Question

Exercises 14-23 deal with the Boolean algebra \(\left\{ {{\bf{0,1}}} \right\}\) with addition, multiplication, and complement defined at the beginning of this section. In each case, use a table as in Example \(8\).

Verify the domination laws.

Short answer

The domination law \(x + 1 = 1,x \cdot 0 = 0\) is verified.

Step-by-step solution

Definition

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

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.

Domination laws

\(\begin{array}{c}x + 1 = 1\\x \cdot 0 = 0\end{array}\)

Using the domination law

x can take on the value of 0 or 1. The Boolean sum is 1 if one of the two elements (or both) are 1.

\(\begin{array}{*{20}{c}}x&{x + 1}\\0&1\\1&1\end{array}\)

The last two columns of the table are identical.

Therefore, you get \(x + 1 = 1\).

Using the domination law

The Boolean product is 1 if both elements are 1.

\(\begin{array}{*{20}{r}}x&{x \cdot x}\\0&0\\1&0\end{array}\)

The last two columns of the table are identical.

Therefore, you get \(x \cdot 0 = 0\).