= Solution
A <forcing atom> is a condition $p\in\mathbb P$ such that any two strengthenings of $p$ are <compatible forcing conditions>:
$$
p\text{ is an atom}\ \Longleftrightarrow\ \forall q,r\leq p\ \exists s\in\mathbb P\ (s\leq q,r).
$$
An <atomless forcing order> has no such condition; equivalently,
$$
\boxed{\forall p\in\mathbb P\ \exists q,r\leq p\ (q\perp r).}
$$
The meaning of $q\perp r$ is that there is no common stronger condition. For an arbitrary <partial order>, an atom need not be a minimal element: the definition concerns compatibility below it. This distinction prevents a minimal-element definition from misclassifying an order with descending but mutually compatible conditions.
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