Question

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

38. Prove that in a Boolean algebra, the law of the double complement holds; that is, x̿ = x for every element \({\bf{x}}\).

Short answer

The given law of double complement x̿ = x is proved.

Step-by-step solution

Definition

The complement of an element: \(\bar 0 = 1\) and \(\bar 1 = 0\)

The Boolean sum \( + \) or \(OR\) is \(1\) if either term is \(1\).

The Boolean product or \(AND\) is \(1\) if both terms are \(1\) .

Identity laws

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

Complement laws

\(\begin{array}{l}x \vee \bar x = 1\\x \wedge \bar x = 0\end{array}\)

Commutative laws

\(\begin{array}{l}x \vee y = y \vee x\\x \wedge y = y \wedge x\end{array}\)

Distributive laws

\(\begin{array}{c}x \vee \left( {y \wedge z} \right) = \left( {x \vee y} \right) \wedge \left( {x \vee z} \right)\\x \wedge \left( {y \vee z} \right) = \left( {x \wedge y} \right) \vee \left( {x \wedge z} \right)\end{array}\)

Using the law of complement

Law of the double complement x̿ = x

To proof: x̿ = x

PROOF

\(y\)is the complement of \(x\)if and only if \(x \vee y = 1\)and\(x \wedge y = 0\).

Note: The complement is also unique, which was proven in one of the previous solutions.

Thus, it needs to show that \(x\) is the complement of \(\bar x\).

Using the commutative and complement law

\(\begin{array}{c}\bar x \vee x = x \vee \bar x = 1\\\bar x \wedge x = x \wedge \bar x = 0\end{array}\)

Thus,\(x\) is the (unique) complement of \(\bar x\).

Therefore, it gets x̿ = x