= Solution
We first justify the <submodularity of matroid rank>, which will also support later arguments. Choose a maximal <matroid independent set> $I_0$ in $A\cap B$, extend it to a maximal <matroid independent set> $I_A$ in $A$, and then to one $I_U$ in $A\cup B$. These extensions exist by successive additions, and their sizes are the respective values of the <rank of a matroid>. No element of $I_U\setminus I_A$ lies in $A$: such an element could have been added to $I_A$, contradicting maximality in $A$. Therefore $I_0\cup(I_U\setminus I_A)$ is an independent <subset> of $B$, giving
$$
r(A\cap B)+r(A\cup B)-r(A)\leq r(B).
$$
Now let $Y\subseteq Z$ and $x\in\sigma(Y)$. If $x\in Z$, then $x\in\sigma(Z)$ by (ii). Otherwise apply <submodularity of matroid rank> to $Y+x$ and $Z$, whose <intersection> is $Y$:
$$
r(Z+x)+r(Y)\leq r(Y+x)+r(Z)=r(Y)+r(Z).
$$
The monotonicity of the <rank of a matroid> supplies the opposite bound $r(Z+x)\geq r(Z)$, so equality holds. Thus $x\in\sigma(Z)$, proving \b[$\sigma(Y)\subseteq\sigma(Z)$].
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