Abelian category in which every double chain is stationary, is an AB5 category? In studying to write an expository paper in representation theory,
I am reading Abelian Categories with Applications to Rings and Modules
 by Popescu and I have not been able to figure out something which
appears to go without saying in the book.
In Section 5.1, Theorem 1.3 (Azumaya) says that, in an AB5 category, decomposition
of an object into indecomposable objects with local endomorphism rings
is unique in the Krull-Remak-Schmidt-Azumaya sense (i.e. up to reordering
and isomorphism).
The following characterization of the condition AB5 is used in the
proof (at the moment, I do not have the original book, but only the
notes I have made while reading, so I might be using a different notation).
Let $\mathscr{A}$ be an abelian category which has arbitrary coproducts.
Let $\left\{ A_{i}\right\} _{i\in I}$ be a set of objects of $\mathscr{A}$
and let 
$$
\iota_{i}^{I}:A_{i}\to\coprod_{i\in I}A_{i}
$$
denote the coprojection for $i\in I$. For
any finite subset $F$ of the set $I$, let $\iota_{f}^{F}$ denote
the coprojection of $A_{f}$ into $\coprod_{f\in F}A_{f}$, and let
$A_{F}$ denote the source of the image (that is the image itself
if an image is thought of as an object and not an arrow) of the canonical
arrow 
$$
u_{F}:\coprod_{f\in F}A_{f}\to\coprod_{i\in I}A_{i}
$$
defined such that 
$$
u_{F}\iota_{f}^{F}=\iota_{f}^{I}
$$
for all $f\in F$. If for every subobject $A$ of the object $\coprod_{i\in I}A_{i}$
holds the equality 
$$
A=\sum_{F\in T}\left(A\cap A_{F}\right),
$$
with $T$ denoting the set of all finite subsets of $I$, it is said
that $\mathscr{A}$ verifies the condition AB5 or that it is an AB5
category.
In the same section, Theorem 1.4 says that, in an Abelian category
in which every double chain is stationary, every object has a unique
(in the Krull-Remak-Schmidt-Azumaya sense) decomposition into finitely many
indecomposable objects with local endomorphism rings.
By double chain, one means
$$
\left(A_{n}{\rightleftarrows}A_{n+1}\right)_{n\in\mathbb{N}},
$$
with $A_{n}$ being objects, $i_{n}:A_{n+1}\to A_{n}$ being monomophisms and $p_{n}:A_{n}\to A_{n+1}$
being epimorphisms. The chain is stationary if there is $n_{0}\in\mathbb{N}$
such that $i_{n}$ and $p_{n}$ are isomorphisms for all $n\geq n_{0}$.
For the proof of the uniqueness part of Theorem 1.4, it is said in
the book that Theorem 1.3 is used. No proof is given, though.
My question: How does one prove that the condition for using Theorem
1.3 is satisfied in Theorem 1.4, namely that a category described
in Theorem 1.4 indeed verifies the aforementioned condition AB5 needed
for Theorem 1.3?
Thank you for your time and attention!
 A: The point seems to be that the double chain condition is strictly stronger than AB5. More precisely, I claim that

An abelian category satisfying the double chain stabilizing condition admits only finite coproducts.

For such categories, note that AB5 is trivially satisfied since then, with the above notation, $I \in T$ and thus, $A=\sum_{F \in T}(A \cap A_F)$ trivially holds.
To show the claim, suppose there is an infinite collection of nonzero objects $\{A_i\}_{i \in I}$ such that $\coprod_{i \in I}A_i$ exists. Fix a countable subset $\{i_n \;|\; n \in \mathbb{Z}_{>0}\}$ of $I$.
Set $B_0=\coprod_{i \in I}A_i$, and let $B_1=\mathrm{coker}(\iota_{i_1}:A_{i_1} \rightarrow\coprod_{i \in I}A_i)$. This is just $\coprod_{i \in I \setminus\{i_1\}} A_i$
(when someone brings morphisms $A_i \rightarrow X, \; i \neq i_1,$ add the zero morphism $A_{i_1} \rightarrow X$ to get a morphism $\coprod_{i \in I}A_i \rightarrow X$ from the universal property, and this in turn factorizes through a map $B_1 \rightarrow X$; the point of this awkwad description is to make sure that we do not use any coproducts that might not exist). Now it's easy to see that $B_0= A_{i_1} \oplus B_1,$ so one has the canonical maps $B_0 \rightleftarrows B_1$ from the biproduct description. Now repeat the process with $B_1$ to get $B_1 \rightleftarrows B_2=\coprod_{i \in I \setminus\{i_1, i_2\}}A_i$, and so on. 
In the end, one has a double chain $(B_n)_n$ that clearly does not stabilize (kernel of each $B_n \rightarrow B_{n+1}$ is $A_{i_{n+1}} \neq 0$). So this contradicts the double chain stabilizing assumption.
