Question

One that is not postulate of Boolean algebra\[\]
A. Commutative \[\]
B. Duality \[\]
C. Associative\[\]
D. Identity Element\[\]

Answer 1

Hint:We recall the definition of fundamental binary operations the logical AND, the logical NOT, logical OR in Boolean algebra. We take two random inputs and check whether they satisfy the properties given in the option commutative, duality, associative, identity element to get the property which is not a postulate. \[\]

Complete step by step answer:
We know that in Boolean algebra the variables assume truth values T or F instead of real or complex number like in elementary algebra. The operations that happen between two truth values in Boolean algebra are called logical operations and they are of 3 types conjunction or the logical AND operation (denoted as $\wedge $ ), disjunction or the logical OR operation (denoted as $\vee $ ) and the negation or the logical NOT operation (denoted as $\neg $) . The logical AND, logical OR are binary operations while the negation is a unitary operation.\[\]

We denote any binary operation in Boolean algebra as $'o'$ and take two variables as inputs $A$ and $B$ that assume truth values T or F. An operation follows is commutative when the order of the inputs does not change the output. We know in Boolean algebra that \[\begin{align}
  & A\wedge B=B\wedge A \\
 & A\vee B=B\vee A \\
\end{align}\]
 So the binary operation $'o'$ is commutative, $AoB=BoA$. We take three inputs $A,B,C$ that assume truth values T or F. An operation follows is associative if we choose any two inputs and operate their result with the third input and get the same result. We know in Boolean algebra that
\[\begin{align}
  & \left( A\wedge B \right)\wedge C=A\wedge \left( B\wedge C \right) \\
 & \left( A\vee B \right)\vee C=A\vee \left( B\vee C \right) \\
\end{align}\]
So the binary operation $'o'$ is associative, $\left( AoB \right)oC=Ao\left( BoC \right)$. The identity element is the element with operation does not change the operand. We know that in logical AND operation the identity element is T and for logical OR operation the identity element is F. So we have,
\[\begin{align}
  & A\wedge T=A \\
 & A\vee F=A \\
\end{align}\]
So identity element exists for $'o'$. So we have commutative, associative, identity element as the postulates or perquisites for ay operation in Boolean algebra. \[\]
The duality principle states that if we replace one $'\wedge '$ with $'\vee '$ or vice-versa in the Boolean expression which is not a prerequisite for operation in Boolean algebra. So duality is not a postulate of Boolean algebra and the correct option is B.\[\]

Note:
 We note that the truth values T and F also can be written as 1 and 0 and similarly the operations $'\wedge '$ as $'*'$ and $'\vee '$ as$'+'$. The other postulates of Boolean algebra are complement $\neg \left( \neg A \right)=A$ and distributive law $A\wedge \left( B\vee C \right)=\left( A\wedge C \right)\vee \left( A\wedge C \right)$.