[proofplan]
The first three properties follow directly from the definition of rank as the maximum size of an independent subset. The only substantive point is submodularity. For that, we choose a basis of $A \cap B$, extend it to a basis of $A$, then extend again to a basis of $A \cup B$; the extra elements added in the last extension must lie in $B \setminus A$, and together with the original basis of $A \cap B$ they form an independent subset of $B$. Counting these sets gives the desired inequality.
[/proofplan]
[step:Bound the rank between zero and the cardinality of the set]
Fix a subset $A \subset E$. Since $\varnothing \subset A$ and $\varnothing \in \mathcal I$, the collection of independent subsets of $A$ is nonempty, so $r(A) \ge 0$. Every independent subset $I \subset A$ satisfies $|I| \le |A|$, and therefore the maximum of these cardinalities satisfies
\begin{align*}
0 \le r(A) \le |A|.
\end{align*}
[/step]
[step:Prove monotonicity by enlarging the ambient set]
Let $A, B \subset E$ satisfy $A \subset B$. If $I \subset A$ and $I \in \mathcal I$, then also $I \subset B$. Thus every independent subset counted in the definition of $r(A)$ is also counted in the definition of $r(B)$. Taking maxima over these two collections gives
\begin{align*}
r(A) \le r(B).
\end{align*}
[/step]
[step:Show that adjoining one element increases rank by at most one]
Fix $A \subset E$ and $e \in E$. Let $J \subset A \cup \{e\}$ be an independent subset with $|J| = r(A \cup \{e\})$. Define
\begin{align*}
J_0 := J \cap A.
\end{align*}
Since $\mathcal I$ is hereditary, $J_0 \in \mathcal I$, and since $J_0 \subset A$, we have $|J_0| \le r(A)$. Also $J \setminus J_0 \subset \{e\}$, so $|J| \le |J_0| + 1$. Therefore
\begin{align*}
r(A \cup \{e\}) = |J| \le |J_0| + 1 \le r(A) + 1.
\end{align*}
[guided]
We want to compare the largest independent subset of $A \cup \{e\}$ with the largest independent subset of $A$. Choose an independent subset $J \subset A \cup \{e\}$ whose cardinality realizes the rank:
\begin{align*}
|J| = r(A \cup \{e\}).
\end{align*}
Now remove from $J$ anything that is not already in $A$ by defining
\begin{align*}
J_0 := J \cap A.
\end{align*}
Because $J_0 \subset J$ and $J$ is independent, the hereditary axiom for matroids gives $J_0 \in \mathcal I$. Since $J_0 \subset A$, the definition of $r(A)$ gives
\begin{align*}
|J_0| \le r(A).
\end{align*}
The only possible element of $J$ not contained in $A$ is $e$, because $J \subset A \cup \{e\}$. Hence $J \setminus J_0 \subset \{e\}$, so at most one element was removed:
\begin{align*}
|J| \le |J_0| + 1.
\end{align*}
Combining the two inequalities gives
\begin{align*}
r(A \cup \{e\}) = |J| \le |J_0| + 1 \le r(A) + 1.
\end{align*}
This proves that adjoining a single element can increase rank by no more than one.
[/guided]
[/step]
[step:Extend bases through $A \cap B$, $A$, and $A \cup B$]
Let $I \subset A \cap B$ be a basis of $A \cap B$, meaning that $I \in \mathcal I$, $I \subset A \cap B$, and $|I| = r(A \cap B)$. By the matroid augmentation axiom, extend $I$ to a basis $J \subset A$ of $A$, so that $I \subset J$, $J \in \mathcal I$, and $|J| = r(A)$. Again by augmentation, extend $J$ to a basis $K \subset A \cup B$ of $A \cup B$, so that $J \subset K$, $K \in \mathcal I$, and $|K| = r(A \cup B)$.
[/step]
[step:Locate the new elements of the final basis inside $B \setminus A$]
We claim that $K \setminus J \subset B \setminus A$. Indeed, let $x \in K \setminus J$. Since $K \subset A \cup B$, either $x \in A$ or $x \in B$. If $x \in A$, then $J \cup \{x\} \subset K$, so $J \cup \{x\}$ is independent by heredity. This contradicts the maximality of $J$ as a basis of $A$, because $J \cup \{x\}$ would be an independent subset of $A$ properly containing $J$. Therefore $x \notin A$, and since $x \in A \cup B$, we must have $x \in B$. Thus $x \in B \setminus A$.
[/step]
[step:Count an independent subset of $B$ to obtain submodularity]
Since $I \subset J \subset K$ and $K \setminus J \subset K$, the set
\begin{align*}
L := I \cup (K \setminus J)
\end{align*}
is a subset of $K$. Hence $L \in \mathcal I$ by heredity. Also $I \subset A \cap B \subset B$ and $K \setminus J \subset B \setminus A \subset B$, so $L \subset B$. Therefore $L$ is an independent subset of $B$, and so $|L| \le r(B)$.
Because $I \subset J$ and $K \setminus J$ is disjoint from $J$, the sets $I$ and $K \setminus J$ are disjoint. Thus
\begin{align*}
|L| = |I| + |K \setminus J|.
\end{align*}
Using $I \subset J \subset K$, we also have
\begin{align*}
|K \setminus J| = |K| - |J| = r(A \cup B) - r(A).
\end{align*}
Since $|I| = r(A \cap B)$, the inequality $|L| \le r(B)$ becomes
\begin{align*}
r(A \cap B) + r(A \cup B) - r(A) \le r(B).
\end{align*}
Rearranging gives
\begin{align*}
r(A) + r(B) \ge r(A \cup B) + r(A \cap B).
\end{align*}
This is the submodularity inequality, and the four stated rank properties follow.
[/step]
