Prüfer domains are Arithmetical rings 
Suppose $R$ be a Prüfer domain. How should I prove that it is an arithmetical ring?

 A: Assuming that by an "arithmetical ring" you mean a ring in which for all ideals $\mathfrak{a}, \mathfrak{b}, \mathfrak{c}$ we have $\mathfrak{a} \cap (\mathfrak{b} + \mathfrak{c}) = (\mathfrak{a} \cap \mathfrak{b}) + (\mathfrak{a} \cap \mathfrak{c})$, then this is one of a standard list of equivalent condition for an integral domain to be a Prüfer domain.  For instance, Theorem 6.6 in the text Multiplicative Ideal Theory (henceforth MIT) by Larsen and McCarthy contains ten such equivalent conditions, the first being Prüfer domain (i.e., nonzero finitely generated ideals are invertible) and the last being arithmetical ring.
The sequence of logical implications among these 10 conditions is one of the most complicated I have seen -- it is rather far from being a ten-cycle!  In fact this result is stated in my commutative algebra notes but I am still working on typing up the proof, which will be the longest in the entire set of notes (at least three pages).  They prove Prüfer implies arithmetical in MIT as follows:
Show that (i) Prüfer $\implies$ (iii): nonzero finitely generated ideals are cancellable.
(iii) $\implies$ (iv): For every prime ideal $\mathfrak{p}$ of $R$, the localization $R_{\mathfrak{p}}$ is a valuation ring.
And then they prove (iv) $\implies$ (x): arithmetical ring.
In fact I see that I have typed up some of this already, so if no one else intervenes to give a better answer, perhaps I will take this opportunity to get this transcription done.
Remark: Mr. Bill Dubuque has written several posts here and elsewhere giving much longer lists of equivalent conditions for a domain to be Prüfer.  To the best of my recollection his posts do not (understandably!) contain proofs of all of these implications, but they probably contain references, some of which may be freely accessible online (as MIT is not).

$\newcommand{\aa}{\mathfrak{a}}$
$\newcommand{\bb}{\mathfrak{b}}$
$\newcommand{\c}{\mathfrak{c}}$
$\newcommand{\pp}{\mathfrak{p}}$
Okay, here it goes.  I will insert another condition on an integral domain $R$.
(iii$'$): if $\mathfrak{a},\mathfrak{b},\mathfrak{c}$ are ideals of $R$ with $\mathfrak{a}$ nonzero and finitely generated and such that $\mathfrak{a} \mathfrak{b} \subset \mathfrak{a} \mathfrak{c}$, then $\mathfrak{b} \subset \mathfrak{c}$.
(i) $\implies$ (iii): Let $\mathfrak{a}$, $\mathfrak{b}$, $\mathfrak{c}$ be ideals of a Prüfer domain, with $\mathfrak{a}$ finitely generated and nonzero and such that $\mathfrak{a} \mathfrak{c} = \mathfrak{b} \mathfrak{c}$.  Then since $\mathfrak{a}$ is invertible, multiply both sides by $\mathfrak{a}^{-1}$ to deduce $\mathfrak{b} = \mathfrak{c}$.
(iii) $\implies$ (iii$'$): If $\aa \bb \subset \aa \c$, then $\aa \c = 
\aa \bb + \aa \c = \aa(\bb+ \c)$; cancelling $\aa$ gives $\c = \bb + \c$, so
$\bb \subset \c$.
(iii$'$) $\implies$ (iv): Let $\pp$ be a prime ideal of $R$.  I will make use of the (much easier!) exercise that an integral domain is a valuation ring iff principal ideals are linearly ordered under inclusion.  So it suffices to show that for any $\frac{a}{s}, \frac{b}{t} \in R_{\pp}$, we have either
$(\frac{a}{s}) \subset (\frac{b}{t})$ or $(\frac{b}{t}) \subset (\frac{a}{s})$.  Since
$\frac{1}{s}, \frac{1}{t} \in R^{\times}$, it is equivalent to show that
$(a) \subset (b)$ or $(b) \subset (a)$: for this we may clearly assume $a,b \neq 0$.  Now
$\langle ab \rangle \langle a,b \rangle \subset \langle a^2, b^2 \rangle \langle a,b \rangle$,
so
$ab = xa^2 + yb^2$ for some $x,y \in R$.  Then
$\langle y b \rangle \langle a,b \rangle \subset \langle a \rangle \langle a,b \rangle$,
so
$\langle yb \rangle \subset \langle a \rangle$.
Put
$yb = au$ for some $u \in R$.  Then
$ab = xa^2 + uab$, or $xa^2 = ab(1-u)$.
Case 1: $u \in \pp$.  Then $1-u \notin \pp$, so $b = a\left(\frac{x}{1-u}\right) \in a R_{\pp}$.
Case 2: $u \notin \pp$.  Then $a = b\left(\frac{y}{v} \right) \in b R_{\pp}$.
(iv) $\implies$ (x): We always have the inclusion of ideals
$\iota: (\aa \cap \bb) + (\aa \cap \c) \subset \aa \cap (\bb + \c)$.  Whether a homomorphism of $R$-modules is an isomorphism (hence the identity, in this case) can be checked "locally", i.e., after pushing forward to $R_{\pp}$ for all maximal ideals $\pp$ of $R$.  Thus we reduce to the case of a valuation ring.  But a valuation ring is a uniserial ring (or chain ring) -- i.e., given any two ideals, one contains the other -- and this makes the identity almost trivial to check: e.g. if $\aa \subset \bb \subset \c$, then $\aa \cap (\bb + \c) = \aa = (\aa \cap \bb) + (\aa \cap \c)$.  The other five cases are no harder.
