Question

Use a table to express the values of each of these Boolean functions.

\(\begin{array}{l}(a)F(x,y,z) = \bar xy\\(b)F(x,y,z) = x + yz\\(c)F(x,y,z) = x\bar y + \overline {(xyz)} \\(d)F(x,y,z) = x(yz + \bar y\bar z)\end{array}\)

Short answer

a) A table of Boolean expression \(F(x,y,z) = \bar xy\) is

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

b) A table of Boolean expression \(F(x,y,z) = x + yz\) is

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

c) A table of Boolean expression \(F(x,y,z) = x\bar y + \overline {(xyz)} \) is

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

d) A table of Boolean expression \(F(x,y,z) = x(yz + \bar y\bar z)\) is

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

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.

(a) Using the Boolean product

\(F(x,y,z) = \bar xy\)

The function has three variables x, y and z. Each of these variables can take on the value of 0 or 1.

Note: \(\bar xy\) represents \(\bar x \cdot y\)

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

(b) Using the Boolean product and sum

\(F(x,y,z) = x + yz\)

The function has three variables x, y and z. Each of these variables can take on the value of 0 or 1.

Note: \(yz\) represents \(y \cdot z\)

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

(c) Using the Boolean product and sum

\(F(x,y,z) = x\bar y + \overline {(xyz)} \)

The function has three variables x, y and z. Each of these variables can take on the value of 0 or 1.

Note: \(\overline {(xyz)} \) represents \(\overline {(x \cdot y \cdot z)} \) and \(x\bar y\) represents \(x \cdot \bar y\)

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

(d) Using the Boolean product and sum

\(F(x,y,z) = x(yz + \bar y\bar z)\)

The function has three variables x, y and z. Each of these variables can take on the value of 0 or 1.

Note: \(yz\) represents \(y \cdot z\) and \(\bar y\bar z\) represents \(\bar y \cdot \bar z\)

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