Question

use the laws in Definition \(1\) to show that the stated properties hold in every Boolean algebra.

Show that in a Boolean algebra, the complement of the element \(0\) is the element \(1\) and vice versa.

Short answer

1 is the complement of 0 and 0 is the complement of 1 so you get \(\bar 0 = 1\),\(\bar 1 = 0\).

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 \( \cdot \) or \(AND\) is \(1\) if both terms are \(1\).

Identity laws

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

Commutative laws

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

Using the commutative and identity law

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

Let you first check that the complement of \(0\) is \(1\).

Using the commutative and identity law

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

Thus,\(1\) is the complement of \(0\)

Therefore, it gets\(\bar 0 = 1\)

Using the identity and commutative law

Next, you need to check that the complement of \(1\) is \(0\).

Using the identity and commutative law

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

Thus,\(0\) is the complement of \(1\).

Therefore, it gets\(\bar 1 = 0\).