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Source. Theorem 12 and its proof, printed pp. 1328–1329 (published PDF).

Statement. Let $\mathcal A=(A_1,\ldots,A_n)$ and $\mathcal B=(B_1,\ldots,B_n)$ be indexed families of subsets of finite $S$. Let $p_i,q_i\ge0$ be integers with

$$\sum_{i=1}^n p_i=\sum_{i=1}^n q_i=N.$$

There is a subset $X\subseteq S$ that is both a $p$-transversal of $\mathcal A$ and a $q$-transversal of $\mathcal B$ if and only if

$$|A(J)\cap B(K)| \ge p(J)+q(K)-N \tag{1}$$

for all $J,K\subseteq[n]$.

Proof. Suppose first that (1) holds. Taking $K=[n]$ gives

$$|A(J)|\ge |A(J)\cap B([n])|\ge p(J),$$

so Theorem 7 gives an $A$ $p$-transversal. Similarly, taking $J=[n]$ gives $|B(K)|\ge q(K)$ and hence a $B$ $q$-transversal. In particular, the replicated transversal matroid $T_p(\mathcal A)$ has rank $N$ and its bases are the $A$ $p$-transversals.

By the rank formula, condition (1) says exactly that

$$r_p(B(K))\ge q(K) \qquad(K\subseteq[n]).$$

Applying Theorem 4 to $\mathcal B$, multiplicity vector $q$, and the matroid $T_p(\mathcal A)$ gives a $B$ $q$-transversal $X$ independent in that matroid. It has

$$|X|=\sum_i q_i=N=r_p(S),$$

so it is a base and hence also an $A$ $p$-transversal.

Conversely, suppose $X$ is both kinds of transversal. It is a base of $T_p(\mathcal A)$ and a $B$ $q$-transversal independent in that matroid. The necessary direction of Theorem 4 gives

$$r_p(B(K))\ge q(K) \qquad(K\subseteq[n]).$$

Applying the rank formula to every $B(K)$ gives (1) for every $J$.

If $n=0$, then $N=0$, both families are empty, and $X=\varnothing$ proves the equivalence. $\square$

The proof establishes individual feasibility before referring to $p$-transversals as bases. It also supplies the cardinality bars and binds the set $X$ omitted in the printed intermediate prose.