Principal ideal in an integral domain

Question

Let $R$ be an integral domain. How to prove that these two conditions are equivalent:

1. $(d) = (x_1, ..., x_n)$
2. $d | x_1, ..., x_n$ and $\exists a_1, ..., a_n \in R, d = x_1a_1+...+x_na_n$

I know that $x | y \Leftrightarrow (x) \geq (y)$, but I find it difficult to complete the proof.

Answer 1

$d\mid x_1,\ldots, x_n\!\iff (d)\supseteq (x_1),\ldots, (x_n)\!\iff (d)\color{#c00} \supseteq \sum_i\, (x_i) =: (x_1,\ldots, x_n)$

$\exists\ a_i\!:\,\ d\, =\, \sum_i a_i x_i\iff\, d\,\in\, \sum_i\, (x_i) \iff (d) \color{#c00}\subseteq \sum_i\, (x_i) $

Answer 2

$(1)$ implies $(2)$

$(d)=(x_1,...,x_n)$ implies that $x_i\in (d)$ and $x_i=a_id$. $(d)\in (x_1,..,x_n)$ implies $d=a_1x_1+...+a_nx_n$.

$(2)$ implies $(1)$. Let $d$ divides $x_i$ implies that $x_i=a_id$ and $x_i\in (d)$ and $(x_1,...,x_n)\subset (d)$.

$d=a_1x_1+...+a_nx_n$ implies that $d\in (x_1,...,x_n)$ and $(d)\subset (x_1,...,x_n)$.