Question

For which values of the Boolean variables \(x,y\), and \(z\) does

\(\begin{array}{l}{\bf{a)x + y + z = xyz?}}\\{\bf{b)x(y + z) = x + yz?}}\\{\bf{c)\bar x\bar y\bar z = x + y + z?}}\end{array}\)

Short answer

1. The given equation \({\bf{x + y + z = xyz}}\) holds for \((0,0,0),(1,1,1)\)
2. The given equation \({\bf{x(y + z) = x + yz}}\) holds for

\((0,0,0),(0,0,1),(0,1,0),(1,0,1),(1,1,0),(1,1,1)\)

1. The given equation has no solution.

Step-by-step solution

Definition

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

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

The Boolean product \( \cdot \) or \(AND\) is \({\bf{1}}\) if both terms are \({\bf{1}}\).

Using the Boolean sum and product

(a)

\({\bf{x + y + z = xyz}}\)

Now create a table for all possible values of \(x,y\) and \(z\).

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

Then note that \({\bf{x + y + z}}\) contains the same value as \({\bf{xyz}}\) in the first and last row.

Hence it corresponds with \({\bf{x = y = z = 0}}\) and \({\bf{x = y = z = 1}}\).

Using the Boolean sum and product

(b)

\({\bf{x(y + z) = x + yz}}\)

Now create a table for all possible values of \(x,y\) and \(z\).

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

Then note that \({\bf{x(y + z)}}\) contains the same value as \({\bf{x + yz}}\) in the rows.

So, it corresponding to \(x,y,z{\bf{ = }}(0,0,0),(0,0,1),(0,1,0),(1,0,1),(1,1,0),(1,1,1)\).

Using the Boolean sum and product

(c)

\({\bf{\bar x\bar y\bar z = x + y + z}}\)

Now create a table for all possible values of \(x,y\) and \(z\).

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

Then note that \({\bf{\bar x\bar y\bar z}}\) never contains the same value as \({\bf{x + y + z}}\).

Therefore, the given equation has no solutions.