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+3 votes

Consider a Hasse Diagram for a Boolean Algebra of Order 3

What can we comment about it? 
How is it successfully able to represent the Boolean Algebra System?
Is there an easy way to check for distributive lattice, or any other properties of a lattice?

Given/Know :
$\langle B, \vee ,\ \cdot \ , \bar{\ \ } \ ,0, 1 \rangle$ is a Boolean Algebra, 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



It is not expected that one should provide a complete answer to all parts of the question. Whatever one can supply to support its answer is welcomed.

asked in Set Theory & Algebra by Veteran (27k points)   | 231 views

3 Answers

+1 vote

According to me::

Suppose we have A={a,b,c}

now if we calculate power set of of A like P(A)={Phi,{a},{b},{c},{ab},{bc},{ca},{abc}};

Then Poset =[P(A);under subset]

Now u can check below hasse diagram that it is both complemented as well as Distributed lattice that's why it will be boolean algebra.One special point is that in boolean algebra number of vertex will be 2^n and edges n*2^n-2.U can check above. 

answered by Loyal (4.1k points)  

correct it total no of  edges in boolean algebra if no of element is n should be ( n * 2n-1

0 votes


It is a bounded, distributive and complemented lattice. So, it must be a boolean algebra

answered by Veteran (49.7k points)  
0 votes

A lattice is called a Boolean Algebra, if it is distributive and complemented( if complement exists for every element of lattice).

So, For $2-order$ Boolean Algebra, it's easy to check for these properties to say whether a lattice is boolean Algebra (or) not, But for lattices for higher order (say $3$, $4$ (or) $20$), it becomes worse to Check for distributive properties and finding complement for each element.

One way here would be eliminating options using the property that Every element has a unique complement in a Boolean Algebra, but the converse is not True.

Any $n-order$ boolean Algebra is isomorphic to any other $n-order$ boolean algebra. And we know that $\color{olive}{\big[ P(A),\subseteq \big]}$ is a boolean Algebra.

So, any boolean Algebra of $\color{green}{order-2}$ and $\color{green}{order-3}$ must be isomorphic to these two. And also note that a Boolean Algebra of order $n$ has $\color{red}{2^n}$ vertices(nodes) and $\color{red}{n*2^{n-1}}$ edges.

So, now coming to the question the above Hasse diagram is isomorphic to the Boolean Algebra of order-3, Hence it is Boolean Algebra.

answered by Veteran (22k points)  
edited by
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