Question

In Exercises 35–42, use the laws in Definition \(1\) to show that the stated properties hold in every Boolean algebra.

Show that in a Boolean algebra, the dual of an identity, obtained by interchanging the \( \vee \)and \( \wedge \) operators and interchanging the elements \({\bf{0}}\) and \(1\) , is also a valid identity.

Short answer

If \(I\) is a valid identity, then the dual of \(I\) is also a valid identity because we note that all Boolean algebra laws come in dual pairs. The dual of Boolean algebra law is also a Boolean algebra law.

Step-by-step solution

Definition

The dual of an identity is obtained by interchanging \( \vee \) and \( \wedge \) operators and interchanging the elements \(0\) and \(1\) .

If \(I\) is a valid identity, then the dual of \(I\) is also a valid identity because we note that all Boolean algebra laws come in dual pairs. The dual of Boolean algebra law is also a Boolean algebra law.

For example, the dual of the first identity law \(x \vee 0 = x\) is the second identity law \(x \wedge 1 = x\).

Boolean algebra laws 

Identity laws

\(\begin{aligned}x \vee 0 = x\\x \wedge 1 = x\end{aligned}\)

Complement laws

\(\begin{aligned}x \vee \bar x &= 1\\x \wedge \bar x& = 0\end{aligned}\)

Commutative laws

\(\begin{aligned}x \vee y &= y \vee x\\x \wedge y &= y \wedge x\end{aligned}\)

Associative laws

\(\begin{aligned}x \vee \left( {y \vee z} \right) &= \left( {x \vee y} \right) \vee z\\x \wedge \left( {y \wedge z} \right) &= \left( {x \wedge y} \right) \wedge z\end{aligned}\)

Distributive laws

\(\begin{aligned}x \vee \left( {y \wedge z} \right) &= \left( {x \vee y} \right) \wedge \left( {x \vee z} \right)\\x \wedge \left( {y \vee z} \right) &= \left( {x \wedge y} \right) \vee \left( {x \wedge z} \right)\end{aligned}\)

Idempotent laws

\(\begin{aligned}x \vee x &= x\\x \wedge x &= x\end{aligned}\)

Hence, the law of the double complement

\(\overline{\overline x} = x\)