Chinese remainder theorem in a Dedekind domain.

Question

I want to prove the following :
Let $a_1,...,a_n$ ideals and $x_1,...,x_n$ elements of a Dedekind domain $A$. Then the system of congruences $x \equiv x_i \pmod{a_i}$ for $i = 1,...,n$ has a solution $x \in A$ iff $x_i \equiv x_j \pmod{a_i + a_j}$ for all $i \neq j$.
This is my try :
$(\Rightarrow)$\ Let $x \in A$ solution of $x \equiv x_i \pmod{a_i}$. So, for all $i \neq j$ : $$ x_i - x_j = (x-x_j) - (x - x_i) \in (a_i + a_j)$$ hence $$ x_i \equiv x_j \pmod{a_i + a_j}$$ $(\Leftarrow)$\ Assume $x_i \equiv x_j \pmod{a_i + a_j}$. Let $$ \phi : A \rightarrow A/a_1 \oplus ... \oplus A/a_n$$ defined by $\phi(x) = (x + a_1,...,x+a_n)$. If I show that $\phi$ is surjective, then consider $x_1,...,x_n \in A$ and $(x_1+a_1,...,x_n +a_n) \in A/a_1 \oplus ... \oplus A/a_n$. Since $\phi$ is surjective, there exists $x \in A$ such that $$\phi(x) = (x_1 + a_1,...,x_n+a_n)$$ i.e. $x \equiv x_i \pmod{a_i}$ for each $i=1,...,n$. There remains to show that $\phi$ is surjective.
$\phi$ is surjective iff for all ideals $p \neq 0$ of $A$, $\phi_p$ ( the localization of $\phi$ under $p$) is surjective. $A$ is a Dedekind domain hence its localization is a DVR. Assume $A$ is a DVR and show that $\phi$ is surjective. Let $m$ the maximal ideal of $A$ hence :
$a_i = m^{s_i}$ for $s_i \in \mathbb{N}$.
WLOG:
$$s_1 \leq s_2 \leq ... \leq s_n$$ hence for all $i < j$
$$ x_i - x_j \in a_i + a_j = m^{s_i} + m^{s_j} \subseteq m^{s_i}$$ i.e. $x_i \equiv x_j \pmod{a_i}$.
In particular $$x_n \equiv x_i \pmod{a_i}$$ for all $i = 1,...,n$. Hence $$\phi(x_n) = (x_1 + a_1,...,x_n+a_n)$$ i.e. $\phi$ is surjective.
Is my proof correct?