Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $(x\wedge y)\vee z = (x\vee z)\wedge(y\vee z)?$

share|improve this question
add comment

1 Answer 1

up vote 4 down vote accepted

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

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.