Why is this characterization of matroids correct? Let $E$ be a finite set with a collection of subsets $\mathcal{I}$ which form an independence complex/abstract simplicial complex. (I.e. a non-empty descending family of sets.)
We require an additional condition to characterize the independent sets belonging to a matroid.

Question: Why is this unconventional axiom for the independent sets of a matroid:

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*If $D_1, D_2 \not\in \mathcal{I}$, but $D_1 \cap D_2 \in \mathcal{I}$, then for every $e \in E$, $(D_1 \cup D_2) \setminus \{ e \} \not\in \mathcal{I}$.

equivalent to more conventional/common axioms for the independent sets belong to a matroid?

Examples of such more conventional axioms include but are not limited to:

*

*For every $I_1, I_2 \in \mathcal{I}$, if $|I_1| < |I_2|$, then there exists $x \in I_2 \setminus I_1$ such that $I_1 \cup \{ x \} \in \mathcal{I}$.

*For every $I_1, I_2 \in \mathcal{I}$, if $|I_2| = |I_1| + 1$, then there exists $x \in I_2 \setminus I_1$ such that $I_1 \cup \{ x \} \in \mathcal{I}$.

*For all $E' \subseteq E$, the maximal independent subsets of $E$ are equicardinal (pure subcomplex).

 A: First suppose $E$ and $\mathcal{I}$ satisfy the "conventional" axioms. We'll prove the contrapositive of the "unconventional" axiom. (We take it as given that the "conventional" axioms are equivalent to each other: that seems easy to check, and from the original question it seems to be known and accepted; the question is just about equivalence of the "unconventional" axiom.)
Let $D_1, D_2 \subseteq E$ such that $D_1 \cap D_2 \in \mathcal{I}$. Suppose there exists an $e \in E$ such that $(D_1 \cup D_2) \setminus \{e\} \in \mathcal{I}$. We will prove that $D_1 \in \mathcal{I}$ or $D_2 \in \mathcal{I}$. By the "conventional" axioms we can keep adding elements of $(D_1 \cup D_2) \setminus \{e\}$ to $D_1 \cap D_2$ until we have an independent set $D'$ that (1) is contained in $D_1 \cup D_2$ (since those are the only elements we are adding), (2) contains $D_1 \cap D_2$ (since we start with that), (3) has the same size as $(D_1 \cup D_2) \setminus \{e\}$. This independent set $D' = (D_1 \cup D_2) \setminus \{e'\}$ where $e' \notin D_1 \cap D_2$. Without loss of generality $e' \notin D_1$; then $D_1 \subseteq D' \in \mathcal{I}$, so $D_1 \in \mathcal{I}$.
Now conversely suppose that $E$ and $\mathcal{I}$ satisfy the "unconventional" axiom. We'll prove the "conventional" axioms.
Say $I_1,I_2 \in \mathcal{I}$ with $|I_2| = |I_1|+1$. If $|I_2\setminus I_1| = 1$, then $I_1 \subseteq I_2$; say $x$ is the one element of $I_2 \setminus I_1$; then $I_1 \cup \{x\} = I_2$, so $I_1 \cup \{x\} \in \mathcal{I}$.
Next suppose $|I_2 \setminus I_1| = 2$. Write $I_2 \setminus I_1 = \{x_1,x_2\}$. We also can write $I_1 \setminus I_2 = \{y\}$. Now let $D_1 = I_1 \cup \{x_1\}$ and $D_2 = I_1 \cup \{x_2\}$. We have $D_1 \cap D_2 = I_1 \in \mathcal{I}$, and also $(D_1 \cup D_2) \setminus \{y\} = I_2 \in \mathcal{I}$. By the contrapositive of the "unconventional" axiom, at least one of $D_1$ or $D_2$ is in $\mathcal{I}$. This proves the conclusion of the "conventional" axiom in this case.
Next suppose $|I_2 \setminus I_1| = 3$. Write $I_2 = (I_1 \cap I_2) \cup \{x_1,x_2,x_3\}$ and $I_1 = (I_1 \cap I_2) \cup \{y_1,y_2\}$. ... I'm stuck here. 
