Consider a Hasse Diagram for a Boolean Algebra of Order 3
Just by using the diagram and defined Boolean Algebra System as :
$\langle B, \vee ,\ \cdot \ , \bar{\ \ } \ ,0, 1 \rangle$ and for any 3 of its arbitrary elements $a, b, c$ in $B$ the following postulates are satisfied:
where, $\vee$ is Boolean Sum
$\cdot$ is Boolean Product
$\bar{\ \ }$ is Complement
How is that Hasse diagram successfully able to define this Boolean Algebra System; How can I see that it is able to do that?
I know that Boolean Algebra is a distributive lattice that satisfies postulates from $(6)$ to $(10)$. Is there an easy way to check that a lattice is distributive?
I know a method to check if it is Distributive by taking all possible combination of elements and see if they satisfy postulates from $(1)$ to $(5)$. Also, I know that these shapes in a lattice destroys its distributive property
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Hasse diagrams describe general posets (partially ordered sets) according to the following rule: $x \prec y$ if there is a directed path from $x$ to $y$, where we direct all edges up. Going the other way, there is an edge from $x$ to $y$ if $x \prec y$ and there is no $z$ satisfying $x \prec z \prec y$.
Given two elements $x,y$, their join $x \lor y$ is an element $z$ such that $x,y \preceq z$ and all $w$ satisfying $x,y \preceq w$ also satisfy $z \preceq w$. The meet $x \land y$ is defined analogously by replacing $\preceq$ with $\succeq$. (Your notation is $\cdot$.)
An element $0$ is the bottom element if $0 \preceq x$ for all $x$. Similarly, an element $1$ is the top element if $x \preceq 1$ for all $y$. In your diagram these elements are $(0,0,0),(1,1,1)$, and you can identify them as the unique elements at the bottom and top.
For an element $x$, a complement $x'$ is an element satisfying $x \land x' = 0$ and $x \lor x' = 1$ (your notation is $\bar{x}$). I'm not sure under what conditions the complement is unique.
This allows you to define a Boolean algebra given a Hasse diagram, assuming that the corresponding poset has joins and meets, top and bottom elements, and complements.
In your case, your Boolean algebra is an algebra of sets, so it is easy to verify all axioms directly. You can also verify them by going over all $a,b,c$ and performing the required computation, but that's better left to a computer.