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Source. Example 3 and Theorem 11, printed pp. 1327–1328 (published PDF).

Statement. Let $(E_d)_{d\in D}$ be a finite partition of $S$, and let $a_d\ge0$ be integers. There is a $p$-transversal $X$ of $\mathcal A=(A_i)_{i\in I}$ satisfying

$$|X\cap E_d|\le a_d \qquad(d\in D) \tag{1}$$

if and only if, for every $J\subseteq I$,

$$|A(J)| -\sum_{d\in D}\bigl(|A(J)\cap E_d|-a_d\bigr)^+ \ge p(J), \tag{2}$$

where $x^+=\max\{x,0\}$.

Proof. Let $\mathcal M$ consist of the subsets $Y\subseteq S$ satisfying

$$|Y\cap E_d|\le a_d \qquad(d\in D).$$

This is a matroid. It is hereditary and contains the empty set. If $Y,Z\in\mathcal M$ and $|Y|<|Z|$, then for some block $E_d$ one has $|Y\cap E_d|<|Z\cap E_d|$. Choose $z\in(Z\cap E_d)\setminus Y$. Since $|Y\cap E_d|<|Z\cap E_d|\le a_d$, the set $Y\cup\{z\}$ remains in $\mathcal M$. Thus the augmentation axiom holds.

For any $W\subseteq S$, a largest independent subset takes $\min\{|W\cap E_d|,a_d\}$ elements from each block. Hence

$$\begin{aligned} r_{\mathcal M}(W) &=\sum_{d\in D}\min\{|W\cap E_d|,a_d\}\\ &=|W|-\sum_{d\in D}\bigl(|W\cap E_d|-a_d\bigr)^+. \end{aligned} \tag{3}$$

A $p$-transversal satisfies (1) exactly when it is independent in $\mathcal M$. Applying Theorem 4 and substituting $W=A(J)$ into (3) gives precisely (2). $\square$

The direct independence definition works even when $a_d>|E_d|$. The source's exact-capacity base description does not cover that endpoint, and its displayed theorem omits the cardinality bars inside the positive part; both points are recorded in source corrections.