What is the condition for "NOT" irreducibility of a non-zero, non-unit element $a$ in an integral domain $R$?

Question

Actually I got a confusion when I am learning the proof the theorem "An integral domain $R$ with ACCP is FD(Factorization Domain)."
The begining of the proof of the theorem goes as follows-
Suppose on contrary, $D$, which is an ID with ACCP, is not FD. Then $\exists$ a non-zero, non-unit element $a\in R$ such that a does not have factorization. Then $a$ is not irreducible. And so $a=a_1 a_2$, where $a_1, a_2\in R$ are non-trivial factors of $a$. At least one of $a_1, a_2$ must not have a factorization, otherwise factorization of $a_1$ and factorization of $a_2$ will produce factorization of $a$. ...(and so on)
But I cannot understand why "$a$ is not irreducible implies the next statements(bolded lines)
I know that the as per definition of irreducible elements-
A non-zero, non-unit element $a\in R$ is said to be irreducible if whenever $a=bc; b,c \in R$ it implies either $b$ is unit or $c$ is unit.
So I think negetion of the above definition would be-A non-zero, non-unit element $a\in R$ is said to be NOT irreducible if $a$ can be expressed as $a=bc$ where none of $b,c$ is unit.
Can anybody clear my confusions?

Answer 1

Yes, "not irreducible", or "reducible", means that it can be expressed as the product of two non-trival (i.e., non-unit) elements. For instance, 6 is reducible in the ring of integers, because neither 2 nor 3 are units, and 6 = 2*3.