= Solution
The two-argument version of <Fn forcing> consists of finite <partial functions>:
$$
\boxed{\operatorname{Fn}(I,J)=\{p:p\text{ is a function},\ \operatorname{dom}p\subseteq I,\ \operatorname{ran}p\subseteq J,\ |\operatorname{dom}p|<\omega\},\qquad q\leq p\ \Longleftrightarrow\ q\supseteq p.}
$$
Thus a stronger condition specifies more values. The empty <partial function> is the greatest condition. Two conditions are <compatible forcing conditions> exactly when they agree on their common domain; if they do, their <set union> is a common strengthening. This is $\operatorname{Fn}(I,J,\omega)$ in the three-argument convention for <Fn forcing>. If $I$ or $J$ is empty, only the empty condition exists. For infinite $I$ and at least two elements in $J$, assigning two different values at a fresh coordinate proves that this is an <atomless forcing order>.
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