The typical definition of a circuit matroid is as follows. A matroid $M$ consists of a finite set of elements $E(M)$ along with a collection of non-empty subsets $C(M)\subseteq 2^M$ (called circuits) that satisfies the following properties:
1) If $X, Y \in C(M)$ and $X \subseteq Y$ then $X = Y$ (i.e. no circuit is properly contained in another circuit).
2) If $X, Y \in C(M)$ and $X \neq Y$, then for any $a \in X \cap Y$ there exists a circuit $Z \in C(M)$ such that $Z \subseteq (X \cup Y) - a$.
I recently came across a paper that replaced the second axiom with a stronger version:
2) If $X, Y \in C(M)$ and $X \neq Y$, then for any $a \in X \cap Y$ and for any $b \in X\setminus Y$ there exists a circuit $Z \in C(M)$ such that $Z \subseteq (X \cup Y) - a$ and such that $b \in Z$.
Based on the presentation in the paper, these definitions of matroid seem to be equivalent. So far, I've been unable to prove to that the first set of axioms implies the second. Any ideas?
I think I found a proof. By strong induction on size of $|X \cup Y|$.
Base case: $|X \cup Y| = 3$ (It is impossible for $|X \cup Y| < 3$ if they satisfy axiom 1 and weak axiom 2). In this case, $X = \{a,b\}$ and $Y= \{a,d\}$, and it is clear that the circuit $Z \subseteq (X\cup Y)-a$ must be the set $\{b,d\}$, which contains the only element of $X \setminus Y$.
Now, assume that weak set of axioms implies the strong set of axioms whenever $|X \cup Y| < k$ for some $k$. Let there be some $X,Y \in C(M)$ with $X \neq Y$ where $|X \cup Y| = k$, and let there be $a \in X \cap Y$ and $b \in X \setminus Y$. By the weak axiom, there exists a circuit $Z \subseteq (X \cup Y) -a$. If $b \in Z$, then we are done, so consider the case that $b \notin Z$.
Since $Z$ is not a proper subset of $X$ (axiom 1), then there exists some $d \in Z$ that is not in $X$. This element $d$ must be in $Y$. Thus, $d \in Z \cap Y$. Note that $a \in Y \setminus Z$. Further note that $(Z \cup Y) \subseteq (X \cup Y) -b$ so that $|Z \cup Y| < |X \cup Y|$, so that by induction we can apply the strong axiom 2 to $Y$, $Z$, $d$ and $a$ instead of $X$, $Y$, $a$ and $b$. Then, there is a circuit $W \subseteq (Z \cup Y) - d$ such that $a \in W$.
Since $b \notin Y$ and $b \notin Z$, then $b \notin W$. Thus, $b \in X \setminus W$. Note that $a \in W \cap X$, and note that $(X \cup W) \subseteq (X \cup Y) -d$ (since $W \subseteq (Z \cup Y) - d$ and $Z \cup Y \subseteq X \cup Y$) so that $|X \cup W| < |X \cup Y|$. Then, by induction, we can apply the strong axiom 2 to $W$ instead of $Y$, which shows that there is a circuit $Z' \subseteq (X \cup W) - a$ such that $b \in Z'$. Since $X \cup W \subseteq X \cup Y$, this completes the proof.