A question concerning supremum in a Boolean ring

Question

In a Boolean $\sigma$-ring, is it true that for every sequence {$x_n$} and y, we have $supx_n\cdot y = sup(x_n\cdot y)$ ?
Background:
I was reading 'Measure Theory' written by Paul R.Halmos (in the book series Graduate Texts in Mathematics), and encountered this question as an exercise.
A Boolean ring is defined as a commutative ring (maybe without a unit) such that every element of it is idempotent, i.e. for every $x$,$x^{2}=x$. A typical example of a Boolean ring is a class of subsets of a given set $X$(denoted by $S$), such that it is closed under the operation of symmetric difference and intersection, and define $E+F=E\triangle F,E\cdot F=E\cap F$. We define $x\leq y$ if $x\cdot y=x$, motivated by the fact that $E\subset F\iff E\cap F = E$. For a subset $X$ of a Boolean ring $R$, $supX$ is defined as an element $x_0$ of $R$, such that for every $x\in X$, $x\leq x_0$, and for every $a\in R$ such that for every $x\in X$, $x\leq a$, we have $x_0\leq a$ (the existence of $supX$ is not assured, but by a simple calculation we have $sup\{x_1,x_2\}=x_1+x_2+x_1\cdot x_2$, so that if $X$ is finite then $supX$ exists). A Boolean $\sigma$-ring is defined as a Boolean ring such that every countable subset of it has a supremum. In the typical example above, if for every countable subclass $\{X_n\}$ of $S$, we have $\cup X_n$ belongs to $S$(and therefore is equal to $sup\{X_n\}$), then $S$ is a Boolean $\sigma$-ring. Now,since $\cup (X_n\cap Y)=(\cup X_n)\cap Y$, I want to prove that in any Boolean $\sigma$-ring, we have $supx_n\cdot y = sup(x_n\cdot y)$.(We adopt the simplification that $supx_n=sup\{x_1,x_2,...\}$)
My progress:
Suppose $supx_n = x$, then it is obvious that $x\cdot y \geq x_n\cdot y$ for every $n$. So we only need to prove the following: if $z\geq x_n\cdot y$ for every $n$ then $z\geq x\cdot y$. In the case that the ring $R$ has a unit $1$, we could proceed as follows: since $z\geq x_n\cdot y$, so $x_n\cdot y\cdot z = x_n\cdot y$, so $x_n\cdot(1+y-y\cdot z)=x_n$. It follows that $x_n \leq 1+y-y\cdot z$ for every $n$, so $x\leq 1+y-y\cdot z$, so $x\cdot(1+y-y\cdot z) = x$, implying $z\geq x\cdot y$. However in case that the ring does not have a unit, this proof breaks.