[proofplan] We prove that two bases cannot have different cardinalities. If one basis were smaller than the other, the matroid exchange axiom would add an element from the larger basis to the smaller independent set while preserving independence. This contradicts the defining maximality of a basis as a maximal independent set. [/proofplan]

[step:Assume one basis has smaller cardinality and apply exchange]Suppose, for contradiction, that $|B_1| < |B_2|$. Since $B_1$ and $B_2$ are bases, both sets belong to $\mathcal I$. By the 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 $B_1 \cup \{e\} \in \mathcal I$.[/step]

[guided]Assume, for contradiction, that the two bases have different sizes in the direction $|B_1| < |B_2|$. Because $B_1$ and $B_2$ are bases of $M$, they are independent sets, so $B_1 \in \mathcal I$ and $B_2 \in \mathcal I$. The exchange axiom for a finite matroid says that whenever $I_1, I_2 \in \mathcal I$ satisfy $|I_1| < |I_2|$, there is an element of the larger independent set outside the smaller one that can be added to the smaller set while preserving independence. Applying this axiom with $I_1 := B_1$ and $I_2 := B_2$, we obtain an element $e \in B_2 \setminus B_1$ such that \begin{align*} B_1 \cup \{e\} \in \mathcal I. \end{align*} This is the key use of the matroid structure: the larger independent set supplies an element that genuinely extends the smaller independent set.[/guided]

[step:Contradict maximality of the smaller basis] The element $e \in B_2 \setminus B_1$ satisfies $e \notin B_1$, so $B_1 \subsetneq B_1 \cup \{e\}$. The preceding step gives $B_1 \cup \{e\} \in \mathcal I$, which means $B_1$ is properly contained in a larger independent subset of $E$. This contradicts the assumption that $B_1$ is a basis, since a basis is a maximal independent subset of $E$. [/step]

[step:Exclude the opposite inequality and conclude equality] The contradiction shows that $|B_1| < |B_2|$ is impossible. Interchanging the roles of $B_1$ and $B_2$ gives the same argument, so $|B_2| < |B_1|$ is also impossible. Since $E$ is finite, the cardinalities $|B_1|$ and $|B_2|$ are natural numbers, and exactly one of $|B_1| < |B_2|$, $|B_1| = |B_2|$, or $|B_2| < |B_1|$ holds. Therefore $|B_1| = |B_2|$. [/step]