Let $S$ be a semigroup, then an ideal $W$ is a subset $W \subseteq S$ such that $sWt \in W$ for all $s,t \in S^{\bullet}$, where $S^{\bullet}$ is $S$ with an additional unit element added if $S$ does not has one (hence forms a monoid). Further we set $s \le t$ iff $S^{\bullet} s S^{\bullet} \subseteq S^{\bullet} t S^{\bullet}$, which is equivalent with $t = xsy$ for some $x,y \in S^{\bullet}$.
The following claim is taken from the book Automata, Languages and Machines (page 291) by Samual Eilenberg:
Let $W$ be an ideal in $S$ and let $s,t \in S$ with either $s$ or $t$ an idempotent. Then $$ s \le t \mbox{ in $S$ and $t \in W$ } \Leftrightarrow s \le t \mbox{ in W}. $$ Proof: Let $s \in t$ in $S$ with $t \in W$. Then $s \in W$ and there exists $x,y \in S^{\bullet}$ such that $s = xty$. If $s^2 = s$, then $$ s = (sx)t(ys) \in W^{\bullet} t W^{\bullet} $$ so $s \le t$ in $W$. If $t^2 = t$, then $$ s = (xt) t (ty) \in W^{\bullet} t W^{\bullet} $$ so $s \le t$ in $W$. The converse is clear. $\square$
Why is the converse clear, if $s \le t$ in $W$, then if $s = xty$ with $x,y \in W$ we have $s \in W$, but not $t \in W$? And even the case $s = t \notin W$ I do not see why this is not possible?
