : Z, rel_prime m n -> exists x : Z, (m * x == 1 [n]). Admitted. 

My question is how to finish the following proof (maybe using the modulo_inv theorem?):

Variables m n : Z. Hypothesis co_prime : rel_prime m n.  Theorem SimpleChineseRemainder :   forall a b : Z, exists x : Z, (x == a [m]) /\ (x == b [n]). 

Here is what I tried, but I don't know whether it is correct or not.

Proof.     intros a b.     exists ((a * n) * (n ^ (-1) mod m) + (b * m) * (m ^ (-1) mod n)).     refine (conj _ _).         (* case : ((a * n) * (n ^ (-1) mod m) + (b * m) * (m ^ (-1) mod n) == a [m]) *)         red.         rewrite Z.add_sub_swap.         apply Z.divide_add_r.          (* case : ((a * n) * (n ^ (-1) mod m) + (b * m) * (m ^ (-1) mod n) == b [n]) *) 

Can anybody provide any suggestions?

回答1:

Code-golfing Anton's answer, I was hoping that ring would be clever enough to use the Eq information, and that the proof would simply be

Theorem SimpleChineseRemainder' a b :  exists x : Z, (x == a [m]) /\ (x == b [n]). Proof.   destruct (rel_prime_bezout _ _ co_prime) as [u v Eq];   exists (a * v * n + b * u * m); split ; [ exists ((b-a)*u) | exists ((a-b)*v)]; ring. Qed. 

Unfortunately it didn't automatically exploit that u * m + v * n = 1 -> u * m = 1 - v * n. So until we have a stronger tactic, I guess that has to be added manually, like so:

Theorem SimpleChineseRemainder' a b : exists x : Z, (x == a [m]) /\ (x == b [n]). Proof.   destruct (rel_prime_bezout _ _ co_prime) as [u v Eq].   exists (a * (v * n) + b * (u * m)); split ; [ exists ((b-a)*u) | exists ((a-b)*v)].   - replace (v*n) with (1-u*m) by (rewrite <- Eq; ring); ring.   - replace (u*m) with (1-v*n) by (rewrite <- Eq; ring); ring. Qed. 

EDIT: The nsatz tactic is able to solve the equation system. However, it introduces a notation for [ ... ] that conflicts with the notation introduced above, and I don't know how to handle that. However, by changing the notation to i.e. [[ ... ]], the proof becomes just two lines:

Require Import Nsatz. Theorem SimpleChineseRemainder' a b :   exists x : Z, (x == a [[m]]) /\ (x == b [[n]]). Proof.   destruct (rel_prime_bezout _ _ co_prime) as [u v Eq];   exists (a * v * n + b * u * m); split ; [ exists ((b-a)*u) | exists ((a-b)*v)]; nsatz. Qed. 

回答2:

Reusing the proof from Wikipedia which is based on Bézout's lemma, we get the following:

From Coq Require Import ZArith Znumtheory. Import Z.  Definition modulo (a b n : Z) : Prop := (n | (a - b)). Notation "a == b [ n ]" := (modulo a b n) (at level 50).  Section SimpleChineseRemainder. Variables m n : Z. Hypothesis co_prime : rel_prime m n.  Theorem SimpleChineseRemainder a b :  exists x : Z, (x == a [[m]]) /\ (x == b [[n]]). Proof.   destruct (rel_prime_bezout _ _ co_prime) as [u v Eq].   exists (a * v * n + b * u * m); split; [| rewrite add_comm in *];   match goal with |- _ == ?c [_] => replace c with (c * 1) at 2 by apply mul_1_r end;   rewrite <-Eq, mul_add_distr_l, !mul_assoc;   now eexists; rewrite add_add_simpl_l_r, <-mul_sub_distr_r. Qed. End SimpleChineseRemainder. 

文章来源: How do I prove the simplified Chinese Remainder Theorem?