Proof. Since $L$ is not modular, by definition it contains elements $x$ , $y$ and $z$ such that $x\leq z$ and $x\lor (y\land z) < (x\lor y) \land z$ . Then the sublattice formed by $y$ , $x\lor y$ , $y\land z$ , $(x\lor y)\land z$ and $x\lor (y\land z)$ is isomorphic to $N_5$ . This is because $y\land z\leq x\lor(y\land z)<(x\lor y)\land z\leq x\lor y$ while $[x\lor(y\land z)]\lor y=x\lor y$ by absorption and similarly $y\land[(x\lor y)\land z]=y\land z$ . Moreover, $x\lor(y\land z)$ covers $y\land z$ since $y\land z=x\lor(y\land z)$ would imply $x\leq y\land z \leq y$ , whence $x\lor y=y$ and $(x\lor y)\land z=y\land z=x\lor(y\land z)$ contrary to our hypothesis. By the same method, $ (x\lor y)\land z=x\lor y $ leads to a contradiction of nonmodularity, so $ x\lor y $ covers $ (x\lor y)\land z $ .