${\displaystyle {}y}$

${\displaystyle {}z}$

${\displaystyle {}x}$

${\displaystyle {}y=y\sqcup 0=y\sqcup (x\sqcap z)=(y\sqcup x)\sqcap (y\sqcup z)=1\sqcap (y\sqcup z)=y\sqcup z\,.}$

Da ebenso ${\displaystyle {}z=y\sqcup z}$ gilt, folgt ${\displaystyle {}y=z}$.