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 \(8\).

20. Verify the first distributive law in Table \({\bf{5}}\) .

Short answer

The given distributive laws \(x + y z = \left( {x + y} \right)\left( {x + z} \right)\) and \(x\left( {y + z} \right) = x y + x z\) are 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.

Using the distributive law

Distributive laws

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

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

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

The last two columns of the table are identical.

Therefore, you get \(x + y z = \left( {x + y} \right)\left( {x + z} \right)\).

Verifying the second law

Note: you had to verify the first distributive law, but the following part contains the verification of the second law.

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

The last two columns of the table are identical.

Therefore, you get \(x\left( {y + z} \right) = x y + x z\).