In finite semigroup, if $ef = f$ for $e,f \in E(S)$ and $e = xy, f = yx, s = sx = se = sf$ then we could not have $fe = e$

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Let $S$ be a finite semigroup. For two idempotents $e, f$ we set $$ f \le e :\Leftrightarrow ef = f $$ Now let $s \in S$ and $e,f \in S$ be two idempotents. Suppose $s = se = sf$ and there exists $x, y\in S$ such that $$ s = sx, e = xy, f = yx $$ (and hence also $s = sy$ as $sy = sxy = se = s$).

Is it true that if $f \le e$ then we also have $e \le f$, i.e. if $f \le e$ and $f \not\le e$ then the above relations between $s,e$ and $f$ could not hold?

I am stuck with this, hope you can help me!

PS: $E(S) = \{ e \in S \mid e^2 = e \}$ is the set of idempotents of $S$.

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The answer is yes. Indeed $e = xy$ and $f = yx$ implies that $e$ and $f$ are $\mathcal{D}$-equivalent. Now, $ef = f$ implies that $f \leqslant_{\mathcal{R}} e$. In a finite semigroup, the conjunction of these two conditions imply $e \mathrel{\mathcal{R}} f$.