how to prove that the dual of a matroid satisfies the exchange property? In the Wikipedia entry for Dual matroid, an early section states

An alternative definition of the dual matroid is that its basis sets are the complements of the basis sets of $M$. The basis exchange axiom, used to define matroids from their bases, is self-complementary, so the dual of a matroid is necessarily a matroid.

There is also a survey paper by James Oxley, a well-known expert on matroids. In particular, on p. 11 of the paper, Oxley states that the dual of a matroid is itself a matroid, but points us to his book for the proof.
I do not have access to the book, and I'm having a hard time seeing why the dual of a matroid satisfies the exchange property in one particular case.
Suppose we have a matroid $M=(S, I)$. Suppose we define $I^\prime$ as the set of all $C^\prime \subseteq S$ such that $S-C^\prime$ contains some maximal subset $C \in I$. Now suppose further that we have $A^\prime \in I^\prime$ and $B^\prime \in I^\prime$, with $|A^\prime| < |B^\prime|$, and $S-A^\prime$ contains some maximal subset $A \in I$, and $S-B^\prime$ contains some maximal subset $B \in I$.
Now if $B^\prime - A^\prime - A \neq \emptyset$, then it is easy to show that for any $x^\prime \in B^\prime - A^\prime - A$, we have 
$$
S - (A^\prime \cup \{x^\prime\} ) \supseteq A,
$$
resulting in $A^\prime \cup \{x^\prime\} \in I^\prime$, and hence the exchange property is satisfied.
Next, if $B^\prime - A^\prime - A = \emptyset$, and $B\cap H=\emptyset$, where $H \equiv (S - B^\prime) - (S - A^\prime - A)$, then it is easy to show that for any $x^\prime \in B^\prime - A^\prime$, we have
$$
S - (A^\prime \cup \{x^\prime\} ) \supseteq B,
$$
and hence the exchange property is again satisfied.
I am stumped regarding the remaining case: if $B^\prime - A^\prime - A = \emptyset$, but $B\cap H \neq \emptyset$. In this case, it seems that for any $x^\prime \in B^\prime - A^\prime$, we are certain that $S - (A^\prime \cup \{x^\prime\} )$ does not contain fully either $A$ or $B$. Hence, we must prove that in this situation, there exists some $x^\prime \in B^\prime - A^\prime$ such that $S - (A^\prime \cup \{x^\prime\} )$ contains some other maximal subset $C \in I$. I have no clue how to do this.
 A: I will alter your notation a bit. Let $M = (S, \mathcal I)$ be a matroid with a set of bases $\mathcal B$, and consider its dual $M^* = (S,\mathcal I^*)$ with $\mathcal I^* = \{C\subseteq S: \exists B_C \in \mathcal B \text{ s.t. }C\cap B_C=\emptyset\}$. We wish to prove that the independent set exchange property holds for $M^*$.
Consider any two sets $X,Y \in \mathcal I^*$ with $|X| > |Y|$. Then $\exists B_X, B_Y \in \mathcal B$ such that $B_X \cap X = \emptyset$ $B_Y \cap Y = \emptyset$. Furthermore, observe that $B_X \cap Y\subseteq Y\setminus X$ and that $(B_X\setminus Y) \cap (X\setminus Y) = \emptyset$. We know that $\exists B\in \mathcal B$ such that $B_X\setminus Y\subseteq B\subseteq S\setminus Y$ because the Steinitz exchange lemma (see below) guarantees its existence. Now, suppose for contradiction that $X\setminus Y\subseteq B$. Then we would have:
\begin{align}
|B_X| &= |B_X\setminus Y| + |B_X\cap Y| \\
&\leq |B_X\setminus Y| + |Y\setminus X| \\
&<|B_X\setminus Y| + |X\setminus Y| \qquad \left(\text{because }|X|>|Y|\right)\\
&\leq |B| \qquad \left(\text{because }(B_X\setminus Y) \cap (X\setminus Y) = \emptyset\right)
\end{align}
This is clearly absurd: $B,B_X \in \mathcal B$, yet $|B_X| < |B|$! It follows that $X\setminus Y\nsubseteq B$. As a result, there must be some $x\in X\setminus Y$ such that $x\notin B$. For such an $x$, we find that $Y\cup\{x\} \in \mathcal I^*$! Hence, the exchange property holds for $M^*$. It does not take much more work to prove that $M^*$ is a matroid.


Steinitz Exchange Lemma: Let $\mathcal B$ be the set of bases of a matroid $M=(E,\mathcal I)$. $\forall A,B \in \mathcal B$ and $a \in A\setminus B$, $\exists b \in B \setminus A$ such that $(A\setminus \{a\})\cup \{b\} \in \mathcal B$.

Proof: Because $A$ is a basis of $M$, $A\setminus \{a\} \in \mathcal I$. By the augmentation property of matroids, $\exists b \in B\setminus (A\setminus \{a\})$ such that $(A\setminus \{a\})\cup \{b\} \in \mathcal I$. Since $a \in A\setminus B$, we know that $b \in B\setminus A$. In addition, $|(A\setminus \{a\})\cup \{b\}| = |A| = |B|$, so we have $(A\setminus \{a\})\cup \{b\} \in \mathcal B$. $\square$
