How do I prove the simplified Chinese Remainder Theorem?
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 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)