I need to prove the following,
If $A$ is a circuit-hyperplane of a matroid $M$, then the relaxation of $M'$ is also a matroid.
So far I have:
Assume $A \in \mathcal C(M)$, since every proper subset of a circuit is independent then, $A-\{x\} \in \mathcal I(M)$. Now let $A-\{x\}= I_1$ and let there be $B_2 \in \mathcal B(M)$.
Hence we have at least two independent sets, $B_2$ and $I_1$, where $|B_2|>|I_1|$. By $\mathcal I3$, there exists $y \in B_2-I_1$ such that $I_1 \cup \{y\} \in \mathcal I(M)$. Additionally, $I_1 \cup \{y\} \in \mathcal B(M)$.
But $I_1 \cup \{y\}= A-\{x\} \cup \{y\}$. Since $A-\{x\} \cup \{y\} \in \mathcal B(M)$ and $\mathcal B(M') = \mathcal B(M)\cup A$, then $\{A-\{x\} \cup \{y\} \in \mathcal B(M')\}$.
Basically I'm trying to prove the strong basis exchange axiom, and I am having some difficulty with the second part of the proof.