[proofplan] We prove that neither basis can have strictly smaller cardinality than the other. If $|B_1|<|B_2|$, the independent-set exchange axiom produces an element of $B_2\setminus B_1$ that can be added to $B_1$ while preserving independence. This contradicts the defining maximality of $B_1$ as a basis. Reversing the roles of the two bases rules out the opposite strict inequality, so finite cardinal comparison gives equality. [/proofplan] [step:Rule out that $B_1$ has smaller cardinality than $B_2$] Assume, for contradiction, that $|B_1|<|B_2|$. Since $B_1$ and $B_2$ are bases, they are independent sets, so $B_1,B_2\in\mathcal I$. By the independent-set exchange axiom for the matroid $M$, applied to the independent sets $B_1$ and $B_2$, there exists an element $e\in B_2\setminus B_1$ such that \begin{align*} B_1\cup\{e\}\in\mathcal I. \end{align*} Because $e\notin B_1$, the inclusion $B_1\subsetneq B_1\cup\{e\}$ is proper. Thus $B_1$ is properly contained in a larger independent set, contradicting that $B_1$ is maximal under inclusion among elements of $\mathcal I$. Therefore $|B_1|<|B_2|$ is impossible. [guided] Suppose, aiming for a contradiction, that the first basis is strictly smaller: \begin{align*} |B_1|<|B_2|. \end{align*} The reason this inequality is useful is that the matroid exchange axiom compares two independent sets of different finite sizes. Since $B_1$ and $B_2$ are bases, they are in particular independent sets, so $B_1,B_2\in\mathcal I$. Applying the independent-set exchange axiom to these two independent sets gives an element \begin{align*} e\in B_2\setminus B_1 \end{align*} such that \begin{align*} B_1\cup\{e\}\in\mathcal I. \end{align*} Now $e\notin B_1$, so adding $e$ genuinely enlarges $B_1$: \begin{align*} B_1\subsetneq B_1\cup\{e\}. \end{align*} But the enlarged set is still independent. This is exactly what the definition of a basis forbids: a basis is a maximal independent set under inclusion, meaning there is no independent set in $\mathcal I$ that properly contains it. Hence the assumption $|B_1|<|B_2|$ contradicts the fact that $B_1$ is a basis. [/guided] [/step] [step:Reverse the roles of the bases and conclude equality] The same argument with $B_1$ and $B_2$ interchanged shows that $|B_2|<|B_1|$ is also impossible. Since $E$ is finite, the cardinalities $|B_1|$ and $|B_2|$ are natural numbers. Exactly one of \begin{align*} |B_1|<|B_2|,\qquad |B_1|=|B_2|,\qquad |B_2|<|B_1| \end{align*} holds. The two strict alternatives have been ruled out, so \begin{align*} |B_1|=|B_2|. \end{align*} This proves that all bases of $M$ have the same cardinality. [/step]