Since \(z \le _{\mathcal L} x\), there exists \(t\in M\) with \(z = t\cdot x\). Using associativity and the relations \(x\cdot u = y\) and \(y\cdot v = x\), we compute

It suffices to verify the following:

(1) Image in the correct \(\mathcal L\)-class. If \(z \, \mathcal L\, x\) then \(z\cdot u \, \mathcal L\, y\). Indeed, writing \(z = t\cdot x\) for some \(t\), we have \(z\cdot u = t\cdot x\cdot u = t\cdot y\), so \(z\cdot u \, \mathcal L\, y\) (equivalently, by compatibility of \(\mathcal L\) with right-multiplication).

(2) Injectivity on the \(\mathcal L\)-class of \(x\). If \(z,w \, \mathcal L\, x\) and \(z\cdot u = w\cdot u\), then \(z=w\). By Lemma 36, \(z\cdot u\cdot v = z\) and \(w\cdot u\cdot v = w\). Hence \(z = z\cdot u\cdot v = w\cdot u\cdot v = w\).

(3) Surjectivity onto the \(\mathcal L\)-class of \(y\). Let \(z \, \mathcal L\, y\). Set \(w := z\cdot v\). Then \(w \, \mathcal L\, y\cdot v = x\), so \(w \, \mathcal L\, x\). Moreover, by Lemma 36 (applied with \(z\) in the \(\mathcal L\)-class of \(y\)),

so \(z\) lies in the image of \(\rho _u\).

(4) \(\rho _u\) and \(\rho _v\) are inverses on the respective \(\mathcal L\)-classes. If \(z \, \mathcal L\, x\), then \(z\cdot u\cdot v = z\) by Lemma 36; if \(z \, \mathcal L\, y\), then \(z\cdot v\cdot u = z\) (the same lemma with the roles of \(x,y\) interchanged).

(5) Preservation of \(\mathcal H\)-equivalence. For \(z,w \, \mathcal L\, x\) one should verify

Using \(\rho _{u,v}\) from Lemma 36 gives \(z\cdot u \, \mathcal R\, z\) and \(w\cdot u \, \mathcal R\, w\), and the \(\mathcal L\)-compatibility from (1) supplies the \(\mathcal L\)-side; a routine transitivity argument then completes the proof. (Details omitted.)

(\(\Rightarrow \)). Assume \(e^2=e\), \(e \, \mathcal L\, x\), and \(e \, \mathcal R\, y\). For idempotents one has the characterizations

Hence \(x e = x\) and \(e y = y\). By multiplicative compatibility of Green’s relations (see 23 and 22), from \(y \, \mathcal R\, e\) we deduce \(x y \, \mathcal R\, x e = x\), and from \(x \, \mathcal L\, e\) we deduce \(x y \, \mathcal L\, e y = y\).

(\(\Leftarrow \)). Assume \((x y)\, \mathcal R\, x\) and \((x y)\, \mathcal L\, y\). Consider the right-translation \(\rho _y(z)=z y\). By Green’s Lemma (37), \(\rho _y\) restricts to a bijection from the \(\mathcal L\)-class of \(x\) onto the \(\mathcal L\)-class of \(x y\). Since \(y \, \mathcal L\, x y\), there exists \(t\) in the \(\mathcal L\)-class of \(x\) with \(t y = y\). From \((x y)\, \mathcal R\, x\) choose \(u\) with \(x y \cdot u = x\). Applying Lemma 36 (with \(x \, \mathcal R\, x y\)) to \(t\) gives \(t y u = t\). Therefore

so \(t\) is idempotent. Moreover \(t \, \mathcal R\, y\) because \(t y = y\), and \(t \, \mathcal L\, x\) since \(t\) was chosen in the \(\mathcal L\)-class of \(x\). Thus there exists an idempotent \(t\) with \(t \, \mathcal L\, x\) and \(t \, \mathcal R\, y\).