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Source. Theorem 2d, printed p. 152 (published PDF).

Statement. With pairwise disjoint independent $J_1,\ldots,J_k$ and $E'=E\setminus\bigcup_iJ_i$ as above, pairwise disjoint bases $B_i$ of $M$ satisfying $J_i\subseteq B_i$ exist if and only if

$$|A|\ge\sum_{i=1}^k \bigl(r(E)-r((E'\setminus A)\cup J_i)\bigr) \qquad(A\subseteq E'). \tag{1}$$

Proof. If such bases exist, $B_i\setminus A$ is an independent subset of $(E'\setminus A)\cup J_i$, since it cannot contain any other seed. Hence

$$|B_i\cap A|\ge r(E)-r((E'\setminus A)\cup J_i).$$

Summing inside $A$ proves necessity.

For sufficiency, first apply (1) to $A=\varnothing$. Each summand $r(E)-r(E'\cup J_i)$ is nonnegative by rank monotonicity, so their sum can be at most zero only if

$$r(E'\cup J_i)=r(E)\qquad(1\le i\le k). \tag{2}$$

Thus the rank obstruction must be checked before constructing a packing of full bases.

As in Theorem 1d, contract $J_i$ and restrict to $E'$. The resulting matroid $M_i$ has

$$r_i(A)=r(A\cup J_i)-r(J_i),\qquad r_i(E')=r(E)-r(J_i),$$

where the second equality uses (2). Substituting these expressions in Theorem 2c on $E'$ gives exactly (1). It yields disjoint bases $K_i$ of $M_i$. Each $B_i=K_i\cup J_i$ is independent in $M$ and has $r(E)-r(J_i)+r(J_i)=r(E)$ elements, so is a base. The unions are pairwise disjoint because the $K_i$ lie in $E'$ and the seeds were disjoint. This proves sufficiency, including $E'=\varnothing$ and rank zero. $\square$