[proofplan] Remove the chosen element $e$ from the basis $B_1$, producing an independent set whose cardinality is one less than that of the basis $B_2$. The matroid augmentation axiom then supplies an element of $B_2$ that can be added while preserving independence. Since $e \notin B_2$, this augmenting element cannot be the removed element $e$, so it lies in $B_2 \setminus B_1$. The resulting independent set has the same cardinality as a basis, hence must itself be a basis. [/proofplan] [step:Remove $e$ from $B_1$ to obtain an independent set smaller than $B_2$] Fix an element $e \in B_1 \setminus B_2$. Define \begin{align*} A := B_1 \setminus \{e\}. \end{align*} Since $B_1 \in \mathcal I$ and $\mathcal I$ is hereditary, $A \in \mathcal I$. Because $e \in B_1$, we have \begin{align*} |A| = |B_1| - 1. \end{align*} Since $B_1$ and $B_2$ are bases of the finite matroid $M$, they have the same cardinality, so \begin{align*} |A| = |B_2| - 1 < |B_2|. \end{align*} [/step] [step:Use augmentation to add an element from $B_2$] Apply the augmentation axiom to the independent sets $A$ and $B_2$. Since $A,B_2 \in \mathcal I$ and $|A| < |B_2|$, there exists an element \begin{align*} f \in B_2 \setminus A \end{align*} such that \begin{align*} A \cup \{f\} \in \mathcal I. \end{align*} [guided] We now use the only genuinely matroid-specific ingredient: the augmentation axiom. The two independent sets to compare are \begin{align*} A := B_1 \setminus \{e\} \end{align*} and $B_2$. The set $A$ is independent because it is a subset of the independent set $B_1$, and $B_2$ is independent because every basis is independent. We also proved \begin{align*} |A| < |B_2|. \end{align*} Therefore the augmentation axiom applies and gives an element \begin{align*} f \in B_2 \setminus A \end{align*} such that \begin{align*} A \cup \{f\} \in \mathcal I. \end{align*} This is exactly the exchange move we want: after deleting $e$ from $B_1$, we can add some element from $B_2$ without destroying independence. [/guided] [/step] [step:Show the augmenting element lies outside $B_1$] We claim that $f \in B_2 \setminus B_1$. From the previous step, $f \in B_2$. Also $f \notin A = B_1 \setminus \{e\}$. Since $e \notin B_2$ and $f \in B_2$, we have $f \neq e$. Thus $f$ is neither in $B_1 \setminus \{e\}$ nor equal to $e$, so $f \notin B_1$. Hence \begin{align*} f \in B_2 \setminus B_1. \end{align*} [/step] [step:Conclude the exchanged set is a basis] Define \begin{align*} B' := (B_1 \setminus \{e\}) \cup \{f\}. \end{align*} By construction, $B' = A \cup \{f\} \in \mathcal I$. Since $f \notin A$, its cardinality is \begin{align*} |B'| = |A| + 1 = |B_1|. \end{align*} Thus $B'$ is an independent set with the same cardinality as the basis $B_1$. It remains only to justify that such an independent set is a basis. Suppose, for contradiction, that $B'$ is not a basis. Since a basis is a maximal independent set, there exists an element $g \in E \setminus B'$ such that \begin{align*} B' \cup \{g\} \in \mathcal I. \end{align*} Then \begin{align*} |B_1| = |B'| < |B' \cup \{g\}|. \end{align*} Applying augmentation to the independent sets $B_1$ and $B' \cup \{g\}$ gives an element $h \in (B' \cup \{g\}) \setminus B_1$ such that $B_1 \cup \{h\} \in \mathcal I$, contradicting the maximality of the basis $B_1$. Therefore $B'$ is a basis, which is the desired conclusion. [/step]