Question

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

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

Short answer

a) The table of the given Boolean function \(F(x,y,z) = \bar z\) is

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

b) The table of the given Boolean function \({\bf{F(x,y,z) = \bar xy + \bar yz}}\) is

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

c) The table of the given Boolean function \(F(x,y,z) = x\bar yz + \overline {(xyz)} \) is

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

d) The table of the given Boolean function \({\bf{F(x,y,z) = \bar y(xz + \bar x\bar z)}}\) is

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

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 definition of complement

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

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

Therefore, the table of the given Boolean function \(F(x,y,z) = \bar z\) is

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

(b) Using Boolean product and sum

\({\bf{F(x,y,z) = \bar xy + \bar yz}}\)

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 \(\bar x \cdot y\) and \(\bar yz\) represents \(\bar y \cdot z\)

Therefore, the table of the given Boolean function \({\bf{F(x,y,z) = \bar xy + \bar yz}}\) is

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

(c) Using Boolean product and sum

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

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

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

Therefore, the table of the given Boolean function \(F(x,y,z) = x\bar yz + \overline {(xyz)} \) is

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

(d) Using Boolean product and sum

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

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 \(\bar y \cdot \bar z\)

Therefore, the table of the given Boolean function \({\bf{F(x,y,z) = \bar y(xz + \bar x\bar z)}}\) is

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