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I heard that the "pentagon lattice" $N_5 = (\{0,a,b,c,1\},\le)$ is not distributive, where $\le$ is the reflexive transitive closure of $\{(0,a),(a,b),(b,1),(0,c),(c,1)\}$. What elements of $N_5$ do not satisfy the distributive law $(x\vee y)\wedge z = (x\wedge z)\vee(y\wedge z)$ or $(x\wedge y)\vee z = (x\vee z)\wedge(y\vee z)?$

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up vote 4 down vote accepted

$a\vee(b \wedge c)=a$, while $(a\vee b)\wedge(a\vee c)=b$

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I've been looking for an example of $(x\vee y)\wedge z = (x\wedge z)\vee(y\wedge z)$ myself. Could you show one? – Pteromys Dec 31 '11 at 12:36
We get from the duality principle $b\wedge(a\vee c) \ne (b\wedge c)\vee(a\wedge c)$. – Pteromys Dec 31 '11 at 13:35

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