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Found in: Page 818

Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

Use a table to express the values of each of these Boolean functions.$$\begin{array}{l}{\bf{a) F(x,y,z) = \bar z}}\\{\bf{b) F(x,y,z) = \bar xy + \bar yz}}\\{\bf{c) F(x,y,z) = x\bar yz + }}\overline {{\bf{(xyz)}}} \\{\bf{d) F(x,y,z) = \bar y(xz + \bar x\bar z)}}\end{array}$$

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

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

See the step by step solution

Step 1: 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.

Step 2: (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

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

Step 3: (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

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

Step 4: (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

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

Step 5: (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

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