Is every prime element of a commutative ring "veryprime"? Let $R$ denote a commutative ring.
Define a function $$\| : R \times R \rightarrow \mathbb{N} \cup \{\infty\}$$
such that $a \| b$ is the number of times $a$ divides $b$ (and include $0$ in $\mathbb{N}$ to allow for the case where $a$ does not divide $b$). Explicitly:


*

*$a \| b$ is the maximum $k \in \mathbb{N}$ satisfying $a^k \mid b$, as long as the set of all $k \in \mathbb{N}$ such that $a^k \mid b$ is bounded.

*$a \| b=\infty$ otherwise.


Then in general, we have
$$d \| ab \geq (d \| a) + (d \| b).$$
Call $p \in R$ veryprime iff $p$ is a non-zero, non-unit that satisfies
$$p \| ab = (p \| a) + (p \| b).$$
It follows that if $p$ is veryprime, then $p$ is prime. (Proof. Suppose that $p$ is veryprime and that $p \mid ab$. Then $(p \| ab) \geq 1$. So $(p \| a) + (p \| b) \geq 1$. Hence $(p \| a) \geq 1$ or $(p \| b) \geq 1$. Hence $p \mid a$ or $p \mid b$.)

Question. Is every prime element of a commutative ring necessarily veryprime? If not, is veryprime equivalent to a more familiar condition?

 A: The ring $\mathbb{Z}/4\mathbb{Z} $ provides a counterexample. 
The  element $2$ in that ring is prime, and
 $2 \mid 2$ while $2^2 \nmid 2$ yet $2^n \mid 2^2 = 0$ for every $n$.
If one excludes zero-divisors, things should work by an inductive argument; it just needs to be possible to cancel by prime elements.
A: Let $p$ be an element with $p\mid ab\implies p\mid a\lor p\mid b$ and that is not a divisor of zero. 
Assume $p\|ab\ne p\|a+p\|b$. Then certainly $p\|a$ and $p\|b$ are both finite. So write $a=p^ka'$, $b=p^mb'$ with $k,m\in\mathbb N_0$ and $p\nmid a'$, $p\nmid b'$.
Then $ab=p^{k+m}a'b'$ and $p\|ab\ne k+m$ means that $p^{k+m+1} \mid ab$, say $ab=p^{k+m+1}c$.
Then $$p^{k+m}(pc-a'b')=0. $$
As $p$ is not a divisor of zero, we conclude $pc=a'b'$, i.e., $p\mid a'b'$ contradicting $p\nmid a'$, $p\nmid b'$.
We conclude that $p\|ab= p\|a+p\|b$.
Now that I've gotten this far in writing this up, I see that quid presented the counterexaple suggested by this finding. At least this elaborates on quid's closing remark. :) 
