<p>Mai exact, trebuie sa fie satisfacuta egalitatea ((a ~1~ {*} M ~1~ {*} x ~1~ + a ~2~ {*} M ~2~ {*} x ~2~ + ... + a ~k~ {*} M ~k~ {*} x ~k~ ) mod n) mod n ~i~ = a ~i~ , <tex>\forall</tex> i <tex>\in</tex> {1, 2, ..., k}.</p>

<p>Datorita faptului ca n ~i~ | n, avem (a ~1~ {*} M ~1~ {*} x ~1~ + a ~2~ {*} M ~2~ {*} x ~2~ + ... + a ~k~ {*} M ~k~ {*} x ~k~ ) mod n ~i~ = a ~i~ . Stim ca M ~i~ = n / n ~i~, <tex>\forall</tex> j <tex>\neq</tex>  i, M ~j~ mod n ~i~ = 0, deci <tex>\forall</tex> j <tex>\neq</tex>  i, (a ~j~ {*} M ~j~ {*} x ~j~) mod n ~i~ = 0. Este suficient sa demonstram ca (a ~i~ {*} M ~i~ {*} x ~i~ ) mod n ~i~ = a ~i~ <tex>\Leftrightarrow</tex>(a ~i~ {*} (M ~i~ {*} x ~i~ ) mod n ~i~ ) mod n ~i~ = a ~i~ . Folosind M ~i~ {*} x ~i~ <tex>\equiv</tex> 1 (mod n ~i~) ajungem la egalitatea evidenta

<p>Datorita faptului ca n ~i~ | n, avem (a ~1~ {*} M ~1~ {*} x ~1~ + a ~2~ {*} M ~2~ {*} x ~2~ + ... + a ~k~ {*} M ~k~ {*} x ~k~ ) mod n ~i~ = a ~i~ . Stim ca M ~i~ = n / n ~i~, <tex>\forall</tex> j <tex>\neq</tex> i, M ~j~ mod n ~i~ = 0, deci <tex>\forall</tex> j <tex>\neq</tex>i, (a ~j~ {*} M ~j~ {*} x ~j~) mod n ~i~ = 0. Este suficient sa demonstram ca (a ~i~ {*} M ~i~ {*} x ~i~ ) mod n ~i~ = a ~i~ <tex>\Leftrightarrow</tex>(a ~i~ {*} (M ~i~ {*} x ~i~ ) mod n ~i~ ) mod n ~i~ = a ~i~ . Folosind M ~i~ {*} x ~i~ <tex>\equiv</tex> 1 (mod n ~i~) ajungem la egalitatea evidenta

a ~i~ = a ~i~ (q.e.d.)</p>
<p>Mai ramane de verificat daca valoarea x este unica. Fie x ^'^ o alta solutie; avem x < n si x ^'^ < n. Daca  x <tex>\equiv</tex> a ~i~ (mod n ~i~ ) si x ^'^ <tex>\equiv</tex> a ~i~ (mod n ~i~ ), <tex>\forall</tex> i <tex>\in</tex> {1, 2, ..., k}, presupunand ca x ^'^ < x, avem x ^'^ - x <tex>\equiv</tex> 0 (mod n ~i~ ), <tex>\forall</tex> i <tex>\in</tex> {1, 2, ... k}.</p>