I just started learning Boolean Algebra and have this homework question
Demonstrate that the pentagon lattice is non-distributive
I know this is non-distributive because $b$ complements $a$ and $c$. So when I check if distributive law holds for $ a \lor (b \land c) = (a \lor b) \land (a \lor c)$. I get
$$ a \lor (b \land c) = (a \lor b) \land (a \lor c)$$ $$ a \lor 0 = (a \lor b) \land (a \lor c)$$ $$ a = (a \lor b) \land (a \lor c)$$ $$ a = 1 \land c $$ $$ a = c $$
Since $ a \neq c$ the pentagon is not distributive. But why is it if I use the other law of distribution that I don't get the same answer? I would think they both would fail if the lattice is non-distributive. What am I doing wrong here?
$$ a \land (b \lor c) = (a \land b) \lor (a \land c)$$ $$ a \land 1 = (a \land b) \lor (a \land c)$$ $$ a = 0 \lor a$$ $$ a = a $$
This is very simple. If try to find the complement of 'b' you will get 'a' and 'c'. And any element in a distributve lattice should have atmost one complement. Distributive property comes from set theory where every element has atmost one complement and if lattice has an element with more than one complement then the lattice will not follow distributive property.
How to find complement? ans: If any element two elements 'a' and 'b' such that a∨b = upper bound of lattice and a∧b = lower bound of lattice, a and b are complements of each other.
what are upper bound and lower bound? Ans: upper bound : an element to which every element relates. Lower bound :this element relates to every element.