Solution
= Solution
Write $r(Y)=r$. Since $y\notin\sigma(Y)$, the one-element increment bound gives $r(Y+y)=r+1$. Since $y\in\sigma(Y+x)$,
$$
r(Y+x+y)=r(Y+x)\leq r+1.
$$
But $Y+y\subseteq Y+x+y$ gives the reverse lower bound $r(Y+x+y)\geq r(Y+y)=r+1$. Hence $r(Y+x+y)=r(Y+y)$, which is precisely \b[$x\in\sigma(Y+y)$]. This is the <matroid closure exchange property>; it follows from one-element <rank of a matroid> increments and does not require exchanging an entire <matroid basis>.