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 \(8\).

19. Verify the associative laws.

Short answer

The given associative law \(x + (y + z) = (x + y) + z,x(yz) = (xy)z\) is proved.

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.

Associative laws

\(\begin{array}{c}{\bf{x + (y + z) = (x + y) + z}}\\{\bf{x(yz) = (xy)z}}\end{array}\)

Using the associative law

x and y can both take on the value of 0 or 1.

The Boolean sum is 1 if one of the two elements (or both) are 1.

\(\begin{array}{*{20}{r}}x&y&z&{y + z}&{x + y}&{x + (y + z)}&{(x + y) + z}\\0&0&0&0&0&0&0\\0&0&1&1&0&1&1\\0&1&0&1&1&1&1\\0&1&1&1&1&1&1\\1&0&0&0&1&1&1\\1&0&1&1&1&1&1\\1&1&0&1&1&1&1\\1&1&1&1&1&1&1\end{array}\)

The last two columns of the table are identical.

Therefore, you get \(x + (y + z) = (x + y) + z\).

Using the associative law

The Boolean product is 1 if both elements are 1.

\(\begin{array}{*{20}{r}}x&y&z&{yz}&{xy}&{x(yz)}&{(xy)z}\\0&0&0&0&0&0&0\\0&0&1&0&0&0&0\\0&1&0&0&0&0&0\\0&1&1&1&0&0&0\\1&0&0&0&0&0&0\\1&0&1&0&0&0&0\\1&1&0&0&1&0&0\\1&1&1&1&1&1&1\end{array}\)

The last two columns of the table are identical.

Therefore, you get \(x(yz) = (xy)z\).