do not need to break any ties at this point, so it has no effect.

Index $3$ is associated with three features. We associate each feature with exactly one of $U_{2}, U_{3},$ and $U_{4}$ (the next three available $U_{k}$). For instance, pair $\{3,4\}$ with $U_{2}$, $\{3\}$ with $U_{3}$, and the other $\{3\}$ with $U_{4}$. Suppose it happens that $U_{3} < U_{2} < U_{4}$. Then the feature $\{3\}$ paired with $U_{3}$ receives label 2 (the next available order-of-appearance label). The feature $\{3,4\}$ receives label 3. And the feature $\{3\}$ paired with $U_{4}$ receives label 4.

Index $4$ has three features, but $\{2,5,4\}$ and $\{3,4\}$ are already labeled. So the only remaining feature, $\{6,4\}$, receives the next available order-of-appearance label: 5. $U_{5}$ is associated with this feature, but since we do not need to break ties here, it has no effect. Indices $5$ and $6$ belong to already-labeled features.

So the features can be listed with order-of-appearance indices as $$\label{eq:ex_A_oa}
    A_{1} = \{2,5,4\}, A_{2} = \{3\}, A_{3} = \{3,4\}, A_{4} = \{3\}, A_{5} = \{6,4\}.$$ Let $\indfao_{n}$ indicate the set of order-of-appearance feature labels for the features to which index $n$ belongs; i.e., if the features are labeled according to order of 