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Source. Theorem 5 and its proof, printed pp. 1325–1326 (published PDF).

Statement. Let $(S,M)$ be a finite matroid with rank $r$, let $\mathcal A=(A_i)_{i\in I}$ be a finite indexed family, and let $k,t\ge0$ be integers. There is a $k$-transversal $X$ of $\mathcal A$ with $r(X)\ge t$ if and only if, for every $J\subseteq I$,

$$k|A(J)|\ge |J|, \tag{6}$$

and

$$r(A(J))\ge |J|+t-|I|. \tag{7}$$

Proof. First take $k=0$. If $I\ne\varnothing$, condition (6) fails on a singleton, and no representative assignment has multiplicity at most zero. If $I=\varnothing$, the only support is $\varnothing$ and has rank zero; condition (7) at $J=\varnothing$ is $0\ge t$, so both sides hold exactly when $t=0$.

Now suppose $k\ge1$. Replace every $x\in S$ by its $k$ labeled copies and put

$$A_i^k=A_i\times[k]\subseteq S^k.$$

By the lift in the definitions, $k$-transversals of $\mathcal A$ are exactly projections of full transversals of $(A_i^k)_{i\in I}$. By Theorem 6,

$$r_{M^k}(Z)=r_M(\pi(Z)) \qquad(Z\subseteq S^k). \tag{8}$$

Suppose first that a $k$-transversal $X$ of rank at least $t$ exists, and lift a witnessing representative assignment to a full transversal $Z$ of $(A_i^k)$. Hall's necessary condition gives

$$\left|\bigcup_{i\in J}A_i^k\right| =k|A(J)|\ge |J|,$$

which is (6). The necessary direction of Perfect's criterion applied in $M^k$ gives

$$r_{M^k}\left(\bigcup_{i\in J}A_i^k\right) \ge |J|+t-|I|.$$

Using (8), its left side is $r_M(A(J))$, proving (7).

Conversely, suppose (6) and (7) hold. Equation (6) is exactly Hall's condition for the copied family $(A_i^k)$, so that family has a full transversal. Equations (7) and (8), together with Perfect's criterion, give a partial-transversal range $Y$ of $(A_i^k)$ satisfying $r_{M^k}(Y)\ge t$. Since the copied family has a full transversal, transversal augmentation extends $Y$ as a set to a full-transversal range $Z$. Rank is monotone, so $r_{M^k}(Z)\ge t$. Its projection is a $k$-transversal $X$, and (8) gives $r_M(X)\ge t$. This proves sufficiency. $\square$

The proof is relative to finite Rado through Perfect's corollary. Hall's theorem and all copied-ground-set identities are linked or proved explicitly.